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physical-chem - Atomic Structure I. Atomic structure A. B....

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Unformatted text preview: Atomic Structure I. Atomic structure A. B. Different model of atomic structure Isotopic mass and relative atomic mass 1. Mass spectrometer a) Determination of atomic mass b) Determination of molecular mass Radioactivity of nuclide 1. Nature of radioactivity a) Penetrating power of radiation 2. Radioactive decay a) Half life b) Use of radioactive isotope Emission spectrum of hydrogen 1. Origin of emission and absorption spectrum 2. Pattern of energy levels of hydrogen 3. Determination of the ionization energy a) Through the convergence limit of Lyman series b) Through Rydberg constant Detail electronic configuration 1. Definition of ionization energy 2. Successive ionization energy 3. Periodicity of First ionization energy a) First I.E. across periods b) First I.E. down a group 4. Quantum numbers a) Energy of electrons in different orbitals 5. Building up of electrons in an atom a) Aufbau principle b) Pauli exclusion principle c) Hund's rule d) Building up of electrons e) Detail electronic configuration of some atoms 6. Electronic configuration of first 36 elements a) Extra stability of half-filled and full-filled subshell 7. Position of s-, p-, d- and f-block elements Shape of s, p and d orbitals Variation in atomic radius Variation in electronegativity C. II. Electronic structure of atoms A. B. III. IV. Atomic orbitals A. A. B. Periodic Table Energetics I. Enthalpy Change (∆H) A. B. C. Definition of a system Definition of enthalpy (H) and enthalpy change (∆H) 1. Constant pressure process a) Work done by a system Standard enthalpy change of formation (∆Hfo) 1. Definition of ∆Hfo a) Standard enthalpy change (∆Ho) b) Standard enthalpy change of formation (∆Hfo) c) Standard enthalpy change of formation of an element 2. Direct determination of ∆Hfo a) Use of simple calorimeter Other enthalpy change (∆H) 1. Examples of enthalpy change a) Enthalpy of combustion b) Enthalpy of atomization c) Enthalpy of neutralization d) Enthalpy of hydrogenation e) Enthalpy of solution f) Enthalpy of reaction Hess’s Law 1. Indirect determination of ∆Hfo a) Hess’s Law (1) (2) Use of energy level diagram / energy cycle Use of equation D. E. II. Bonding Energy A. B. C. Bond energy 1. Definition of bond energy 2. Determination of bond energy term (bond enthalpy) Strenght of covalent bond 1. Relationship between bond length and bond energy 2. Factors affecting the bond strength Estimation of enthalpy change by bond energy Electron affinity Lattice energy 1. Born-Haber cycle a) Determination of lattice energy by Born-Haber cycle 2. Factors affecting the value of lattice energy Stoichiometry of ionic compounds Limitation of enthalpy change (∆H) Entropy (S) and entropy change (∆S) Gibbs free energy change (∆G) III. Energetics of formation of ionic compounds A. B. C. IV. Gibbs free energy change (∆G) A. B. C. Chemical Bonding Page 1 Chemical Bonding I. Shape of molecule A. Lewis structure of molecule 1. Oxidation no. of individual atoms in a molecule 2. Formal charge of individual atoms in a molecule 3. Limitation of octet rule Valence Shell Electron Pair Repulsion Theory (VSEPRT) 1. Octet expansion of elements in Period 3 2. Position of lone pair 3. 3-dimensional representation of molecular shape Hybridization Theory 1. Overlapping of atomic orbital 2. Hybridization theory 3. Examples a) sp hybridizaion b) sp2 hybridizaion c) sp3 hybridizaion d) sp3d hybridizaion e) sp3d2 hybridizaion 4. Structure and shape of hydrocarbons a) Characteristic of Carbon-Carbon bond b) Shape of hydrocarbons (1) (2) (3) Saturated hydrocarbons Unsaturated hydrocarbons Aromatic hydrocarbons (a) Delocalization of π-electrons (b) Stability of benzene B. C. D. Molecular Orbital Theory II. Ionic bonding A. B. C. Strength of ionic bond - lattice energy X-ray diffraction 1. Electron density map of sodium chloride Periodicity of ionic radius 1. Definition of radius 2. Periodicity of ionic radius 3. Size of isoelectronic particles Chemical Bonding Page 2 III. Covalent bonding A. B. C. H2+ ion Electron diffraction Covalent radius 1. Definition of covalent radius 2. Addivity of covalent radius 3. Breaking down of addivity in covalent radius and bond energy a) Resonance (1) (2) (3) Resonance structure of benzene Resonance structure of nitrate ion Rules in writing resonance structures Delocalization energy b) D. Breaking down of addivity in bond enthalpies (1) Dative covalent bond 1. Examples of H3N→BF3 and Al2Cl6 a) H3N→BF3 b) Al2Cl6 IV. Bonding intermediate between ionic and covalent A. B. Differences between ionic bond and covalent bond Incomplete electron transfer in ionic compound 1. Electron density map of LiF comparing with those of NaCl and H2 2. Difference among lattice energies of Na, Ag and Zn compounds a) Lattice energies of sodium halide, silver halide and zinc sulphate b) Bonding intermediate between covalent and ionic 3. Polarization of ionic bond a) Fajans’ Rules in polarization of ionic bond Electronegativity 1. Definition of electronegativity 2. Pauling scale of electronegativity Polarity in covalent bond 1. Deflection of a liquid jet by an electric field 2. Dipole moment a) Vector quantity of dipole moment b) Polarity of molecule c) Factors affecting dipole moment (1) (2) (3) Inductive effect (I) Mesomeric effect / resonance effect (R) Presence of lone pair C. D. V. Metallic bonding A. B. C. D. Electron sea model of metal Strength of metallic bond Melting and boiling of metal Strength of ionic bond, covalent bond and metallic bond Intermolecular forces Intermolecular Forces I. Van der Waals’ forces A. B. Discovery of van der Waals' forces Origin of van der Waals' forces 1. Induced dipole-induced dipole attractions 2. Dipole-dipole interactions 3. Dipole-induced dipole attractions Relative strength of different origins of van der Waals' forces Nature Strength Solubility and hydrogen bond Intramolecular hydrogen bond Examples of hydrogen bonding 1. Ethanoic acid dimer 2. Simple hydrides - CH4, NH3, H2O and HF 3. Structure and physical properties of ice 4. Biochemical importance of hydrogen bond a) DNA b) Protein Strength of van der Waals' forces and hydrogen bond C. II. Hydrogen bond A. B. C. D. E. F. III. Relationship between structures and properties States of Matter I. Solid state A. Metallic structure 1. Hexagonal close packing 2. Cubic close packing - face centred cubic 3. Tetrahedral holes and octahedral holes 4. Body-centred cubic structure Giant ionic structure Covalent giant structure 1. Allotrope of carbon a) Diamond b) Graphite 2. Silicon(IV) oxide Molecular crystal 1. Iodine – face centred cubic 2. Carbon dioxide (Dry ice) – face centred cubic Gas laws 1. Charles' law 2. Boyle's law 3. Avogadro's law 4. PVT surface Ideal gas equation (Ideal gas law) 1. Molar volume Determination of molecular mass 1. Experimental determination of molar mass of a gas 2. Experimental determination of molar mass of a volatile liquid Dalton's law of partial pressure 1. Mole fraction 2. Relationship between partial pressure and mole fraction B. C. D. II. Gaseous state A. B. C. D. Chemical Kinetics Page 1 Chemical Kinetics I. Rates of chemical reactions A. B. Definition of rate of reaction 1. Unit of rate of reaction Measuring the rate of reaction 1. Different approaches a) Constant amount approach b) Constant time approach 2. Interpretation of physical measurements made in following a reaction a) Volume of gas formed b) Colorimetric measurement Rate Law or rate equation 1. Differential form and integrated form of rate law 2. Graphical presentation of reaction rate 3. Order of reaction a) Experimental determination of order of reaction (1) (2) By measuring the rates of reaction at different reactant concentration By plotting graph of ln rate versus ln [X] II. Order of reaction A. B. Examples of different reaction 1. First order reaction a) Half life (t½) b) Carbon-14 dating c) Examples of calculation 2. Second order reaction 3. Zeroth order reaction Collision theory 1. Distribution of molecular speed (Maxwell-Boltzman distribution) 2. Effect of temperature change on molecular speeds 3. Collision theory and activation energy a) Arrhenius equation Determination of activation energy 1. By measuring the rate of reaction at different temperature 1 2. By plotting graph of ln k versus T 1 3. By plotting graph of ln rate versus T III. Collision theory and activation energy A. B. Chemical Kinetics Page 2 IV. Energy profile of reaction A. Transition state 1. Order of reaction leads to interpretation of reaction mechanism at molecular level a) Multi-stages reaction (1) (2) Alkaline hydrolysis of 2-chloro-2-methylpropane Reaction between bromide and bromate(V) in acidic medium B. b) Rate determining step 2. Energetic stability and kinetic stability Effect of catalyst 1. Characteristics of catalyst 2. Theory of catalyst a) Homogeneous catalysis b) Heterogeneous catalysis 3. Application of catalyst a) Use of catalyst in Haber process and hydrogenation b) Catalytic converter c) Enzymes Concentration Temperature Pressure Surface area Catalyst Light V. Factors influencing the rate of reaction A. B. C. D. E. F. Chemical Equilibria Page 1 Chemical Equilibria I. Nature of equilibrium A. B. Dynamic nature Examples of equilibrium 1. Bromine water in acidic and alkaline medium 2. Potassium dichromate in acidic and alkaline medium 3. Hydrolysis of bismuth chloride Equilibrium law 1. Introduction to different equilibrium constant (K) – Kc, Kp, Kw, Ka, Kb, KIn, Kd 2. Equilibrium constant (Kc and Kp) 3. Degree of dissociation (α) 4. Concentration of solid 5. Determination of equilibrium constant (Kc) a) Esterification b) Fe3+(aq) + NCS-(aq) d FeNCS2+(aq) 6. Relationship between rate constant and equilibrium constant Effect of change in concentration, pressure and temperature on equilibria (Le Chaterlier's principle) 1. Concentration 2. Pressure 3. Temperature ∆H a) Equation : ln K = constant T 4. Effect of catalyst on equilibria 5. Examples of calculation C. D. Chemical Equilibria Page 2 II. Acid-base Equilibria A. Acid-base Theory 1. Arrhenius definition 2. Bronsted-Lowry definition 3. Lewis definition Strength of acid 1. Leveling effect Dissociation of water 1. Ionic product of water pH and its measurement 1. Definition of pH 2. Temperature dependence of pH 3. Measurement of pH a) Use of pH meter (1) Calibration of pH meter Colour of indicator B. C. D. b) E. Use of indicator (1) Strong and weak acids/bases 1. Measuring of pH or conductivity of acid / base 2. Dissociation constant (Ka and Kb) a) Dissociation of polybasic acid (1) Charge effect 3. F. G. H. 4. Buffer 1. Principle of buffer action a) pH of an acidic buffer b) pH of an alkaline buffer 2. Calculation of buffer solution Theory of Indicator Acid-base titration 1. Difference between equivalence point and end point 2. Titration using pH meter 3. Titration using indicator a) Choosing of indicator 4. Thermometric titration 5. Conductimetric titration a) Strong acid vs Strong base b) Weak acid vs Strong base c) Weak acid vs Weak base Calculation involving pH, Ka and Kb a) Relationship between Ka and Kb (pKa and pKb) b) Relationship between pH, pOH and pKw c) Some basic assumptions applied Experimental determination of Ka Chemical Equilibria Page 3 III. Redox Equilibria A. Redox reaction 1. Balancing of redox reaction a) By combining balanced half-ionic equation (1) (2) Steps of writing balanced half-ionic equation Combining half-ionic equations B. C. b) By the change in oxidation no. 2. Faraday and mole 3. Calculation of mass liberated in electrolysis Electrochemical cells 1. Measurement of e.m.f. a) Potentiometer b) High impedance voltmeter / Digital multimeter 2. Use of salt bridge a) Requirement of a salt bridge b) An electrochemical cell does not need salt bridge 3. Cell diagrams (IUPAC conventions) Electrode potentials 1. Standard hydrogen electrode 2. Relative electrode potential (Standard reduction potential) a) Meaning of sign and value 3. Electrochemical series 4. Use of electrode potential a) Comparing the oxidizing and reducing power b) Calculation of e.m.f. of a cell c) Prediction of the feasibility of redox reaction (1) Disproportionation reaction D. E. Secondary cell and fuel cell 1. Lead-acid accumulator 2. Hydrogen-oxygen fuel cell Corrosion of iron and its prevention 1. Electrochemical process of rusting 2. Prevention of rusting a) Coating b) Sacrificial protection c) Alloying Phase Equilibria I. One component system A. Water 1. Phase diagram of water 2. Isotherm of water 3. PVT surface of water Carbon dioxide 1. Phase diagram of carbon dioxide 2. Isotherm of carbon dioxide Ideal system 1. Raoult's law 2. Composition of the liquid mixture comparing with composition of the vapour 3. Boiling point-composition diagram a) Definition of boiling point b) Boiling point-composition diagram c) Fractional distillation Non-ideal solution 1. Deviation from Raoult's law a) Positive deviation b) Negative deviation c) Enthalpy change of mixing 2. Elevation of boiling point and depression of freezing point by an involatile solute 3. Boiling point of two immiscible liquid a) Steam distillation Partition of a solute between two phases 1. Solvent extraction a) Calculation 2. Paper chromatography a) Rf value b) Separation of amino acids B. II. Two components system A. B. III. Three components system A. Atomic Structure I. Atomic structure A. B. Different model of atomic structure Isotopic mass and relative atomic mass 1. Mass spectrometer a) Determination of atomic mass b) Determination of molecular mass Radioactivity of nuclide 1. Nature of radioactivity a) Penetrating power of radiation 2. Radioactive decay a) Half life b) Use of radioactive isotope Emission spectrum of hydrogen 1. Origin of emission and absorption spectrum 2. Pattern of energy levels of hydrogen 3. Determination of the ionization energy a) Through the convergence limit of Lyman series b) Through Rydberg constant Detail electronic configuration 1. Definition of ionization energy 2. Successive ionization energy 3. Periodicity of First ionization energy a) First I.E. across periods b) First I.E. down a group 4. Quantum numbers a) Energy of electrons in different orbitals 5. Building up of electrons in an atom a) Aufbau principle b) Pauli exclusion principle c) Hund's rule d) Building up of electrons e) Detail electronic configuration of some atoms 6. Electronic configuration of first 36 elements a) Extra stability of half-filled and full-filled subshell 7. Position of s-, p-, d- and f-block elements Shape of s, p and d orbitals Variation in atomic radius Variation in electronegativity C. II. Electronic structure of atoms A. B. III. IV. Atomic orbitals A. A. B. Periodic Table I. Atomic structure Unit 1 Page 1 Topic Reference Reading I. Atomic Structure 1.1, 1.3 Chemistry in Context, 3rd Edition, ELBS pg. 2–4, 57–58 Physical Chemistry, 4th Edition, Fillans pg. 3–5, 10–17 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 6–12 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 1, 2,–4, 54–62 Different model of atomic structure Mass spectrometer I. Atomic Structure Unit 1 Syllabus Notes A. Different model of atomic structure Dalton’s model Thomson’s model Rutherford’s model Bohr’s model Wave model Orbital model Dalton’s model (1808) Atom is a solid, indivisible sphere. Thomson’s model (1899) Atom is made of electrons evenly spread throughout positive mass. Rutherford’s model (1911) Atom is made of electrons and nucleus where most of the space is empty. Bohr’s model (1913) The electron is moving around the nucleus in certain circular orbits. Wave model (1924) The electron is moving around the nucleus in a wave like motion. Orbital model (1927) The exact position of the electron cannot be predicted or measured, an electron density cloud is used to describe the probability of finding an electron in certain region. N.B. Model is just as good as what you want it to explain. B. Isotopic mass and relative atomic mass Isotopic mass – The relative isotopic mass of a particular isotope of an element is the mass of one atom of that isotope on the Carbon-12 scale. (a mass has no unit). Relative atomic mass – The relative atomic mass of an element is the weighted average of the isotopic masses of its natural isotopes on the Carbon-12 scale. Carbon-12 scale – A scale in which Carbon-12 isotope is chosen as the standard and 12 units is assigned to it by definition. The masses of other atoms are determined by comparing with this standard. I. Atomic structure 1. Mass Spectrometer Unit 1 Page 2 A device in which particles are ionized and the accelerated ions are separated according to their mass to charge (m/e) ratio. Construction of Mass Spectrometer Heater – to vaporize the sample, either element or molecule. Ionization chamber – gas molecules are bombarded with fast speed electrons to form positive ions. M(g) + e- (fast) → M+(g) + 2e- Accelerating plate / Negatively charged plate – to accelerate the beam of positive ions across the electric field. Magnet / Magnetic field – to deflect the ion beam, the magnetic field is adjusted so that ions of a particular mass are focused onto the ion-detector Detector – detect the signal and pass it on to a recorder. Vacuum pump – to evacuate the apparatus. This allow the positive ions to travel in straight lines without collision and prevents presence of other substances in the spectrometer that will ionize and affect the result Knowing the magnetic field strength(B), Accelerating voltage(V) and the radius of the path(r), the mass of the ion can be calculated. (Calculation not required.) Important uses of Mass Spectrometer 1. Measurement of isotopic masses of an element. 2. Measurement of molecular mass. 3. Identification of compounds - the various m/e signals and peak heights are processed by computer. 4. Determination of ionization energies. Mass Spectrum A spectrum is obtained with a mass spectrometer, in which a beam of ion is arranged in order of increasing mass to charge (m/e) ratio. The detector current is directly proportional to the relative abundance of the ions reaching the detector. So by using a mass spectrum of a sample of an element, the relative atomic mass of the element can be determined. I. Atomic structure Unit 1 Page 3 a) Determination of relative atomic mass from mass spectrum (Not to scale) The four isotopes of lead are lead-204, lead-206, lead-207 and lead-208. The height of the peak at 204 = 0.07" The height of the peak at 206 = 1.81" The height of the peak at 207 = 1.66" The height of the peak at 208 = 2.81" Total height of the four peaks = 0.07" +1.81" + 1.66" + 2.81" = 6.35" Since, the height of the peak ∝ the relative abundance The relative atomic mass of lead = 204 × 0.07" 1.81" 1.66" 2.81" + 206 × + 207 × + 208 × 6.35" 6.35" 6.35" 6.35" = 207(207.19) b) Determination of molecular mass Another important use of mass spectrometer is to determine the mass of a molecule. Similar to an atom, a molecule will be ionized upon bombardment with fast moving electrons. Once a molecule is ionized, it becomes energetic and unstable. It tends to break up into several fragments. Since the fragment always has a smaller mass than the original molecule, the peak on the mass spectrum with the highest (m/e) ratio corresponds to the original molecule. For example, the following is the spectrum of a sample of methane CH4. Ionization Fragmentation Fragment m/e 12 CH4 + e- → CH4+ + 2eCH4+ → CH3+ + H and CH4+ 16 12 CH3+ → CH2+ + H etc. 12 CH3+ 15 12 CH2+ 14 12 CH+ 13 C+ 12 Besides the peak with m/e ratio 16, a very small peak (about 1% of that of 16) with m/e ratio 17 is found. It is caused by the existence of trace amount of C-13 isotope in the nature. I. Atomic structure Carbon-12 scale Unit 1 Page 4 After the invention of mass spectrometer, structure of numerous number of organic molecules have been studied. Because all organic compound contains carbon, it would be more convenient if carbon is used as the standard of reference in mass spectrum. As a result, in 1961, C-12 was chosen as the standard for atomic mass and carbon-12 scale is established. In carbon-12 scale, by definition, 12 g of C-12 contains 1 mole of C-12 atoms. Since the definition is solely arbitrary, there is no wonder why Avogadro number is not a whole number. Indeed the experimental determination of the more precise value of Avogadro number is still the interest of many scientists. Glossary orbital – Loosely speaking, it is used to describe the geometrical figure which describes the most probable location of an electron. More accurately, it is used to denote the allowed energy level for electrons. Carbon–12 scale (relative) isotopic mass (relative) atomic mass Past Paper Question 90 2A 3 a 96 1A 1 a i ii iii 98 2A 4 c 99 2A 1 a i 90 2A 3 a 3a Give an account of the use of a mass spectrometer for determining the relative masses of particles. The modern method of determining atomic masses uses the mass spectrometer. 6 C 2 marks the sample (an element) is bombarded by electrons to form positively charged ions. the ions are accelerated by the electric field between plates A and B. the ions travel along until deflected by the magnetic field the magnetic field is adjusted so that ions of a particular mass are focused into the ion-detector knowing the magnetic field strength, accelerating voltage and the radius of the path, the atomic mass can be calculated. - the mass spectrometer responds to the mass-to-charge (m/e) of the ionized species 4 marks Many candidates did not point out that the mass spectrometer responds to the mass to charge ratio. Some candidates did not know the positively charged ions are accelerated by the electric field. I. Atomic structure Unit 1 96 1A 1 a i ii iii 1a i Write down the number of neutrons, protons and electrons in one atom of carbon-12, 12C, and in one atom of carbon-13, 13C . 12 C 6n, 6p, 6e ½ mark 13 C 7n, 6p, 6e ½ mark ii The isotopic mass of 12C is 12.000 atomic mass unit (a.m.u.). Calculate the mass, in kg, of 1 mole of 12C atoms. (1 a.m.u. = 1.6605 × 10-27 kg; Avogadro constant, L = 6.0221 × 1023 mol-1) Mass of 1 mole of 12C = 12.000 × 1.6605 × 10-27 × 6.0221 × 1023 1 mark = (0.0119996) = 0.0120 kg 1 mark (Accept answers which could round off to 0.012) iii The following data were obtained from the mass spectrum of a carbon-containing compound : Ion Mass / a.m.u. Relative intensity 12 + C 12.000 100.00 13 + C 13.003 1.12 Using the above data, calculate the relative atomic mass of carbon. 12.000 × 100.00 + 13.003 × 1.12 1 mark relative atomic mass = (100.00 + 1.12) 1214.57 = 101.12 = 12.011 1 mark (Accept answers which could round off to 12.01) (Deduct ½ mark if unit is given) C Well answered in general, though some candidates did not give the unit, kg, as required. Many candidates wrongly assigned the unit a.m.u. to the relative atomic mass of carbon. 98 2A 4 c 4c Sketch the expected mass spectrum for a gas sample having the composition : N2 78%, O2 21% and CO2 1%. (You only need to consider the major isotope of each element.) 99 2A 1 a i 1a i Rubidium occurs naturally in two isotopic forms. The table below lists the mass and relative abundance of each isotope. Isotope Mass / a.m.u. Relative abundance 85 Rb 84.939 72.15% 87 Rb 86.937 27.85% (I) Suggest an experimental method to detect the isotopes of rubidium and state how the relative abundance of each isotope can be obtained. (II) Calculate the relative atomic mass of rubidium. Page 5 1 2 2 3 I. Atomic structure Unit 2 Page 1 Topic Reference Reading I. Atomic Structure 1.2 Chemistry in Context, 3rd Edition ELBS pg. 83–89, 93–95 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 13–20 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 76–81 Nature of radioactivity Radioactive decay C. Radioactivity of nuclide Unit 2 Syllabus Notes 1. Nature of radioactivity There are totally 109 elements, but owing to the presence of the isotope, there are over 1600 different nuclei. Only 238 nuclei are stable, all others exhibit some kind of radioactivity. They eject proton, neutron and electron or emit strong electromagnetic radiation. The process is called radioactive decay and the elements are said to be radioactive. a) Penetrating power of radiation There are 3 types of radioactivity. Emission α particle β particle γ radiation Nature He nuclei (2p + 2n) fast moving electrons electromagnetic wave Relative velocity 1/20 velocity of light 3 – 99% velocity of light 100% velocity of light Range in air a few cm a few m a few km Relative penetrating power 1 100 10000 1 1 Penetrating power is ∝ mass ; ∝ charge ; ∝ velocity Emission α particle β particle γ radiation Relative mass 4 units 1 / 1840 unit – Relative charge +2 -1 0 Relative effect of electric and magnetic field small deflection large deflection no deflection Effect of magnetic field Effect of electric field (The magnetic field is going into the paper.) 1 The deflection is ∝ charge and ∝ mass All of the three kinds of radioactivity have the ability to ionize the air, so a Geiger-counter can be used to measure the activity of the radioactive material. I. Atomic structure 2. Radioactive decay Unit 2 Page 2 In a nuclear reaction, unlike ordinary chemical reaction, nuclei are involved instead of electrons. In the reaction, 1. one element can be changed to another element by radioactive decay, atomic fission or atomic fusion. 2. a large amount of energy is released in the process, the energy is from the binding energy in the nucleus. The process in which an element changes to another element in nuclear reaction is known as transmutation. Alpha Decay This is a decay common to most of the radioactive isotopes of elements with Z > 82 (Pb). (high p: n ratio) e.g. a) b) c) 238 92 226 88 224 88 U → 4 He + 234 Th 2 90 4 222 Ra → 2 He + 86 Rn 220 Ra → 4 He + 86 Rn 2 When an element undergoes alpha decay, its mass no. will be reduced by 4 and the atomic no. will be reduced by 2. Note: i) the charge of the particle, e.g. the charge on He nucleus, is usually omitted for simplification. ii) the mass number and the atomic number in all nuclear reaction must be conserved. Beta Decay This is a decay common to most radioactive isotopes of elements with Z < 82. (low p: n ratio) e.g. a) b) c) 14 6 108 47 1 0 C → 0 1 e + 14 N − 7 Ag → 0 1 e + 108 Cd − 48 0 1 n → −1 e + 1 H When an element undergoes beta decay, its mass no. will remain unchanged and the atomic no. will increase by 1. Gamma Radiation Unlike alpha and beta decay, an element emitting gamma radiation will not change to another element. It will only change from a high energy(excited) state to an low(stable) energy state. e.g. a) 234 90 Th* → γ–ray + 234 90 Th Most of the heavy isotope will undergo multiple decays to form a series. The U-238 isotope can decay through multiple stages to a stable Pb-206 isotope I. Atomic structure a) Half life of radioactive decay Unit 2 Page 3 Half life - the time taken by a given amount of radioactive isotope to decay to half of the original amount. e.g. Radioactive Isotopes Uranium-238 Carbon-14 Polonium-214 Half-life 4.5 × 109 years 5.7 × 103 years 1.5 × 10-4 seconds The stability of an isotope is directly proportional to its half life. For example, if there are 100 Uranium-238 isotopes, it will take 4.5 x 109 years to decay to half of the original amount (50 Uranium-238 isotopes). The rate of radioactive decay, unlike rate of chemical reaction, only depends on the amount of isotopes present. The changes in external factors such as temperature and pressure have no effect on it. Rate of radioactive decay The rate of the decay decreases as the decay proceeds because the rate of decay is directly proportional to the amount of isotope remains. This forms an exponential decay curve. After 5 half-lives, the fraction remaining will be = ½×½×½×½×½ = 0.03125 = 3.125% I. Atomic structure Artificial Transmutation of elements Unit 2 Page 4 Artificial transmutation reactions are usually achieved by bombarding an element with a stream of particles, especially neutrons, protons, alpha particles and other small nuclei. e.g. Uranium-238 can be transmutated to Uranium-239 by bombarded with an neutron. i.e. 238 92 U + 1 n → 239 U 0 92 239 92 239 93 The unstable Uranium-239 then decays in two stages by emitting β-particles. i.e. a) b) U → 239 Np + 0 1 e 93 − 239 Np → 94 Pu + 0 1 e − 14 7 14 6 N in air, is transmutated to 14 1 14 1 i.e. 7 N + 0 n → 6 C + 1 H In nature the, e.g. a) b) c) b) 1. 2. 3. 4. 5. 6. 27 13 27 13 27 13 24 Al + 1 n → 11 Na + 4 He 0 2 27 Al + 1 n → 12 Mg + 1 p 0 1 28 Al + 1 n → 13 Al + γ-ray 0 C by the neutron from the cosmic ray (fast moving neutron). In some cases, if the speeds of intruding neutrons are different, it will cause different modes of nuclear reaction. Use of radioactive isotope Treatment of cancer Study of metabolic pathways. Thickness gauges and empty packet detection. Archaeological and geological dating. Sterilizing surgical instruments. Detection of leakage. (e.g. water pipes) Glossary nuclide - A type of atom specified by its atomic number, atomic mass, and energy state, such as carbon 14. penetrating power radioactive decay half-life I. Atomic structure Unit 2 Page 5 Past Paper Question 91 2A 1 a i ii 93 1A 1 a i ii 95 2A 2 a i 99 1A 2 b i ii 91 2A 1 a i ii 1a The radioactive decay of i 238 92 U may be represented by 238 92 U→ 206 82 Pb + α + β. ii C State the nature of α and β particles and compare their penetrating power. α particles are bare helium nucleus with two positive charge and four units of mass. β particles are fast electrons, which is negatively charged. The penetrating power of β particles is 100 times stronger than α particles. Balance the above equation. 238 206 4 0 92 U → 82 Pb + 8 2 He + 6 −1 e Many candidates incorrectly stated that α particles are helium atoms. Some candidates did not emphasize that the electrons in β particles are moving very fast. 3 1 mark 1 mark 1 mark 2 2 marks 93 1A 1 a i ii 1a Name the particles X and Y in the following nuclear reactions, and gives their charge and mass. Compare the action of a magnetic field upon the paths of X and Y 229 225 i X is _______________ 90 Th → X + 88 Ra ii X is / helium nucleus / α particle 225 225 Y is _______________ 88 Ra → Y + 89 Ac Y is / fast moving electron / β particle. The path of α and β particles are deflected/bent by a magnetic field but they are deflected in opposite direction. OR by a diagram 1 mark 1 mark 1 mark 1 mark 4 The magnetic field is going into the paper. Bent motion Correct direction 95 2A 2 a i 228 2a Ra decays by the emission of β particles. The half-life for the decay is 6.67 years. 88 1 mark 1 mark i C Write a balanced equation to represent the decay of 228 88 228 88 Ra . 1 mark 1 Ra → 228 89 Ac + 0 −1 e Many candidates did not make use of the Periodic Table given in the question paper to find the symbol (Ac) for the product of the decay, and hence they arrived at incorrect answers. 99 1A 2 b i ii 2b Radioactive decay of radium (Ra) in rocks and soil produces radioisotopes of radon (Rn). One of the isotopes 222 86 Rn decays to give 222 86 218 84 Po with a half-life of 3.82 days. 222 86 i ii Write the nuclear equation for the decay of Explain why Rn . Rn is considered hazardous to health. II. Electronic Structure of atoms Unit 1 Page 1 Topic Reference Reading II. Electronic Structure of atoms 2.1 Chemistry in Context, 3rd Edition ELBS pg. 69–71 Physical Chemistry, 4th Edition, Fillans pg. 63–74 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 31–36 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 64–68 Emission spectrum of hydrogen Determination of ionization energy II. Electronic Structure of atoms A. Emission spectrum of hydrogen Unit 1 Syllabus Notes Emission Spectroscopy White light from tungsten lamp consists of a continuous range of wavelengths of visible light. The white light can be separated into different colour light by a prism and forms a continuous spectrum of visible light, like a rainbow. But some lights, for example, the light from the sodium lamp or green light from the flame test of copper, are not composed of a continuous range of wavelengths, instead they are only mixtures of a limited number of wavelengths. In this case, the spectrum is known as a line spectrum. 1. Origin of emission Emission spectrum i. At room temperature, nearly all the atoms in a given sample of substance are in the ground state, the electrons occupy the orbital of lowest energy. The atom is in the ground state because all electrons in the atoms are having the lowest possible energy. ii. Energy is given to the atoms by heating the element with Bunsen flame or hitting with electrons in discharge tube. The electrons takes up the energy and are promoted to a higher energy state, excited state. iii. Since the electrons in excited state are not stable, they tends to fall back to a lower energy state to dissipate their extra energy in form of radiation. Each transition involves a definite amount of energy, called a quantum. iv. Different transition has different energy and each transition corresponds to a single line in the emission spectrum. The greater the number of electrons making a particular transition, the more intense the spectral line is. This gives coloured emission lines on dark background. v. The relationship between the wavelength(λ) and frequency(υ) of any electromagnetic radiation is given by c where c is the velocity of light λ= υ vi. And, the relationship between energy of radiation is given by the equation ∆E = hυ where h is Planck’s constant and υ is frequency II. Electronic Structure of atoms 2. Unit 1 Page 2 Pattern of energy levels of hydrogen Balmer series i. The emission line spectrum of hydrogen atom consists of many lines, some of them fall in the visible region and some of them fall in the invisible region. ii. The visible part of the spectrum was first studied by the scientist Balmer. This series of spectrum lines was named after him as Balmer series. iii. The red line, the line with longest wavelength in the visible region, corresponds to the transition of an electron from the third energy level (n=3) to the second energy level (n=2). It gives out a quantum of light, called photon. With increasing frequencies, the lines get closer and closer until they merge, this is because the difference between energy levels gets smaller as the frequency increases. Relationship between emission spectrum and energy levels iv. The point, where the spectral lines merge, corresponds to the electron transition between energy levels with n = ∞ and n = 2. v. At n = ∞, the electron has effectively left the atom. That means the atom has been ionized. II. Electronic Structure of atoms Unit 1 Page 3 Complete Hydrogen Emission Spectrum i. Besides Balmer series, the complete spectrum is composed of many series. The origin of other series is similar to that of Balmer series, the only difference is the destination of the electrons. ii. Bohr obtained an expression for the frequencies of the lines in the spectrum. 1 1 υ = cR H ( 2 − 2 ) where RH is Rydberg constant n1 n 2 c is velocity of light n1 and n2 are integers, called principal quantum number, with n2 greater than n1. n1 : the principal quantum number of the energy level to which the electron returns. n2 : the principal quantum number of the energy level from which the electron leaves. II. Electronic Structure of atoms 3. Unit 1 Page 4 Determination of the ionization energy The ionization energy of hydrogen can be determined by measuring the frequency the convergence limit (υ∞→1) of Lyman series or by determining the Rydberg constant through the frequency of a specific line. a) Through the convergence limit of Lyman series Ionization of hydrogen = ∆E = h υ∞→1 However, it is quite difficult to determine the convergence limit accurately as the lines are getting close together when approaching the convergence limit and the spectrum becomes a bright continum beyond the limit. Therefore, the limit is usually determined graphically. In Lyman series, there is a convergence limit in the line spectrum. It represents the point where the difference in energy levels diminished. ii. The frequencies of the first lines in Lyman series are measured. The differences in frequencies between two adjacent lines are calculated. e.g. ∆υ1 = (υ2 - υ1), ∆υ2 = (υ3 - υ2) ...iii. A graph of υ against ∆υ is plotted. iv. The curve is extrapolated, and the υ at y-intercept, where ∆υ = 0, will correspond to the transition of electron from n = ∞ to n = 1. v. The Ionization energy can be calculated by I.E. = hυ i. b) Through Rydberg constant For example, the Rydberg constant can be determined by measuring the frequency of the first line of Lyman series (υ2→1). Thereafter, by applying the Ryberg equation, the ionization energy of hydrogen can be calculated. i. The Rydberg constant is calculated from the first line in the Lyman series. υ2→1 = cRH ( 2 − υ∞→1 = cRH ( 2 − 1 1 1 1 1 ) 22 1 ) ∞2 ii. The frequency corresponding to the ionization process is calculated by iii. The ionization energy(I.E.), the energy transition from the ground state and the continuum for one electron is Ionization energy(I.E.) = h υ∞→1 = hcRH where L is the Avogadro's constant The ionization energy per mole = LhcRH Glossary emission spectrum Paschen series absorption spectrum Rydberg equation Planck’s constant Balmer series Lyman series II. Electronic Structure of atoms Unit 1 Page 5 Past Paper Question 90 2A 2 c 95 1A 1 b i ii 97 2A 2 a 99 1A 2 a i ii 90 2A 2 c 2c Give a brief account for the Balmer series in the spectrum of the hydrogen atom. It is a series of lines resulting from photons emitted or absorbed by the hydrogen atom with wavelength in the visible region. These spectral lines are the result of transitions of the electron from one energy level to another 2 marks with the lower energy level with quantum number n1 = 2. 1 1 1 The wavelengths of this series satisfies the formula, = RH (22 - n 2) 1 mark λ 2 where RH is the Rydberg constant n2 = 3,4,5 ... λ = wavelength 1 mark C Many candidates gave the correct formula for the relationship between wavelength and quantum numbers, but only a few defined their symbols. 95 1A 1 b i ii 1b i What can you deduce from the fact that the spectral lines in the atomic emission spectrum of hydrogen are not equally spaced ? The energy differences between electron shells in a hydrogen atom are not the same (converge). 1 mark ii In the atomic emission spectrum of hydrogen, the convergence limit for the Lyman series occurs at 3.275 x 1015 Hz. Calculate the ionization energy of hydrogen, in kJ mol-1 (Planck constant, h = 6.626 x 10-34 J s; Avogadro constant, L = 6.023 x 1023 mol-1) I.E. = hυ 1 mark = (6.26 × 10-34) × (3.275 × 1015 × 6.023 x 1023) 1 mark = 1.307 × 106 Jmol-1 / 1307 kJmol-1 C Many candidates were unable to point out that the energy differences between two energy levels are not the same and hence the spectral lines in the emission spectrum are not equally spaced. Many candidates neglected to give the answer to (ii) in kJmol-1. 97 2A 2 a 2a Describe and account for the characteristics of the emission spectrum of atomic hydrogen. 99 1A 2 a i ii 2a i Outline the origin of atomic emission. ii Briefly explain why each series of atomic emission lines of hydrogen converges at the lower wavelength end. (3 marks) 4 1 2 7 II. Electronic Structure of atoms Unit 2 Page 1 Topic Reference Reading II. Electronic Structure of atoms Unit 2 2.2–2.3 Modern Physical Chemistry ELBS pg. 26–33, 39–41 Chemistry in Context, 3rd Edition ELBS pg. 71–78 Physical Chemistry, 4th Edition, Fillans pg. 74–77, 80–84, 88–95 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 37–42, 49–53 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 38–42, 64, 68–71, 73–74, 170, 173–175 Detail electronic configuration B. Detail electronic configuration 1. Definition of first ionization energy Syllabus Notes X(g) → X+(g) + eFirst ionization energy – Energy required to remove an electron from a free atom in gaseous state. Normally measured in electron volts (eV) or kJmol-1. (1eV = charge of electron × 1V = 1.60 × 10-19 C × 1V = 1.60 × 10-19 J) 2. Successive ionization energy (presence of electron shells) The information from successive ionization energy can be used to deduce the electronic configuration of an atom. The ionization of Potassium (K) can take place in steps 1st ionization K(g) → K+(g) + e2nd ionization K+(g) → K2+(g) + e3rd ionization K2+(g) → K3+(g) + ei. ∆HI1 = 419 kJmol-1 ∆HI2 = 3051 kJmol-1 ∆HI3 = 4412 kJmol-1 The value of successive ionization energy increases as more electrons are removed from the atom. This is because extra energy is needed to removed an electron from an ion of increasing positive charge. ii. The 2nd I.E. is much higher than 1st I.E. It can be observed from the graph that the I.E.s of K atom can be divided into 4 groups. 1 in the first group, 8 in the second group, 8 in the third group and remaining 2 in the fourth. iii. This indicates that a K atom has four electron shells and electronic configuration of 2,8,8,1. II. Electronic Structure of atoms 3. Unit 2 Page 2 Periodicity of First ionization energy (presence of electron sub-shell) a) First I.E. across periods First ionisation Energy of the first 38 elements i. The 1st I.E. increases roughly across a period due to the increase in nuclear charge. ii. Besides the main trends, some irregularities are found within a period. This suggests the presence of subshells (s, p, d and f sub-shells) within an electron shell (n = 1, 2, 3 ...) iii. In period 2 and 3, the increasing trends can further be broken up into 3 parts, in a 2-3-3 manner. This suggest the presence of subshells. iv. In period 4, from K to Kr, the trend can be broken into 4 parts, in a 2-10-3-3 manner. This suggest the presence of another set of subshells. b) First I.E. down a group 53 0 510 Li 170 0 16 0 0 F 2400 2200 He Ne First I.E. / kJmol-1 490 4 70 4 50 430 4 10 390 3 70 3 50 First I.E. / kJmol-1 150 0 14 0 0 13 0 0 12 0 0 110 0 10 0 0 900 Cl Br I First I.E. / kJmol-1 Na 2000 18 0 0 16 0 0 14 0 0 12 0 0 10 0 0 K Rb Cs Ar Kr Xe 1st I.E. of group I i. 1st I.E. of group VI 1st I.E. of group 0 On down a group, the 1st I.E. decreases as the number of electron shells or the size of the atom increases. The attraction between the outermost electron and the nucleus decreases with increasing atomic size. II. Electronic Structure of atoms Unit 2 Page 3 A prevailing explanation for these variation in ionization energy is that different electrons occupy different orbitals in an atom. Since different orbitals have different energies, the electron in different orbital shows difference in energy. A more energetic electron would be removed more readily and has a lower ionization energy. 4. Quantum numbers To facilitate the discussion of the detail electronic configuration of an atom, a set of four quantum numbers is assigned to each electron in an atom. Quantum numbers is the I.D. no of the electron in an atom. The four quantum numbers are Principal quantum number (n) Subsidiary quantum number (l) Magnetic quantum number (m) Spin quantum number (s) specify the principal energy level. specify the shape of the orbital. specify the direction of the orbital. specify the spin of the electron. Allowed value Any natural number Depends on n, in the range of 0, 1, 2 .. (n-1) Depends on l, in the range of -l to +l Either +½ or -½ Electrons with different subsidiary quantum numbers can be denoted by the letters s, p, d and f, or s orbital, p orbital, d orbital and f orbital. l = 0, s 1, p 2, d 3 f a) Energy of electrons in different orbitals i. ii. iii. iv. The principal quantum no. (n) determines the principal energy level of the orbital. With the same principle quantum no. (n), the energy of orbitals increases in following order, s < p < d < f. The energy of orbitals is independent of magnetic quantum number (m) and spin quantum number (s). The orbitals are labelled according to the values of n, l and m. e.g. The orbitals with n = 2, l = 1 and m = 1, 0 and -1 are called 2px, 2py and 2pz orbitals. II. Electronic Structure of atoms 5. Unit 2 Page 4 Building up of electrons in an atom a) Aufbau principle In building up the electronic configuration of an atom in its ground state, the electrons are placed in the orbitals in order of increasing energy b) Pauli exclusion principle – c) In any atom, no two electrons can have all four quantum numbers the same. Hund’s rule (Rule of maximum multiplicity) Electrons in degenerate orbitals will occupy each orbital singly before electron pairing takes place. To generalize the three rules, the following procedure of filling the orbitals shown be observed i. Of the available orbitals, the added electron will always occupy the one with the lowest energy. (Aufbau principle) ii. Each orbital may hold only two electrons, and they must have opposite spin. (Pauli exclusion principle) iii. Where a number of orbitals of equal energy are available, the added electron will go into empty orbital, keeping electron spins the same, before spin-pairing occurs. (Hund's rule) According to the Aufbau principle, the electrons will fill the available orbitals with lowest energy first. The order of energy levels need not to be exactly parallel to the principal quantum. When the atom becomes heavy, the high energy levels tend to overlap each other. For example, the energy of 4s orbital is lower than that of 3d orbital. II. Electronic Structure of atoms Unit 2 Page 5 The order of energy of the orbitals can be predicted by the pattern shown on the left. The order of the energy levels mainly depends on the attraction between the orbital and the nucleus. The stronger the attraction, the lower will be the energy of the orbital. (because of lower potential energy) This once again depends on the spatial distribution of the orbital. For example, 3s orbital has lower energy than 3p orbital because i. Distance from the nucleus 3s orbital is closer to the nucleus than 3p orbital, thus the 3s orbital will experience more attraction. ii. Penetration of 3s orbital Spatial distribution of 3s orbital has 2 small peaks close to the nucleus and make it experience more attraction. Spatial distribution of 3s and 3p orbitals Moreover, from the fact that 4s orbital is less energetic than 3p orbital. The effect of penetration of penetration must be more significant than the effect of distance and shielding effect. iii. Shielding / Screening effect Besides attraction with the nucleus, 3p electrons also experience repulsion will Kshell, L-shell and 3s electrons. This is known as shielding / screening effect of inner electrons. Note : The completely filled shell can be denoted by K, L, M, N according to the principal quantum no Shielding effect and effective nuclear charge Owing to shielding effect, the electrons in an atom experience attraction from the nucleus less than the actual nuclear charge. The effective nuclear charge is always less than actual nuclear charge. There is two kinds of shielding effect Primary shielding effect This is stronger than the secondary shielding effect. This is the shielding effect of the inner electrons on the outer electrons from the nucleus. i.e. electrons in shells of lower principal quantum number are more effective shields than electrons in shells of higher quantum number and s-orbital electrons is more effective in shielding than p orbital electrons. Secondary shielding effect The shielding effect of the electrons/repulsion between electrons in the same orbital. This is much weaker than primary shielding effect. II. Electronic Structure of atoms d) Building up of electrons i. Unit 2 Page 6 The quantum no. of the electron in hydrogen atom in ground state l = 0, m = 0, s = +½ n = 1, 1s 1 electron in box 1s1 Electron with spin quantum no. +½ is represented by an upward arrow. ii. The quantum no. of the two electrons in helium atom in ground state n = 1, n = 1, 1s l = 0, l = 0, m = 0, s = +½ m = 0, s = -½ 2 electron in box 1s2 iii. The quantum no. of the three electrons in lithium atom in ground state n = 1, n = 1, n = 2, 1s 2s 1s22s1 l = 0, l = 0, l = 0, m = 0, s = +½ m = 0, s = -½ m = 0, s = +½ 00 n = 1, n = 1, n = 2, n = 2, n = 2, n = 2, 1s 2s l = 0, l = 0, l = 0, l = 0, l = 1, l = 1, 2p m = 0, s = +½ m = 0, s = -½ m = 0, s = +½ m = 0, s = -½ m = 1, s = +½ m = 0, s = +½ iv. The quantum no. of the six electrons in carbon atom in ground state 0 0 000 e) i. 1s 1s22s22p2 Detail electronic configuration of some atoms The electronic configuration of argon atom in ground state 2s 2p 3s 3p 1s22s22p63s23p6 0 0 000 0 000 II. Electronic Structure of atoms Unit 2 Page 7 ii. The electronic configuration of potassium atom in ground state 1s 2s 2p 3s 3p 4s 1s22s22p63s23p64s1 0 0 000 0 000 0 1s 2s 2p 3s 3p 3d iii. The electronic configuration of scandium atom (Z=21) in ground state 4s 0 0 000 0 000 00000 0 1s22s22p63s23p63d14s2 Note : or [Ar]3d14s2 or K L 3s23p63d14s2 Writing of electronic configuration follows the principal quantum no., not the order of placing electrons iv. The electronic configuration of germanium atom (Z=32) in ground state 1s 2s 2p 3s 3p 3d 4s 4p 0 0 000 0 000 00000 0 000 1s22s22p63s23p63d104s24p2 6. 1 2 3 4 5 6 7 8 9 10 11 12 or [Ar]3d104s24p2 or K L M 4s24p2 Electronic configuration of first 36 elements H He Li Be B C N O F Ne Na Mg 1 s1 1s2 1s22s1 1s22s2 1s22s22p1 1s22s22p2 1s22s22p3 1s22s22p4 1s22s22p5 1s22s22p6 1s22s22p63s1 1s22s22p63s2 13 14 15 16 17 18 19 20 21 22 23 24 Al Si P S Cl Ar K Ca Sc Ti V Cr 1s22s22p63s23p1 1s22s22p63s23p2 1s22s22p63s23p3 1s22s22p63s23p4 1s22s22p63s23p5 1s22s22p63s23p6 1s22s22p63s23p64s1 1s22s22p63s23p64s2 1s22s22p63s23p63d14s2 1s22s22p63s23p63d24s2 1s22s22p63s23p63d34s2 1s22s22p63s23p63d54s1 25 26 27 28 29 30 31 32 33 34 35 36 Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 1s22s22p63s23p63d54s2 1s22s22p63s23p63d64s2 1s22s22p63s23p63d74s2 1s22s22p63s23p63d84s2 1s22s22p63s23p63d104s1 1s22s22p63s23p63d104s2 1s22s22p63s23p63d104s24p1 1s22s22p63s23p63d104s24p2 1s22s22p63s23p63d104s24p3 1s22s22p63s23p63d104s24p4 1s22s22p63s23p63d104s24p5 1s22s22p63s23p63d104s24p6 a) Extra stability of half-filled and full-filled subshell It can be observed that Cr (Z=24) and Cu (Z=29) do not follow the general trend. Instead of [Ar]3d44s2, Cr has the electronic configuration [Ar]3d54s1 Instead of [Ar]3d94s2, Cu has the electronic configuration [Ar]3d104s1 Extra stability is associated with i. half-filled orbital and ii. completely/full-filled orbital. This is because i. a half-filled or full-filled d orbitals mean an even distributed electrons in an atom which make the electron repulsions weaker. ii.4s2 is less favorable than 4s1 as pairing up of the 4s electrons creates greater repulsion between the 2 electrons. This also accounts for the anomaly of first I.E. of Be and N in period 2. Be has a full-filled 2s orbital and N has a half-filled 2p orbital. II. Electronic Structure of atoms Unit 2 Page 8 7. Position of s-, p-, d- and f- block elements According to the electronic structure of elements, the periodic table can be divided into different blocks. s-block elements : Group IA to Group IIA e.g. Na 1s22s22p63s1 p-block elements : Group IIIA to Group VIIA e.g. N 1s22s22p3 d-block elements : Transition metal e.g. Mn 1s22s22p63s23p63d54s2 f-block elements : Lanthanide and Actinide e.g. Pr 1s22s22p63s23p63d104s24p64d104f35s25p6 Across the s-block, electron is being added into the s sub-shell. While moving across, p-, d- and f- block, the electron is added into the p, d and f sub-shell. Glossary First ionization energy electron shell/orbital electron sub-shell/sub-orbital quantum number Aufbau principle Pauli exclusion principle Hund's rule degenerate orbitals – orbitals with the same energy penetration shielding / screening effect effective nuclear charge electron in box half-filled orbital full-filled orbital s-block p-block f-block transition metal 90 2B 4 a 92 2A 1 a ii iii 93 1A 2 a v 95 1A 1 a i 96 1A 1 c i ii 97 2A 2 b i ii iii 98 1A 1 a d-block Past Paper Question 95 2B 5 a iv 98 2B 8 a 95 2B 6 a 90 2B 4 a 4 Account for the following observations. 4a The first ionization energies of the second row elements are in the order Li < B < Be < C < O < N < F < Ne. Increase in effective nuclear charge with full explanation, i.e. no. of protons increase 1 mark From Li to Ne, electrons are added in the same principal shell, the shielding effect involved is only a secondary one 1 mark filled 2s orbital; stability 1 mark Extra stability, Be > B 1s22s2 half-filled electronic configurations; half-filled stability 1 mark N > O 1s22s22p3 C Some candidates failed to mention or explain the abnormally high ionization potentials of Be and N. 4 II. Electronic Structure of atoms Unit 2 Page 9 92 2A 1 a ii iii 1a ii Arrange the following elements in order of increasing first ionization energy : F, Ne and Na. Explain your order. 2½ Na < F < Ne 1 mark Na has the least I.E. because there is a single 3s-electron which is not strongly attracted towards the nucleus and losing it would accomplish the completed K- and L- shell status. It is for this reason that an electron in Ne requires the highest I.E. To lose an electron means that the F atom would move further away from the K- and Lshell status and this requires a relatively high I.E. 1½ mark iii Describe and explain the variation of the successive ionization energies of potassium. 3 There is a gradual increase in the successive ionization energies in general because the nucleus becomes effectively more positively-charged as each electron is being lost. The increase is greatest when the 2nd electron are being ionized, suggesting the 2nd electron is closer to the nucleus than the first. Similar greater but less abrupt I.E.’s occur when the 10th and 18th electrons are being ionized. 3 marks 93 1A 2 a v 2a Answer the following questions by choosing in each case one of the species listed below, putting it in the box and giving the relevant equation(s). Al(s), AlCl3(s), AlO2-(aq), Na(s), CO32-(aq), Cu2+(aq), P4O10(s), S(s), S2O32-(aq), Zn2+(aq) v Which species has the following ionization energies: 1st 500; 2nd 4600; 3rd 6900; 4th 9500 kJmol-1? Na(s) 1 mark 1 95 1A 1 a i 1a i The first four successive ionization energies of an element A are 578, 1817, 2746 and 10813 kJ mol-1 respectively. 1 To which group in the Periodic Table does A belong ? Group III / IIIA / 3 / 3A 1 mark 95 2B 5 a iv 5a The table below lists some properties of the alkali metals. Standard electrode Melting point / ºC Element Atomic Ionic radius / nm First ionization potential / V radius / nm energy / kJmol-1 Li 0.123 0.060 520 -3.04 180 Na 0.157 0.095 495 -2.71 98 K 0.203 0.133 418 -2.92 64 Rb 0.216 0.148 403 -2.93 39 Cs 0.235 0.169 374 -2.95 29 iv Explain the trend in the first ionization energy from Li to Cs. The atomic radius increases as the group is descended hence the outer-shell electrons become less strongly held 2 marks and it is easier to remove an e- from the gaseous atom of the element. C iv the effective attractive forces on the electron is greatest in the case of the smallest first atom. 95 2B 6 a 6a Write the electronic configuration of a titanium atom in its ground state. 1s22s22p63s23p63d24s2 3d 4s [Ar] 11000 2 (accept any correct form of electronic configuration) 1 mark 2 II. Electronic Structure of atoms Unit 2 Page 10 96 1A 1 c i ii 1c Explain why i the first ionization energy of oxygen is greater than that of sulphur; 1 S has 1 more e- shell / has a large atomic size than O, ½ mark ½ mark the attraction experienced by the outermost electron is smaller hence S has a smaller I.E. (DO NOT accept explanations in terms of effective nuclear charge; deduct ½ mark for any misconception) C The question was poorly answered. Many candidates incorrectly attributed the difference in ionization energies to effective nuclear charge and/or screening effect. In fact, the effective nuclear charge experienced by the outermost shell electrons of sulphur was greater than that of oxygen. Such answers reflected the fact that many candidates did not fully understand the concepts associated with screening effect and effective nuclear charge. ii the first ionization energy of oxygen is smaller than that of fluorine. 1 F has 1 more proton in the nucleus, ½ mark the effect of increase in nuclear charge has outweighed the screening effect of the additional electron hence F has a greater I.E. ½ mark Or, Effective nuclear charge experienced by outermost e- of atom increases across a period. ∴ outermost e- of O is more easily removed than that of F. 1 mark C Many candidates pointed out that the outermost shell electron of fluorine experiences a greater effective nuclear charge. But they explained their answers rather vaguely. This again showed that they did not grasp these concepts well but just produced their answers by citing some terms learnt in their lessons. Some candidates incorrectly explained the difference in ionization energies between O and F in terms of electronegativity. 97 2A 2 b i ii iii 2b i What is an electron shell in an atom ? ii For each of the electron shells with principal quantum numbers 1, 2 and 3 list the subshells. iii Deduce the maximum number of electrons in each of the electron shells in (ii). 98 1A 1 a 1a Arrange the following chemical species in the order of increasing ionization enthalpy. Explain your arrangement. N+, N and O 98 2B 8 a 8a Which of the following, (I) or (II), is the ground state electronic configuration of chromium atom ? Explain your answer. 3d 4s (I) [Ar] (II) [Ar] 6 3 2 11111 1 11110 2 III. Atomic orbitals Page 1 Topic Reference Reading III. Atomic orbitals Modern Physical Chemistry ELBS pg. 41–42 Physical Chemistry, 4th Edition, Fillans pg. 78–80 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 39–40 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 72 2.4 Shape of atomic orbitals III. Atomic orbitals In the orbital model of atom, the electrons are moving in a wave like motion. Because of its extremely small mass, the position of individual electron cannot be located precisely by any phyical mean, an electron density diagram is therefore used to described the distribution of electrons. Atomic orbital is the region within which there is a high probability, i.e 90%, of finding an electron. Different orbital have different shape, size and energy. A. Shape of s, p and d orbitals s orbital s-orbital is spherical in shape. The electron density at the nucleus is zero. For 1s orbital of hydrogen, the electron density (probability density of finding a electron) increases as the distance from the nucleus increases and reaches a maximum at a distance equal to Bohr’s radius. And then, the electron density decreases as the distance from the nucleus increases. Syllabus Notes p orbital p-orbital is dump bell in shape. The three p orbitals are perpendicular to each other. The three porbitals have the same energy (degenerate). d orbital There are 5 degenerate d-orbitals. dxy, dxz and dyz orbitals lie on xy, xz and yz plate respectively. dx2-y2 orbital lies on xy plate with the lobes are directing to the x and y axes. dz2 orbital looks like pz orbital but with a ring at the centre. Glossary probability density degenerate s orbital p orbital d orbital III. Atomic orbitals Page 2 Past Paper Question 91 2A 3 c 92 1A 2 c 96 1A 1d 91 2A 3 c 3c Give a brief account of the electron density of the 1s orbital of a hydrogen atom. 1s orbital is spherical in shape, the variation is exactly the same in any direction from the nucleus. At a certain distance from nucleus, there is a maximium probability of electron density. (The most probable distance of the electron from the nucleus of the hydrogen atom is equal to the Bohr’s radius.) The electron density is zero at the nucleus and at infinite distance from the nucleus. 4 C Most candidates were able to give the correct shape of the 1s orbital, however, they failed to describe the variation of electron density as a function of distance from the nucleus. 92 1A 2 c 2c Give the shape of a 1s-orbital and comment on its physical significance. 1s-orbital is spherical. ½ mark Physical Significance ½ mark - The probability of finding an electron at various points in space is called an orbital. - For a give 1s-orbital, the region of highest probability of finding the electron, say 90% of the charge distribution is pictorially represented by a single surface along which the probability of finding the electron is constant. ½ mark - They should be thought of as shapes with fuzzy and indistinct edges, like a cloud. ½ mark C Only a handful of candidates were able to explain clearly the meaning of the spherical shape of the 1s-orbital and its physical significance, i.e. the most simple description of the surface of the sphere represents the probability of finding, say 90 %, of the 1s-electron. A majority of the candidates did not make the important point that orbitals should be thought of as shapes with fuzzy and indistinct edges, like clouds in reality. 96 1A 1d 1d Sketch the pictorial representation of a p orbital and indicate the location of the nucleus in your diagram. 2 1 nucleus (or along any axis) Many candidates did not do what the question asked for and drew three p-orbitals instead of one. 1 mark C IV. Periodic Table Page 1 Topic Reference Reading IV. Periodic Table Modern Physical Chemistry ELBS pg. Chemistry in Context, 3rd Edition ELBS pg. 37-41 Physical Chemistry, 4th Edition, Fillans pg. 85–88 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 111–112, 334–335 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 170–171, 175–176 2.5 Variation in atomic radii Variation in electronegativity IV. Periodic Table A. Variation in atomic radius Covalent radius – Van der Waals’ radius – Half of the distance between two nuclei in an element where the atoms are chemically combined together. Half of the distance between two nuclei in an element where the atoms are not chemically combined together. Syllabus Notes If not specified, atomic radius is normally referring to covalent radius. However, for those atoms where formation of chemical bond is impossible, atomic radius will be referring to van der Waals’ radius. a) Factor affect the size of atomic radius proton to electron ratio (p/e) – An increase in this ratio causes an increase in effective nuclear charge, resulting a contraction of the electron cloud of the ion. The ionic radius of a cation is, of course, smaller than the atom itself. Across a period – Atomic size decreases as atomic number increases. All electrons go into the same quantum shell and the increase in nuclear charge is not effectively shielded. Effective nuclear charge increases across a period and has a greater attraction on the outermost electrons drawing them closer to the nucleus. Down a group – Atomic size increases as no. of electron shells increases. The electron in the outermost shell is effectively shielded from the nucleus by the inner electrons. IV. Periodic Table Page 2 Variation of atomic radii B. Variation in electronegativity Electronegativity is a measure of tendency of an atom in a stable molecule to attract electrons within a bonds. Therefore, it is closely related to the effective nuclear charge experienced by the bonding electrons. The scientist Pauling computed a scale of electronegativity according to bond dissociation energies, ionization energy and electron affinity values of an atom. In Pauling scale, 4 is assigned to the most electronegative atom, fluorine. Pauling electronegativity values of some elements (The shaded ones are more electronegative than C) H 2.1 C 2.5 Si 1.8 Ge 1.8 Sn 1.8 Li 1.0 Na 0.9 K 0.8 Rb 0.8 Cs 0.7 Be 1.5 Mg 1.2 Ca 1.0 Sr 1.0 Ba 0.9 B 2.0 Al 1.5 _ _ N 3.0 P 2.1 As 2.0 Sb 1.9 O 3.5 S 2.5 Se 2.4 Te 2.1 F 4.0 Cl 3.0 Br 2.8 I 2.5 In general, electronegativity values increase from left to right across each period and decrease down each group. Noble gases do not has any value of electronegativity because they do not form any stable molecule at all. Glossary atomic radius covalent radius van der Waals radius proton to electron ratio electronegativity IV. Periodic Table Page 3 Past Paper Question 89 2B 4 d ii 92 2A 3 b i ii 95 2B 5 a I 89 2B 4 d ii 4d Comment on the relative oxidizing properties of ii the non-metal C, N, O and F. C<N<O<F - increase in electron affinity / electronegativity from C to F - decrease in atomic size - 2p electrons are less effectively shielded - increase in nuclear charge - F is particularly reactive because of the weak F–F bond due to lone pair-lone pair repulsion 3 1 mark 2 marks 92 2A 3 b i ii 3b i Define the covalent radius of an atom. 1 The covalent radius is defined as on-half the distance between two atoms of the same kind held together by covalent bond. 1 mark ii State and explain the trends in the covalent radius on going down any group and going across a short period of the 2 periodic table. Going down any group of the periodic table the covalent radius increases ½ mark because the elements become larger as the no. of electron increases. ½ mark Going across any period the covalent radius decreases ½ mark because the nuclear charge increases. ½ mark 95 2B 5 a i 5a The table below lists some properties of the alkali metals. Standard electrode Melting point / ºC Element Atomic Ionic radius / nm First ionization potential / V radius / nm energy / kJmol-1 Li 0.123 0.060 520 -3.04 180 Na 0.157 0.095 495 -2.71 98 K 0.203 0.133 418 -2.92 64 Rb 0.216 0.148 403 -2.93 39 Cs 0.235 0.169 374 -2.95 29 i Explain why the atomic radii increase from Li to Cs. As the group is descended, the next atom has another shell of electrons around the nucleus, therefore larger the atomic radius. 1 mark C as the group is descended, the 'next' atom has one more shell of electrons around the nucleus; 1 Energetics I. Enthalpy Change (∆H) A. B. C. Definition of a system Definition of enthalpy (H) and enthalpy change (∆H) 1. Constant pressure process a) Work done by a system Standard enthalpy change of formation (∆Hfo) 1. Definition of ∆Hfo a) Standard enthalpy change (∆Ho) b) Standard enthalpy change of formation (∆Hfo) c) Standard enthalpy change of formation of an element 2. Direct determination of ∆Hfo a) Use of simple calorimeter Other enthalpy change (∆H) 1. Examples of enthalpy change a) Enthalpy of combustion b) Enthalpy of atomization c) Enthalpy of neutralization d) Enthalpy of hydrogenation e) Enthalpy of solution f) Enthalpy of reaction Hess’s Law 1. Indirect determination of ∆Hfo a) Hess’s Law (1) (2) Use of energy level diagram / energy cycle Use of equation D. E. II. Bonding Energy A. B. C. Bond energy 1. Definition of bond energy 2. Determination of bond energy term (bond enthalpy) Strenght of covalent bond 1. Relationship between bond length and bond energy 2. Factors affecting the bond strength Estimation of enthalpy change by bond energy Electron affinity Lattice energy 1. Born-Haber cycle a) Determination of lattice energy by Born-Haber cycle 2. Factors affecting the value of lattice energy Stoichiometry of ionic compounds Limitation of enthalpy change (∆H) Entropy (S) and entropy change (∆S) Gibbs free energy change (∆G) III. Energetics of formation of ionic compounds A. B. C. IV. Gibbs free energy change (∆G) A. B. C. I. Enthalpy Change (∆H) Unit 1 Page 1 Topic Reference Reading I. Enthalpy Change (∆H) Chemistry in Context, 3rd Edition ELBS pg. 165–166 Physical Chemistry, 4th Edition, Fillans pg. 99 – 101 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 204 – 206 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 146–147 3.0–3.1 Definition of a system Enthalpy and enthalpy change I. Enthalpy Change (∆H) Unit 1 Syllabus Notes The discovery of detailed electronic structure of atom serves as a very good example in how does a chemist work. Nobody have seen the atom directly. The electronic structure is only the result of deduction from some macroscopic observations (ionization energies and successive ionization energies). The energy change associated with a reaction provides another useful tools to investigate the chemistry involved. A. Definition of a system In the study of energetics, the region or particular quantity of matter of interest is defined as the system. The surroundings are defined as everything other than the system. For example, the flask, gas syringe and everything inside them is defined as the system. The atmosphere and everything outside the flask and syringe is defined as the surroundings. B. Definition of enthalpy (H) and enthalpy change (∆H) The enthalpy of a substance, sometimes called its heat content, is an indication of the total energy content of the substance. The absolute enthalpy (H) of a substance depends on i. potential energy inherent in the electrical and nuclear interactions of the constituent particles (chemical energy), ii. kinetic energy possessed by the atoms and sub-atomic particles (kinetic energy) The absolute enthalpy (H) of a substance is not measurable since it is still not possible to measure all the interaction between the sub-atomic particles in an atom. Instead, we can measure the enthalpy change (∆H) of a system. Enthalpy change (∆H) can be arbitrarily defined as the heat change (q) at constant pressure. I. Enthalpy Change (∆H) 1. Constant pressure process Unit 1 Page 2 Consider the reaction between magnesium and hydrochloric acid : Mg ( s ) + 2 HCl( aq ) → MgCl2 ( aq ) + H 2 ( g ) Initial stage 1. Starting from room temperature (25ºC), we allow the magnesium and hydrochloric acid to react. Immediate after reaction 2. Immediately after the reaction, the temperature has increased due to heat energy released in the reaction. Final stage 3. We let the system cool back to room temperature (25ºC). We measure all the heat transferred between system and surroundings in order to return the system to 25ºC, this is the enthalpy change for the reaction. By law of conservation of energy, H1 and H2 must be equal. When the system is cooled down to 25 ºC from 50 ºC, some energy is lost to the surrounding in form of heat energy(q). By definition, ∆H = Hfinal - Hinitial = H3 - H1 = q Since H1 is larger than H3 in this exothermic reaction, it can be seen that ∆H of an exothermic reaction is always negative. Strictly speaking, according to the definition, in order to determine the enthalpy change, the amount of heat exchanged (q) between the system and the surrounding have to be determined. Since the quantity of the heat change (q) is difficult to be measured, it would be more convenient if we assume that all the heat energy evolved in the above example is absorbed by the system. Then the amount of the heat evolved can be calculated from the heat capacity of the system. ∆H = heat change (q) = - Heat capacity of the system × temperature change of the system (∆T) N.B. In an exothermic reaction, ∆T is positive but q and ∆H are negative, so a negative sign is added in the expression. I. Enthalpy Change (∆H) a) Work done by a system Unit 1 Page 3 Upon mixing of magnesium and dilute hydrochloric acid in the flask, hydrogen gas evolves. The volume of the system increase and the system has done a work against the atmospheric pressure. Work done is defined as Force(F) × Distance(d) [Pressure(P) × Area(A)] × [Volume change(∆V) ÷ Area(A)] Pressure(P) × Volume change(∆V) P∆V ∆V = which Vfinal - Vinitial = = = is positive if the volume of the system is increasing, and is negative if the volume of the system is decreasing. If the volume is kept constant in this case, the pressure will no longer be a constant. More heat will evolve in this reaction since the work done against atmospheric pressure is now exhibited in form of heat. Therefore, if the pressure of the system is not kept constant, according to the definition of ∆H, the heat change (q) measured would not be the same as the value of ∆H. Glossary system surrounding enthalpy change (∆H) heat change / transfer work done I. Enthalpy Change (∆H) Unit 2 Page 1 Topic Reference Reading I. Enthalpy Change (∆H) 3.2 Modern Physical Chemistry ELBS pg. 215–222 Chemistry in Context, 3rd Edition ELBS pg. 165 – 169 Physical Chemistry, 4th Edition, Fillans pg. 102 – 104 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 208 – 210 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 148–150 Standard enthalpy change of formation (∆Hfo) Determination of enthalpy change Other enthalpy change C. Standard enthalpy change of formation (∆Hfo) i. Standard state Unit 2 Syllabus Notes For the basis of comparison and tabulation of data, a standard state has been arbitrarily defined. All the reactants and products i. in the pure state if they are crystalline substances or liquids ii. at 1 atmosphere pressure if they are gases iii. at a concentration of 1M if they are dissolved in a solvent iv. at a specified temperature, usually, but not necessarily, 298K (25ºC) The condition of 1 atm and a specified temperature is called standard condition. ii. Meaning of superscripts o º = = = = = standard pure substance excited electronic state transition state, activated complex infinite dilution (with respect to all solutes present) * t ∞ 1. Definition of ∆Hfo a) Standard enthalpy change (∆Ho) The enthalpy change (∆H) for a reaction which occurs under standard conditions is called the standard enthalpy change. It is represented by the symbol ∆Ho. Note : The superscript o does not contain the information of the temperature. To be specific, the standard enthalpy change at 298K should be written as ∆Ho(298K). If the temperature is not specified, by convention, it is assumed to be 298K. e.g. i. 2H2(g) + O2(g) → 2H2O(g) ii. 2H2(g) + O2(g) → 2H2O(l) ∆H = -484 kJmol-1 ∆H = -572 kJmol-1 It can be seen that the value of enthalpy change depends on the physical state of the reactant and product. Therefore, in writing of equation, the physical state of individual species must be specified. I. Enthalpy Change (∆H) Unit 2 Page 2 b) Standard enthalpy change of formation (∆Hfo) Standard enthalpy change of formation (∆Hfo)– The enthalpy change which occurs when 1 mole of a compound is formed, under standard conditions, from its elements in their standard state. e.g. i. Standard enthalpy of formation of carbon dioxide ∆Hfo(298K) = -393.5 kJmol-1 C(graphite) + O2(g) → CO2(g) ii. Standard enthalpy of formation of ethane ∆Hfo(298K) = -52.3 kJmol-1 2C(graphite) + 2H2(g) → C2H4(g) Standard enthalpy of formation give us a rough ideal on the energetic stability of the compound respect to its constituent elements. A more negative standard enthalpy formation means more energetically stable. c) Standard enthalpy change of formation of an element According to the definition, standard enthalpy change of formation of an element, e.g. oxygen, is the enthalpy change associated with the following change, O2(g) → O2(g). Actually, there is no change, the standard enthalpy change of formation of an element is always zero. 2. Direct determination of ∆Hfo Some of the enthalpy of formation can be determined directly by measuring the heat change in direct synthesis of the compound, e.g. ∆Hfo[CO2(g)], ∆Hfo[SO2(g)]. a) Use of simple calorimeter A simple calorimeter can be used to determine the amount of heat absorbed or released in a reaction. For an exothermic reaction, the temperature will raise. For an endothermic reaction, the temperature will fall. If the heat capacity of the system is known. The heat energy evolved or absorbed can be calculated from the temperature change. I. Enthalpy Change (∆H) Unit 2 Page 3 e.g. Determination of enthalpy of reaction of Zn(s) + 2Ag+(aq) → Zn2+ + 2Ag(s) An excess of zinc powder was added to 50.0 cm3 of 0.100M AgNO3(aq) in a polystyrene cup. Initially, the temperature was 21.10ºC and it rose to 25.40ºC. Calculate the enthalpy change for the reaction : Zn(s) + 2Ag+(aq) → Zn2+ + 2Ag(s) Assume that the density of the solution is 1.00 gcm-3 and its specific heat capacity is 4.18 kJkg-1K-1. Ignore the heat capacity of the metals. i. Since the polystyrene cup is an insulator, its heat capacity is assumed to be zero and there is no energy exchanged between system and surroundings. ii. All the chemical energy released in the reactions transformed into heat energy which raises the temperature of the solution. iii. As there is no lost of energy to the surroundings, the total energy change in the system is zero. ∆H, due to reaction (at constant T) + change in heat energy of solution = 0 iv. ∆H + mcp∆T = 0 where m is mass of the solution cp is the specific heat capacity of the solution under constant pressure ∆T is the temperature change in K v. 50.0 ∴ ∆H = -mcp∆T = - 1000 kg × 4.18 kJkg-1K-1 × 4.30 K = -0.899 kJ vi. The value -0.899 kJ corresponds to 0.00500 mole of silver ions. In order to obtain the value for 2 mole of silver ions as specified by the equation, 2 mole The enthalpy change using 2 mole of Ag+ ions = -0.899 kJ × 0.00500 mole = -360 kJ vii. Zn(s) + 2Ag+(aq) → Zn2+ + 2Ag(s) ∆H = -360kJmol-1 viii. Strictly speaking, ∆H cannot be written as ∆Ho in this question because the conditions of the experiment were not standard, but the values of ∆H and ∆Ho would be very close. ix. Usually, a graph is plotted and extrapolated to determined the change in temperature, ∆T. I. Enthalpy Change (∆H) Unit 2 Page 4 D. Other standard enthalpy change (∆Ho) 1. Examples of enthalpy change a) Standard enthalpy of combustion ∆Hco It refers to the complete combustion of 1 mole of the substance under standard conditions. e.g. C(graphite) + O2(g) → CO2(g) ∆Hco[C(graphite)] = -393.5 kJmol-1 b) Standard enthalpy of atomization ∆Hoat It refers to the formation of 1 mole of gaseous atoms from the element in the defined physical state under standard conditions. e.g. ∆Hoat[C(graphite)]= +715 kJmol-1 i. C(graphite) → C(g) ii. ½H2(g) → H(g) ∆Hoat[½H2(g)]= +218 kJmol-1 For a compound, the value would be the enthalpy change involved in atomization of 1 mole of the compound being concerned. ∆Hoat[CO2(g)] = +1606 kJmol-1 e.g. CO2(g) → C(g) + 2O(g) c) Standard enthalpy of neutralization ∆Honeutralization It is the heat change when an acid and a base react to form 1 mole of water under standard condition. ∆Honeutralization = -57.1 kJmol-1 e.g. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l) ∆Honeutralization = -55.2 kJmol-1 Neutralization of a weak acid is found to be less exothermic than that of a strong acid because energy is required to ionize the molecules of weak acid first. d) Standard enthalpy of hydrogenation ∆Hohydro It is the heat change when 1 mole of an unsaturated compound is converted to saturated compound by reaction with gaseous hydrogen at 1 atm. ∆Hohydro = -238.8 kJmol-1 e.g. CH2CH=CHCH2(g) + 2H2(g) → CH3CH2CH2CH3(g) e) Standard enthalpy of solution ∆Hosol’n It is the heat change when 1 mole of a substance is dissolved at 1 atm in a stated amount of solvent or infinite amount of solvent. e.g. LiCl (s) + 100H 2 O (l) → LiCl (aq,100H O) ∆Hosol’n = -35 kJmol-1 i. 2 ii. LiCl( s ) → LiCl( aq ) H2O( l ) ∆Hosol’n = -37.2 kJmol-1 f) Standard enthalpy of reaction ∆Horxn It is the heat change in a reaction at 1 atm between the number of moles of reactants shown in the equation for the reaction. ∆Horxn = -845.6 kJmol-1 e.g. 2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s) Glossary standard enthalpy of formation standard state calorimeter enthalpy of atomization enthalpy of neutralization enthalpy of combustion I. Enthalpy Change (∆H) Unit 2 Page 5 Past Paper Question 91 2A 1 b ii 92 2A 1 b i 99 1B 7 b 91 2A 1 b ii 1b ii The enthalpy of neutralization of ethanoic acid with aqueous sodium hydroxide is -55.2 kJ mol-1 while that of hydrochloric acid is -57.3 kJ mol-1. Account for the difference in these two values. The enthalpy of neutralization of a strong acid (HCl) by a strong base (NaOH) is -57.3 kJ mol-1. The enthalpy of neutralization of ethanoic acid with NaOH is -55.2 kJ mol-1 because ethanoic acid is a weak acid and is only 2 marks slightly ionized, and the difference is due to the enthalpy of ionization. 92 2A 1 b i 1b i Define the standard enthalpy of formation of a compound, using CH3OH(l) as an illustration. The standard enthalpy of formation of a compound is the standard enthalpy change that occurs when one mole of the compound is made from its constituent elements under standard conditions (298 K and 1 atmospheric pressure). e.g. C(s) + 2H2(g) + ½O2(g) → CH3OH(l) 1 mark Ci Many candidates did not include temperature and pressure in their definition of standard enthalpy of formation. 99 1B 7 b 7b In an experiment to determine the enthalpy change of neutralization, a polystyrene foam cup was used as a calorimeter. When a solution of an acid was poured into a solution of an alkali in the calorimeter, the temperature rise was recorded by a thermometer which also served as a stirrer. State THREE sources of error in the result obtained in such an experiment. 2 1 I. Enthalpy Change (∆H) Unit 3 Page 1 Topic Reference Reading I. Enthalpy Change (∆H) 3.3 Modern Physical Chemistry ELBS pg. 222–226 Chemistry in Context, 3rd Edition ELBS pg. 170–174 Physical Chemistry, Fillans pg. 62–75 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 210 – 213 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 150–157 Hess’s Law Applications of Hess’s Law E. Hess’s Law 1. Indirect determination of ∆Hfo Unit 3 Syllabus Notes Not ∆Hfo of all compounds can be determined directly. This is because i. Not all compounds can be synthesized directly from its constituent elements at standard state. e.g. CH4(g) ii. In some reactions, the extent of the reaction is difficult to control. e.g. NH3(g) iii. Presence of side products / side reactions. 4Na(s) + O2(g) → 2Na2O(s), 2Na(s) + O2(g) → Na2O2(s) For the ∆Hfo of the compound which cannot be determined directly, law of conservation of energy (Hess’s Law) can be used to determine its value. a) Hess’s Law Hess's Law The change in enthalpy accompanying a chemical reaction is independent of the pathway between the initial and final states. The enthalpy change for a reaction is constant and is the sum of enthalpy changes for the intermediate steps. The standard enthalpy change of the reaction, C(s) + ½O2(g) → CO(g), cannot be determined directly since the extent of the reaction is difficult to control, inevitably some C(s) will remain and some CO2(g) will be present. But, the following 2 data can be determined experimentally by calorimetry, ∆Ho = -394 kJmol-1 C(s) + O2(g) → CO2(g) CO(g) + ½O2(g) → CO2(g) ∆Ho = -283 kJmol-1 I. Enthalpy Change (∆H) Unit 3 Page 2 (1) Use of energy level diagram / energy cycle i. Energy Cycle By Hess's Law, the heat change accompanying Route 1 should equal to that of Route 2. ∴ ∆H1 = ∆H2 + ∆H3 ∆H2 = ∆H1 - ∆H3 ∴ ∆Hfo[CO(g)] = ∆H2 = -394 kJmol-1 - (-283 kJmol-1) = -111 kJmol-1 ii. Energy Level Diagram To represent the relative enthalpy of the substance involved, an energy level diagram may be used. Steps of drawing an energy level diagram 1. Draw a line to represent the enthalpy of 1 mole of graphite and 1 mole of oxygen at standard condition. 2. The combustion of 1 mole of graphite produces CO2(g) and releases 394 kJ. 3. The combustion of CO(g) gives the same product, 1 mole of CO2(g). The energy released is 283 kJ. 4. The gap represents the heat of formation of CO(g). Steps of constructing energy cycle / energy level diagram 1. 2. 3. Write down an equation according to the definition of the problem. Write down the common reference to which both reactant and product are related. In the above example, it is CO2(g). Complete the cycle / diagram. In general, there are three popular common reference used in construction of energy cycles. They are i. combustion products ii. constituent elements in their standard states iii. constituent atoms N.B. I. Enthalpy Change (∆H) (2) Use of equation Unit 3 Page 3 The standard heat of reaction can also be determined if the standard heat of formation of the reactants and products are given. The generalized statement of this approach in the calculation of ∆Horxn is ∆Horxn = Σ∆Hfo(products) - Σ∆Hfo(reactants) e.g. C(s) + O2(g) → CO2(g) CO(g) + ½O2(g) → CO2(g) ∆Hfo = -394 kJmol-1 ∆Hfo = -283 kJmol-1 If you want to determine the heat of formation of CO(g), C(s) + ½O2(g) → CO(g), following method may be used. For the reaction, CO(g) + ½O2(g) → CO2(g), ∆Horxn = -283 kJmol-1 ∆Horxn -283 kJmol-1 -283 kJmol-1 ∆Hfo[CO(g)] Note : = Σ∆Hfo(products) - Σ∆Hfo(reactants) = {∆Hfo[CO2(g)]} - {∆Hfo[CO(g)] + ∆Hfo[½O2(g)]} = (-394 kJmol-1) - (∆Hfo[CO(g)] + 0 kJmol-1) = (-394 kJmol-1) - (-283 kJmol-1) = -111 kJmol-1 By definition, ∆Hfo of any element, e.g. O2, is zero. Or simply, by doing some manipulations on the equations. C(s) + O2(g) → CO2(g) CO(g) + ½O2(g) → CO2(g) ∆Hfo = -394 kJmol-1 ∆Hfo = -283 kJmol-1 i. Reverse the second equation and the sign of ∆Hfo. ii. Add the two equations together. ∆Hfo = -394 kJmol-1 C(s) + O2(g) → CO2(g) CO2(g) → CO(g) + ½O2(g) ∆Hfo = +283 kJmol-1 --------------------------------------------------------------------------------------------C(s) + O2(g) + CO2(g) → CO2(g) + CO(g) + ½O2(g) --------------------------------------------------------------------------------------------∆Hfo = -111 kJmol-1 C(s) + ½O2(g) → CO(g) Glossary Hess’s Law energy level diagram energy cycle I. Enthalpy Change (∆H) Unit 3 91 2A 1 b i Page 4 Past Paper Question 90 1A 2 b 91 1B 4 b i ii iii 92 2A 1 b ii 93 2A 3 b i ii 94 2A 2 a i ii 95 2A 1 b 97 2A 2 c 98 2A 2 c i ii 90 1A 2 b 2b Show how ∆Hof for CuSO4·5H2O(s) can be determined using the following data: ∆Hosoln CuSO4·5H2O(s) = + 8 kJ mol-1 ; ∆Hof CuSO4(s) = - 773 kJ mol-1 -1 ∆Hosoln CuSO4(s) = - 66 kJ mol ; ∆Hof H2O(l) = - 286 kJ mol-1 aq + Cu(s) + S(s) + 9/2 O2(g) + 5H2(g) ∆Hf CuSO4(s) + 5 ? ∆Hf H2O(l) aq + CuSO4(s) + 5H2O(l) ∆Hsoln CuSO4(s) ∆Hf CuSO4?H 2O(s) CuSO4?H 2O(s) + aq ∆Hsoln CuSO4?H 2O(s) CuSO4(aq) + 5H2O(l) 3 C - 773 kJ Cu(s) + S(s) + 2O2(g) → CuSO4(s) 5H2(g) + 2½O2(g) → 5H2O(l) 5 × - 286 kJ aq + CuSO4(s) → CuSO4(aq) - 66 kJ CuSO4(aq) + 5H2O(l) → CuSO4·5H2O(s) + aq -8 kJ ------------------------------------------------------------------------------------------Cu(s) + S(s) + 5H2(g) + 4½O2(g) → CuSO4·5H2O(s) ∴∆Hfo for CuSO4·5H2O(s) = (-773) + 5(-286) + (-66) - 8 = -2277 kJ mol-1 2 marks for working (cycle/equations and expressions) 1 mark for correct answer (sign, numeric and units) Deduct ½ mark for wrong units. Only about half of the candidates did this numerical problem correctly. Many errors arose as a result of a poorly constructed Born-Haber cycle. Very few candidates used the more analytical approach of setting up equations. 91 1B 4 b i ii iii 4b You are required to determine indirectly the enthalpy of formation of calcium carbonate(s). You are give the reagents: calcium, calcium carbonate, a strong acid, and water; the enthalpies of formation of carbon dioxide(g) and water(l) are available. i Give equations of suitable reactions for this determination. CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) Ca(s) + 2HCl(aq) → CaCl2(aq) + H2(g) 2 marks (missing or incorrect physical state -½; wrong coefficient -½; inappropriate acid e.g. H2SO4 and HNO3 -½) ii Name the major pieces of apparatus needed for this determination. Calorimeter (beaker or plastic cup, stirrer, temperature bath, insulation, electric heater) ½ mark thermometer or thermocouple ½ mark measuring cylinder, burette or pipette ½ mark balance ½ mark iii Suggest one experimental difficulty in the direct determination of the enthalpy of formation of calcium carbonate from its elements. 3 Ca(s) + C(s) + O2(g) → CaCO3(s) 2 Side reactions: 2Ca(s) + O2(g) → 2CaO(s); C(s) + O2(g) → CO2(g) 2 marks for any one explanation: - extent of reaction cannot be controlled - side reactions - directly combustion of Ca can be violent - cannot react or cannot react to form CaCO3 (1 mark only) C This section showed that the candidates were weak in applying general chemical knowledge to experiments. Most candidates scored marks by pointing out a very common piece of equipment and giving a partial explanation of the experimental difficulty. Of those who gave the equations with correct species, many failed to give balanced coefficients and states of the species. Many incorrect explanations like "heat loss to surroundings" and "large heat of formation" were given for (iii). 2 2 2 I. Enthalpy Change (∆H) Unit 3 91 2A 1 b ii 1b i Calculate the enthalpy of formation of NaCl(s) from the following data : Reaction ∆Ho/kJmol-1 NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) -57.3 H2(g) + ½O2(g) → H2O(l) -285.9 ½H2(g) + ½Cl2(g) → HCl(g) -92.3 HCl(g) + aq → HCl(aq) -71.9 Na(s) + ½O2(g) + ½H2(g) + aq → NaOH(aq) -425.6 NaCl(s) + aq → NaCl(aq) +3.9 ∆Ho = -425.6 kJ mol-1 Na(s) + ½O2(g) + ½H2(g) + aq → NaOH(aq) NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) ∆Ho = -57.3 kJ mol-1 H2O(l) → H2(g) + ½O2(g) ∆Ho = +285.9 kJ mol-1 ½H2(g) + ½Cl2(g) → HCl(aq) ∆Ho = -92.3 + (-71.9) = -164.2 kJ mol-1 NaCl(aq) → NaCl(s) + aq ∆Ho = -3.9 kJ mol-1 1 mark ∆Ho = (-425.6) + (-57.3) + (+285.9) + (-164.2) + (-3.9) = -365.1 kJ mol-1 (no unit -½) 3 marks C Some candidates were not able to give a correct energy cycle and hence failed to obtain the correct answer. Some weaker candidates did not even give state symbols in the chemical equations. 92 2A 1 b ii 1b ii Given the following thermochemical data at 298K : Standard enthalpy of combustion of CH3OH(l) -726.6 kJmol-1 -393.5 kJmol-1 Standard enthalpy of formation of CO2(g) -285.8 kJmol-1 Standard enthalpy of formation of H2O(l) Calculate the standard enthalpy of formation of CH3OH(l) at 298K. ∆Ho = -726.6 kJmol-1 CH3OH(l) + 3 2 O2(g) → CO2(g) + 2H2O(l) ∆Ho = Σnp(∆Hfo)p - Σnr(∆Hfo)r ∆Ho = (∆Hfo(CO2) + 2∆Hfo(H2O) - (∆Hfo(CH3OH) + 3 2 ∆Hfo(O2)) Page 5 4 3 Since ∆Hfo(O2) is zero. ∴ -726.6 = (-393.5 + 2 × (-285.8)) - (∆Hfo(CH3OH) + 0) ∴ ∆Hfo(CH3OH) = -238.5 kJ mol-1 missing sign(-1), missing unit(-½) 93 2A 3 b i ii 3b Given the following thermochemical data at 298 K: -1 -1 Compound ∆Hocombustion /kJ mol ∆Hoformation /kJ mol correct steps 2 marks correct answer 1 mark i ii C cyclopropane (g) -2091 — propene (g) -2058 — propane (g) -2220 — water (l) — -285.8 Calculate the enthalpy change involved in the hydrogenation of cyclopropane to propane. 9 (1) ∆(g) + 2 O2(g) → 3H2O(l) + 3CO2(g) ∆H = -2091 kJmol-1 (2) CH3CH2CH3(g) + 5O2(g) → 4H2O(l) + 3CO2(g) ∆H = -2220 kJmol-1 (3) H2(g) + ½O2(g) → H2O(l) ∆H = -285.8 kJmol-1 1 mark (1) + (3) - (2) ∆(g) + H2(g) → CH3CH2CH3(g) 1 mark ∆H = -2091 + (-285.8) - (-2220) = -156.8 kJmol-1 2 marks Calculate the enthalpy change involved in the conversion of cyclopropane to propene. Comment on the relative stabilities of cyclopropane and propene. 9 (4) H2C=CH–CH3(g) + 2 O2(g) → 3CO2(g) + 3H2O(l) ∆H = -2058 kJmol-1 (1) - (4) ∆(g) → H2C=CH–CH3(g) 1 mark -1 ∆H = -2091 - (-2058) = -33 kJmol 1 mark 1 mark The conversion from cyclopropane to propene is exothermic, hence propene is more stable because in the structure of cyclopropane there is squeezing of bond angles / ring is highly strained / poor overlapping of atomic orbital. 1 mark Weaker candidates produced wrong answers because they used the wrong signs and some could not correlate the fact that the cyclopropane molecule was highly strained, and that it would release the excess energy upon conversion to propene, resulting in an exothermic reaction. 4 4 I. Enthalpy Change (∆H) Unit 3 94 2A 2 a i ii 2a Given the following thermochemical data. Reaction ∆Ho98 / kJ mol-1 2 C (graphite) + 2H2(g) → CH4(g) - 75.0 C (graphite) + O2(g) → CO2(g) - 393.5 H2(g) + ½O2(g) →H2O(l) - 285.9 i Calculate the enthalpy change ∆Ho98 for the reaction CH4(g) + 2O2(g) → 2H2O(l) + CO2(g). 2 2 marks ∆Ho98 = -393.5 + 2(-285.9) - (-75.0) = - 890.3 kJmol-1 2 ii The enthalpy change ∆Ho98 is - 801.7 kJ mol-1 for the following reaction. 2 CH4(g) + 2O2(g) → 2H2O(g) + CO2(g) Calculate the enthalpy change of vaporization of water at 298K. ∆H = - 801.7 kJ mol-1 CH4(g) + 2O2(g) → 2H2O(g) + CO2(g) - CH4(g) + 2O2(g) → 2H2O(l) + CO2(g) ∆H = - 890.3 kJ mol-1 –––––––––––––––––––––––––––––––––––––––––––––––––––––– ∆H = + 88.6 kJ mol-1 1 mark 2H2O(l) → 2H2O(g) -1 ∆Hovaporization(H2O) = + 44.3 kJ mol 1 mark C Many candidates gave +88.6 kJ, instead of + 44.3 kJ mol-1 for the enthalpy change of vaporization of water. 95 2A 1 b 1b You are provided with the following thermochemical data: Reaction ∆Ho298/kJmol-1 + Ag(s) + aq → Ag (aq) + e +105.56 1 3 -207.36 2 N2(g) + 2 O2(g) + aq + e → NO3 (aq) 1 -167.15 2 Cl2(g) + aq + e → Cl (aq) 1 Ag(s) + 2 Cl2(g) → AgCl(s) -127.07 Calculate the standard enthalpy change for the reaction AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq). 1 mark The reaction AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq) is equivalent to Ag+(aq) + Cl-(aq) → AgCl(s) ∆H = Σ∆Hf(products) - Σ∆Hf(reactants) 1 mark = -127.07 - (+105.56) - (-167.15) 2 mark = -65.48 kJmol-1 (Accept any correct method using an energy cycle.) C Some candidates could not construct the energy cycle correctly. Mistakes in the numerical calculation were common. 97 2A 2 c 2c Given the following thermochemical data at 298 K: = -393.5 kJ mol-1 Standard enthalpy change of formation of CO2(g) = -285.8 kJ mol-1 Standard enthalpy change of formation of H2O(l) Standard enthalpy change of combustion of CH3CH2OH(l) = -1336.9 kJ mol-1 calculate the standard enthalpy change of formation of CH3CH2OH(l) at 298 K. Page 6 2 2 4 4 I. Enthalpy Change (∆H) Unit 3 98 2A 2 c i ii 2c Both H2(g) and CH3OH(l) are possible fuels for powering rockets. Their combustion reactions are shown below. H2(g) + ½O2(g) → H2O(g) CH3OH(l) + 1½O2(g) → CO2(g) + 2H2O(g) Compound molar mass / g ∆Hfo / kJmol-1 44 -394 CO2(g) 18 -242 H2O(g) 32 -239 CH3OH(l) For each of the above reactions, calculate the enthalpy change at 298 K per kg of the fuel-oxygen mixture in the mole ratio as indicated in the stoichiometric equation The effectiveness of a fuel can be estimated by dividing the enthalpy change per kg of the fuel-oxygen mixture in its combustion reaction by the average molar mass of the product(s) in g. Deduce which of the above two fuels is more effective in powering rockets. Page 7 i ii II. Bonding Energy Unit 1 Page 1 Topic Reference Reading II. Bonding Energy 3.4.1 Modern Physical Chemistry ELBS pg. 229–233 Chemistry in Context, 3rd Edition ELBS pg. 174–180 Physical Chemistry, Fillans pg. 76–83 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 214 – 215 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 157–159 Definition of bond enthalpies Determination of bond enthalpies II. Bonding Energy This is the principles of energetics and bonding energy again. Unit 1 Syllabus Notes Energy change always involves with bond formation and bond breaking. Usually, the bond formation process evolves energy, i.e. exothermic and bond breaking process absorbs energy, i.e. endothermic. ii. Bonding is just a balance between attractions and repulsions. If the attractions between atoms are stronger that the repulsions, the bond will form. Otherwise, the bond will be weak or will not form at all. A. Bond energy 1. Definition of bond energy i. In discussion of bond energy, two definitions have to be made clear. Bond Dissociation Energy, D: The energy required to break one mole of a particular covalent bonds in molecules. e.g. Cl–Cl bond in Cl2 molecule In contract, enthalpy of atomization is the energy required to produce one mole of gaseous atom, for a diatomic molecule, it is half of D. Also known as average bond energy or bond enthalpy, is an average value of bond dissociation energies for a given type of bond. e.g. E(C–H) is the average value of bond dissociation energies of C-H bond in all kinds of hydrocarbon. Not all C–H bonds are equivalent, even in the same molecule. For example, C–H bond at position 1 is stronger than the bond at position 2. The presence of the electronegative oxygen makes the C–H bonds become more polar and stronger. Bond Energy Terms, E(X–Y): 2 HC HH HH 1 C OH By convention, if not specified, bond energy is referring to bond energy term (average bond energy). Enthalpy changes for four successive dissociation reactions : Reaction CH4(g) CH3(g) + H(g) CH3(g) → CH2(g) + H(g) CH2(g) → CH(g) + H(g) CH(g) → C(g) + H(g) ∆Ho(298K) kJmol-1 +425 +470 +416 +335 Since dissociation energy refers to a particular bond, it is important to specify the process involved. The standard enthalpy change for the reaction, CH4(g) → C(g) + 4H(g), ∆Hoat[CH4(g)] = (+425) + (+470) + (416) + (+335) = +1646 kJmol-1 The bond energy terms, E(C–H), in CH4(g) = +1646 kJmol-1 ÷ 4 = 413 kJmol-1 In contrast with the bond dissociation energy, bond energy term does not refer to one particular bond but it is the average bond energy of a particular type of bond. II. Bonding Energy Unit 1 Bond C–H C–C C–Cl E(X-Y) kJmol-1 413 347 346 Page 2 After the collection of a number of bond dissociation energies for a particular type of bond, the bond energy term can be determined. 2. Determination of bond energy term (bond enthalpy) Bond energy term can be determined by i. thermochemical methods (by Hess’s Law) ii. spectroscopic methods (not required) iii. electron impact methods (not required) Determination of E(C=O) in CO2 by using an energy level diagram E(C=O) should be half of the enthalpy of reaction of CO2(g) → C(g) + 2O(g), ∆Hoat[CO2(g)] The enthalpy changes of the first three stages can be determined experimentally. Stage 1 Atomization of graphite ∆Hoat[C(graphite)] = +715 kJmol-1 C(graphite) → C(g) Stage 2 Atomization of oxygen ∆Hoat[O2(g)] = +498 kJmol-1 O2(g) → 2O(g) Stage 3 Enthalpy of formation of carbon dioxide ∆Hfo[CO2(g)] = -393 kJmol-1 C(graphite) + O2(g) → CO2(g) Stage 4 Enthalpy of atomization of carbon dioxide ∆Hoat[CO2(g)] CO2(g) → C(g) + 2O(g) By Hess’s Law ∆Hoat[CO2(g)] = - ∆Hfo[CO2(g)] + ∆Hoat[C(graphite)] + ∆Hoatm[O2(g)] = - (-393 kJmol-1) + 715 kJmol-1 + 498 kJmol-1 = +1606 kJmol-1 E(C=O) = ∆Hoat[CO2(g)] ÷ 2 = +1606 kJmol-1 ÷ 2 = +803 kJmol-1 Glossary bond energy bond energy terms (average bond energy) enthalpy of atomization bond dissociation energy Past Paper Question II. Bonding Energy Unit 2 Page 1 Topic Reference Reading II. Bonding Energy 3.4.2–3.4.3 Modern Physical Chemistry ELBS pg. 229–233 Chemistry in Context, 3rd Edition ELBS pg. 174–180 Physical Chemistry, Fillans pg. 76–83 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. Strength of covalent bond Factors affecting the bond strength B. Strength of covalent bond 1. Relationship between bond length and bond energy Unit 2 Syllabus Notes The bond strength of a bond can be measured by the bond energy. The stronger the bond, the more energy will be required to break the bond. For identical atoms, the bond strength is inversely proportional to the bond length. The stronger the bond, the closer will be the nuclei drawn together. Therefore, bond strength can be served as a indicator of bond strength. Bond C–C C=C 2. Bond length / nm 0.154 0.134 Bond energy +347 kJmol-1 +612 kJmol-1 Factors affecting the bond strength i. atomic size ii. number of bonding electrons / bond order iii. bond polarity i. atomic size Bond Bond length/nm F–F 0.142 Cl–Cl 0.199 Br–Br 0.228 I–I 0.266 Bond energy +158 kJmol-1 +242 kJmol-1 +193 kJmol-1 +151 kJmol-1 As the size of the atoms gets bigger, the bonding electrons are further from the nucleus and the shielding effect of the electron is getting more significant. Therefore, the strength of the bond decreases with increasing atomic size. Fluorine shows anomaly in group VII, this is because of the repulsion between lone pairs on the two atoms which weakens the bond. However if F is bonded to an atom with no lone pair, owing to the small size of F, the bond would be very strong. This explains why F four a larger variety of compounds that the other elements because the compound formed is energetically more stable than the others. For the same kind of bond, e.g. C–C bond, the bond energy increases with increasing no. of bonding electron. Bond F–H Cl–H Br–H I–H Bond length/nm 0.092 0.127 0.141 0.161 Bond energy +568 kJmol-1 +432 kJmol-1 +366 kJmol-1 +298 kJmol-1 ii. number of bonding electrons / bond order Bond Bond energy / kJmol-1 C–C +346 C=C +610 C≡C +837 N–N +163 N=N +410 N≡N +945 II. Bonding Energy iii. bond polarity Bond Difference in E.N. N–H 0.9 O–H 1.4 F–H 1.9 Unit 2 Page 2 Bond energy +388 +463 +562 Polarization of the bond gives the bond certain ionic character. The gives the bond certain ionic attraction on top of covalent bond. bond polarity Glossary Past Paper Question bond length 90 2B 4 c 91 2B 4 c i 95 2B 4 a ii 98 1A 1 b ii bond order molecular orbital theory 95 2B 4 c I 90 2B 4 c 4 Account for the following observations. 4c Dinitrogen tetraoxide and hydrogen peroxide are both unstable to heat. N2O4 d 2NO2 H2O2 → 2HO· or homolytic bond fission/state the products weak N–N and O–O bond strengths due to lone pair-lone pair repulsions in O2 and long N–N bond in N2 3 1 mark 1 mark 1 mark 91 2B 4 c i 4c i Why is the Cl-Cl bond stronger than the F-F or Br-Br bonds? 2 We expect that the X-X bond energy decreases on descending the group because the distance between the nuclear protons and the bond-pair electron increases, so the attractive force decreases. This is so for Cl2 and Br2. 1 mark The F-F bond is weaker than that of Cl2 due to non-bonding electron repulsion between lone pairs, on the F nuclei. 1 mark Ci The abnormally weak F–F bond strength was attributed to 'internuclear repulsion', to 'the small size of the fluorine atom', or to 'electron repulsion' and 'lone-pair repulsion', and few answers cited non-bonding electron repulsion between the two fluorine atoms. Some candidates thought that the Cl–Cl bond is stronger than the Br–Br bond because 'chlorine is more electronegative' or because 'the effective nuclear charge of Br is smaller'. Two points were required for the second part of the answer : a mention of the change in distance between nuclear protons and shared (bond-pair) electrons, and the consequent change in attractive force. 95 2B 4 a ii 4a Explain the following facts: ii The bond dissociation energy of F2 is less than that of Cl2. Unusually short F–F distance leads to high repulsion due to lone pair electrons of fluorine atoms. Therefore F–F bond is weaker than expected. 2 marks C Many candidates did not know that the small bond dissociation energy of F2 is due to the strong repulsion of lone electron pairs of the fluorine atoms in the F2 molecule. 95 2B 4 c i 4c Consider the data given below for the hydrogen halides and answer the questions that follow. Bond dissociation energy / kJmol-1 Standard enthalpy change of formation / kJmol-1 H–F -269.4 +562 H–Cl -92.8 +430 H–Br -36.8 +367 H–I +26.1 +298 i Explain briefly the trend in the bond dissociation energy of the hydrogen halides. Trend : Bond dissociation energy decreases with increasing molecular mass of hydrogen halides. 1 mark Explanation : The H–X bond lengths increase as the atomic radii of the halogens increase. The longer the bond, the weaker it is and the bond dissociation energy is smaller. 2 marks C Some candidates did not know that the H–X bond lengths increase as the atomic radii of the halogens increase from F to I and that the longer the bond, the weaker it is and the smaller is the bond dissociation energy. 98 1A 1 b ii 1b ii Explain why the carbon-oxygen bond lengths in CO and CO2 are different. 2 3 2 II. Bonding Energy Unit 3 Page 1 Topic Reference Reading II. Bonding Energy 3.4.4 Modern Physical Chemistry ELBS pg. 229–233 Chemistry in Context, 3rd Edition ELBS pg. 174–180 Physical Chemistry, Fillans pg. 76–83 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 157–158 Estimation of bond enthalpies and bond length C. Estimation of enthalpy change by bond energy Unit 3 Syllabus Notes With the atomization energies and Hess's Law, bond energy terms can be used for estimation of enthalpy changes. Since energy is required in bond breaking and energy is released in bond formation, the overall enthalpy can be estimated by the equation, ∆H = energy required - energy released = ΣE(bond broken) - ΣE(bond formed). i. Estimation of Enthalpy of formation of methane, CH4(g) E(C–H) ∆Hoat[C(graphite)] ∆Hoat[½H2(g)] +413 kJmol-1 +715 kJmol-1 +218 kJmol-1 Value estimation by bond energy terms ∆Hfo[CH4(g)] = energy required - energy released = (∆Hoat[C(graphite)] + 4 × ∆Hoat[½H2(g)]) - (4 × E(C–H)) = (715 + 4 × 218) - (4 × 413) = -65 kJmol-1 Value determination by Hess’s Law ∆Hco[C(graphite)] ∆Hco[H2(g)] ∆Hco[CH4(g)] -393.5 kJmol-1 -285.8 kJmol-1 -890.3 kJmol-1 ∆Hfo[CH4(g)] = (∆Hco[C(graphite)] + 2 × ∆Hco[H2(g)]) - ∆Hco[CH4(g)] = (-393.5 + 2 × (-285.8)) - (-890.3) = -74.8 kJmol-1 It can be seen that the value estimated by bond energy terms differs from the value determined by Hess’s Law. This is because bond energy terms is only the average bond energy and not the actual value of the bond being concerned. Therefore, the value from Hess’s Law is always preferred. II. Bonding Energy Unit 3 Page 2 ii. Estimation of Enthalpy of formation of 1-chloroethane, CH3CH2Cl(g), by bond energy terms E(C–H) E(C–C) E(C–Cl) ∆Hoat[C(graphite)] ∆Hoat[½H2(g)] ∆Hoat[½Cl2(g)] +413 kJmol-1 +347 kJmol-1 +346 kJmol-1 +715 kJmol-1 +218 kJmol-1 +122 kJmol-1 ∆Hfo[CH3CH2Cl(g)] = energy required - energy released = (2 × ∆Hoat[C(graphite)] + 5 × ∆Hoat[½H2(g)] + ∆Hoat[½Cl2(g)]) - (5 × E(C–H) + E(C–C) + E(C–Cl)) = (2 × 715 + 5 × 218 + 122) - (5 × 413 + 347 + 346) = -116 kJmol-1 iii. Use of bond energy term to predict the heat of reaction For the reaction, H–H(g) + Cl–Cl(g) → 2H–Cl(g) E(H–H) E(Cl–Cl) E(H–Cl) +435.9 kJmol-1 +243.4 kJmol-1 +432.0 kJmol-1 ∆H = energy required - energy released = ΣE(bond broken) - ΣE(bond formed) = (E(H–H) + E(Cl–Cl)) - (2 × E(H–Cl)) = (435.9 + 243.4) - (2 × 432.0) = -184.7 kJmol-1 (experimental value 184.6 kJmol-1) Glossary Past Paper Question 96 2A 1 a i 96 2A 1 a i ii 1a For the hydrogenation of buta-1,3-diene, H2C=CH–CH=CH2(g) + 2H2(g) → CH3CH2CH2CH3(g) the experimental molar enthalpy change is -239 kJmol-1. i Estimate the molar enthalpy change for the above hydrogenation using the bond energy terms below : 2 Bond Bond energy term / kJmol-1 H–H 436 C–H 413 C–C 346 C=C 611 1 mark ∆H = -2EC–C - 4EC–H + 2EC–C + 2EH–H or = -2(346) - 4(413) + 2(611) + 2(436) (1 mark) 1 mark = -250 kJmol-1 (0 marks for omitting the eve sign; deduct ½ mark for no units) III. Energetics of formation of ionic compounds Page 1 Topic Reference Reading III. Energetics of formation of ionic compounds 3.5.1–3.5.2 Modern Physical Chemistry ELBS pg. 69–70, 74, 226–228 Chemistry in Context, 3rd Edition ELBS pg. 180–182 Physical Chemistry, Fillans pg. 84–90 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 216 – 218 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 85–86, 160–162 Lattice energy Electron affinity and magnitude of electron affinity Determination of lattice energy by Born-Haber cycle III. Energetics of formation of ionic compounds A. Electron affinity Electron affinity, ∆Heo or EA – the enthalpy change which occurs when one mole of electrons are added to one mole of neutral atom in the gaseous state to form one mole of gaseous ions under standard conditions., i.e. A(g) + e- → A-(g). Syllabus Notes The value of the electron affinity is depending on the attraction between the incoming electron and the nucleus and the shielding effect offered by the existing electrons. i.e. the electron affinity of O, O(g) + e- → O-(g) ∆Heo = -142 kJmol-1 i.e. the electron affinity of O- (or second electron affinity of O) O-(g) + e- → O2-(g) ∆Heo = +791 kJmol-1 The first EA of O is exothermic because the attraction between the incoming electron and the nucleus is atronger than the repulsion between the incoming electron and the existing electrons. The second EA of O is endothermic because in the presence of the extra electron, the extra electron charge will offer a stronger repulsion with the incoming electron. Furthermore, the electron cloud of O- will expand and the incoming electron will occupy a position further from the nucleus which weakens the attraction with the nucleus. B. Lattice energy Lattice energy, ∆Holat – the enthalpy change which occurs when one mole of the ionic compound is formed, as a crystal lattice, from its constituent gaseous ions. The value of lattice energy is directly proportional to the strength of the ionic bond. ∆Holat = -788.5 kJmol-1 e.g. Na+(g) + Cl-(g) → NaCl(s) Since lattice energy always associates with bond formation, it must have a negative value, i.e. exothermic. The lattice energy cannot be determined directly by experiment, we construct an energy diagram called BornHaber cycle to determine the lattice energy. 1. Born-Haber cycle Born-Harber cycle – A thermodynamic cycle derived by application of Hess’s Law. Commonly used to calculate lattice energies of ionic solids and average bond energies of covalent compound. III. Energetics of formation of ionic compounds a) Determination of lattice energy by Born-Haber cycle Generalized Born-Haber cycle for an ionic compound Detail Born Haber cycle of sodium chloride Page 2 Transformation of elements in their standard states to gaseous ions, ∆H1 ∆Hoat[Na(s)] = +108 kJmol-1 ∆Hio[Na(g)] = +500 kJmol-1 ∆Hoat[½Cl2(g)] = +121 kJmol-1 ∆Heo[Cl(g)] = -364 kJmol-1 ∆H1 = ∆Hoat[Na(s)] + ∆Hio[Na(g)] + ∆Hoat[½Cl2(g)] + ∆Heo[Cl(g)] = 108 + 500 + 121 + (-364) = 365 kJmol-1 Standard heat of formation of the ionic crystal lattice, ∆H2 ∆H2 = ∆Hof[NaCl(s)] = -411 kJmol-1 Lattice energy, ∆Holat = ∆Holat[NaCl(s)] Finally, by Hess’s Law ∆H1 + ∆Holat[NaCl(s)] = ∆H2 ∆Holat[NaCl(s)] = ∆H2 − ∆H1 = (-411) - (+365) = -776 kJmol-1 3. Factors affecting the value of lattice energy Note: ∆Hio – ionization energy, I.E. ∆Heo – electron affinity, E.A. The magnitude of the lattice energy depends on i. The packing efficiency of the crystal lattice – small and similarity in size of ions result in better packing and more negative lattice energy ii. The charge density of the constituent ions – higher the charge density of the ion, stronger will be the electrostatic attraction and the lattice energy will be more negative. 1 Lattice energy ∝ r + r + r +, r - : radii of cation and anion III. Energetics of formation of ionic compounds C. Stoichiometry of ionic compounds Page 3 The calculation of the lattice energies and use of Born-Haber cycle can also be used to explain why group I, II and III metal always form M+, M2+ and M3+ ion respectively. e.g. Born-Haber cycle for Ca+Cl-, Ca2+(Cl-)2 and Ca3+(Cl-)3 The lattice energies are calculated based on the mathematical model. By summing up the terms, the enthalpy of formation of the 3 hypothetical ionic crystal can be calculated. ∆Hfo = -155 kJmol-1 Ca(s) + ½Cl2(g) → Ca+Cl-(s) 2+ Ca(s) + Cl2(g) → Ca (Cl )2(s) ∆Hfo = -763 kJmol-1 Ca(s) + 2 Cl2(g) → Ca3+(Cl-)3(s) 3 ∆Hfo = +1356 kJmol-1 The enthalpy of formation of CaCl2(s) is most exothermic among the three. This explain why calcium chloride has the empirical formula CaCl2 because CaCl2(s) is the most stable one among the three. The octet rule only helps to predict the formation of Ca2+ ion but it doesn't offer explanation. The rationale behind the formation of Ca2+ ion is solely from an energetic point of view. Glossary lattice energy electron affinity Born-Haber cycle intermediate type of bond III. Energetics of formation of ionic compounds Page 4 Past Paper Question 92 1A 3 d 94 1A 1 a i ii iii 96 2B 5 a i ii 98 1A 1 c i 99 2A 1 a ii 92 1A 3 d 3d Briefly describe and explain the change in the lattice energy of Group I chlorides on descending the group. The lattice energy decreases in magnitude on descending the group. NaCl 771; CsCl 645 kJ mol-1 1 mark 2 marks This is due to increasing size of cation results in smaller attraction between cation and anion. 3 94 1A 1 a i ii iii 1a i Write an equation to represent the change related to the second electron affinity of oxygen. 1 1 mark O-(g) + e- → O2-(g) ii The first and second affinities of oxygen are -142 kJmol-1 and +791 kJmol-1 respectively. Explain why they have 2 opposite signs. Addition of an e- to the O(g) is exothermic Q attraction between the nucleus and the incoming e- outweighs the 1 mark repulsion between the e- in O atom and the incoming e-. After addition of an e-, the size of O expands, hence the attraction for another electron is weakened, and the repulsion between the O-(g) ion and the e- makes the second electron affinity an endothermic process. 1 mark iii Explain why all the inert gases have positive first electron affinities. 1 Screening effect of the stable p6 configuration of inert gases (s2 in the case of He) makes the addition of an extra ean unfavourable process ∴ inert gases have +ve first E.A. 1 mark C It was surprising to find that many candidates could not distinguish between electron affinity and electronegativity and hence they performed badly in this part. 96 2B 5 a i ii 5a The first electron affinities of chlorine and bromine are -364 kJmol-1 and -295 kJmol-1 respectively. i What is the meaning of 'electron affinity' ? Electron affinity is the molar enthalpy change ½ mark ½ mark for the following process at standard state/conditions X(g) + e- → X-(g) Or ½ mark electron affinity of an element X is the enthalpy change 1 mark when 1 mole of X-(g) is formed from gaseous atoms of X ½ mark under standard conditions. C Most candidates did not give a precise definition for 'electron affinity'. They did not point out that the electron affinity of an element X is the molar enthalpy change when atoms of X in the gaseous state take up electrons to form X-(g) ions under standard conditions. ii Explain why the first electron affinity of chlorine is more negative than that of bromine. Cl has a smaller size / a smaller no. of electron shells 1 mark ∴ Cl can exert a stronger attraction on an extra electron 1 mark 98 1A 1 c i 1c The theoretical lattice enthalpies of NaCl(s) and AgCl(s), and the experimental lattice enthalpy of AgCl(s) are given in the table below. Compound Theoretical lattice enthalpy Experimental enthalpy NaCl(s) - 770 ? AgCl(s) - 833 - 905 i Calculate the experimental lattice enthalpy of NaCl(s) using the following thermochemical data. ∆Ho/kJ mol-1 108 Atomization enthalpy of Na(s) 495 First ionization enthalpy of Na(g) 239 Bond dissociation enthalpy of Cl2(g) - 349 Electron affinity of Cl(g) - 411 Enthalpy change of formation of NaCl(s) 2 2 3 III. Energetics of formation of ionic compounds 99 2A 1 a ii 1a ii Using the thermochemical data given below, calculate the lattice enthalpy of rubidium chloride. ∆Ho298 / kJ mol-1 Rb(s) → Rb(g) 82 Rb(g) → Rb+(g) + e403 Cl2(g) → 2Cl(g) 242 Cl(g) + e → Cl (g) -348 2Rb(s) + Cl2(g) → 2RbCl(s) -861 Page 5 IV. Gibbs free energy change (∆G) Page 1 Topic Reference Reading IV. Gibbs free energy change (∆G) 3.0 Modern Physical Chemistry ELBS pg. Chemistry in Context, 5th Edition ELBS pg. 339–350 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. Assignment Reading Syllabus Notes Gibbs free energy change IV. Gibbs free energy change (∆G) (not required in A-Level) So far, we have applied the concept of energetic to different chemical systems such as ionization energy, bond energy and lattice energy. However, in certain instances, changes in enthalpy are only part of the driving force for the change. Some processes take place spontaneously even though they are endothermic. i.e. ∆H > 0. The observation seems to be contradictory, the products of an endothermic change is less stable than the reactants. A. Limitation of enthalpy change (∆H) For example, melting is an endothermic change. Ice melts readily at room temperature. Although the heat content of water is higher than ice, does that mean ice is more stable than water ? If ice is more stable than water, why does it change to water ? Therefore, enthalpy change ∆H, is not a very good description of the relatively stability of systems. ∆H Water melting process Ice B. Entropy (S) and entropy change (∆S) Consider another example, the free expansion of a balloon in vacuum. Since no work is done in the expansion, the temperature will remain constant, i.e. ∆H = 0. The size of the balloon increases and more spaces are available for the gaseous molecules to travel. Therefore, the arrangement and the movement of the molecules are getting more random. ree exp ansion in vacuum f→ The expansion is spontaneous which means that the bigger balloon is more stable than the smaller balloon though the ∆H is 0. It can be generalized that the state with a high degree of disorderness is always favourable in nature. The scientist called the disorderness in arrangement of a system – entropy S and the change in the disorderness – entropy change ∆S. In the above example, the arrangement of the molecules becomes more random and the entropy of the system increases. i.e. ∆S > 0. IV. Gibbs free energy change (∆G) C. Gibbs free energy change (∆G) Page 2 In order to address the limitation of enthalpy change ∆H in describing the relative stability of different systems. The scientist J. Willard Gibbs defined another function – Gibbs free energy change ∆G. Gibbs free energy change is defined as ∆G = ∆H - T∆S where ∆H is enthalpy change T is temperature in Kelvin ∆S is entropy change For example, ice has a more ordered packing that water. Thus, melting is a process associated with an increase in disorderness i.e. ∆S > 0. Enthalpy change (∆H) of melting = 6010 kJmol-1 Entropy change (∆S) of melting = 22.0 kJmol-1 Temperature -10ºC (263.15K) 0ºC (273.15K) 10ºC (283.15K) ∆H of melting 6010 kJmol-1 6010 kJmol-1 6010 kJmol-1 -T∆S of melting -5790 kJmol-1 -6010 kJmol-1 -6230 kJmol-1 ∆G of melting 220 kJmol-1 0 kJmol-1 -220 kJmol-1 Obviously, whether ice will melt or freeze is depending on temperature. At any temperature above 0ºC, ice tends to melt. At any temperature below 0ºC, water tends to freeze. Moreover, the value of ∆G is always smaller than zero. All spontaneous changes take place in the direction of decrease in free energy. i.e. ∆G < 0. However, free energy change ∆G only indicates the energetic stability of a system. Whether the change will take place is also depending on the kinetic stability of the system which will be discussed in the chapter of rate of reaction. Glossary Past Paper Question spontaneous reacton disorderness Gibbs free energy change (∆G) entropy (S) entropy change (∆S) Chemical Bonding Page 1 Chemical Bonding I. Shape of molecule A. Lewis structure of molecule 1. Oxidation no. of individual atoms in a molecule 2. Formal charge of individual atoms in a molecule 3. Limitation of octet rule Valence Shell Electron Pair Repulsion Theory (VSEPRT) 1. Octet expansion of elements in Period 3 2. Position of lone pair 3. 3-dimensional representation of molecular shape Hybridization Theory 1. Overlapping of atomic orbital 2. Hybridization theory 3. Examples a) sp hybridizaion b) sp2 hybridizaion c) sp3 hybridizaion d) sp3d hybridizaion e) sp3d2 hybridizaion 4. Structure and shape of hydrocarbons a) Characteristic of Carbon-Carbon bond b) Shape of hydrocarbons (1) (2) (3) Saturated hydrocarbons Unsaturated hydrocarbons Aromatic hydrocarbons (a) Delocalization of π-electrons (b) Stability of benzene B. C. D. Molecular Orbital Theory II. Ionic bonding A. B. C. Strength of ionic bond - lattice energy X-ray diffraction 1. Electron density map of sodium chloride Periodicity of ionic radius 1. Definition of radius 2. Periodicity of ionic radius 3. Size of isoelectronic particles Chemical Bonding Page 2 III. Covalent bonding A. B. C. H2+ ion Electron diffraction Covalent radius 1. Definition of covalent radius 2. Addivity of covalent radius 3. Breaking down of addivity in covalent radius and bond energy a) Resonance (1) (2) (3) Resonance structure of benzene Resonance structure of nitrate ion Rules in writing resonance structures Delocalization energy b) D. Breaking down of addivity in bond enthalpies (1) Dative covalent bond 1. Examples of H3N→BF3 and Al2Cl6 a) H3N→BF3 b) Al2Cl6 IV. Bonding intermediate between ionic and covalent A. B. Differences between ionic bond and covalent bond Incomplete electron transfer in ionic compound 1. Electron density map of LiF comparing with those of NaCl and H2 2. Difference among lattice energies of Na, Ag and Zn compounds a) Lattice energies of sodium halide, silver halide and zinc sulphate b) Bonding intermediate between covalent and ionic 3. Polarization of ionic bond a) Fajans’ Rules in polarization of ionic bond Electronegativity 1. Definition of electronegativity 2. Pauling scale of electronegativity Polarity in covalent bond 1. Deflection of a liquid jet by an electric field 2. Dipole moment a) Vector quantity of dipole moment b) Polarity of molecule c) Factors affecting dipole moment (1) (2) (3) Inductive effect (I) Mesomeric effect / resonance effect (R) Presence of lone pair C. D. V. Metallic bonding A. B. C. D. Electron sea model of metal Strength of metallic bond Melting and boiling of metal Strength of ionic bond, covalent bond and metallic bond I. Shape of molecule Unit 1 Page 1 Topic Reference Reading Syllabus Notes I. Shape of molecule 4.0–4.1 Modern Physical Chemistry ELBS pg. 84–86 Organic Chemistry, Solomons, 6th Edition pg. 8–14 Lewis structure Chemical Bonding I. Shape of molecule Unit 1 The shape of a molecule can be predicted and explained by several different theory i. Valence Shell Electron Pair Repulsion Theory – repulsions among electron centres (bond pair and lone pair electrons) have to be minimized. ii. Hybridization Theory – overlapping of hybridized atomic orbitals iii. Molecular Orbital Theory – overlapping of atomic orbitals to form molecular orbital A. Lewis structure of molecule Before the shape of a molecule can be derived, it would be easier to draw the Lewis structure (cross-dot diagram) of the molecule first. Steps of drawing Lewis structure i. Determination of central atom Not all atoms can be the central atom in a molecule. e.g. in H2O, H can only be the peripheral atom since it can form only 1 bond. ii. Determination the no. of the bonds have to be formed. Count the electron available Count the number of valence electrons provided by each atom. If the molecule is negatively charged, add extra no. of electrons accordingly. If the molecule is positively charged, subtract the no. of electrons accordingly. Count the electron required Count the total no. of electrons required if the outermost shells of all the atoms have to be filled. e.g. in NH3, there should be 8 electrons for N and 2 electrons for each H atom. Determine the no. of bonds The total no. of bond pairs can be calculated by the following method. Assuming each atom in the molecule needs the max. no. of e- to fill its valence shell e.g. 2 for H, 8 for O or N. Formation of each bond can save the use of 2 electrons. no. of bond pair = (no. of electron required - no. of electron available) ÷ 2 no. of bond pair = ((2 × 3 + 8) - 8) ÷ 2 = 3 e.g. NH3 no. of bond pair = ((8 × 4) - 24) ÷ 2 = 4 e.g. NO3iii. Drawing the skeleton of the molecule Within a covalent molecule, there should be at least 1 bond pair between 2 atoms. iv. Put the rest of the electrons into the diagram until octet is fulfilled and include the charge if any. I. Shape of molecule Examples NH3 i. Central atom – N ii. N 3H e- available N 3H e- required Unit 1 Page 2 5 1×3 8 8 2×3 14 iii. iv. HNH H Total 3 N–H Remaining 8 6 2 HNH H 3 N–H 1 lone pair Total 6 2 8 no. of bond = (14 - 8) ÷ 2 = 3 NH4+ i. Central atom – N ii. N 4H +1 (charge) e- available N 4H e- required 5 1×4 -1 8 8 2×4 16 iii. iv. H HNH H Total 4 N–H Remaining iii. 8 8 0 H HNH H 4 N–H Total iv. 8 8 + no. of bond = (16 - 8) ÷ 2 = 4 NO3i. Central atom – N ii. N 3O -1 e- available N 3O e- required 5 6×3 1 24 8 8×3 32 ONO O Total 3 N–O Remaining 24 6 18 ONO O 1 N=O 2 N–O 8 lone pairs Total 4 4 16 24 no. of bond = (32-24)÷2 = 4 1. Oxidation no. of individual atoms in a molecule Oxidation no. – The difference between the number of electrons associated with an atom in a compound as compared with an atom of the element. Oxidation no. is related to the difference in electronegativity of the atoms in the molecule. It is assumed that the bonding electrons will be attracted to the more electronegative atom from the less electronegative. i.e. the more electronegative atom will has more electrons than if neutral, results in negative oxidation no. e.g. CO2 Oxygen is more electronegative than carbon, the 4 electrons in the double bond is attracted towards the oxygen. Oxidation no. of oxygen = no. of electron in neutral O atom - no. of electrons on O in CO2 = 6 - 8 = -2 Oxidation no. of carbon = no. of electron in neutral C atom - no. of electrons on C in CO2 = 4 - 0 = +4 Oxygen is more electronegative than nitrogen OCO e.g. NO3- - ONO O O.N. of the doubly bonded oxygen = no. of e- in neutral O atom - no. of e- on O in NO3= 6 - 8 = -2 O.N. of the singly bonded oxygen = no. of e- in neutral O atom - no. of e- on O in NO3= 6 - 8 = -2 O.N. of nitrogen = no. of e- in neutral N atom - no. of e- of N in NO3= 5 - 0 = +5 2. Formal charge of individual atoms in a molecule I. Shape of molecule Unit 1 Page 3 Unlike oxidation no., formal charge is independent of difference in electronegativity. Formal charge of an atom is calculated by assuming that the bond pair electrons are shared equally between the two atoms bonded together. e.g. CO2 Formal charge of oxygen = no. of e- in neutral O atom - no. of e- associated with O in CO2 =6-6=0 Formal charge of carbon = no. of e- in neutral C atom - no. of e- associated with C in CO2 =4-4=0 Formal charge of the doubly bonded oxygen = no. of e- in neutral O atom - no. of e- associated with O in NO3=6-6=0 Formal charge of the singly bonded oxygen = no. of e- in neutral O atom - no. of e- associated with O in NO3= 6 - 7 = -1 Formal charge of nitrogen = no. of e- in neutral N atom - no. of e- associated with N in NO3= 5 - 4 = +1 Non-zero formal charge is usually indicated on the Lewis structure. Similar to oxidation no., the sum of formal charges equals the overall charge carried by the molecule. To minimize the fuzziness of calculation, some formal charges of some atoms can be memorized. OCO e.g. NO3- + ONO O N.B. The positive sign and negative sign must be accompanied for all oxidation no. and formal charges. I. Shape of molecule 3. Limitation of octet rule Unit 1 Page 4 The forth-mentioned method of drawing Lewis structure is only limited to the elements in Period 2. This is because, for principal quantum number 2, there are only 2s, 2px, 2py and 2z orbitals available. Each orbital can accommodate 2 electrons and makes up to a total of 8. Though the largest no. of valence electron would be 8, the no. can be less than 8. For example, in NO2, O N O there are totally ( 5 + 6 + 6 ) = 15 valence electrons which is an odd number. It is not possible for all atoms to have an octet structure. Actually, there are 8 electrons about each O atom and only 7 eletrons about N. However, for period 3 elements, there are 3d orbitals available which is capable to accommodate 10 more electrons. This is why P can form both PCl3 and PCl5 but N can only form NCl3. This phenomenon is called expansion of octet (or expansion of coordination sphere) and will be explained in the chapter of hybridization theory. For those which does not obey octet rule, e.g. PCl5, SO2, another method have to be used to draw the Lewis structure. It is assumed that all valence electrons are capable to form bond with peripheral atoms. And the peripheral will form the required number of bond to achieve octet. e.g. in PCl5, P has 5 valence electrons and each Cl is capable to form 1 single bond. Therefore, the Lewis structure will become Cl Cl Cl P Cl + Cl . e.g. in SO2, S has 6 valence electrons and each O is capable to form 1 double bond. Therefore, the Lewis structure will become O S O . e.g. in SO42- ion, S has 6 valence electrons ; 2 O atoms with 0 formal charge each forming 2 bond ; 2 O atoms with O O -1 formal charge each forming 1 bond. Therefore, the Lewis structure will become O- S O- . e.g. XeF4 is a noble gas compound not obeying octet rule. Xe has 8 outermost electrons, each F is capable to form 1 bond. As a result, 4 electrons about Xe will remains as the lone pairs. Therefore, the Lewis structure will F F become F Xe F . However, for some structures do not obey octet rule, sometimes memorization is required. Glossary Past Paper Question 6Aa Lewis structure 92 2B 6 Aa i Oxidation no. 92 2B 6 Ba i formal charge 92 2B 6 Aa i i Write the formula in each case of ONE compound or ion in which nitrogen has the oxidation state +5, +3, -1 and -3. NO3-, NO2-, NH2OH, NH3 ½ mark each 2 92 2B 6 Ba i 6Ba i Write the formula in each case of ONE compound or ion in which sulphur has the oxidation state -2, +2, +4 and +6. S2-, S2O32-, SO32-, SO42½ mark each 2 I. Shape of molecule Unit 2 Page 1 Topic Reference Reading I. Shape of molecule Modern Physical Chemistry ELBS pg. 84–86 Physical Chemistry, Fillans pg. 126–143 Organic Chemistry, Solomons, 6th Edition pg. 9–14 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 125–128, 131–132 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 90–92 4.2 Shape of molecule Valence Shell Electron Pair Repulsion Theory B. Valence Shell Electron Pair Repulsion Theory (VSEPRT) The shape of a molecule was related to the number of electrons in the outer shell of the central atom. 1. 2. 3. Unit 2 Syllabus Notes Electron-pairs, whether in bonding orbitals or lone-pair orbitals, arrange themselves in space so as to minimize their mutual repulsions. The repulsion between lone-pair/lone-pair is greater than that between lone-pair/bond-pair which is, greater than that between bond-pair/bond-pair. Where there are more than four pairs of electrons to consider. the interactions where the electron-pairs make an angle greater than 90° at the central atom are ignored. No. of electron centre in the outer shell of the central atom 2 3 Shape of the framework linear trigonal planar 120º BF3 Bond angle Example 180º BeCl2 4 tetrahedral 109º28' CH4 NH3 H2O 5 trigonal bipyramidal 120º 90º PF5 SF 4 ClF3 6 octahedral 90º SF6 IF5 7 pentagonal bipyramidal 72º 90º IF7 Note: An electron centre can either be a lone pair, single bond, double bond or triple bond. I. Shape of molecule 1. Unit 2 Octet expansion of elements in Period 3 Page 2 Elements in period 3 and onwards are capable to use the low lying energy d-orbitals available to form more bonds. Therefore, the element in period 3 can exceed the octet by expansion of its coordination sphere. This would be further explained in hybridization theory. 2. Position of lone pair Since lone pair electrons, comparing with bonding pair, is closer to the central atom and has greatest repulsion with other electron centres, only repulsions with the lone pair need to be considered. e.g. in SF4 If the lone pair occupies the equatorial position, there would be only 2 90º lone pair / bond pair repulsions. If the lone pair occupies the axial position, there would be 3 90º lone pair / bond pair repulsions. The 120º repulsion is neglected as it is much weaker than 90º repulsion. For the molecule with 5 electron centres, the lone pair always occupies the equatorial position to minimize the repulsion. Steps of using VSEPR theory to determine the shape of a molecule. 1. 2. 3. 4. Draw the Lewis structure. Count the number of electron centre around the central atom and choose the corresponding framework. Put the electron centres (with the atom if applicable) on the vertex of the framework so that the repulsion among them are minimized. Determine the bond angle. In the determination of the shape of a molecule, only the geometry of the atoms are considered. The position of the lone pair is not counted. 3. 3-dimensional representation of molecular shape a bond coming out of the paper a bond going into the paper a bond on the paper a lone pair Usually two bonds are used to defined the plane of the paper and then the other bonds are added with respect to this plane. I. Shape of molecule Example BF3 F Unit 2 Lewis structure F B F Page 3 Shape of molecule trigonal planar F F B F no. of electron centre 3 single bonds Total : 3 4 single bonds Framework chosen Bond angle ∠ F–B–F : 120 º CH4 H HC H H tetrahedral H H C H H ∠ H–C–H : 109½º Total : 4 3 single bonds 1 lone pair Total : 4 NH3 HNH H trigonal pyramidal H NH H ∠ H–N–H : ≈ 107º H2O HOH 2 single bonds 2 lone pair Total : 4 angular / bent / V-shaped H O H ∠ H–O–H : ≈ 105º Although all CH4, NH3 and H2O has 4 electron centres, they possess slightly different bond angle, 109½º, 107º and 105º respectively. 109½º is the angle for a perfect tetrahedral. However, the repulsion with lone pair is greater than the repulsion with bond pair. The presence of a lone pair on N squeezes the bond angle to 107º, the two lone pairs on O squeezes the bond angle even further to 105º Example PCl5 Lewis structure Cl Cl Cl P Cl no. of electron centre 5 single bonds Total : 5 Framework chosen Shape of molecule trigonal bipyramidal Cl Cl P Cl Cl Cl Bond angle ∠ Cl–P–Cl : 120º × 3 ∠ Cl–P–Cl : 90º × 6 Cl SF 6 F F S F F F F 6 single bonds Total : 6 4 single bonds octahedral F F F S F F F ∠ F–S–F : 90º NH4+ H HNH H + tetrahedral H + NH H H ∠ H–N–H : 109½º Total : 4 2 single bonds 2 lone pairs Total : 4 NH2- HNH angular / bent / V-shaped HN H ∠ H–N–H : ≈ 105º CO2 OCO 2 double bonds Total : 2 linear OCO ∠ O=C=O : 180º SO 2 O S O 2 double bonds 1 lone pair Total : 3 angular / bent / V-shaped O S O ∠ O=S=O : ≈ 120º I. Shape of molecule Unit 2 Oxidation no. 91 2B 5 e 93 1A 3 a ii formal charge Valence shell electron pair repulsion theory Page 4 Glossary Past Paper Question Lewis structure octet expansion 91 1A 3 d i ii 92 1A 3 g i ii 93 1A 1 b i ii 94 1A 1 c i ii 95 1A 2 e i ii 96 1A 2 b i ii 97 1A 3 b i ii iii 98 1A 3 a i ii iii 99 2A 3 c i 98 1A 4 a i 98 2A 1 c i 91 1A 3 d i ii 3d Draw the molecular shapes of i PCl5(g) Cl Cl P Cl Cl Cl 1 1 mark 1 ii SF4(g) F S F F F 1 mark 91 2B 5 e 5e Why is the bond angle in NF3 smaller than that in NH3? (Valence Shell Electron Pair Repulsion approach) Bond pair electrons nearer to F than H since F is more electronegative ∴ bond-pair/bond pair repulsion is smaller in NF3 than in NH3. HNH H N FFF 2 ½ mark ½ mark 1 mark or (Hybridization approach) The primary scheme of hybridization of N is sp3 the p orbital is more directional than s orbital and have been attracted strongly towards the electronegative F as a result the N in NF3 has higher s-character than N in NH3 Bond angle ∝ s-character e.g. Ideal bond energy of following hybridization: sp = 180º, sp2 = 120º, sp3 = 109½º 92 1A 3 g i ii 3g Draw diagrams showing the shapes of the following molecules. Indicate the lone pairs (if any) on each central atom. i ICl2 1½ Cl I Cl Linear shape 1 mark Lone pairs ½ mark XeOF4 1 e- only ii 1½ O note: Xe is [Kr]4d105s25p6 Square-pyramidal1 mark Lone pair ½ mark F Xe F F F I. Shape of molecule Unit 2 Page 5 93 1A 1 b i ii 1b For each of the following molecules, draw a three-dimensional structure and state the molecular geometry. i SiF4 F F Si F F 2 tetrahedral F 2 marks 2 2 marks ii OF2 O F bent / V-shaped 93 1A 3 a ii 3a Consider the following compound, X: ii Suggest expected values for the bond angles α, β and γ. α: 108º-110º β: 178º-182º γ: 118º-122º 1½ ½ mark each 94 1A 1 c i ii 1c For each of the following molecules, draw a three-dimensional structure showing the positions of the bond electron pairs and lone electron pairs (if any). In each case, state the molecular geometry and whether the molecule possesses a non-zero dipole moment. i BF3 2 F F B F 1 mark ½ mark ½ mark 2 ii trigonal / triangular planar no dipole moment ClF3 F F Cl F T-shaped Possesses a net dipole [0 mark if shape of ClF3 is described as triangular planar] 1 mark ½ mark ½ mark 95 1A 2 e i ii 2e For each of the following species, draw a three-dimensional structure showing the bond electron pairs and lone electron pairs of the central atom. State the shape of the species in each case. i ICl4Cl Cl I Cl Cl 1½ ii Square planar SCl2 S Cl Cl (1 mark, no charge -½ mark, no lone pair -½ mark) ½ mark 1½ 1 mark ½ mark V-shaped / bent / angular 96 1A 2 b i ii I. Shape of molecule Unit 2 2b For each of the following chemical species, draw a three-dimensional structure showing the bond electron pairs and lone electron pair(s) of the central atom underlined. State the shape of the species in each case. i ClO3- Page 6 1½ O O Cl O 1½ mark 1½ ii pyramidal (Deduct ½ mark for omitting the negative charge) NOF N F O 1½ mark angular or V-shaped (Deduct ½ mark for missing out the lone pair) (1 mark for 3-D structure; ½ mark for the shape) 97 1A 3 b i ii iii 3b For each of the following sulphur-containing chemical species, state its shape and the oxidation state of sulphur. i H2S ii SO2 iii SO4298 1A 3 a i ii iii 3a For each of the nitrogen-containing chemical species below, state its shape and the oxidation state of nitrogen. i NO2+ ii NH3 iii NO398 1A 4 a i 4a Alcohol E has the structure CH3CH(OH)C2H5. i Draw a three-dimensional representation of E. 98 2A 1 c i 1c i Draw the three-dimensional structure of BF3. 99 2A 3 c i 3c Consider the hydrides of three Period 3 elements : SiH4, PH3 and H2S i For each hydride, draw a three-dimensional structure showing the bond electron pairs and lone electron pair(s), if any, of the central atom. 3 1 1 1 1 4 I. Shape of molecule Unit 3 Page 1 Topic Reference Reading I. Shape of molecule Modern Physical Chemistry ELBS pg. 84–86 Physical Chemistry, Fillans pg. 126–143 Organic Chemistry, Solomons, 6th Edition pg. 26–37 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 133–140 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 419 4.3 Hydridization Theory C. Hybridization Theory Actually, the electrons in the molecule are not stationary, they undergo wave like motion. The valence shell electron pair repulsion theory is not good enough to explain this. Therefore, the scientist developed another theory, hybridization theory. Unit 3 Syllabus Notes Questions that cannot be answered by VSEPRT 1. electrons are not stationary 2. the strength of the 2 bonds in a double bond are not the same. 3. No. of unpaired e- available for bond formation. 4. shape of some molecules 5. strength of some bonds 1. Overlapping of atomic orbital Instead of fixed electron centre, covalent bond is now viewed as an overlapping of atomic orbitals. In hydrogen, two hydrogen 1s atomic orbitals are overlapped to form a covalent bond. The two 1s orbitals interfere with each other constructively. This causes a higher electron density along the internuclear axis and create the attraction to bring the two atoms together. There are two kinds of overlaps i. head-on overlap ii. sideway overlap Sideway overlap (Laterally overlap) of the orbital give π (pi) bond. Head-on overlap (End to End overlap) of the orbital give σ (sigma) bond. Since the bond strength of a covalent bond is directly proportional to the area of overlap, in most of the case, the σ bond will be stronger than the π bond. Also, the π electrons, unlike σ electrons, are not concentrated along the internuclear axis. In large atoms, π bonds are not favorable because, being removed from the line between the centres of the atoms, the π bond rapidly weakens as the size of the atom increases. e.g. C=C double exists but Si=Si double bond doesn't exist. I. Shape of molecule Unit 3 Page 2 σ bond and π bond can be distinguished by the shape of their cross section. σ bond has a cylindrical cross section about the bond axis but π bond does not has a cylindrical cross section. Orbital representation of one σ bond and two π bonds ` 2. Hybridization theory Cross section Hybridization means mixing. Hybridization theory addresses many questions that cannot be explained by Valence Shell Electron Pair Repulsion Theory. i. electrons are not stationery, they undergo wave like motion. ii. The strength of C=C (612 kJmol-1) is not twice of the strength of C–C (347 kJmol-1) bond. i.e. the two bonds in double bond are not equvalent. iii. There are not enough unpaired electrons in a C atom in ground state (1s22s22p2) for the formation of 4 single bonds. iv. The four single bonds in CH4 are equivalent and the shape is tetrahedral. Consider CH4 molecule, experimental findings tell us that the 4 C–H bonds are equivalent and the molecule is tetrahedral A C atom in ground state has only 2 unpaired electron in 2 different 2p orbitals which can only form 2 single bond. If one electron is promoted from 2s orbital to 2p orbital, you will get 4 unpaired electrons for bond formation. However, if the 4 unpaired electrons overlap with 1s electron of the 4 H atoms independently, the bond formed will not have equivalent strength. This is because the 4 unpaired electrons are not equivalent. One is a 2s electron and 3 are 2p electrons. Furthermore, the bond angle will not be 109½º any more. The angle between different p orbitals are only 90º. Therefore, scientist proposed that the 4 unpaired electrons interfere with each other first (i.e. mixed with each other) to form 4 equivalent orbitals before any bond is formed. Because the new orbitals are formed from one s orbital and three p orbital. They are called sp3 hybridized atomic orbitals or sp3 hybrid orbitals. 1s hybridized C* sp3 excited state C * 1s ground state C 2s 2p 2 2 110 1s 2s 2p 2 1 111 2 1111 It can be seen that the main purpose of developing hybridization theory, or any new theory, is to explain the experiemental findings. I. Shape of molecule Unit 3 Page 3 The process of combining the orbitals is called hybridization. Note : Hybridized atomic orbitals = ∑ atomic orbitals 4- r t l oba s i no. of hybridized atomic orbitals = no. of atomic orbitals used hybridization excited state hybridized atomic orbital overlapping of orbitals (bond formation) In some of the hybridization processes, electron is promoted to the higher orbital first. e.g. from 2s to 2p orbital in C. This requires energy. Since bond formed will be stronger with hybridization, this will be paid off by the energy released during bond formation. Since hybridization is just a kind of interference, only the orbitals with similar energy can be hybridized. i.e. 2s orbital cannot hybridize with 3s orbital promotion ground state ∆H (more stable product) Hybridized orbital sp sp 2 sp 3 3 3 sp d or dsp (not required) sp3d2 or d2sp3 (not required) * Table needed to be remembered. 3. Examples No. of electron pairs (bond & lone pairs) 2 3 4 5 6 Geometrical arrangement Linear Trigonal planar Tetrahedral Trigonal bipyramidal Octahedral Typical examples BeCl2, C2H2, CO2 BeF3, C2H4 CH4, NH3, H2O PCl5 Fe(CN)63+,SF6 a) sp hybridization e.g. C2H2 (linear) At first, one 2s electron is promoted to 2p orbital. One 2s orbital and one 2p orbital undergo hybridization to form two hybridized orbitals. sp hybridized orbitals consists of two lobes with linear geometry. One sp hybridized orbital overlaps with the 1s orbital of H to form a C–H σ bond. Another sp hybridized orbital overlaps with the sp hybridized orbital of another C to form a C–C σ bond. The two p orbitals on one C overlap with the two p orbitals of another C laterally to form 2 C–C π bond. It can be seen that not all the three C–C bond are equivalent. Because of the cylindrical symmetry of the σ bond, the C–H bond is free to rotate. Rotation of the C≡C bond will break the two π bonds, therefore C≡C bond is not free to rotate. ground state C 1s 2s 2p 2 2 110 1s 2s 2p * promotion→ excited state C 2 1 111 1s sp 2p ybridization h → hybridized C * 2 11 11 Orbital representation of ethyne (H–C≡C–H) I. Shape of molecule b) sp2 hybridization e.g. BF3 (trigonal planar) 1s ground state B 2s 2p Unit 3 Page 4 1s 2s sp2 2p 2p 2 2 100 promotion→ hybridization excited state B* 1s 2 1 110 2 111 0 → hybridized B* The three sp2 hybridized orbitals overlap with 2p orbitals of the three fluorine to form 3 σ bonds. sp2 hybridized orbitals consists of three lobes with trigonal planar geometry. On B, there is still an empty 2p orbital and it does not fulfill octet. BF3 is an electron deficiency species and ready to accept electron from other molecule. e.g. C2H4 (trigonal planar) 1s ground state C 2s 2p one 2s and two 2p orbitals sp2 hybridized orbitals 1s 2s sp2 2p 2p 2 2 110 promotion→ excited state C* * 1s 2 1 111 2 111 1 ybridization h → hybridized C Unlike BF3, the unhybridized p orbital of the C at right angle to the plane has an unpaired electron. The p orbitals from the two C overlap laterally to form a π bond. Since the π bond is not free to rotate, ethene has a planar structure. I. Shape of molecule c) sp3 hybridization Unit 3 Page 5 e.g. CH4 (tetrahedral) 1s ground state C 2s 2p 1s 2s sp3 2p 2 2 110 promotion→ excited state C* 1s 2 1 111 2 1111 * ybridization h → hybridized C sp3 hybridized orbitals consists of four lobes with tetrahedral geometry directed to the four vertices of a tetrahedron. CH4 e.g. NH3 (trigonal pyramidal) 1s ground state N 2s 2p ybridization h → hybridized N 1s sp3 2 2 111 2 1112 One of the sp3 hybridized orbital is occupied by an electron pair. This is the lone pair electrons of the molecule. NH4 e.g. H2O (angular) 1s ground state O 2s 2p ybridization h → hybridized O 1s sp3 2 2 211 2 1122 In CH4 molecule, being symmetrical, adopts the bond angle of regular tetrahedron 109º29'. In NH3 molecule, the bond angle is decreased by 2.2º (106º45'). In H2O molecule, the bond angle is decreased by 5.0º (104º31') from the tetrahedral value. The tend can be predicted by the valence shell electron pair repulsion theory which state that the lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion and a lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion. H2O It can be observed that the bond angle is related to the percentage of s character in the hybridized orbital. sp – 180º (50% s character), sp2 – 120º (33% s character), sp3 – 109½º (25% s character) The higher the s character, the larger will be the bond angle. I. Shape of molecule Unit 3 Page 6 d) sp3d hybridization (not required in A-Level) e.g. PF5 (trigonal bipyramidal) 3s 3p P [Ne] 3d 3s 3p sp3d 3d 3d 2 111 00000 → promotion P * [Ne] 1 111 10000 0000 * ybridization h → P [Ne] 11111 The hybridization process involves the promotion of an electron from 3s orbital to 3d orbital. The central atom, P, does not obey octet rule, this is known as expansion of the octet. This is impossible for elements in period 2, as there is no low lying energy d orbital available. i.e. Energy required to promote 2p electrons to 3s orbital will be large that cannot be paid off in the bond formation process. e) sp3d2 hybridization (not required in A-Level) e.g. SF6 (octahedral) [Ne] 3s23p4 3s S [Ne] 3p 3d 3s 3p sp3d2 3d 3d 2 211 00000 promotion→ S * [Ne] [Ne] 1 111 11000 111111 000 ybridization h → S * The six sp3d2 hybrid orbitals are identical. They overlap with the p orbitals of the six fluorine atoms to form a perfect octahedral. e.g. XeF4 (square planar) [Kr] 4d105s25p6 5s Xe [Kr] 4d 10 5p 5d 5s 5p sp3d2 5d 5d 2 222 00000 → promotion hybridization Xe [Kr] 4d 10 2 211 11000 221111 000 F F Xe F F → Xe [Kr] 4d 10 Not all noble gases can form compound. As the principle quantum number increases, the energy gap between the orbitals gets smaller. For heavy noble gas atom, electron from lower energy level may be promoted to higher energy level to yield unpaired electrons for bonding. I. Shape of molecule Unit 3 Page 7 Glossary Lewis structure octet expansion excited state Oxidation no. formal charge Valence shell electron pair repulsion theory Hybridization theory σ(sigma) bond π(pi) bond ground state I. Shape of molecule Unit 3 90 2B 6 a i iii v Page 8 Past Paper Question 90 2B 4 d 94 2B 4 c 95 1A 2 a 98 2B 5 a i 90 2B 4 d 4 Account for the following observations. 4d Oxygen usually forms compounds with oxidation numbers -2, -1 and 0, whereas sulphur can form compounds with many more oxidation numbers ranging from -2 to +6. Sulphur has empty low lying energy 3d orbital - to expand its coordination sphere, thus capable of forming compounds with a wide range of oxidation number. 1 mark Oxygen has a much higher electronegativity than S and therefore it usually acts as an oxidant. 1 mark Oxygen cannot expand its octet/has no low lying energy orbital for bonding. 1 mark 90 2B 6 a i iii v 6a The following species are either impossible to prepare or very unstable. Explain, in each case, why this is so. i NCl4 Nitrogen cannot have 9e- in its valence shell and cannot expand its octet because N has no low lying energy orbitals for bonding. 1½ mark iii ArCl2 Ar does not form compounds because of the stability of close-shell electronic configuration. 1½ mark 2v [PO4] P cannot have oxidation state higher than +5. 1½ mark 3 1½ 1½ 1½ 94 2B 4 c 4c What is the highest oxidation state of iodine? Give an example of an iodine-containing compound in which iodine 2 is in this oxidation state. Hightest O.S. of I = +7 1 mark 1 mark Example : IF7, IO495 1A 2 a 2a Explain why phosphorus can form PCl3 and PCl5, while nitrogen can form only NCl3. P has low-lying empty 3d orbital / energy level, promotion of e- to 3d, gives 5 unpaired electrons. It is capable to form 5 covalent bond. 1½ mark In N, there is no low-lying d orbitals, excitation of e- is not possible. ∴ It can form only 3 covalent bonds. ½ mark 98 2B 5 a i 5a Consider the following compound F. CH3 a b c F CH CH C CH d 2 5 i Give the hybridization states of the carbon atoms, a, b, c and d. I. Shape of molecule Unit 4 Page 1 Topic Reference Reading I. Shape of molecule 4.4 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 14–23, 33–34, 122–129 Organic Chemistry, Solomons, 6th Edition pg. 52–59, 61–65 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 110–111, 141–143 The unique nature of carbon Structure and Shape of hydrocarbons 4. Structure and shape of hydrocarbons Unit 4 Syllabus Notes a) Characteristic of Carbon–Carbon bond i. Bond strength Comparing with other single bond, C–C bond is the one with the highest bond strength. Therefore, it is capable to form a stable long chain molecule. This characteristic is called catenation of C atom. Bond C–C N–N F–F Cl–Cl Br–Br I–I Bond energy +346 kJmol-1 +163 kJmol-1 +158 kJmol-1 +242 kJmol-1 +193 kJmol-1 +151 kJmol-1 ii. Ability to form multiple bond Furthermore, C is capable to form relative stable double bond and triple bond. Bond Bond energy / kJmol-1 C–C +346 C=C +610 C≡C +837 This make a lot of different kinds of connection possible in the organic molecule. There is over millions of different organic compounds identified and may be billions not identified. b) Shape of hydrocarbons (1) Saturated hydrocarbons In saturated hydrocarbons, all C atoms are tetrahedrally bonded. The arrangement can be viewed as the repulsion of the four bonding electron pairs or the result of sp3 hybridization. Furthermore, since σ bond has cylindrical (circular) cross section and the area doesn't change when the bond rotates, all σ bond are free to rotate. Shape of methane Shape of ethane I. Shape of molecule (2) Unsaturated hydrocarbons Rigidity of double and triple bond – Unit 4 Page 2 Rotation of the C=C and C≡C bonds will break the one π bond and two π bonds respectively, therefore C=C and C≡C bond are not free to rotate. Shape of ethene (H2C=CH2) (3) Aromatic hydrocarbons By X-ray diffraction, it was found that the six C–C bonds in benzene are equivalent and with the bond length intermediate between single C–C bond and double C=C bond. Shape of ethyne (H–C≡C–H) Type of bond Single C–C Benzene C–C Double C=C Triple C≡C Bond length / nm 0.154 0.139 0.134 0.120 Kekulé structure According to energetic studies, the actual benzene is more stable that the Kekulé structure. This is due to the delocalization of the π electrons / resonance of π electrons (a) Delocalization of π-electrons 2s C ground state [He] 2p 2 110 2s 2p → promotion C excited state [He] 1 111 sp 2 p hybridization → C sp2 hybrid atomic orbital [He] 111 1 The six carbon atoms undergo sp2 hybridization. One lobe of each C sp2 hybrid orbital overlaps with lobe of another C to form C–C σ bond. Another lobe overlap with 1s orbital of H to form C–H σ bond. The p orbitals overlap above and below the plane laterally to form an extensive π orbital. This gives benzene a planar structure which guarantees maximium p orbital overlapping. I. Shape of molecule (b) Stability of benzene Unit 4 Page 3 Chemical properties of benzene, hexane, cyclohexane and cyclohexene H H H H Benzene Reaction bromine in CCl4 (in dark) Hexane Benzene No reaction. Br2 decolorizes slowly only in the presence of Fe catalyst with evolution of HBr. (Electrophilic substitution) No reaction. No reaction. Hexane No reaction. Upon heating, Br2 decolorizes slowly with evolution of HBr. (Radical substitution) No reaction. No reaction. H H H H H H HH H H H H H H H HH H H HH HH H Cyclohexane Cyclohexane Similar to hexane. (Radical substitution) Cyclohexene Cyclohexene Decolorize immediately. (Electrophilic addition) acidified potassium manganate(VII) KMnO4(aq)/H+(aq) neutral potassium manganate(VII) KMnO4(aq) alkaline potassium manganate(VII) KMnO4(aq)/OH-(aq) No reaction. No reaction. No reaction. No reaction. No reaction. Purple manganate(VII) decolorize immediately. (Oxidation) Purple manganate(VII) turns to black ppt. of manganese(VI) oxide, MnO2(s). (Oxidation) Purple manganate(VII) turns to green manganate(VI), MnO42(aq), immediately. (Oxidation) The chemical properties of benzene is so different from ordinary alkene e.g.(cyclohexene). It resembles the properties of alkane instead. This is due to the extra stability from delocalization of π electrons. For example, its reaction with bromine is a substitution reaction (electrophilic substitution) instead of addition reaction (electrophilic addition) in other typical alkenes. This will be explained with more details in the section of reaction mechanism. Glossary catenation electrophilic substitution electrophilic addition reaction mechanism I. Shape of molecule Unit 4 Page 4 Past Paper Question 93 1A 3 a i 94 2C 9 b i 96 1A 3 c i ii 97 1A 4 c I 99 2A 3 b 93 1A 3 a i 3a Consider the following compound, X: i Give the hybridization state of the carbon atoms indicated by a, b and c. a: sp3 b: sp c: sp2 1½ ½ mark each 94 2C 9 b i 9b Ethene and chloroethene can undergo polymerization to give polyethene (PE) and polyvinyl chloride (PVC) respectively. PVC is more rigid and durable than PE, but incineration of PVC causes a more serious pollution problem. i Describe the bonding and shape of the ethene molecule in terms of the type and spatial arrangement of the orbitals 3 involved. There are 4 C–H bonds and 1 C=C bond in an ethene molecule. The C’s in ethene are sp2 hybdrized 1 mark The C–H bond is formed by overlapping of a sp2 hybrid orbital with an atomic orbital of H. The C=C bond consisted of a s bond and a p bond. These bonds are formed by the overlapping of two sp2 hybride orbitals and of the two unhybridized p orbitals on C’s respectively. 1 mark H H C C H H σ bond π bond Since for sp2 hybridization, the hybrid orbitals are arranged on a triangular plane. ∴ Ethene molecule is planar and the C–C–H bond angles are 120º 96 1A 3 c i ii 3c Consider the following structure : 1 mark i ii Give the hybridization states of the carbon atoms a, b and c. a: sp3, b sp, c: sp2 Suggest expected values for the bond angles α, β and γ. a : 108-110º, b : 180º, g : 119-121º (Deduct ½ mark for missing out the degree notation) 1½ 3 × ½ mark 1½ 3 × ½ mark 97 1A 4 c i 4c i Give the hybridization state of the carbon atoms and the bond angles in ethene. 99 2A 3 b 3b Briefly describe the bonding in benzene and give TWO pieces of supportive evidence. 3 I. Shape of molecule Unit 5 Page 1 Topic Reference Reading Syllabus Notes I. Shape of molecule Modern Physical Chemistry ELBS pg. 84–86 Organic Chemistry, Solomons, 6th Edition pg. 23–26 4.5 Molecular Orbital Theory D. Molecular Orbital Theory (Not required in A-Level) According to hybridization theory, all electrons in O2 should be paired up. 1s ground state O 2s 2p 1s sp2 2p Unit 5 2 2 211 → hybridized O hybridization 2 221 1 O undergoes sp2 hybridization. The hybridized sp2 orbital of the two O overlaps with each other to form a σ bond. The unhybridized 2p orbitals of the overlaps laterally to form a π bond. This approach fails to explain why liquid O2 is paramagnetic, a phenomenon arisen from unpaired electron. Scientist developed another theory called molecular orbital theory to explain this. For example, in H2 molecule, the two 1s orbitals can interfere constructively to create a bonding molecular orbital or destructively to create a anti-bonding molecular orbital. In bonding molecular orbital, the electron density is concentrated along the internuclear axis, thus tends to bring the two hydrogen atoms together. In anti-bonding molecular orbital, the electron density is concentrated on the two sides of the molecule, also there is a node(an area with zero electron density) between the two hydrogen atoms. This tends to break the two hydrogen atoms apart. As a result, bonding orbital has lower energy than antibonding orbital. Similar to Aufbau principle in building up of electrons, the molecular orbital with lower energy will be filled first. Each molecular orbital can accommodate 2 electrons. Electron pair in bonding orbital will have opposite spin and electron pair in anti-bonding orbital will have similar spin. N.B. ∑ atomic orbital = molecular orbital = no. of molecular orbital no. of atomic orbital I. Shape of molecule Molecular orbitals of oxygen Unit 5 Page 2 When the two oxygen atoms is approaching each other, the atomic orbitals start to overlap. 1s orbital is not involved in the overlapping since it is inner orbital. The overlappings of 2s, 2px, 2py and 2pz give rise to 6 different energy levels. The 12 electrons fill up the molecular orbitals according to the energy. 2 unpaired electron can be found in the 2 degenerate anti-bonding π* 2p molecular orbital. This explain why oxygen is paramagnetic. No. of electrons in bonding orbital = 10 No. of electrons in anti-bonding orbital = 6 Bond order = (no. of e- in bonding orbital - no. of e- in anti-bonding orbital) ÷ 2 = (10 - 6) ÷ 2 =2 Glossary Past Paper Question Molecular orbital theory paramagnetism II. Ionic bonding Page 1 Topic Reference Reading II. Ionic bonding 4.7 Modern Physical Chemistry ELBS pg. 64–66 Chemistry in Context, 3rd Edition ELBS pg. 133, 192–195 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 99–104, 145–146, 149–151 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 83–86, 115–118, 125–127, 171–173 Strength of ionic bond X-ray diffraction Atomic and ionic size II. Ionic bonding Like all other kinds of bonding, ionic bond is a result of balance between attraction and repulsion. Syllabus Notes There are electrostatic attractions among oppositely charged particle and repulsion among similarly charged particle. The ions will come to equilibrium when the attractive and repulsive forces balance one another an the ions will stay in positions in the crystal lattice. The structure of ionic compound is called giant ionic structure or giant ionic lattice. II. Ionic bonding Different ionic lattices Page 2 CsCl - Simple cubic r+ r- = 0.933 NaCl - Face-centered cubic (Octahedral) r+ r- = 0.524 Two factors govern the packing of ions to form giant lattice: Each ion tends to have the highest no. of neighbours of opposite charge, as this would increase the lattice stability. The number of nearest neighbours is called the coordination umber of the central ion. ii.. Relative sizes of the ions. The arrangement of the ions in an ionic crystal is determined by the ionic radii ratio (r+/r-): Coordination no. 8 6 4 Shape Simple cubic Face centered cubic (Octahedral) Face-centered cubic (Tetrahedral) r+/r0.73 and above 0.41 - 0.73 0.22 - 0.41 Example CsCl NaCl ZnS i. Ionic bonding is non-directional in nature. An ion is surrounded by many oppositely charged ions in all directions, i.e. no particular orientation is favoured. Characteristics of ionic compounds 1. 2. 3. 4. Good electrolytes. Hard solids because the inter-ionic forces within an ionic crystal are usually strong. High melting point. Soluble in water and insoluble in organic solvents. A. Strength of ionic bond - lattice energy By definition, mMn+(g) + nXm-(g) → MmXn(s) ∆Holat Lattice energy can be served as a good estimate for the strength of the ionic bond. It has a typical value range from 600–1000 kJmol-1. II. Ionic bonding B. X-ray diffraction The structure of an ionic crystal can be determined by a technique called X-ray diffraction. When X-rays strike a crystal, they are diffracted by the electrons in the atoms or ions. The larger the atom, the more electrons it possesses and the brighter the spot will be on the diffraction pattern. From the pattern of spots on the X-ray film, the distribution of electron in the crystal can be calculated. Points of equal density in the crystal are joined by contours giving an electron density map. Page 3 1. Electron density map of sodium chloride From the electron density map, it can be seen that electrons are concentrated near the nuclei in ionic compounds. In NaCl, ions are not perfectly spherical but tend to expand to fill up any empty space present, until their charge clouds experience repulsion from that of the neighbouring ions. The electron density between the ions fall almost to zero. Electron density map of sodium chloride C. Periodicity of ionic radius 1. Definition of ionic radius Ionic radius – the effective radius of ions in crystal lattices. It is the distance measured from the centre of the ion to the region where the electron density is zero. 2. Periodicity of ionic radius Similar to atomic radius, the size of an ion is mainly governed by the strength of the effective nuclear charge. On top of this, it is also affected by the presence of other ions. If an ion is present in a surrounding with few ions around, its electron cloud would tend to expand to fill up the space, until it is repelled by the electron clouds of other ions Size of cation It is always smaller than the size of the corresponding atom as electron is removed. The proton to electron ratio increases. The nuclear charge will have a greater effect on the fewer remaining electrons, which makes the electron cloud contracts. Size of anion It is always bigger than the size of the corresponding atom as electron is added. The additional electron enters the highest energy level without an increase in nuclear charge. This causes greater repulsion among the electrons. The above discussion applies to simple ion only. Certainly, a polyatomic ion is much larger than a simple ion because it contains a larger no. of atoms. II. Ionic bonding 3. Size of isoelectronic particles Page 4 O2- ion, F- ion, Na+ ion, Mg2+ ion are said to be isoelectronic because they contain the same no. of electron and have the same electronic configuration, – 2,8. Glossary Past Paper Question Therefore, their sizes show the order of O2- > F- > Na+ > Mg2+. electrovalent bond coordination no. X-ray diffraction isoelectronic 92 1A 3 b 95 2B 5 a ii iii 97 1A 1 c electron density map ionic radius 92 1A 3 b 3b The size of three anions is in the order: Br- < H- < I-. Briefly explain why H- is smaller than I-, but larger than Br-. I- has more e- shells ([Kr]5s25p6) than H- (1s2). 1 mark 1 mark H- is large due to electron repulsion between 2 electrons in 1s orbital. 2 95 2B 5 a ii iii 5a The table below lists some properties of the alkali metals. Standard electrode Melting point / ºC Element Atomic Ionic radius / nm First ionization potential / V radius / nm energy / kJmol-1 Li 0.123 0.060 520 -3.04 180 Na 0.157 0.095 495 -2.71 98 K 0.203 0.133 418 -2.92 64 Rb 0.216 0.148 403 -2.93 39 Cs 0.235 0.169 374 -2.95 29 ii Explain why the atomic radius is larger than the ionic radius in each element and why the ratio, atomic radius : 3 ionic radius is greatest in the case of Li. The atom has one more electron shell than the cation. 1 mark In M+, the nuclear charge outweighs the screening effect of the electrons and the electron cloud of M+ experiences a stronger attraction. Therefore, the ionic radii are smaller than the atomic radii. 1 marks Atomic radius : ionic radius ratio is highest in the case of Li, because it has the smallest size. In Li+, the eexperience the strongest attraction ∴ contraction in size is greatest. 1 mark iii Explain why the ionic radius of K+ is larger than that of Ca2+, although they have the same electronic 2 configuration. Electronic configuration of K+ and Ca2+ are both 1s22s22p63s23p6. 1 mark Both number and arrangement of electrons are the same in two cations. Ca2+ is doubly charged, while K+ is singly charged, the outermost e- in Ca2+ experiences a stronger attraction (effective nuclear charge). Hence the radius of 1 mark Ca2+ is smaller. 97 1A 1 c 1c Arrange, with explanation, the following chemical species in the order of decreasing size. F, O and O- 3 III. Covalent bonding Page 1 Topic Reference Reading III. Covalent bonding 4.6.0–4.6.2 Modern Physical Chemistry ELBS pg. 78–82, 112–113 Chemistry in Context, 3rd Edition ELBS pg. 101–105 Organic Chemistry, Solomons, 5th Edition pg. 15–18, 506–509 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 114–115 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 86–90, 92–94 H2+ ion Additivity of covalent radius Breaking down of additivity due to resonance Dative covalent bond III. Covalent bonding A. H2+ ion For simplicity, H2+ ion can be used as the simplest model of covalent bond. In H2+ ion, only 1 electrons and 2 protons need to be considered. Syllabus Notes There are only 3 forces involved. 1 repulsion term between the 2 protons and 2 attraction terms between the electron and 2 protons. The bond length (equilibrium distance between two nuclei) depends on the balance of the repulsion and the attraction. Electron density map for the H 2 ion + The two positive nuclei are bond together by sharing the negative electron cloud. This arrangement leads to a lower potential energy than if the electron cloud were not shared. And a lower P.E. results in greater stability. Electron density is not uniform on the electron density map, but is greater along the internuclear axis. The high concentration of electron between two positive nuclei serves to bind the nuclei together - covalent bond. The attraction is even greater with the neutral H2 molecule, where two electrons are available for bonding. This is revealed in the decrease in the internuclear distance and increase in bond dissociation energy. + H2 H2 Internuclear distance (nm) 0.104 0.074 Bond Dissociation energy (kJmol-1) 257 435 III. Covalent bonding B. Electron diffraction Page 2 The electron density map of hydrogen cannot be obtained by X-ray diffraction / X-ray crystallography because hydrogen is volatile and cannot be crystallized easily. And also, the electron density of the hydrogen atom is too low to diffract the X-ray. Another similar method called electron diffraction is used to determine the structure of a gaseous molecule. When a beam of electrons is passed through a gas or vapour at low pressure. The nuclei of atoms in gas molecule scatter (diffract) the electrons. From the diffraction pattern, the distance between the nuclei and the structure of the molecule can be determined. C. Covalent radius 1 Definition of covalent radius Covalent radius – half the distance between two atoms of the same kind held together by covalent bond. 2. Additivity of covalent radius If there is a molecule A–B, the interatomic distance A–B is approximately equal to the arithmetic mean of the A–A and B–B distance. A− B = A− A + B−B 2 and because of this simple relationship, it is possible to obtain interatomic distance of certain compound by simple addition of covalent radii. e.g. Average bond length of C–C bond : 0.154 nm : 0.074 nm Bond length of H–H bond in H2 0.154 nm + 0.074 nm By calculation, bond length of C–H bond = = 0.114 nm 2 Average bond length of C–H bond determined by experiment : 0.108 nm The two values rather agree with each other. 3. Breaking down of additivity in covalent radius and bond energy This simple additive relationship of covalent radii was found to fail for certain molecules. e.g. benzene. The reason is the molecules are actually resonance hybrid of two or more possible structures. There are compounds whose bond length differ from calculated values. e.g. benzene and nitrate ion. Compound benzene C=C C–C N–O N=O Calculated 0.134 nm 0.154 nm 0.136 nm 0.115 nm Bond lengths (nm) Experimental 0.139 nm 0.121 nm NO3− All bond lengths are found to be equivalent. This is due to the delocalization of electrons / reasonance of electrons. III. Covalent bonding a) Resonance / Delocalization of electrons (1) Resonance structure of benzene In benzene molecule, six carbon atoms are joined together in a ring by overlap of carbon sp2 hybrid orbitals, forming C–C σ bonds. C–H σ bonds are formed by further overlap with hydrogen 1s orbitals. the p-orbitals of each carbon atoms overlap with each other to formal a new molecular orbital. The electrons are delocalized in a system of π orbitals above and below the ring. This make all the six C–C bonds equivalent. Delocalization of electron is also called resonance of electrons. Two Lewis structures are used to represent the real structure. Page 3 or Resonance structures of benzene Representation of hybrid Since one π bond is shared between two C–C bonds, the bond order of the C C bond is said to be 1½ . (2) Resonance structure of nitrate ion Usually, formal charge is included in the resonance structure and the hybrid structure to see the distribution of the electron. Since one π bond is shared among three N–O bonds, the bond order of the N O bond is said to be 12 . N.B. Resonance structures are only representations on paper. They do not exist. The real structure of the molecule cannot be represented by any individual resonance structure. Furthermore, the electrons are not shifting forth and back in the structure. Instead, they are evenly distributed in the structure. Orbital representation is always the better way to represent the actual structure. sp2 O sp2 O N O sp2 sp2 (3) Rules in writing resonance structures i. ii. iii. iv. Resonance structures exist only on paper. In writing resonance structures, only electrons are allowed to move. The actual molecule or ion will be better represented by a hybrid of these structures. The energy of the actual molecule is lower than the energy that might be estimated for any contributing structure. III. Covalent bonding b) Breaking down of additivity in bond enthalpies The presence of delocalization of π electrons is further confirmed by the breaking down of addivitiy in bond enthalpies. Page 4 H H H C C C C H C C H H or H H H H Kekulé structure of benzene H H or German scientist Kekulé proposed that benzene is a ring compound with alternating C–C single bond and C=C double bond joining the carbon atoms together. To verify the validity of the structure, some reactions of benzene are studied. e.g hydrogenation Hydrogenation of benzene H H H H Benzene H + 3H2(g) H (l) H H H HH H H H H HH H Cyclohexane Estimation of enthalpy of hydrogenation by bond energy terms Bond broken 3 C=C 3 H–H Bond formed 3 C–C 6 C–H E(C–H) E(H–H) E(C–C) E(C=C) +413 kJmol-1 +436 kJmol-1 +347 kJmol-1 +612 kJmol-1 ∆H = energy required - energy released = ΣE(bond broken) - ΣE(bond formed) = (3 × E(C=C) + 3 × E(H–H)) - (3 × E(C–C) + 6 × E(C–H)) = (3 × 612 + 3 × 436) - (3 × 347 + 6 × 413) = -375 kJmol-1 (experimental value -210 kJmol-1) The calculated value is so differed from the experimental value, less heat is released than the calculated value. The difference between these values indicates that the actual structure of benzene is more stable than Kekulé structure by (-210) - (-375) = 165 kJmol-1 III. Covalent bonding (1) Delocalization energy This extra stability is attributed to the delocalization of the bonding electrons over all 6 carbon atoms. Page 5 H The double bond electrons are not localized at fixed positions, instead, they delocalized around the ring and make the 6 C–C bonds equivalent. This distribution minimizes the repulsion among the electrons and give a more stable structure. 165 kJmol-1 (delocalization energy) -375 kJmol-1 H H H H Resonance structure of benzene -210 kJmol-1 H H or The breakdown of the additivity rule demonstrates the importance of bond energy terms in understanding the chemical bonding in benzene. III. Covalent bonding D. Dative covalent bond Dative covalent bond – Page 6 The linkage of two atoms by a pair of electrons, both electrons being provided by one of the atoms. The dative covalent bond is also known as coordinate bond. The formation of a dative bond requires an unbonded(lone) pair of electrons in one atom (donor) and an empty orbital in another atom (acceptor). 1 Examples of H3N→BF3 and Al2Cl6 a) H3N→BF3 Boron trifluoride does not possess an octet of electrons; this is a suitable compound with which to introduce dative covalency. BF3 gas reacts with NH3 gas to give a white solid with the composition NH3BF3. 1s ground state B 2s 2p 1s 2s sp2 2p 2p 2 2 100 → promotion hybridization excited state B* 1s 2 1 110 2 111 0 → hybridized B* B undergoes sp2 hybridization and uses the three sp2 hybridized orbitals to overlap with 2p orbitals of fluorine. BF3 has a planar structure and a vacant 2p orbital. The 2p orbital accepts the lone pair from NH3 and the hybridization of B changes from sp2 to sp3. b) Al2Cl6 In gaseous state, two AlCl3(g) molecule dimerizes to form Al2Cl6(g) molecule. 3s 3p 3s 3p Al [Ne] 2 100 promotion→ hybridization Al* [Ne] 1 110 * sp 2 3p → hybridized Al [Ne] 111 AlCl3(g) has a planar structure similar to BF3(g). The vacant 3p orbital accepts lone pair electrons on chlorine atom to form dative bond. 0 Cl Al Cl Cl Cl Al Cl Cl Glossary Octet rule electron diffraction additivity of covalent radius resonance delocalization delocalization energy dative covalent bond / coordinate bond hybrid structure III. Covalent bonding Page 7 92 2B 5 a iii Past Paper Question 92 2A 3 b iii 95 1A 1 c 96 2A 1 a i ii 98 2A 1 c ii 92 2A 3 b iii 3b iii The covalent radius of carbon is 0.077 nm. The measured carbon-carbon bond length in benzene is 0.139 nm. Estimate the carbon-carbon bond length in ethane. Explain any difference in the carbon-carbon bond lengths in these two molecules. 1 mark C–C bond length of CH3–CH3 = 2 × 0.077 = 0.154 nm C–C bond in ethane is a single bond which is longer than the carbon-carbon bond in benzene. In benzene, it is 1 + 1 mark because of delocalization electrons in the ring. The average bond order is larger than 1. 92 2B 5 a iii 5a iii Draw resonance electronic structures of CO and CO2, showing lone pair electrons and charges where appropriate. 3 2 CO OCO CO CO Any 2 structures C Any 2 structures ½ mark each Very few candidates were able to draw resonance structures for CO and CO2. The charges on atoms in these structures are neglected in many textbooks, but the students should be able to deduce these by "book-keeping" of the lone pair and bond pair electrons. OCO OCO 95 1A 1 c 1c Account for the fact that the carbon-oxygen bond lengths in CO, CO2 and CO32- are 0.113, 0.116 and 0.129 nm respectively. CO CO -+ 3 ½ mark ½ mark ½ mark ½ mark Bond order = 3 / the bond formed between C & O is triple bond. It is the shortest. CO2 OCO Bond order = 2 / the bond formed between C & O is double bond. CO32- It is a resonance hybrid of the following : O -O C O-O -O C O O OC O- C ½ mark 1 Bond order = 13 (intermediate between C–C and C=C bond). It is the longest. ½ mark This question was not well answered. The answer required descriptions or diagrams of the electronic structures of the three compounds and the deduction of the bond orders. It is more meaningful to show lone pairs of electrons, and formal charges in these electronic structure diagrams. Many candidates could not give the correct structure for CO, and few described the resonance of the CO32- structures. III. Covalent bonding 96 2A 1 a i ii 1a For the hydrogenation of buta-1,3-diene, H2C=CH–CH=CH2(g) + 2H2(g) → CH3CH2CH2CH3(g) the experimental molar enthalpy change is -239 kJmol-1. i Estimate the molar enthalpy change for the above hydrogenation using the bond energy terms below : 2 Bond Bond energy term / kJmol-1 H–H 436 C–H 413 C–C 346 C=C 611 1 mark ∆H = -2EC–C - 4EC–H + 2EC–C + 2EH–H or = -2(346) - 4(413) + 2(611) + 2(436) (1 mark) 1 mark = -250 kJmol-1 (0 marks for omitting the eve sign; deduct ½ mark for no units) ii Explain why the estimated value differs from the experimental value. 2 Buta-1,3-diene is a conjugated system / delocalization of π electrons / resonance / 1 mark mesomerism occurs in buta-1,3-diene. or Page 8 H H C H or H2C CH CH CH2 H C C H (1 mark) + H2C CH CH CH2 etc. C H (1 mark) The molecule is thus stabilized. ∴ energy released in hydrogenation is less than that in non-conjugated system. 1 mark (Award 1 mark for the explanation: the bond energy given is only the average bond energy / bond energy of a covalent bond depends on its bonding environment) Very few candidates related the difference in ∆H values to the delocalization of π electrons in the conjugated system. C 98 2A 1 c ii 1c ii BF3 reacts with NH3 to form an adduct, BF3·NH3. Account for the formation of the adduct and draw its threedimensional structure. IV. Bonding intermediate between ionic anc covalent Page 1 Topic Reference Reading IV. Bonding intermediate between ionic and covalent 4.8 Chemistry in Context, 3rd Edition ELBS pg. 111–113 Organic Chemistry, Solomons, 6th Edition pg. 38–41 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 111–114 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 96–99, 162 Intermediate type of bond Incomplete electron transfer in ionic compound Electronegativity Polarization of ionic bond Polarity in covalent bond IV. Bonding intermediate between ionic and covalent A. Differences between ionic bond and covalent bond The similarity in nature of ionic and covalent bonds is reflected by the same order of magnitude of lattice energies (600-1000 kJmol-1) and covalent bond energies (300-1000 kJmol-1). i. Distribution of electrons In case of ionic solid, in certain region, the electron density between the ions fall almost to zero. In case of covalent molecule, the electron density is highest between the bonding atoms and decreases outwards. The electron density is concentrated along a line between 2 nuclei. Syllabus Notes ii. Ionic compounds are formed by complete transfer of electron. Covalent molecules are formed by sharing of electron among the bonding nuclei. iii. In ionic compounds, oppositely charged ions attract electrostatically to one another. In covalent compounds, the positive charged nuclei also attract electrostatically the electrons shared by the atoms involved. Moreover, both of them are electrostatic in nature. B. Incomplete electron transfer in ionic compound 1. Electron density map of LiF comparing with those of NaCl and H2 NaCl (nearly purely ionic) LiF (intermediate between ionic and covalent) H2 (purely covalent) All the bonding arise from the interaction between the electrons and the nuclei. Different arrangements of the electrons and atoms will give different kind of bonding. From the electron density map, one can see that electrons are concentrated near the nuclei in ionic compounds while there are substantial electron density between the nuclei of hydrogen atoms. In the LiF, its electron density map shows that it is somewhere between a typical ionic compound and covalent compound. IV. Bonding intermediate between ionic anc covalent 2. Difference among lattice energies of Na, Ag and Zn compounds Page 2 Besides the Born-Haber cycle, the lattice energies can also be calculated in terms of electrostatic interactions among ions. The mathematical treatment of the calculation is beyond the scope of the syllabus, but it is worthy to note the basic assumptions and the limitation of the model. Further reading about the calculation of lattice energy can be found on pg. 71–73 of Modern Physical Chemistry ELBS. Assumptions of the mathematical model for calculation: i. The geometrical arrangement of the ions within the crystal and the interionic distances are known. These can be determined by a method called X-ray diffraction. ii. The bonding within the crystal is completely ionic and all ions are spherical in shape, i.e. no sharing of electron. a) Lattice energies of sodium halide, silver halide and Zinc sulphide Compound NaCl NaBr NaI AgCl AgBr AgI ZnS Theoretical lattice energy -777 kJmol-1 -731 kJmol-1 -686 kJmol-1 -768 kJmol-1 -759 kJmol-1 -736 kJmol-1 -3427 kJmol-1 Experimental lattice energy from Born-Haber cycle -781 kJmol-1 -742 kJmol-1 -699 kJmol-1 -890 kJmol-1 -877 kJmol-1 -867 kJmol-1 -3615 kJmol-1 Percentage error 0.5 % 1.5 % 1.9 % 15.9% 15.5% 17.8% 5.5% b) Bonding intermediate between covalent and ionic In sodium halide, the theoretical values calculated from the mathematical model close to the values determined by the Born-Haber cycle. This is because the difference in electronegativity between sodium and halogen is very large, thus the electron transfer is almost complete and shows no evidence of electron sharing. This agrees with the basic assumption of the model. The great discrepancy of between the theoretical values and the experimental values of silver halides and zinc sulphide implies that our model for calculation is not realistic. Owing to the small size of the silver ion which polarizes the large halide ion easily, the bond possesses some covalent characters. The experimental values also suggest that the bonds are stronger than the expected. This is also an evidence of the presence of bonding intermediate between covalent and ionic. IV. Bonding intermediate between ionic anc covalent The discrepancies implies the presence of an appreciable proportion of covalent bonding. Page 3 Partial covalent properties is due to incomplete electron transfer in ionic compound and this is called polarization of ions. From spectroscopic studies of the vapours of alkali metal halides, the internuclear distance in these molecule is less than in the corresponding ionic bond. Halide LiBr LiI Crystal 0.275 0.300 Internuclear separation (nm) Vapour 0.217 0.239 This implies a stronger bond than in the crystal. This is caused by a higher concentration of electron density between the nuclei than in the crystal. This arises from distortion of the electron cloud of one ion or both from a spherical distribution. Polarization of an ion representing a transition from ionic bonding to covalent. 3. Polarization of ionic bond When an ionic bond is formed between the cation X+ and anion Y-, owing to its charge, the X+ ion would tend to attract the outer electrons in the charge cloud of Y-. This causes a displacement of the outer electrons of Y- back to X+, so that the outer electron of X is not completely transferred to Y. The ionic bond formed is said to have some covalent character. a) Fajans’ Rules in polarization of ionic bond i. The polarizing power of a cation is increased as (a) the size is reduced and (b) the charge is increased. ii. The polarizability of anion increases as (a) the size increases and (b) the charge is increased. iii. Covalent character is stronger for cations which do not have a noble gas electron arrangement (d electrons have poor shielding properties). Thus Na+ is less polarizing than Cu+. i.e. Ions of transition metal are usually much smaller than the ions of main group metals) IV. Bonding intermediate between ionic anc covalent C. Electronegativity 1. Definition of electronegativity Page 4 Electronegativity – A measure of tendency of an atom in a stable molecule to attract electrons within a bond. In H–Cl molecule, the bonding electrons are much closer to the chlorine than to the hydrogen; the chlorine being the more electronegative atom. 2. Pauling scale of electronegativity There are several method to measure the electronegativity of an element. One of the commonly used method is Pauling scale of electronegativity. Pauling defined the electronegativity of an atom as the power of that atom in a molecule to attract electrons. In contrast with electron affinity, it is a measure of the power of an single gaseous atom to attract electrons. In Pauling scale, 4 is assigned to the most electronegative atom, fluorine. Pauling electronegativity values of some elements (The shaded ones are more electronegative than C) H 2.1 C 2.5 Si 1.8 Ge 1.8 Sn 1.8 Li 1.0 Na 0.9 K 0.8 Rb 0.8 Cs 0.7 Be 1.5 Mg 1.2 Ca 1.0 Sr 1.0 Ba 0.9 B 2.0 Al 1.5 _ _ N 3.0 P 2.1 As 2.0 Sb 1.9 O 3.5 S 2.5 Se 2.4 Te 2.1 F 4.0 Cl 3.0 Br 2.8 I 2.5 In general, electronegativity values increase from left to right across each period and decrease down each group. D. Polarity in covalent bond 1. Deflection of a liquid jet by an electric field A dry glass rod rubbed with polythene develops a positive charge. An ebonite rod rubbed with fur develops a negative charge A number of liquids are tested Liquid showing a marked deflection Water Trichloromethane Propanone Ethoxyethane Nitrobenzene Cyclohexene Ethanol Liquid showing no or small deflection Tetrachloromethane Benzene Cyclohexane IV. Bonding intermediate between ionic anc covalent The zero or very small deflection with tetrachloromethane is due to the symmetry of the molecule in which the four dipoles counteract each other exactly. Molecules which do not possess a permanent dipole moment are nevertheless polarised to some extent when placed in an electrical field. 2. Dipole moment Page 5 If two charges of equal magnitude but opposite in sign, +q and -q, are separated by a distance d, then the system is said to have a dipole moment (µ) of magnitude give by: = = Dipole moment (µ) charge × distance between the +ve and -ve centres q×d This dipole moment is usually expressed in terms of Debye (D). 1 D ≈ 3.34 × 10-30 Cm. For example, in HCl, Cl attracts electrons more strongly than H does, the dipole moment of the HCl is 1.1 D. Cl is said to exert a negative inductive effect. a) Vector quantity of dipole moment Dipole moment is a vector quantity. For a molecule with more than one polar bond, the dipole moment of the molecule is given by the vector sum of the dipole moments caused by the various polar bonds. In asymmetric molecules (e.g. HCl, CH3CH2Cl), the center of the positive charge and the center of the negative charge do no coincide and a permanent dipole moment results. b) Polarity of molecule The overall distortion of charge in molecules, which results from unequal sharing of electrons, is known as polarization. It represents the departure of the bond from purely covalent, it introduces some ionic character into the bond. Polarization of ions represent the existence of some covalent character in the ionic bonding. Polarization of a covalent bond represents the existence of some ionic character in the covalent bond. IV. Bonding intermediate between ionic anc covalent c) Factors affecting dipole moment Page 6 The dipole moment of a molecule is affected by i. inductive effect – a substituent effect on an organic compound due to the permanent polarity or polarizability of groups. This is transmitted through σ bond. ii. mesomeric effect / resonance effect – delocalization of electron via π bond. iii. presence of lone pair. (1) Inductive effect (I) This refers to the displacement of electron cloud mainly due to difference in electronegativity. -I effect : electron withdrawing +I effect : electron donating (2) Mesomeric effect / resonance effect (R) This refers to the stabilization due to delocalization of electron through π bond.. -R effect : electron withdrawing +R effect : electron donating For example, in benzenamine, i. N is more electronegative than C and N exerts a -I effect. ii. by resonance, lone pair on N is donated to the benzene ring and N exerts a +R effect. NH2 NH2 NH2 H N H The final dipole moment would be the sum of the two effects. N.B. Benzene ring can serves as a reservoir of electrons in resonance. If it is attached to a +R group, it will behave as a -R group. On another hand, if it is attached to a -R group, it will behave as a +R group. (3) Presence of lone pair Lone pair can be considered as a negative centre and contributes a dipole moment. For example, experimentally, NH3 has a stronger dipole moment (1.46 D) than NF3 (0.24D). But 1. The difference in electronegativity between N(3.0) and F(4.0) is greater than that between N(3.0) and H(2.1). 2. The NF3 has a bond angle smaller than NH3 has. According to these two factors only, NF3 should has a stronger dipole moment than NH3. But in NF3, the lone pair imposes a dipole moment opposing the dipole moment of N–F, this results in a much weaker resultant dipole moment. Furthermore, the direction of the dipole moment cannot be determined because we do not known which is stronger. IV. Bonding intermediate between ionic anc covalent Page 7 Glossary Past Paper Question polarization of ion polarizing power polarizability Fajans rules electronegativity Pauling scale dipole moment vector inductive effect mesomeric effect / resonance effect 97 2A 1 a i ii iii 98 1A 1 c ii 97 2A 1 a i ii iii 1a i Explain the terms 'dipole' and 'dipole moment', using HBr as an example. ii Explain why the dipole moment of HF is greater than that of HI. iii State the effect of an electric field on molecules of the following compounds and explain the effect in terms of dipole moment. Cl Cl Cl 7 Cl 98 1A 1 c ii 1c The theoretical lattice enthalpies of NaCl(s) and AgCl(s), and the experimental lattice enthalpy of AgCl(s) are given in the table below. Compound Theoretical lattice enthalpy Experimental enthalpy NaCl(s) - 770 ? AgCl(s) - 833 - 905 ii Why is there such a large difference between the theoretical and experimental lattice enthalpies of AgCl(s) ? 1 V. Metallic bonding Page 1 Topic Reference Reading V. Metallic Bonding 4.9 Modern Physical Chemistry ELBS pg. 74–76 Chemistry in Context, 3rd Edition ELBS pg. 137 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 146–147 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 121–123 Metallic bonding Electron Sea model Delocalization model V. Metallic bonding A. Electron Sea model of metal The metal atoms are closely packed (high coordination number) to form a crystal lattice. Valence electrons of individual atoms are contributed to form a sea of electrons. The lattice of positive ions is then immersed in the sea of electrons. The attraction arises from the electrostatic attraction among the positive ions, mobile electrons and positive ions. Syllabus Notes General properties of metal i. ii. iii. iv. Thermal conductivity – the presence of mobile electrons Electrical conductivity – the presence of mobile electrons Malleability and ductility – non-directional electrostatic attraction between nuclei and mobile electrons Shiny lustre – The mobile electron is excited and re-emits the energy in form of lustre. From another point of view, metallic bond may be considered an delocalization of valence electrons from each atom over the entire metallic crystal. (Further reading about band theory of metallic bonding can be found in Modern Physical Chemistry ELBS pg. 158–159) B. Strength of metallic bond Strength of a metallic bond depends on i. packing efficiency of the crystal Transition metal atoms are much smaller than the atoms in Group 1A and 2A. ii. availability of the valence electron Group 1A metal has 1 valence e-. Group 2A metal has 2 valence e-. Transition metal use 3d e- on top of 2 4s e-. iii. effective nuclear charge In general, the metallic bond strength increases with increasing effective nuclear charge on the valence electrons. However, if the effective nuclear charge is too high, this will lower the availability of valence electrons and the bond strength will drop. e.g. Hg has a first I.E. 10.44 V comparing with 5.14 V of Na. V. Metallic bonding C. Melting and boiling of metal Upon melting, only small amount of the metallic bonds are broken. Majority of the metallic bond have to be broken upon boiling. This causes a big gap between the melting point and boiling point of metal. e.g. Na (371K/1156K), Fe (1809K/3135K) Page 2 In molecular substance, most of the attractions among the molecules have been broken upon melting. This results in a smaller gap between the melting point and boiling point. e.g. N2 (63K/77K), I2 (387K/458K) D. Strength of ionic bond, covalent bond and metallic bond Ionic bond (non-directional) Covalent bond (directional) Metallic bond (non-directional) 0 1000 Estimated by Lattice energy Bond energy terms Atomization energy 2000 Range (kJmol-1) -780 (NaCl) – -3791(MgO) 158 (E(F–F)) – 945.4 (E(N≡N)) 107.3 (Na) – 514.2 (V) 3000 4000 kJmol -1 I onic bond Covalent Metallic Glossary Past Paper Question malleability 92 1A 2 d ii iii 98 1A 3 b i ii ductility lustre 92 1A 2 d ii iii 2d ii Describe the bonding in metallic crystals. ‘Free electron’ model or ‘Mobile electron’ description. iii Of the three energy ranges in kJmol-1 given below: 5 - 100 200 - 700 800 - 1500, which is the most likely energy range for the change M(s) → M(g), where M is a metal? The best answer out of the three is 200-700 kJ mol-1. 1½ 1½ mark ½ ½ mark 98 1A 3 b i ii 3b Sketch the trends for the properties mentioned in (i) and (ii) below, and account for the trend in each case. i melting point of the alkali metals, Li, Na and K 2 ii boiling point of the Period 3 elements, Na, Mg and Al 2 V. Metallic bonding Page 3 Intermolecular forces Intermolecular Forces I. Van der Waals’ forces A. B. Discovery of van der Waals' forces Origin of van der Waals' forces 1. Induced dipole-induced dipole attractions 2. Dipole-dipole interactions 3. Dipole-induced dipole attractions Relative strength of different origins of van der Waals' forces Nature Strength Solubility and hydrogen bond Intramolecular hydrogen bond Examples of hydrogen bonding 1. Ethanoic acid dimer 2. Simple hydrides - CH4, NH3, H2O and HF 3. Structure and physical properties of ice 4. Biochemical importance of hydrogen bond a) DNA b) Protein Strength of van der Waals' forces and hydrogen bond C. II. Hydrogen bond A. B. C. D. E. F. III. Relationship between structures and properties I. Van der Waals' forces Page 1 Topic Reference Reading I. Van der Waals’ forces 4.10.1 Modern Physical Chemistry ELBS pg. 93–95 Chemistry in Context, 3rd Edition ELBS pg. 114–117 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 116–118 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 99–103 van der Waals' forces Introduction to intermolecular forces and intramolecular forces Study of chemistry can be divided into 2 main categories : 1. 2. Chemical properties of substances (e.g. stability and reactivity) These are depending on the intramolecular forces i.e. forces within a molecule – covalent bond. Physical properties of substances (e.g. melting point, boiling point and solubility) These are depending on the intermolecular forces i.e. forces among the molecule. e.g. van der Waals'' forces and hydrogen bond. Intermolecular forces Syllabus Notes There are two type of intermolecular forces : van der Waals'' forces and hydrogen bond. v an der Waals' forces Hydrogen b ond induced dipole-induced dipole attractions dipole-dipole interactions dipole-induced dipole attractions I. Van der Waals’ forces A. Discovery of van der Waals' forces van der Waals, Johannes Diderik (van der vals), 1837–1923, was a Dutch physicist. He discovered that real gas doesn't obey ideal gas equation perfectly. In ideal gas equation, it is assumed that there is no interaction among the gas molecules and the gas molecules occupy no volume. Ideal gas law : PV = nRT where P is the pressure measured V is the volume measured n is the number of mole of gas molecule R is the universal gas constant T is the temperature in Kelvin Instead, the real gas follows another equation named after him, van der Waals' equation or real gas equation. an2 van der Waals' equation : ( P + V2 ) (V - nb) = nRT where a and b are called van der Waals' constant. a is a measurement of the attraction among the particles and b is a measurement of the volume of the gas molecule. I. Van der Waals' forces The values of the empirical constants a and b are depending on the nature of the gas. Gas He N2 CO2 CH3CH2CH2CH3 NH3 a / atmdm6mol-2 0.0341 1.39 3.63 147 4.19 b / dm3mol-1 0.0237 0.0391 0.0427 0.123 0.0372 Page 2 It can be seen that the attractions among the molecules and the size of the molecules increase from He to butane. The weak forces of mutual attraction is called van der Waals forces. B. Origin of van der Waals' forces van der Waals' forces have 3 origins i. induced dipole-induced dipole attractions ii. dipole-dipole (permanent dipole-permanent dipole) interactions iii. dipole-induced dipole attractions 1. Induced dipole-induced dipole attractions induced dipole-induced dipole attractions is a kind of electrostatic attraction arises from fluctuation of electron cloud. Even at 0 K, electrons in a molecule are in constant motion. Slight relative displacement of electrons in a molecule will give rise to a temporary electrical dipole. It will induce another temporary dipole on a neighbouring molecule and an attractive force will be established between the two. Induced dipole-induced dipole attractions also has many other names i. London force – it was first explained by scientist London in 1930. ii. Dispersion force – it is caused by the unsymmetrical dispersion of electron cloud. iv. Temporary induced dipole-temporary induced dipole attractions – dipole is not permanent. Strength of induced dipole-induced dipole attractions depends on the polarizability of the electrons involved. i. larger molecular size ⇒ larger electron cloud ⇒ higher polarizability ⇒ stronger induced dipoleinduced dipole attractions A larger molecule means a larger electron cloud and higher polarizability, this gives strong induced dipoleinduced dipole attractions. In general, an unbranched molecule has a larger size than branched molecule. Butane (b.p. -0.5ºC) has a higher boiling point than 2-methylpropane (b.p. -12ºC). ii. small effective nuclear charge ⇒ more diffuse electron cloud ⇒ higher polarizability ⇒ stronger induced dipole-induced dipole attractions Fluorine is highly electronegative with high effective nuclear charge on the valence electrons. CF4 (m.m. 88) has a greater molecular mass than benzene (m.m. 78) but its boiling point (-128ºC) is lower than that of benzene (80ºC). Teflon, poly(tetrafluoroethene), is used to make the non-sticky surface of frying pan. The high electronegativity, hence low polarizability of fluorine atom results in very small induced dipole-induced dipole attractions. I. Van der Waals' forces 2. Dipole-dipole interactions (permanent dipole-permanent dipole interactions) Page 3 There is dipole-dipole interactions among all polar molecules. The oppositely charged ends tend to attract each other and the similarly charged ends tend to repel each other. The molecules will arrange in a way with lowest potential energy for the system. This means maximum amount of attractions. 3. Dipole-induced dipole attractions Like a temporary dipole, a polar molecule can also induce another temporary dipole on another molecule. The attractions caused by this mean is called dipole-induced dipole attractions. C. Relative strength of different origins of van der Waals' forces Molecule Ar (non-polar) N2 (non-polar) CH4 (non-polar) CO (polar) HCl (polar) induced dipole-induced dipole attractions 100 % 100 % 100 % ≈ 100 % 81 % dipole-dipole interactions (permanent dipole-permanent dipole interactions) nil nil nil 0.005 % 15 dipole-induced dipole attractions nil nil nil 0.08 % 4% Although dipole-dipole interactions and dipole-induced dipole attractions are very weak comparing with induced dipole-induced dipole attractions, polar molecule always has a higher m.p. and b.p. than a non-polar molecule with comparable molecular mass. This is because on top of dipole-dipole interactions and dipole-induced dipole attractions, there is always induced dipole-induced dipole attractions among polar molecules. Glossary Past Paper Question van der Waals’ forces dipole-dipole interactions 91 2A 2 a 95 2B 4 a i 98 1A 1 b i temporary dipole induced dipole-induced dipole attractions dipole-induced dipole attractions polarizability 91 2A 2 a 2a Arrange the following substances in order of increasing boiling points: C2H5Cl, CH4 and C2H6. Explain your order by comparing the relative magnitudes and nature of the intermolecular forces. CH4 < C2H6 < C2H5Cl 1 mark C2H5Cl has one highly polarizable Cl atom and has a permanent dipole moment, while C2H6 is less polarizable than C2H5Cl and has no permanent dipole moment. Thus C2H5Cl is expected to have stronger intermolecular 1 mark forces than C2H6, and a higher boiling point. Both CH4 and C2H6 are non-polar but the molecular size and surface area of C2H6 is larger than those of CH4, so 2 marks C2H6 will have a stronger van der Waals’ forces. 4 I. Van der Waals' forces 95 2B 4 a i 4a Explain the following facts: i The boiling points of the halogens increase as the group is descended. Boiling points of halogens depend on the strength of their intermolecular forces (van der Waal’s forces) which is 1 mark related to the strength of the instantaneous dipole of the molecule. Descending the group, with the increase in relative molecular size, the strength of the instantaneous dipole increases. Hence more energy is needed for the boiling of the higher halogens. 1 mark 98 1A 1 b i 1b i An iodine molecule can be represented by the diagram below, with each dot '·' representing an atomic nucleus. Page 4 2 2 (I) Using one or more diagrams of this kind, illustrate your understanding of the two terms 'covalent radius' and 'van der Waals' radius'. (II) Account for the difference between the covalent radius and van der Waals' radius for iodine. II. Hydrogen bond Page 1 Topic Reference Reading II. Hydrogen bond 4.10.2 Modern Physical Chemistry ELBS pg. 90–92 Chemistry in Context, 3rd Edition ELBS pg. 117–126 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 118–123 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 103–112 Hydrogen bond Strength of hydrogen bond Intramolecular hydrogen bond II. Hydrogen bond A. Nature Hydrogen bond can be considered as a kind of exceptionally strong dipole-dipole interactions. When a hydrogen is bonded to a very electronegative element e.g. N, O and F, the bond pair electrons would be attracted towards the latter one. The hydrogen atom without the bonding electron will become a naked proton. It is then attracted by the lone pair of another very electronegative element. naked proton Syllabus Notes δ+ H F δ- δ+ H F δ- intermolecular hydrogen bond bond pair very close to the electronegative atom Since the charge separation is large, this will give a strong dipole. The electrostatic attraction between the naked proton and the lone pair is called hydrogen bond. The size of proton, N, O and F are small, this facilitates close approach of two dipoles. Necessary condition for the formation of hydrogen bond. 1. A hydrogen atom attaching to a very electronegative atom i.e. N, O, F. 2. A lone pair on a very electronegative atom i.e. N, O, F. II. Hydrogen bond B. Strength i. Loneliness of the proton Page 2 If the hydrogen is bonded to a more electronegative atom, the covalent bond will be more polarized. This causes the proton to be more isolated and the hydrogen bond would be stronger. ii. Availability of the lone pair on the electronegative atom CH3 CH3 C OH CH3 2-methylpropan-2-ol (b.p. 82.4 °C, m.p. 25.6 °C) Steric effect – pentan-1-ol (b.p. 138.1 °C, m.p. -78.9 °C) CH3 CH2 CH2 CH2 OH If the lone pair is sterically hindered by bulky groups around it, it would be less available to form hydrogen bond. Though both melting point and boiling point are depending on the attraction among the particles, their value may or may not parallel to each other. e.g. 2-methylpropan-2-ol (b.p. 82.4 °C, m.p. 25.6 °C), pentan-1-ol (b.p. 138.1 °C, m.p. -78.9 °C) Arrangement of particles Ordered Attraction among the particles High Random Medium Random Low C. Solubility and hydrogen bond The solubility of the molecule is parallel to its ability to attract with the water molecule. this is related to its ability to form hydrogen bond with water molecules. N.B. In certain extent, The solubility is depending on the polarity rather than the size of the molecule. e.g. glucose (C6H12O6) is soluble in water but methane (CH4) is insoluble in water. II. Hydrogen bond D. Intramolecular hydrogen bond Page 3 Besides intermolecular hydrogen bond, hydrogen bond may be formed within a single molecule, i.e. intramolecular hydrogen bond. Boiling point of nitrophenol : H O +N -O H O +N -O O O H O +N -O O 2-nitrophenol (214ºC) 3-nitrophenol (290ºC) 4-nitrophenol (279ºC) 2-nitrophenol has lower boiling point and solubility than the other two isomer. Intramolecular hydrogen bond is formed within the molecule. The intramolecular hydrogen bonding prevents formation of intermolecular bonding hydrogen bond between molecules. Intramolecular hydrogen bonding is impossible in 3- and 4-nitrophenol. E. Examples of hydrogen bonding 1. Ethanoic acid In gas phase or organic solvent, ethanoic acid dimerizes to form a dimer. The presence of an eight-membered ring is confirmed by electron diffraction studies. O H3C C OH O HO C CH3 In aqueous solution, the molecules of a carboxylic acid link up with water molecules rather than form dimers. 2. Simple hydrides - CH4, NH3, H2O and HF CH4 91 109 NH3 195 240 H2O 273 373 HF 190 293 melting point (K) boiling point (K) There is no hydrogen bond among CH4 molecule. 1 per molecule in NH3 and HF. 2 per molecule in H2O. Furthermore, boiling point is a better indicator of strength of hydrogen bond than melting point because on top of attractions, melting process involves breaking of crystal lattice. Therefore, melting point is also depending on the packing of the crystal. Consider the boiling point of group V, VI and VII hydrides. The boiling point of the first member is much higher than the expected because of the presence of hydrogen bond on top of van der Waals' forces. The boiling points of group IV hydrides follow the general trend of increasing boiling point with increasing molecular size. This is because of the presence of van der Waals' forces only. Boiling point of Group IV, V, VI and VII hydrides II. Hydrogen bond 3. Structure and physical properties of ice Page 4 Each water is capable to form two hydrogen bonds. Including the two covalent bond, oxygen is tetrahedrally coordinated. In ice, the tetrahedrally coordinated oxygen atoms fit into a nonplanar, six-membered ring of water molecules associated through hydrogen bonding. The ring structure extends in a 3-dimensional way. This gives ice an open structure and lower density than water. When ice melts, the energy absorbed breaks some of the intermolecular associations and the hexagonal crystal lattice collapses. The density of water increases with increasing temperature up to 4ºC as the hexagonal structure continues to collapse. 4. Biochemical importance of hydrogen bond a) DNA (Deoxyribonucleic acids) Hydrogen bonding plays a crucial role in the structures of many biochemical substances, such as deoxyribonucleic acids (DNA). The two helical nucleic acid chains are held together by hydrogen bonds. These hydrogen bonds are formed between specific pairs of bases on the DNA chains, such as between the adenine unit in one chain and a thymine unit in the other; or between a guanine unit in one chain and a cytosine unit in the other. II. Hydrogen bond b) Protein Protein are made up from α-amino acid molecules. The –NH2 group condenses with –COOH group to form a peptide link, ( O H CN Page 5 ). The polypeptide forms a helix structure with the amide groups joined together by hydrogen bonds. RO H2N C H α-amino acid O HRO H HRO H HR H C OH CNC CNC CNC chain of polypeptide F. Strength of van der Waals’ forces and hydrogen bond H-bonds have the strength in the range 5–40 kJmol-1. van der Waals’ forces have roughly the same strength. For a H-bonded molecule, there are both hydrogen bond and van der Waals’ force. Therefore, a H-bonded substance always has a higher boiling point than those not H-bonded. non-polar molecule polar molecule molecule with H-bond induced-induced induced-induced dipole-dipole induced-induced dipole-dipole H-bond Range (kJmol-1) -780 (NaCl) – -3791(MgO) 158 (E(F–F)) – 945.4 (E(N≡N)) 107.3 (Na) – 514.2 (V) Ionic bond (non-directional) Covalent bond (directional) Metallic bond (non-directional) Estimated by Lattice energy Bond energy terms Atomization energy Comparing with ionic bond, covalent bond and metallic bond, both H-bond and van der Waals' forces are very weak. This explains why only 100ºC is required to boil water but 8000ºC is required to decompose water into hydrogen and oxygen atoms. O H H weak hydrogen bond and van der Waals' forces O strong covalent bond H O H H H Glossary hydrogen bond steric effect intramolecular hydrogen bond open structure DNA double helix structure amino acid peptide link amide group polypeptide II. Hydrogen bond Page 6 92 2B 6 Bc i 94 2A 2 b i 98 2B 6 a i 92 2C 7 d 94 2B 4 b Past Paper Question 92 2B 6 Ac i 93 1A 3 d 94 1A 2 d i ii iii 97 1A 1 d i ii 98 1A 2 b 99 1A 1 b iii 6Ac 92 2B 6 Ac i i Account for the difference in boiling points between NH3 and PH3. PH3 molecules - van der Waals intermolecular forces NH3 molecules - additional hydrogen bonding 2 1 mark 1 mark 92 2B 6 Bc i 6Bc i Account for the difference in boiling points between H2O and H2S. Intermolecular hydrogen bonding in H2O but not H2S H2S van der Waals forces and permanent dipole-permanent dipole interactions only 2 1 mark 1 mark 92 2C 7 d 7 A carboxylic acid P, with a relative molecular mass less than 100, contains C, 55.8%; H, 7.0%; and O, 37.2% by mass . An attempt to convert P to its methyl ester Q by prolonged refluxing of P with methanol in the presence of aqueous H2SO4 gave the desired ester Q but with much of the starting material P unchanged. (Relative atomic masses : H 1.0; C 12.0; O 16.0) 7d Which has a higher boiling point, the carboxylic acid or its methyl ester? Explain your answer. 3 Carboxylic acid has higher b.p. than its ester ½ mark Hydrogen bonding 1 mark O RCO H O RCO or lone pair on oxygen bonded to acidic proton Need more energy to break off H-bonding. ∴higher b.p. H 1 mark ½ mark 93 1A 3 d 3d The boiling point and density of butan-1-ol are 117.9 ºC and 0.81 gcm-3 respectively, and those of its isomer 2methylpropan-2-ol are 82.2 ºC and 0.79 gcm-3 respectively. Account for these differences. Alcohol molecules are held together by hydrogen bond. ½ mark Methylpropan-2-ol is a tertiary alcohol and butan-1-ol is a primary alcohol. ½ mark Due to steric interaction, 3º alcohol forms weaker intermolecular hydrogen bonds so its boiling point is lower. 1 mark Further, the molecules are less closely packed and therefore results in lower density. ½ mark 94 1A 2 d i ii iii 2d i Explain the term “hydrogen bonding”. H atom is situated between 2 small / electronegative atoms. 1 mark 1 mark One of these has a lone pair of electrons / bond energy in the range of 2 – 10 kJmol-1. ii Draw a diagram of the structure of a compound which has hydrogen bonds. Indicate the hydrogen bond(s) clearly. A actual example of intra / intermolecular hydrogen bond. The example must show X–HLY or XLHLX 1 mark iii Explain why (I) the boiling point of CH4 is lower than that of SiH4, and (II) the boiling point of NH3 is higher than that of PH3. (I) Boiling points is expected to increase with number of electrons in the molecule (CH4 → SiH4) due to increasing van der Waal’s forces. 1 mark (II) Boiling point of NH3 higher than the expected (i.e. >PH3) due to increased intermolecular attraction (hydrogen bonding) 1 mark 2 2 1 2 II. Hydrogen bond Page 7 94 2A 2 b i 2b Account for each of the following: i Concentrated H3PO4 has a high viscosity. 2 In H3PO4, the intermolecular force is H-bond, ½ mark and each H3PO4 molecule is capable of forming 3 H-bonds with its neighbours. With strong intermolecular forces, 1 mark H3PO4 has a high viscosity. O P OH OH OH ½ mark 94 2B 4 b 4b Explain why hydrogen fluoride is a liquid at room temperature and atmospheric pressure while the other hydrogen 2 halides are gases. The anomalous boiling point of HF is due to extensive H-bonding between HF molecules and other hydrogen halides have no hydrogen bondings among the molecules. 1 mark H F H F H F H F H F H F H F (diagram showing H-bonds in HF) 97 1A 1 d i ii 1d Explain why i the boiling point of HF is higher than that of HCl; ii the boiling point of HI is higher than that of HBr. 1 mark 2 98 1A 2 b 2b Which compound, H2O or F2O, would you expect to have a higher boiling point ? Explain your answer. 98 2B 6 a i 6a Consider the structures of the two synthetic polymers shown below. CH2CH2 poly(ethene) n HN(CH2)6NHCO(CH2)4CO nylon-6.6 n 2 i Suggest an explanation for the fact that nylon-6,6 has a higher tensile strength than poly(ethene). 99 1A 1 b iii 1b Account for each of the following : iii At 273 K, ice has a smaller density than water. III. Relationship between structures and properties Page 1 Topic Reference Reading Syllabus Notes III. Relationship between structures and properties 4.0.1.0 4.11 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 109 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 402–403, 468–469 The relationship between structures and properties of materials In chemistry, chemists intend to explain the chemical and physical properties of substances in term of particles and the interaction among the particles. All properties can be explained according to the following framework. particles → attractions among particles → arrangement of particles → structure → properties Example Zn(s) Particles Zn atom Attractions among particles strong metallic bond due to delocalized electron weak metallic bond due to less delocalized electron strong ionic bond Arrangement of particles hexagonal close packing random arrangement in liquid face centered cubic Structure giant metallic structure liquid at room temperature giant ionic structure Physical properties high thermal and electrical conductivities; hard; malleable; ductile; dense; high m.p. & b.p.; insoluble in any solvent except Hg(l) high thermal and electrical conductivities; relatively low m.p. & b.p. low thermal and electrical conductivities; only conducts electricity in molten state or aqueous state; soluble in water low thermal and electrical conductivities; insoluble in water low thermal and electrical conductivities; relatively low m.p. and b.p.; high heat capacity Chemical properties reducing agent very weak reducing agent quite inert Hg(l) NaCl(s) Hg atom Na+ ion & Cl- ion AgCl(s) H2O(l) Ag+ ion & Cl- ion H2O molecule H2(g) H2 molecule ionic bond with very strong covalent character strong O–H covalent bond and relatively weak intermolecular forces among molecules (Hydrogen bond and van der Waals' forces) strong H–H covalent bond; extremely weak intermolecular forces among molecules (van der Waals' forces only) cubic structure H2O is angular in shape; random arrangement in liquid; open structure in ice H–H is linear in shape; random arrangement in gas giant ionic structure simple molecular structure / molecular crystal in ice simple molecular structure quite inert quite inert; only decomposes into H and O atom at 8000K reducing agent low thermal and electrical conductivities; extremely low m.p. and b.p.; low heat capacity III. Relationship between structures and properties Example Graphite Particles C atom Attractions among particles strong covalent bond with delocalized electrons between C atoms; weak van der Waals' forces between hexagonal layers strong covalent bond between C atoms strong covalent bond between C, H and O; weak intermolecular forces among molecules Arrangement of particles C atoms are arranged in hexagonal layers with large gap between layers C atoms are arranged tetrahedral network glucose molecule has a ring structure Structure giant covalent structure Physical properties high thermal and electrical conductivities; extremely high m.p. and b.p.; brittle; insoluble in any solvent Page 2 Chemical properties reducing agent Diamond C atom giant covalent structure simple molecule structure Glucose C6H12O6 molecule low thermal and electrical conductivities; high m.p. and b.p.; extremely hard; insoluble in any solvent low thermal and electrical conductivities; moderate m.p. and b.p.; soluble in water reducing agent chars upon heating; a weak reducing agent. Glossary particle bonding / attraction arrangement structure properties III. Relationship between structures and properties Page 3 Past Paper Question 90 2A 3 c 91 1A 2 d iii 93 2B 4 b iv 94 2C 9 b iv 99 1A 1 b i 90 2A 3 c 3c Arrange the following substances in order of increasing melting point: NaF, F2, HF. Explain your order in terms of the bonding involved. Melting involves the overcoming of forces between the units of which the solid is made. Two of the given solid HF, F2 would be expected to form molecular solids; the other, NaF would be expected to form ionic solid. The coulombic forces holding an ionic solid together are strong, hence NaF would be expected to have high melting point. 1 mark The attraction between F2 molecules is quite weak and due only to van der Waals forces would be expected to have a low melting point. 1 mark Even though HF forms molecular solid but because of the presence of hydrogen bonding between H and neighbouring F atom, which can hold the molecules together more strongly and therefore, expected to have much 1 mark higher melting point than F2. 1 mark F2(-219.6ºC) < HF(-89.1ºC) < NaF(993ºC) 91 1A 2 d iii 2d iii Explain why the electrical conductivities of iron and of caesium chloride are different in the solid state. In CsCl, the ions are held together by strong coulombic forces and cannot be easily displaced. Because of lack of mobility of ions, conductivity is low. In iron, the iron atoms occupy the BCC sites and electrons are delocalized. Conductivity is high because electrons can move. 1 mark 93 2B 4 b iv 4b Account for each of the following facts: iv There is a significant difference in the melting point between silicon (m.p. 1410 ºC) and lead (m.p. 328 ºC). Silicon has a giant covalent network structure. C.N. = 4 Lead has a close packed metallic structure. C.N. = 12 1 mark The Si–Si in silicon is strong compared with the metallic bond in lead and in the liquid state of lead is residual metallic bond remained. ∴ silicon has a higher melting point than lead. 1 mark 94 2C 9 b iv 9b Ethene and chloroethene can undergo polymerization to give polyethene (PE) and polyvinyl chloride (PVC) respectively. PVC is more rigid and durable than PE, but incineration of PVC causes a more serious pollution problem. iv Explain why PVC is more rigid than PE. The structure of PVC can be represented by 4 1 2 1 The attraction between polymer chains of PVC is dipolar attraction while that between PE polymer is van der Waal's force which is weaker when compared with the former. Hence PVC is more rigid and more durable than PE. 1 mark 99 1A 1 b i 1b Account for each of the following : i At 298 K and 1 atm pressure, carbon dioxide is a gas whereas silicon dioxide is a solid. States of Matter I. Solid state A. Metallic structure 1. Hexagonal close packing 2. Cubic close packing - face centred cubic 3. Tetrahedral holes and octahedral holes 4. Body-centred cubic structure Giant ionic structure Covalent giant structure 1. Allotrope of carbon a) Diamond b) Graphite 2. Silicon(IV) oxide Molecular crystal 1. Iodine – face centred cubic 2. Carbon dioxide (Dry ice) – face centred cubic Gas laws 1. Charles' law 2. Boyle's law 3. Avogadro's law 4. PVT surface Ideal gas equation (Ideal gas law) 1. Molar volume Determination of molecular mass 1. Experimental determination of molar mass of a gas 2. Experimental determination of molar mass of a volatile liquid Dalton's law of partial pressure 1. Mole fraction 2. Relationship between partial pressure and mole fraction B. C. D. II. Gaseous state A. B. C. D. I. Solid State Unit 1 Page 1 Topic Reference Reading I. Solid state 4.12.1 Modern Physical Chemistry ELBS pg. 154–155 Chemistry in Context, 3rd Edition ELBS pg. 134–136 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 147–149 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 45, 119–121 Close packing structure Body centred cubic structure I. Solid state Unit 1 Syllabus Notes A. Metallic structure In metal crystal, the metal atoms tends to pack with each other closely and has a high coordination no. Coordination no. – the number of atoms closest to a particular atom. There are two kinds of close packing - hexagonal close packing and cubic close packing. 1. Hexagonal close packing (74% of the space is filled) Some of the metal have hexagonal close packing (h.c.p.). e.g. Cobalt, Titanium and Zinc In hexagonal close packing, each atom is surrounded by 12 atoms, i.e. coordination is 12. 6 on the same layer, 3 on the top and 3 at the bottom. Since the orientation of the first and third layer are the same, this is also called abab... pattern. Unit cell – smallest portion of the crystal which contains all the fundamentals (information) without repetition. 1 12 × 6 1 2×2 3×1 No. of atoms in an unit cell 12 at the corner 2 on the face 3 in the cell =2 =1 =3 =6 Total no. of atoms in an unit cell I. Solid State 2. Unit 1 Cubic close packing - face centred cubic (74% of the space is filled) Page 2 Cubic close packing (c.c.p.) is also called face centre cubic packing (f.c.c.). Examples are Copper, Nickel and Calcium The coordination no. of any atom in the structure is also 12. The first layer has a different orientation with the third layer of atoms. This is called abcabc... pattern. By looking at a four layers unit cell, 1 atom at the first layer, 6 at the second layer, 6 at the third layer and 1 at the fourth layer, a face centred cubic unit cell can be constructed. No. of atoms in an unit cell 8 at the corner 6 on the face 1 8×8 1 6×2 =1 =3 =4 Total no. of atoms in an unit cell I. Solid State 3. Unit 1 Tetrahedral holes and octahedral holes Page 3 Although the crystal is closely packed, there is still some spaces between the atoms. They are called holes. There are two types of holes – tetrahedral hole and octahedral hole Octahedral hole Octahedral hole is surrounded by 6 atoms. It is the space between two layers of triangularly arranged atoms. From another angle, it can be seen that the six atoms are arranged in a form of octahedron. Tetrahedral hole Tetrahedral hole is surround by 4 atoms. The four atoms are arranged in a form of tetrahedron. It has a smaller size than octahedral hole. 4. Body-centred cubic structure (68% of the space is filled) e.g. Sodium, Potassium and Chromium Body-centred cubic (b.c.c.) structure is not a close packing structure. Only 68% of the space is filled comparing with 74% in the two close packing structures. In body-centred cubic structure, the coordination no. of any atom is 8. No. of atoms in an unit cell 8 at the corner 1 at the centre 1 8×8 1×1 =1 =1 =2 face centred cubic Total no. of atoms in an unit cell Glossary coordination no. tetrahedral hole hexagonal close packing cubic close packing octahedral hole body centred cubic unit cell I. Solid State Unit 1 90 1A 2 c 91 1A 2 d i 92 1A 2 d i 97 1A 1 a i ii Page 4 Past Paper Question 90 1A 2 c 2c Compare and contrast the two close packing atomic arrangements in metals. Metals having the hexagonal close packing has the abab... arrangement for atoms in different layers. Metals having the face-centred cubic close packing has the abcabc... arrangement for atoms in different layers. 2 marks 1 mark Both arrangements have the same packing efficiency. or Atoms in both close packing arrangements all have the co-ordination number of 12. 91 1A 2 d i 2d i Iron has a body-centred structure. Draw a unit cell representation of iron. 3 1 1 mark 92 1A 2 d i 2d The arrangement of atoms in metals can be described by the close-packing of spheres. i Which close-packed structure does abcabcabc... describe? Indicate on the diagram below one tetrahedral hole (marking it T) and one octahedral hole (marking it O). 1½ Cubic closest packed structure. ½ mark ½ mark for O and ½ mark for T 97 1A 1 a i ii 1a At room temperature, iron has a body-centred cubic structure. i Draw the unit cell representation of iron. ii Deduce the number of atoms in one unit cell of iron. 2 I. Solid state Unit 2 Page 1 Topic Reference Reading I. Solid state 4.12.2 Modern Physical Chemistry ELBS pg. 159–162, 165–166 Chemistry in Context, 3rd Edition ELBS pg. 138–146 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 149–151 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 124–126 Giant ionic structure B. Giant ionic structure Ionic crystal is the result of a balance between attractions and repulsions. Unit 2 Syllabus Notes If the attraction forces and the repulsion forces are not balanced, the structure will be unstable. The ions will be rearranged and another more stable structure will be resulted. This depends on the relative size of the ions in the ionic crystal. Stable Limiting radius Unstable Stable The arrangement of the ions in an ionic crystal is determined by the ionic radii ratio (r+/r-): Coordination no. 8 6 4 Shape Simple cubic Face centered cubic (Octahedral) Face-centered cubic (Tetrahedral) r+/r0.73 and above 0.41 - 0.73 0.22 - 0.41 Example CsCl NaCl ZnS Caesium chloride Simple cubic + Cs ions occupy the centre of the cubes formed by 8 Cl- ions no. of Cs+ ions in the unit cell 1 at the centre 1×1=1 Total no. of Cs+ ions = 1 no. of Cl- ions in the unit cell 1 4 at the corner 4 × 4 = 1 Total no. of Cl- ions = 1 Sodium chloride face-centred cubic (octahedral) Na+ ions occupy the octahedral holes formed by 6 Cl- ions no. of Na+ ions in the unit cell 1 at the centre 1×1=1 1 12 on the edge 12 × 4 = 3 Total no. of Na+ ions = 4 no. of Cl- ions in the unit cell 1 6 on the face 6×2=3 1 8 at the corner 8 × 8 = 1 Total no. of Cl- ions = 4 Coordination no. of Na+ ion = 6 Coordination no. of Cl- ion = 6 Zinc blende (Zinc sulphide) face centred cubic (tetrahedral) Zn2+ ions occupy half of the tetrahedral holes formed by 4 S2- ion no. of Zn2+ ions in the unit cell 4 in the cell 4×1=4 Total no. of Zn2+ ions = 4 no. of S2- ions in the unit cell 1 6 on the face 6×2=3 1 8 at the corner 8 × 8 = 1 Total no. of S2- ions = 4 Coordination no. of Cs+ ion = 8 Coordination no. of Cl- ion = 8 Coordination no. of Zn2+ ion = 4 Coordination no. of S2- ion = 4 I. Solid state Unit 2 Page 2 Both the no. of ions in the unit cell and ratio of coordination no. can be used to determine the empirical formula of the ionic compound. The structure of an ionic crystal is labeled according to the positions of the formula units, not the position of individual ions. The position of each formula unit is called a lattice point. Take CsCl as the example, consider each pair of Cs+ ion and Cl- ion as a lattice point. The lattice points are arranged in a simple cubic manner i.e. 8 lattice point occupy the 8 corners of a cube. Glossary Past Paper Question giant ionic structure 94 1A 1 b i ii 99 2A 2 c i ii zinc blende 94 1A 1 b i ii 1b The crystal structure of a compound AxBy can be described as a simple cubic lattice of A atoms with B atoms at the middle of all the edges. i What is the empirical formula of this compound? 1 No. of atom A per unit-cell = 8 × 8 = 1 1 No. of atom B per unit-cell = 12 × 4 = 3 ∴ Empirical formula of the compound is AB3. 1 mark ii What are the coordination numbers of an A atom and a B atom respectively? Coordination number of A = 6 ½ mark Coordination number of B = 2 ½ mark 99 2A 2 c i ii 2c i Consider the unit cell of calcium fluoride shown below: 1 1 ii (I) (II) (I) (II) State the respective coordination numbers of each calcium ion and each fluoride ion. Describe the lattice of calcium ions and that of fluoride ions. Draw the unit cell of caesium chloride. Describe the lattice of caesium ions and that of chloride ions in caesium chloride. I. Solid state Unit 3 Page 1 Topic Reference Reading I. Solid state 4.12.3 Modern Physical Chemistry ELBS pg. 159–162, 165–166 Chemistry in Context, 3rd Edition ELBS pg. 138–146 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 152–154 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 46, 123–124 Giant covalent structure C. Giant covalent structure 1. Allotrope of carbon Unit 3 Syllabus Notes Allotropy – The existence of more than one physical form of the element in the same state. e.g. graphite and diamond, red phosphorus and white phosphorus, oxygen and ozone. Carbon has two common allotropes – graphite and diamond. The two allotropes have different physical properties but similar chemical properties. This can be explained by the structures of the two allotropes. Properties appearance hardness density electrical conductivity melting point a) Diamond Each carbon is sp3 hybridized. The sp3 hybrid orbitals are overlapped to form strong σ covalent bond ⇒ hard, high density, high melting point and boiling point. The bonding is extended to form a 3-dimensional network. Diamond colourless, transparent crystal hardest natural substance known 3.5 gcm-3 does not conduct 3550 ºC Graphite black, shiny solid brittle 2.2 gcm-3 conducts in one direction 3730 ºC b) Graphite Each carbon is sp2 hybridized. The sp2 hybrid orbitals are overlapped to form strong σ covalent bond. The p orbitals perpendicular to the carbon layer are overlapped to form π bond. ⇒ higher melting point than diamond. This gives a hexagonal layer structure. There are only weak van der Waals’ forces between the layers. ⇒ brittle and low density The π electrons are delocalization above and below the layer in a direction parallel to the layer. ⇒ conducts electricity in one direction, shiny appearance I. Solid state 2. Silicon(IV) oxide Unit 3 Page 2 Quartz / Sand / silicon(IV) oxide / silicon oxide / silicon dioxide (SiO2) The structure of quartz is similar to that of diamond. Each silicon atom is joined to 4 oxygen atoms and each oxygen atom is then connected to another silicon atom. The silicon and oxygen are joined together by strong covalent bond. Upon melting, a lot of energy is required to break the Si–O covalent bond. Therefore, silicon(IV) oxide has a relatively high melting point 1610ºC. Glossary Past Paper Question allotrope 90 2B 4 b 92 2A 3 a i ii 98 2A 2 a i ii hexagonal layer structure 90 2B 4 b 4 Account for the following observations. 4b SiO2 is a solid with a high melting point, whereas CO2 is a gas at room temperature. SiO2 - three dimensional polymeric solid with very strong Si-O bonds. (Si-O bond must be stated; with strong bonds only - 0 mark) CO2 - exists as discrete molecules with only weak intermolecular van der Waals’ forces. 4 2 marks 2 marks 92 2A 3 a i ii 3a i Describe the bonding and intermolecular forces in ice and in SiO2 solid. Ice consists of covalent H2O molecules held together by hydrogen bonding. Quartz, SiO2 solid, is an infinite three dimensional network(giant structure) solid in which each Si atom is covalently bonded in a tetrahedral arrangement to four O atoms. 2 marks ii What type of interactions must be overcome to melt these solid? Hydrogen bonding must be overcome in order to melt ice. In order for SiO2(s) to melt, some of the strong Si-O covalent bonds have to be broken. 2 marks 98 2A 2 a i ii 2a The structures of the two allotropes of carbon, diamond and graphite, are shown below. 2 2 6 i ii Comment on the three different carbon-carbon distances as indicated in the above structures. With reference to the above structures, explain why diamond is hard whereas graphite is soft enough to be used as lubricant. I. Solid state Unit 4 Page 1 Topic Reference Reading I. Solid state 4.12.4 Modern Physical Chemistry ELBS pg. 159–162, 165–166 Chemistry in Context, 3rd Edition ELBS pg. 138–146 Physical Chemistry, Fillans pg. 212–220 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 151–152 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 47, 130–132 Molecular crystal Unit 4 Syllabus Notes D. Molecular crystal Molecular crystal consists of molecules held in lattice sites by weak intermolecular forces such as van der Waals’ forces or hydrogen bonds. The structure of molecular crystals are determined by the shape of the basic molecular units which tend to pack together as efficient as possible. 1. Iodine – face centred cubic 2. Carbon dioxide (Dry ice) - face centred cubic Glossary Past Paper Question molecular crystal II. Gaseous state Unit 1 Page 1 Topic Reference Reading II. Gaseous state 0.5.3.1–0.5.3.2 ILPAC 9 (2nd ed.), John Murray, pg 7–19 Modern Physical Chemistry (4th ed.), Bell & Hyman Ltd., pg 143–144 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., pg 155 Unit 1 Assignment A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., pg 163–165, 167 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 137–139 Reading Syllabus Charles' law Molar volume Boyle's law Avogadro's law Ideal gas equation (Ideal gas law) Molar volume of gases II. Gaseous state A. Gas Laws 1. Charles’ Law Notes For a given mass of gas, the volume is directly proportional to the absolute temperature (Kelvin scale) at a given pressure. V ∝ T 2. Boyle’s Law For a given mass of gas, the volume is inversely proportional to the pressure at 1 a given temperature. V ∝ P 3. Avogadro’s Law At constant pressure and temperature, the volume of a gas is directly proportional to the number of gaseous particle. V ∝ N or V ∝ n N : no. of particle, n : no. of mole of particle II. Gaseous state 4 PVT surface Unit 1 Page 2 PVT relationship of an ideal gas can be represented by a 2-dimensional surface. PVT surface for an ideal gas B. Ideal gas equation (Ideal gas law) By combining the three gas laws, an ideal gas equation can be derived. nT 1 By combining V ∝ T, V ∝ P and V ∝ n ⇒ V ∝ P PV Therefore, nT = constant The constant is called gas constant, R and the relationship becomes PV = nRT Unit P : 1 atm = 101325 Nm-2 (Pa) = 760 mmHg V : 1 m3 = 1000 dm3 = 1000000 cm3 n : no. of mole T : must be in Kelvin (K) R : 8.314 JK-1mol-1 = 82.05 cm3atmK-1mol-1 II. Gaseous state 1. Molar volume Unit 1 Page 3 Since the volume occupied by a gas is mainly space whose size is depending on the motion of the particles only, all gases have the similar molar volume. i.e. the molar volume is independent of the nature of the particles. It is found that the molar volume of any gas at 0 ºC (273 K) and 1 atm is about 22.4 dm3. The condition 0 ºC and 1 atm is called standard temperature and pressure (stp) or standard condition. Gas Helium Hydrogen Carbon dioxide Formula He H2 CO2 Molar volume at stp, Vm / dm3mol-1 22.396 22.432 22.262 By using the ideal gas equation, the molar volume of a gas at different condition can also be determined. For example, the condition 25ºC (273K) and 1 atm is called room temperature and pressure (rtp) or room condition. The molar volume at stp can be determined by using ideal gas equation. PV = nRT P1V1 P2V2 n1T1 = n2T2 1 atm × V2 1 atm × 22.4 dm3 1 mole × 273 K = 1 mole × 298 K V2 = 24.5 dm3 Since the value is only an approximate value, usually the molar volume of a gas is assumed to be 24 dm3 at rtp. Glossary Charles' law absolute temperature (Kelvin scale) Boyle's law Avogadro's law ideal gas equation gas constant / universal gas constant molar volume standard temperature and pressure (stp) / standard condition room temperature and pressure (rtp) / room condition 94 2A 1 a i PVT surface Past Paper Question 94 2A 1 a i 1a Gas container A and B each contain an ideal gas at low pressure and 298 K. The volume of container A is twice that of container B, but the number of moles of ideal gas contained in A is only half of that in B. i Calculate the ratio of the gas pressure in the two containers. PAVA = nART 1 mark PBVB = nBRT PA n A V B 1 mark P B = n B ( V A) 11 1 = 2 (2 ) = 4 1 PA = 4 PB or P B = 4 PA 1 mark 3 II. Gaseous State Unit 2 Page 1 Topic Reference Reading II. Gaseous State 0.5.3.3 ILPAC 9 (2nd ed.), John Murray, pg 21–25 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., pg 158–160 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 143–144 Unit 2 Assignment A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 168, 180–181 Reading Syllabus Notes Determination of molecular mass C. Determination of molecular mass Although molecular mass of a substance is a microscopic properties which cannot be measured directly, ideal gas equation provides a link between molecular mass of a gas and some measurable quantities. PV = nRT where P is the pressure of the gas V is the volume of the gas n is the number of mole of gaseous particle R is the gas constant T is the temperature in Kelvin m The quantity of no. of mole is related to the molar mass by the relationship, n = M , where m is the mass of the m gas and Mm is the molar mass of the gas. Since molar mass is numerically the same as molecular mass, molecular mass can be determined by measuring the mass, pressure, volume and temperature of a gas. PV = nRT m PV = M (RT) m where m is the mass of the gas Mm is the molar mass of the gas m Mm = PV (RT) or m ρ Mm = P (RT) where, ρ is the density of the gas = V II. Gaseous State 1. Unit 2 Experimental determination of molar mass of a gas Page 2 By determining the mass, pressure, volume and temperature of a gas (e.g. CO2(g)), the molar mass of the gas can be determined. In the following experiment, a volumetric flask is used to measure the volume and the mass of the gas. Measured values Mass of volumetric flask filled with air Mass of volumetric flask filled with CO2 Mass of volumetric flask filled with water Room temperature Atmospheric pressure = = = = = 47.993 g 47.998 g 152.8 g 26 ºC 757 mmHg Given values Density of water at 26ºC = 1.00 gcm-3 Density of air at 26ºC and 757 mmHg = 0.00118 gcm-3 Gas constant (R) = 0.0821 atmdm3K-1mol-1 Calculation Molar Mass of CO2 = mass of CO2 × gas constant × temp. in K mRT = PV pressure of CO2 × volume of CO2 Note ! Your should keep all the units in the intermediate steps. Mass of CO2 Mass of water = = Mass of volumetric flask filled with water - Mass of volumetric flask (filled with air) (the mass of air is neglected in this step) 152.8 g - 47.933 g = 104.9 g 104.9 g mass of water Volume of the flask = density of water = 1.00 gcm-3 = 104.9 cm3 Mass of air = Volume of flask × density of air = 104.9 cm3 × 0.00118 gcm-3 = 0.124 g Mass of empty volumetric flask Mass of CO2 = Mass of volumetric flask filled with air - Mass of air = 47.933 g - 0.124 g = 47.809 g = mass of volumetric flask filled with CO2 - mass of empty volumetric flask = 47.998 g - 47.809 g = 0.189 g Pressure of CO2 = atmospheric pressure Volume of CO2 = volume of the flask mass of CO2 × gas constant × temp. in K mRT = PV pressure of CO2 × volume of CO2 0.189 g × 0.0821 atmdm3K-1mol-1 × (273 + 26)K = 44.4 gmol-1 105.0 757 3 760 atm × 1000 dm Molar Mass of CO2 = = II. Gaseous State 2. Unit 2 Experimental determination of molar mass of a volatile liquid Page 3 The molar mass of a volatile liquid cannot be determined directly according to ideal gas equation, the liquid must be vaporized first. By injecting a known mass of volatile liquid into a heated gas syringe, the molar mass of the volatile liquid can be determined using the ideal gas equation. Molar Mass of the volatile liquid = = mass of the liquid × gas constant × temp. in K pressure of the vapour × volume of the vapour mRT PV The mass of the liquid is measured by a hypodermic syringe and a balance. It is determined by comparing the mass of the hypodermic syringe before and after injection. The hypodermic needle is equipped with a self-sealing rubber to ensure no loss of vapour from the gas syringe. The volume of the vapour is measured by the volume displaced by the vapour in the heated gas syringe. Precautions : 1. 2. 3. The needle of hypodermic syringe must be long enough to inject the liquid into the inner part of gas syringe where the temperature is higher. This ensures complete vaporization of liquid. The temperature and volume readings must become steady before any record is taken. To ensure accurate injection of liquid, the hypodermic syringe should be rinsed with the liquid first and any bubbles in the syringe should be expelled. Keep the syringe horizontal and avoid touching the piston or warming the barrel with you hand, either of which could result in loss of liquid. II. Gaseous State Unit 2 Measured values Mass of hypodermic syringe and liquid before injection Mass of hypodermic syringe and liquid after injection Temperature of the vapour Atmospheric pressure Volume of air in gas syringe before injection Volume of air and vapour in gas syringe after injection Given values Gas constant (R) = 0.0821 atmdm3K-1mol-1 1 atm = 760 mmHg Calculation m PV = nRT = M RT m Page 4 = = = = = = 13.189 g 12.920 g 100 ºC 762 mmHg 5.0 cm3 66.0 cm3 66.0 - 5.0 (13.189 -12.920 g) 762 3 × 0.0821 atmdm3K-1mol-1 × (100 + 273) K Mm 760 atm × ( 1000 dm ) = Mm = 135 gmol-1 Mass of liquid used = mass of hypodermic syringe before injection - mass of hypodermic syringe after injection = 13.189 g - 12.920 g = 0.269 g Temperature of vapour = temperature of the steam jacket = 100 ºC = 373 K 762 Pressure of vapour = atmospheric pressure = 760 atm = 1.00 atm Volume of vapour = Volume of air and vapour in gas syringe - Volume of air in gas syringe = 66.0 cm3 - 5.0 cm3 = 61.0 cm3 = 0.0610 dm3 mass of the liquid × gas constant × temp. in K mRT pressure of the vapour × volume of the vapour = PV 0.269 g × 0.0821 atmdm3K-1mol-1 × 373 K = 135 gmol-1 1.00 atm × 0.0610 dm3 self-sealing rubber steam jacket Molar Mass of the volatile liquid = = Glossary Past Paper Question molar mass 95 2A 3 a i 97 2A 1 b 99 1A 1 a i ii gas syringe hypodermic syringe 95 2A 3 a i 3a i Derive an expression for the molar mass M of an ideal gas, in terms of its density d, pressure P, and absolute temperature T. PV = nRT ½ mark m n=M where m is the mass of dry air ½ mark m PV = M (RT) m d M = PV (RT) = P (RT) 1 mark Ci Generally well answered. However, some candidates did not express M in terms of d, P and T explicitly. 97 2A 1 b 1b At 360 K and 101 kPa pressure, the vapour produced by 0.226 g of a volatile liquid occupies 85.0 cm3. Assuming that the vapor behaves ideally, calculate the molar mass of the liquid. (1 kPa = 1 × 103 Nm-2) (Gas constant, R = 8.31 JK-1mol-1) 2 II. Gaseous State Unit 2 99 1A 1 a i ii 1a At 433 K, 0.569 g of aluminium chloride produces a vapour having a volume of 96.0 cm3 and a pressure of 80.0 kPa. i Assuming that the aluminium chloride vapour behaves as an ideal gas, calculate its molar mass. (1 kPa = 1 × 103 Nm-2) ii Deduce the molecular formula of aluminium chloride under the above conditions and suggest a structure consistent with this formula. Page 5 II. Gaseous State Unit 3 Page 1 Topic Reference Reading II. Gaseous State 4.13.4 ILPAC 9 (2nd ed.), John Murray, pg 33–34 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., pg 156 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 140 Unit 3 Assignment A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 169 Reading Syllabus Notes Dalton's law of partial pressure Relationship between partial pressure and mole fraction D. Dalton’s law of partial pressure Dalton’s law of partial pressure – The total pressure (PTotal) exerted by a mixture of gases is equal to the sum of the partial pressure (p) of each constituent gas. PTotal = p1 + p2 + p3...pn The partial pressure is defined as the pressure the gas would exert if it was contained in the same volume as that occupied by the mixture. 1. Mole fraction Mole fraction is the percentage of a particular component in a mixture in term of number. e.g. For a mixture containing 0.2 mole of oxygen, 0.3 mole of hydrogen and 0.3 mole of carbon dioxide, the mole fraction oxygen in the mixture can be calculated by 0.2 no. of mole of oxygen molecules mole fraction of oxygen = total no. of mole of molecules = 0.2 + 0.3 + 0.3 = 0.25 It is different from percentage by mass. In the above example, 0.2 mole of oxygen weighs 6.4 g, 0.3 mole of hydrogen weighs 0.6 g and 0.3 mole of carbon dioxide weighs 13.2 g. Percentage by mass of oxygen = 6.4 g mass of oxygen total mass = 6.4 g + 0.6 g + 13.2 g = 0.32 = 32% II. Gaseous State 2. Unit 3 Relationship between partial pressure and mole fraction Page 2 According to the ideal gas equation, the pressure of a gas is directly proportional to the no. of gaseous particle or the no. of mole of particle present. P ∝ N or P ∝ n where N is the total no. and n is the no. of mole, they are related by the expression N = nL where L is Avogadro's constant. By combining with Dalton's law of partial pressure, the partial pressure exhibited by individual component in a gaseous mixture should also be proportional to the no. of mole of that kind of particle. 1 : PT ∝ n 2 : pi ∝ ni where PT is the total pressure and n is the total no. of mole of particles. pi is the partial pressure of a component and ni is the no. of mole of that component Combining 1 & 2, pi ni PT = n = mole fraction of individual component in the mixture The mole fraction of individual component can be calculated from the partial pressure pi and the total pressure PT. Glossary Past Paper Question Dalton's law of partial pressure 91 1A 2c 95 2A 3 a ii 96 2A 2 a ii partial pressure mole fraction percentage by mass 91 1A 2c 2c A container holds a gaseous mixture of nitrogen and propane. The pressure in the container at 200ºC is 4.5 atm. At -40ºC the propane completely condenses and the pressure drops to 1.5 atm. Calculate the mole fraction of propane in the original gaseous mixture. ntRT1 = P1V1 (1) (2) nt (1 - x)RT2 = P2V1 where nt is total number of moles, x is mole fraction of propane (2) ÷ (1) T2 P2 1.5 473.15 (1 - x) T = P (1 - x) = 4.5 · 233.15 = 0.067646 x = 0.3235 1 1 1 mark for correct final answer (0.315 - 0.324) 2 marks for intermediate answers (partial pressure, partial volume, no. of mole), equations and method. 95 2A 3 a ii 3a ii At 98.6 kPa and 300 K, the density of a sample of dry air is 1.146 g dm-3. Assuming that dry air contains only nitrogen and oxygen and behaves ideally, calculate the composition of the sample. (Gas constant, R = 8.31 J K-1 mol-1) Molar mass of dry air 1.146 gdm-3 1.146 × 103 gm-3 = 98.6 kPa × 8.31 JK-1mol-1 × 300 K = 98.6 × 103 Nm-2 × 8.31 JK-1mol-1 × 300 K = 28.98 gmol-1 1 mark Let mole fraction of N2 be x 28.02 x + 32.0 (1 - x) = 28.98 1 mark x = 0.759 Composition of dry air 1 mark 75.9% N2 and 24.1% O2 C ii A number of candidates made mistakes in the numerical calculation and in the units used. 96 2A 2 a ii 2a ii At 298 K 1.0 dm3 of N2 at 0.20 Pa pressure was mixed with 2.0 dm3 of O2 at 0.40 Pa in a 4.0 dm3 container. Assuming that both N2 and O2 behave ideally, calculate the pressure of the gaseous mixture at 298 K. In the mixture ½ mark partial pressure of N2 = ¼ × 0.20 Pa = 0.05 Pa ½ mark partial pressure of O2 = ½ × 0.40 Pa = 0.20 Pa pressure of mixture = (0.05 + 0.20) Pa = 0.25 Pa 1 mark 3 3 2 Chemical Kinetics Page 1 Chemical Kinetics I. Rates of chemical reactions A. B. Definition of rate of reaction 1. Unit of rate of reaction Measuring the rate of reaction 1. Different approaches a) Constant amount approach b) Constant time approach 2. Interpretation of physical measurements made in following a reaction a) Volume of gas formed b) Colorimetric measurement Rate Law or rate equation 1. Differential form and integrated form of rate law 2. Graphical presentation of reaction rate 3. Order of reaction a) Experimental determination of order of reaction (1) (2) By measuring the rates of reaction at different reactant concentration By plotting graph of ln rate versus ln [X] II. Order of reaction A. B. Examples of different reaction 1. First order reaction a) Half life (t½) b) Carbon-14 dating c) Examples of calculation 2. Second order reaction 3. Zeroth order reaction Collision theory 1. Distribution of molecular speed (Maxwell-Boltzman distribution) 2. Effect of temperature change on molecular speeds 3. Collision theory and activation energy a) Arrhenius equation Determination of activation energy 1. By measuring the rate of reaction at different temperature 1 2. By plotting graph of ln k versus T 1 3. By plotting graph of ln rate versus T III. Collision theory and activation energy A. B. Chemical Kinetics Page 2 IV. Energy profile of reaction A. Transition state 1. Order of reaction leads to interpretation of reaction mechanism at molecular level a) Multi-stages reaction (1) (2) Alkaline hydrolysis of 2-chloro-2-methylpropane Reaction between bromide and bromate(V) in acidic medium B. b) Rate determining step 2. Energetic stability and kinetic stability Effect of catalyst 1. Characteristics of catalyst 2. Theory of catalyst a) Homogeneous catalysis b) Heterogeneous catalysis 3. Application of catalyst a) Use of catalyst in Haber process and hydrogenation b) Catalytic converter c) Enzymes Concentration Temperature Pressure Surface area Catalyst Light V. Factors influencing the rate of reaction A. B. C. D. E. F. I. Rates of chemical reactions Page 1 Topic Reference Reading I. Rates of chemical reactions 5.1 A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 298–300 Assignment Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 397–399, 401–403 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 352–354, 356–358 Reading Syllabus Definition of rate of reaction Unit of rate of reaction Measuring of rate of reaction I. Rates of chemical reactions Notes A. Definition of rate of reaction In every reaction, reactant is converted to product. Reactant → Product Rate of reactions is reflected by either rate of appearing of product or rate of disappearing of reactant. Rate of reaction = change in amount of product = time taken change in amount of reactant time taken The amount can be measured in term of mass, volume or any physical quantities related to amount. If the reaction is performed in solution, the amount can also be expressed in term of concentration. Rate of reaction = ∆ [product] = ∆t ∆ [reactant] ∆t [ ] : concentration Note : ∆ [reactant] ∆t because rate of reaction is always positive ∆ [reactant] but the quantity is negative. ∆t negative sign is added for the i. ii. iii. iv. v. vi. concentration temperature catalyst surface area pressure light Rate of a reaction depends on : 1. Unit of rate of reaction If the rate of reaction is expressed as the change in concentration of reactant or product with respect to time, ∆ [product] mol dm-3 , the unit of rate of reaction would be concn time-1 (e.g. , moldm-3s-1) s ∆t I. Rates of chemical reactions B. Measuring the rate of reaction By definition, rate is the change in certain quantity with respect to time. For example, speed is the change in distance with respect to time. Rate = 1. ∆ amount ∆ time Speed = ∆ distance ∆ time Page 2 Different approaches i. constant amount approach ii. constant interval approach There are two approaches to measure the rate, a) Constant amount approach Constant amount approach – Like Formula-1 car racing, a fixed no. of laps is set. The one who finishes all the laps first would be the fastest one and would be the winner. e.g. Rate of disproportionation of thiosulphate in acidic medium Sodium thiosulphate and hydrochloric acid solution is put into a beaker. In acidic medium, thiosulphate decomposes into yellow suspension of sulphur. Once the cross on the paper is masked by the yellow suspension, the time is noted. As constant amount of sulphur precipitate is needed to mask the cross and the time taken is noted, the rate of reaction can be estimated. Rate of reaction = Amount of sulphur precipitate formed time taken 1 Rate ∝ time taken Different concentration of thiosulphate and acid can be used to study the kinetic of the reaction. I. Rates of chemical reactions b) Constant interval approach Page 3 Constant interval approach –Like Daytona 24 hours car racing, a fixed interval e.g. 24 hours is set. The one who can finish more laps in 24 hours would be the winner. If the amount of a reactant or a product can be traced at constant intervals throughout the reaction, the rate of reaction can also be determined. For example, the reaction of disproportionation of thiosulphate, S2O32-(aq) + 2H3O+(aq) → SO2(g) + S(s) + 3H2O(l), can also be followed by titrating a fixed amount (e.g. 25.00 cm3) of reacting mixture at regular interval (e.g. 2 minutes) with standard I2(aq) solution using starch indicator. I2(aq) + 2S2O32-(aq) → 2I-(aq) + S4O62-(aq) Before titration, the disproportionation reaction must be quenched (stopped) first. This reaction is catalyzed by acid, therefore, quenching can be done by neutralizing the acid with excess alkali, e.g. sodium hydrogencarbonate (a weak alkali is preferred) or by sudden cooling of the reaction mixture. - d [S2O32-] Rate = dt The rate of reaction at any particular instant can be calculated from the slope. Rate = - slope It is found that the reaction is first order with respect to S2O32-(aq) and zero order with respect to H3O+(aq). Rate = - d [S2O32-] = k [S2O32-] dt In another example, for the catalytic decomposition of H2O2(l) in the presence of MnO2(s) catalyst, it can be quenched by an acidic medium where acid is the negative catalyst to the reaction. The concentration of the H2O2(l) can be determined by titrating with standard MnO4-(aq). 5H2O2(aq) + 2MnO4-(aq) + 6H+(aq) → 2Mn2+(aq) + 8H2O(l) + 5O2(g) I. Rates of chemical reactions 2. Interpretation of physical measurements made in following a reaction Page 4 Besides titration, there are many other methods to follow a reaction. There are many quantities proportional to the amount of a chemical in the reaction mixture. e.g. volume of gas, colour, conductivity and optical activity a) Volume of gas formed In the reaction between magnesium and hydrochloric acid, hydrogen gas is evolved. Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) 1 d[HCl(aq)] d[MgCl2(s)] d[H2(g)] = = dt Rate = - 2 dt dt The rate of reaction can be traced by monitoring the volume of the hydrogen gas evolved. N.B. mass All pure solids have a constant density or concentration, equals volume . Therefore, the rate of a reaction is usually depending on the size of the solid particle rather than the amount of the solid present. A graph of volume of hydrogen against time b) Colorimetric measurement If one of the reacting substances or products has a colour, the intensity of this colour will change during the reaction. This method relies on the measurement of intensity of colour, therefore, it is called colorimetry. An example will be the disappearance of the colour of bromine during the oxidation of methanoic acid by bromine: Br2(aq) + HCO2H(aq) → 2Br-(aq) + 2H+(aq) + CO2(g) The change in colour intensity could be followed by a photoelectric device in a colorimeter, as shown in the following figure. Since the colour intensity is directly proportional to the concentration of Br2(aq), therefore, Rate ∝ d (colour intensity of the solution) dt constant time approach initial rate of reaction Glossary rate of reaction constant amount approach quenching colorimetry I. Rates of chemical reactions Page 5 Past Paper Question 93 1A 2 a iii 95 2A 2 b i 96 1B 4 d i ii iii 97 2A 3 a i ii iii 98 2A 3 a iii 99 1B 7 a 93 1A 2 a iii 2a Answer the following questions by choosing in each case one of the species listed below, putting it in the box and giving the relevant equation(s). Al(s), AlCl3(s), AlO2-(aq), Na(s), CO32-(aq), Cu2+(aq), P4O10(s), S(s), S2O32-(aq), Zn2+(aq) iii Which species forms a pale yellow precipitate with dilute hydrochloric acid? S2O32-(aq) + 2H+(aq) → S(s) + SO2(g) + H2O(l) 2 marks S2O32-(aq) C Many candidates picked the wrong chemical species for the inorganic reactions in parts (i) to (iv). 95 2A 2 b i 2b In an aqueous solution, the decomposition of hydrogen peroxide in the presence of manganese(IV) oxide can be represented by the following equation: i 2 C 2H2O2(aq) 2 → 2H2O(l) + O2(g) For a given amount of manganese(IV) oxide, outline how you would use a chemical method to determine the concentration of hydrogen peroxide, at different times, in the reaction mixture. Withdraw a known volume of the reaction mixture ½ mark ½ mark and add to excess dilute H2SO4, 1 mark titrate H2O2 against standard MnO4-. 5H2O2 + 2MnO4- + 6H+ → 2Mn2+ + 8H2O + 5O2 This question was poorly answered. Candidates should brush up their skills in answering questions which involve the description of experimental procedures. MnO 2 96 1B 4 d i ii iii 4d In an experiment to determine the concentration of H2O2 contained in a rainwater sample, 5.0 cm3 of the sample were mixed with an excess of a certain transition metal complex solution, giving a coloured mixture. The absorbance of the mixture was measured by a colorimeter and was found to be 0.0273. When 5.0 cm3 of 1.46 × 10-6 M H2O2 (instead of the rainwater sample) were treated in the same way, an absorbance of 0.0398 was recorded. i Calculate the concentration of H2O2 in the rainwater sample assuming that concentration is directly proportional to 2 absorbance. 1.46 × 10-6 concn. of H2O2 in the sample = 0.0398 × 0.0273 1 mark -6 -6 1 mark = 1.00 × 10 M / 1.0 × 10 M (Deduct ½ mark for answers given in 1 × 10-6 M or 10-6 M) ii Why is the method of titration not suitable for the determination of the concentration of H2O2 in the rainwater 1 sample ? The concentration of H2O2 in rainwater is too low to be determined by titrimetric method. 1 mark iii Why is it not suitable to collect the rainwater sample for this experiment in an iron container ? Fe can be oxidized by H2O2 / transition metal ions can catalyse the decomposition of H2O2. 1 1 mark I. Rates of chemical reactions 97 2A 3 a i ii iii 3 The reaction of iodine with propanone in acidic solutions can be represented by the following equation I2(aq) + CH3COCH3(aq) → CH2ICOCH3(aq) + H+(aq) + I-(aq) 3a i The progress of the reaction can be monitored by a titrimetric method. Outline the experimental procedure. ii State how the initial rate of the reaction can be determined from the titrimetric results. iii Suggest another method to monitor the progress of the reaction. 98 2A 3 a iii 3a The table below lists the rate constants, k, at different temperatures, T, for the first order decomposition of a dicarboxylic acid, CO(CH2CO2H)2, in aqueous solution : CO(CH2CO2H)2(aq) → CH3COCH3(aq) + 2CO2(g) T/K 273 293 313 333 353 k / s-1 2.46 × 10-5 4.75 × 10-4 5.76 × 10-3 5.48 × 10-2 ? iii Suggest a method to monitor the progress of the reaction. 99 1B 7 a 7a In a chemical kinetics experiment, samples of the reaction mixture are removed at regular time intervals for titrimetric analysis. Suggest TWO methods by which the reaction in the samples removed can be stopped or stowed down. Page 6 6 9 II. Order of reaction Page 1 Topic Reference Reading II. Order of reaction 5.2 Advanced Practical Chemistry, John Murray (Publisher) Ltd., 96–99, 103–106 A-Level Chemistry Syllabus for Secondary School (1995), Curriculum Development Council, 197–198 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 403–410 Experiment – Determination of the order of reaction of disproportionation of thiosulphate in acidic medium Assignment A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 302–308 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 358–365 Reading Syllabus Notes Order of reaction Rate Law II. Order of reaction A. Rate Law or rate equation Rate Law or rate equation – A mathematical expression relates the concentration of a species with the rate of reaction or time. There are three different forms of rate law Examples of rate law (ordinary form) e.g. H2(g) + Br2(g) → 2HBr(g) Rate = k’[H2][Br2] ½ [HBr] 1+k’’( [Br ] ) 2 i. ordinary form ii. differential form iii. integrated form e.g. 2NO(g) + O2(g) → 2NO2(g) Rate = k[NO]2[O2] e.g. 2H2(g) + 2NO(g) → 2H2O(g) + N2(g) Rate = k[NO]2[H2] N.B. The rate equation has no direct relationship with the stochiometric coefficient. It can only be determined by experiment. According to collision theory, the rate of reaction always increases with concentration of reactant. concentration of reactant ↑ ⇒ rate of reaction ↑ In some reactions, the rate is directly proportional to the concentration. i.e. double the concentration of a reactant, only double the rate of reaction. But in some reactions, double the concentration of a reactant may cause 4 folds increase in reaction rate. The relationship needs not be linear. e.g. H2(g) + I2(g) → 2HI(g) For the above reaction, it is found experimentally that Rate ∝ [H2] Rate ∝ [I2] Rate ∝ [H2][I2] Rate = k [H2][I2] Rate law or rate equation (an overall second order reaction) The reaction is first order with respect to H2(g) The reaction is first order with respect to I2(g) k, the proportionality constant, is known as rate constant which depends on temperature only. II. Order of reaction Page 2 Rate has the unit (mole dm-3 s-1) and concentration has the unit (mole dm-3). Therefore, the unit of rate constant, k, depends on actually what the rate equation is. e.g. Rate In the rate equation, Rate = k [H2][I2], k = [H ][I ] 2 2 mole dm-3 s-1 -1 3 -1 unit of k = = mole dm s mole dm-3 · mole dm-3 e.g. Rate For the reaction, 2H2(g) + 2NO(g) → 2H2O(g) + N2(g), Rate = k[NO]2[H2], k = [NO]2[H ] 2 mole dm-3 s-1 -2 6 -1 unit of k = = mole dm s (mole dm-3)2 · mole dm-3 1. Differential form and Integrated form of rate law (The derivation of the integrated form of rate law is not required in A-Level) Rate of reaction = ∆ [product] = ∆t ∆ [reactant] ∆t By using notation of calculus Rate of reaction = d [product] - d [reactant] = dt dt (differential form of rate law) For the reaction, H2(g) + I2(g) → 2HI(g) Rate = - d [H2] - d [I2] 1 d [HI] dt = dt = 2 dt - d [H2] dt = k[H2][I2] and Rate = k[H2][I2] ∴ Rate = Keep the [I2] constant by using excess I2 where k’ = k [I2], the reaction is first order with respect to H2(g) - d [H2] = k’[H2] dt d[H2] - [H ] = k’dt 2 f ⌠- d[H2] = k' ⌡dt ⌠ ⌡ [H2] 0 [H2]0 [H2] - [ ln [H2] ][H ] = k' t 20 [H2] [H2] is the concentration at time = t [H2] 0 is the concentration at time = 0 - (ln [H2] - ln [H2] 0) = k’t [H2] 0 ln [H ] = k’t 2 at t = 0 at t = ∞ or [H2] 0 k't [H2] = e (Integrated form of rate law) [H2] 0 = [H2] [H2] 0 [H2] = ∞ ⇒ [ H 2] = 0 II. Order of reaction 2. Graphical presentation of reaction rate Page 3 For the reaction H2O2(aq) + 2I-(aq) + 2H+(aq) → 2H2O(l) + I2(aq) The completeness of the reaction can be monitored by following the concentration of one of the reactant or product. ∆ [product] = ∆t ∆ [reactant] ∆t Rate of reaction = - Using notation of calculus Rate = d [I2] - d [H2O2] 1 - d [I-] 1 - d [H+] = 2 dt = 2 dt dt = dt Note the rate of disappearing of I-(aq) is twice of the rate of disappearing of H2O2(aq). Although H2O(l) is one of the product, it is also the solvent. Because it is in large excess, its concentration would be rather constant. For the reaction of decomposition of NO2(g) at 300 ºC 2NO2(g) → 2NO + O2(g) 1 d [NO] d [O2] 1 - d [NO2] = 2 dt = dt Rate = 2 dt ∆ [product] ∆t Notation of calculus is preferable to or - ∆ [reactant] since the former one reflects the rate ∆t of reaction at a particular instant instead of average rate over a period of time. II. Order of reaction 3. Order of reaction Page 4 Order of individual reactant – Overall order of reaction – The order of a reaction with respect to a given reactant is the power of that reactant’s concentration in the experimentally determined rate equation. The (overall) order of the reaction is the sum of the powers of the concentration terms in the experimentally determined rate equation. e.g. 2H2(g) + 2NO(g) → 2H2O(g) + N2(g) Rate = k[NO]2[H2] The reaction is second order with respect to NO(g) The reaction is first order with respect to H2(g) The overall order of the reaction is 3 or the reaction is a third order reaction. A zeroth order means that the rate of reaction is independent of the concentration of that reactant. A first order means that the rate of reaction is directly proportional to the concentration of that reactant. A second order means that the rate of reaction is directly proportional to the square of the concentration of that reactant. In general, the higher the order of a reactant, the more dependent will be the rate of reaction on the concentration of that reactant. II. Order of reaction a) Experimental determination of order of reaction (1) By measuring the rates of reaction at different reactant concentrations Suppose a reaction X+Y→Z has the rate law rate = k[X]a[Y]b Page 5 The order of reaction with respect to X (i.e. a) can be determined by measuring the rate of reaction with different concentration of X, while the concentration of Y is kept constant. rate1 = k[X]1a[Y]b rate2 = k[X]2a[Y]b (1) ÷ (2) [X]1 a rate1 k[X]1a[Y]b rate2 = k[X]2a[Y]b = ( [X]2 ) [X]1 rate1 ln rate = a ln ( [X] ) 2 2 (1) (2) [X]1 rate1 a = ln rate ÷ ln ( [X] ) 2 2 (2) By plotting graph of ln rate versus ln [X[ The order of reaction with respect to X can also be determined by graphical method. Since more than 2 readings are used to determine the order, the order determined will be more accurately than the above method. For the same reaction X + Y → Z rate = k[X]a[Y]b rate = k'[X]a ln rate = ln (k'[X]a) ln rate = ln k' + a ln [X] According to the above equation, the order of a reaction (i.e. a) with respect to a reactant (i.e. X) can be determination by measuring the rates of reaction (i.e. rate) with different concentrations of the reactant (i.e. [X[). The order of reaction can be obtained from the slope of the graph with (ln rate) plotting against ln [X]. Keep [Y] constant by using excess Y, since b is also a constant. [Y]b will be a constant. II. Order of reaction B. Examples of different reaction 1. First order reaction Page 6 First order reaction – the rate of reaction is directly proportional to the concentration of the reactant. Since the rate of decay is directly proportional to the amount of isotope remains. This form an exponential decay curve. After 5 half-lives, the fraction remaining will be = ½×½×½×½×½ = 0.03125 = 3.125% Radioactive decay is said to be a first order reaction, since the rate is proportional to the number of radioactive isotopes present. a) Half life (t½) The rate of radioactive decay reaction is directly proportional to the abundance of the radioactive isotope. Radioactive decay is so special that the rate of reaction is even independent of temperature. This is the only reaction which is independent of temperature. Radiation ∝ No. of isotope ∝ Rate of decay Rate of decay ∝ I -d I Rate = k I = dt By integration according to time. I0 ln I = kt or I0 kt I =e I0 : abundance of radioactive isotope at time = 0 I : abundance of radioactive isotope at time = t By definition, half life (t½), is the time taken by a given amount of radioactive isotope to decay to half of the original amount. I0 ln I = kt 2I ln I = kt½ = ln 2 N.B.: when t = t½ then I0 = 2 I ln 2 0.693 t½ = k = k I : abundance of radioactive isotope ∝ no. of isotope It can be seen that the half life (t½) of a first order reaction is related to the decay constant (k). II. Order of reaction b) Carbon-14 dating In nature the, 14 7 Page 7 N in the air, is transmutated to 14 6 C by the neutron from the cosmic ray. N +1 n → 6 C +1 H 0 1 14 i.e. 14 6 14 7 C isotope is radioactive. At the same time, it undergoes beta decay which is a first order reaction. 14 6 C → 7 N + 01 e − 14 6 14 As a result of this simultaneous formation and decay of 14 6 C , the atmosphere contains a constant concentration of 14 6 C . The 14 6 C is present as 14 6 CO2 . This 14 6 CO2 enters plants via photosynthesis and animals via the food chain. C . However, when the animal 14 6 Therefore, all living things have a constant proportion of their carbon in form of or plant dies, replacement of 14 6 C ceases while decay of 14 6 C continues. It is assumed that the concentration of 14 6 CO2 in atmosphere remains unchanged throughout million of years. By comparing the content of C in archaeological specimens with that in similar living materials living at the present time, it is possible to estimate the age of the specimens. c) Example of calculation 14 6 Half life (t½) of C = 5730 years If 32 counts per minute per gram of carbon is emitted by living organisms, while a fossilized bone gives 8 counts per minute per gram of carbon, calculate the age of the fossil. The number of counts is proportional to the abundance of the radioactive isotope. I0 ln I = kt 32 ln 2 ln 8 = t · t ½ ln 2 ln 4 = 5730 · t t= ln 4 · 5730 = 2 × 5730 = 11460 years ln 2 and ln 2 0.693 t½ = k = k II. Order of reaction 2. Second order reaction Page 8 Second order reaction – the rate of reaction is directly proportional to the square of the concentration of the reactant. Consider the reaction A(l) + B(l) → C(l) + D(l) which is first order with respect to A, but second order with respect to B. Rate = k[A][B]2 if we keep A in large excess, then [A] will remain a constant throughout the reaction. Rate = k’[B]2 the reaction becomes second order with respect to B, where k’ = k [A] Rate = k’[B]2 = d[B] - k’dt = [B]2 [B] d[B] ⌠ ⌡- k' dt = ⌠ [B]2 ⌡ 0 [B]0 t 1 - k' t = - [B] [B] [B]0 - d[B] dt 1 1 - k' t = - [B] - - [B] 0 1 1 - k' t = [B] - [B] 0 1 1 [B] = k' t + [B]0 1 By plotting a graph [B] versus t, the slope will equal to k’. Since k’ = k [A], the rate constant (k) can be determined. Half life (t½) of second order reaction 1 1 [B] = k' t + [B]0 1 1 ½[B]0 = k' t½ + [B]0 1 1 k' t½ = ½[B] - [B] 0 0 2 1 1 k' t½ = [B] - [B] = [B] 0 0 0 1 t½ = k' [B] 0 N.B. In contrast to the constant half life of a first order reaction, half life of a second order reaction is depending on the initial concentration of the reactant. 1 at t = t½, [B] = 2 [B] 0 II. Order of reaction 3. Zeroth order reaction Page 9 Zeroth order reaction – the rate of reaction is independent of the concentration of the reactant. Actually, there is no reaction with an overall order equals zero. But zero order reaction with respect to individual reactant is commonly found in catalytic reaction. In synthesis of ammonia by Haber process, N2(g) + 3H2(g) d 2NH3(g), it is found that the reaction is zero order with respect to both N2(g) and H2(g). This is because Fe catalyst is used in the synthesis. The availability of the surface of Fe is the limiting factor of the reaction. Both N2(g) and H2(g) are in large excess comparing with Fe catalyst. For a general zero order reaction, A → product Rate = k[A]0 = k(1) = k Rate = k = - d[A] dt - kdt = d[A] t 0 [A] [A]0 [A] [A]0 ⌠-k dt = ⌠d[A] ⌡ ⌡ -kt = [[A]] - kt = [A] - [A]0 [A] = - kt + [A]0 Half life of zero order reaction [A] = - kt + [A]0 1 2 [A] 0 = - kt½ + [A] 0 1 1 kt½ = [A] 0 - 2 [A] 0 = 2 [A] 0 [A] 0 t½ = 2 k Glossary order of reaction rate law / rate equation rate constant differential form of rate law integrated form of rate law overall order of reaction half-life carbon-14 dating decay constant II. Order of reaction Page 10 Past Paper Question 90 1A 2 d i 91 2A 1 a iii 92 1A 2 a i ii 93 2A 2 b i ii 94 1B 4 e i ii iii 95 1A 1 d i ii 96 1A 1 b i ii 97 2A 3 b i ii c 98 2A 2 b i ii 99 2A 3 a i ii 92 2A 2 a ii iii 95 2A 2 a ii 96 2A 3 c i ii 98 2A 3 a ii 95 2A 2 b ii 90 1A 2 d i 2d Consider the reaction A(l) + B(l) → C(l) + D(l) which is first order with respect to A, but second order with respect to B. i Give a rate expression for this reaction 1 2 Rate = k[A][B] 1 mark 91 2A 1 a iii 1a The radioactive decay of that all 206 82 238 92 U may be represented by 238 92 238 92 U→ 206 82 Pb + α + β. 206 82 iii The age of a mineral containing comparison with that of mole ratio of Note : 206 82 238 92 U and 206 82 Pb can be estimated from the mole ratio, Pb : 238 U . It is assumed 4 92 Pb has arisen only by the decay of the uranium and that all the subsequent decays are small in U which has a half-life of 4.51 × 109 years. Calculate the age of the mineral if the I0 = kt where I0 is the initial I Pb : 238 U is 0.231 : 1.000. 92 The integrated form of a first order reaction my be expressed as ln C concentration and I is the concentration at time t. The decay constant k can be calculated from the half-life. 2 ln 1 = 1.5369 × 10-10 year-1 1½ mark k= 4.51 × 109 The ratio of Pb : U = 0.231 : 1.000 ½ mark The amount of U at the beginning (I0) = 1.000 + 0.231 = 1.231 The amount of U at time t (I) = 1 ½ mark ln1231 . = 1.35 × 109 years (no / wrong unit -½) (3 to 4 sig. fig. if not -½) 1½ mark ∴t= 15369 × 10 −10 . Many candidates were not able to establish the correct ratio of I0/I, and hence failed to obtain the correct answer. 92 1A 2 a i ii 2a For the first order reaction, A k → product , the integrated form of the rate equation is [A] = [A] 0 e − kt , where [A] and [A] 0 are the concentrations of A at time = t and time = 0 respectively. i Starting from this equation, derive the relationship between the half-life t½ of the reaction and the rate constant k. 1 The half life of the reaction is defined by the concentration condition, [A] = [A]0/2. ½ mark Substituting into the integrated rate law give [A]0/2 = [A]0 e-kt. ½ mark This is followed by taking the natural logarithm to give ln(0.5) = -kt½. Therefore, t½ = 0.693/k. Without using the half-life method, show how you would determine the rate constant k from a set of experimental 2 measurements of concentration at various times. From the integrated rate equation, an experimental straight line plot can be obtained by taking the ln of the ½ mark integrated rate equation, ln[A] = ln[A]0 - kt. Which can be rearranged to give an equation in the form of a straight line (y = mx +b) 1 mark ln [A] = -kt + ln[A]0 ½ mark Therefore, k can be determined from the slope of the plot ln[A] vs t. A good example to demonstrate that most candidates lack the most basic mathematical application / manipulation skills. No knowledge of calculus was required in answering this question. The candidates should have the ability to take In and to recognize equations in the form of a straight line. Few candidates were able to score full marks in this question. ii C II. Order of reaction 92 2A 2 a ii iii 2a In acid solution, chlorate(V) ions, ClO3-, slowly oxidize chloride ions to chlorine. The following kinetic data are obtained at 25°C. [Cl-]/mol dm-3 [H+]/mol dm-3 Initial rate/mol dm-3 [ClO3-]/mol dm-3 -5 0.08 0.15 0.20 1.0 × 10 0.08 0.15 0.40 4.0 × 10-5 0.16 0.15 0.40 8.0 × 10-5 0.08 0.30 0.20 2.0 × 10-5 ii Determine the order of the reaction with respect to each reactant. Since the rate of reaction 1 mark Rate = k[ClO3-]l [Cl-]m [H+]n n −5 l m n 1.0 × 10 ( 0.08) ( 015) ( 0.2) . 1 2 1 = ⇒ = ⇒ − log 4 = n log ⇒ n = 2 4 4 2 4.0 × 10 −5 ( 0.08) l ( 015) m ( 0.4) n . Similarly l = 1 and m = 1. with working max. 3 marks without working: all correct 2 marks or wrong 0 mark Page 11 3 iii Determine the rate constant at this temperature. Rate = k[ClO3-][Cl-][H+]2 mol dm −3 s −1 10 × 10 −5 . rate k= = = 2.08 × 10 −2 mol −3 dm9 s −1 2 +2 ( 0.08)( 015)( 0.2) ( mol dm −3 ) 4 . [ClO 3 ][Cl ][H ] C Many candidates gave wrong units for the rate constant. 2 1 mark 1 mark no unit -½ mark 93 2A 2 b i ii 2b Given the following data for the reaction at 298 K 2C + 3D + E → P + 2Q -3 -3 -3 -3 -1 Experiment [C] / mol dm [D] / mol dm [E] / mol dm initial rate / mol dm s -3 1 0.10 0.10 0.10 3.0 ? 10 -2 2 0.20 0.10 0.10 2.4 ? 10 -3 3 0.10 0.20 0.10 3.0 ? 10 -2 4 0.10 0.10 0.30 2.7 ? 10 i Deduce the rate law of the above reaction. Rate = k[C]x[D]y[E]z 3.0 × 10-3 1 rate 1 0.10 = ∴x=3 1 mark Since rate 2 = (0.20 )x = 2.4 × 10-2 8 rate 1 0.10 y 3.0 × 10-3 ∴y=0 1 mark rate 3 = (0.20 ) = 3.0 × 10-3 = 1 -3 1 rate 1 0.10 z 3.0 × 10 ∴z=2 1 mark rate 4 = (0.30 ) = 2.7 × 10-2 = 9 Rate = k[C]3[D]0[E]2 = k[C]3[E]2 1 mark ii Calculate the rate constant. Using results of experiment 1 3.0 × 10-3 = k (0.10)3(0.10)2 2 marks k = 300 (mol dm-3)-4 s-1 (1 mark for answer, 1 mark for correct unit) C Some candidates could not do the calculation of the rate constant, due to their unfamiliarity in handling exponential values. 4 2 II. Order of reaction Page 12 94 1B 4 e i ii iii 4e In an experiment to determine the rate constant at 298 K for the decomposition of sodium thiosulphate by a large excess of dilute hydrochloric acid, the time, t, taken for a certain amount of sulphur to appear was measured. Under these conditions: Rate = k’[S2O32-]. i Write a balanced equation for the reaction between hydrochloric acid and sodium thiosulphate. 1 S2O32-(aq) + 2H+(aq) → H2O(l) + SO2(g) + S(s) or Na2S2O3 + 2HCl → 2NaCl + S + SO2 + H2O 1 mark ii Describe how you would carry out this experiment, indicating the measurements you would take. 4 Make standard [S2O32-] solution of different molarity and place equal volume of these solutions in conical flasks. 1½ mark ½ mark Measure temperature of solution / set thermostat to 298 K ½ mark Add a fixed volume of acid to Na2S2O3 (using measuring cylinder) Start stop clock ½ mark Note time, t, when a black “X” marked on a cardboard becomes invisible when looking through the flask. 1 mark iii What quantities would you plot on a graph for the determination of the rate constant k’ at 298 K? 1 1 -1 -3 21 or 0 mark t / s and [S2O3 (aq)] / mol dm C Many candidates have previous experience of this experiment, but nonetheless their answers were poor. The temperature control was omitted in most cases, and many candidates did not give a method for measuring the time taken for the formation of a certain amount of sulphur. Some utilized titration, while others used a colorimeter or filtering and weighing to quantitate sulphur, but none of these answers were acceptable. In (iii), candidates did not know that rate is inversely proportional to reaction time. 95 1A 1 d i ii 1d The iodination of propanone is catalysed by hydrogen ions. The overall equation is: CH3COCH3(aq) + I2(aq) → CH3COCH2I(aq) + HI(aq) Using four mixture B, C, D and E, the progress of the reaction was followed by colorimetric measurement. The results are tabulated below. Composition by volume of mixture /cm3 Mixture propane water 1.00 M HCl 0.05 M I2 in KI Initial rate / mol dm-3 s-1 B 10.0 60.0 10.0 20.0 4.96 × 10-6 C 10.0 50.0 10.0 30.0 5.04 × 10-6 D 5.0 65.0 10.0 20.0 2.45 × 10-6 E 10.0 65.0 5.0 20.0 2.47 × 10-6 i Determine the effects of the changes in concentration of each of the reactants (iodine and propanone) and the catalyst (hydrochloric acid) on the reaction rate. Write the rate expression for the reaction. I 2, use B and C : rate independent of [I2] ½ mark propane, use B and D : rate ∝ [propanone] ½ mark H +, use B and E : rate ∝ [H+] ½ mark + ½ mark rate = k [H ] [CH3COCH3] ii For mixture B, calculate the rate constant for the reaction at the temperature of the experiment. (Density of CH3COCH3 = 0.789 gcm-3) 100 × 0.789 rate = 4.96 × 10-6 = k [0.1] [ 58.078 ] 1 mark -6 58.078 × 4.96 × 10 k = 0.1 × 100 × 0.789 = 3.65 × 10-5 mol-1dm3s-1 1 mark (½ mark for numerical answer; ½ mark for unit) C Many candidates were unable to deduce the quantitative relation between reaction rate and change in concentration of the reactants, merely stating that 'rate increases with increased concentration'. In some cases, the experimental nature of the rate results was not appreciated, so that the order with respect to iodine was calculated to be non-zero or fractional. Some candidates had difficulty in calculating the concentrations of the reactants from the data given. 2 2 II. Order of reaction 95 2A 2 a ii 228 2a Ra decays by the emission of β particles. The half-life for the decay is 6.67 years. 88 Page 13 ii 228 228 A sample containing 0.50 g of 88 Ra is kept in a closed container. Calculate the mass of 88 Ra remaining after 5 3 years. (The decay of a radioisotope can be described by the equation, Nt = N0e-kt, where N0 and Nt respectively represent the initial number and number at time t of the radioisotope, and k is the decay constant.) ln 2 k = t½ = 0.104 year-1 1 mark 0.5 ln N = 0.104 (5) 1 mark t 1 mark Nt = 0.297 g 95 2A 2 b i ii 2b In an aqueous solution, the decomposition of hydrogen peroxide in the presence of manganese(IV) oxide can be represented by the following equation: i ii 2H2O2(aq) 2 → 2H2O(l) + O2(g) For a given amount of manganese(IV) oxide, outline how you would use a chemical method to determine the concentration of hydrogen peroxide, at different times, in the reaction mixture. Withdraw a known volume of the reaction mixture ½ mark ½ mark and add to excess dilute H2SO4, 1 mark titrate H2O2 against standard MnO4-. 5H2O2 + 2MnO4- + 6H+ → 2Mn2+ + 8H2O + 5O2 How would you use the results obtained in (i) to show graphically that the decomposition is first order with respect to H2O2 and to find the rate constant of the decomposition ? Plot [H2O2] against t to obtain the following graph MnO 2 4 2 marks ln 2 for first order reaction t½ = constant and k = t ½ OR Plot ln [H2O2] against t to obtain a straight line 2 marks 1 mark [H2O2] ln [H O ] = - kt 2 20 ⇒ ln[H2O2] - ln[H2O2]0 = -kt 2 marks C for first order reaction, slope = -k. 1 mark This question was poorly answered. Candidates should brush up their skills in answering questions which involve the description of experimental procedures. 96 1A 1 b i ii 1b Carbon-14, 14C, is radioactive, emitting β-particles. i Write an equation for the decay of 14C. 14 6 1 1 mark 14 12 C→ 0 −1 e+ 14 7 N or 14 6 C→β+ 14 7 N ii A charcoal sample from the ruins of an ancient settlement was found to have a C/ C ratio 0.60 times that found in living organisms. (1) Explain why the 14C/12C ratio in the charcoal sample is smaller than that in living organisms. (2) Given that the half-life for the decay of 14C is 5730 years, calculate the age of the charcoal sample. (Note : The integrated form of the rate expression for radioactive decay can be represented by the following 3 II. Order of reaction equation : N0 0.301 log10 ( N ) = t t t ½ where N0 is the initial number of radioactive nuclei; Nt is the number of radioactive nuclei at time t; t½ is the half-life for the decay) (1) In the piece of charcoal intake of C stops; decay of 14C continues while 12C remains constant ∴ 14C / 12C ratio drops 1.0 0.301 t (2) log 0.6 = 5730 ⇒ t = 4223 ± 3 year (Deduct ½ marks for wrong /no unit) Page 14 ½ mark ½ mark 1 mark + 1 mark 96 2A 3 c i ii 3c The following equation represents the decomposition of N2O5(g) : 2N2O5(g) → 4NO2(g) + O2(g) The progress of the above decomposition can be followed by measuring the partial pressures of N2O5(g), PN O , at 25 different times, t. The table below lists the results of such an experiment : t / minute PN O / kPa 2 5 0 13.3 100 10.9 200 8.9 300 7.2 400 5.9 550 4.4 800 2.7 1250 1.1 3 i By plotting a suitable graph, find the order of the decomposition. t 0 100 200 300 400 550 800 1250 2.59 2.39 2.19 1.97 1.77 1.48 0.99 0.10 ln(PN2O5) 1.12 1.04 0.95 0.86 0.77 0.4 0.43 0.04 or log(PN2O5) Plotting the following graph 2 marks ii (1 mark for the straight line, 1 mark for correct labelling of the axes) the reaction is 1st order Using your graph in (i), determine the rate constant of the decomposition. From the plot of ln(PN2O5) against t, - slope = k = 1.99 × 10-3 minute-1 (2.2 × 10-3 – 1.8 × 10-3 min-1) (1 mark for numerical answer, ½ mark for unit) 1 mark 2 1 mark 1 mark II. Order of reaction 97 2A 3 b i ii c 3 The reaction of iodine with propanone in acidic solutions can be represented by the following equation I2(aq) + CH3COCH3(aq) → CH2ICOCH3(aq) + H+(aq) + I-(aq) 3b The following initial rates and initial concentrations were obtained in an experiment at 298 K: Initial concentration / mol dm-3 -3 -1 Initial rate / mol dm s [I2(aq)] [CH3COCH3(aq)] [H+(aq)] 3.5 × 10-5 2.5 × 10-4 2.0 × 10-1 5.0 × 10-3 -5 -4 -1 3.5 × 10 1.5 × 10 2.0 × 10 5.0 × 10-3 -4 -4 -1 1.4 × 10 2.5 × 10 4.0 × 10 1.0 × 10-2 -5 -4 -1 7.0 × 10 2.5 × 10 4.0 × 10 5.0 × 10-3 i Deduce the rate equation for the reaction. ii Calculate the rate constant for the reaction at 298 K. 3c Suppose that the reaction takes place in a buffer solution of pH 4. On the basis of your results in (b) deduce the half-life of the reaction at 298 K. Page 15 5 4 98 2A 2 b i ii 5 2b Potassium-40 is radioactive and decays to give the stable isotope, argon-40. The half-life of the decay is 1.27 × 109 years. i In a rock sample, the ratio of 40K to 40Ar is 1 to 9. Estimate the age of the rock sample. ii The above method of estimation is based on several assumptions. One of the assumptions is that all 40Ar present in the rock sample is derived from the decay of 40K. Give one other assumption relating to 40Ar. 98 2A 3 a ii 3a The table below lists the rate constants, k, at different temperatures, T, for the first order decomposition of a dicarboxylic acid, CO(CH2CO2H)2, in aqueous solution : CO(CH2CO2H)2(aq) → CH3COCH3(aq) + 2CO2(g) T/K 273 293 313 333 353 k / s-1 2.46 × 10-5 4.75 × 10-4 5.76 × 10-3 5.48 × 10-2 ? ii Estimate the rate constant of the reaction at 353 K and hence calculate the half-life of the reaction at the same temperature. 99 2A 3 a i ii 3a Consider the following data for the reaction A + B → products Initial concentration / mol dm-3 [A] [B] Initial rate / mol dm-3s-1 4.0 × 10-2 4.0 × 10-2 6.4 × 10-5 -2 -2 8.0 × 10 4.0 × 10 12.8 × 10-5 -2 -2 4.0 × 10 8.0 × 10 6.4 × 10-5 For this reaction, deduce its rate equation, calculate the rate constant, and 9 i ii III. Collision theory and activation energy Unit 1 Page 1 Topic Reference Reading III. Collision theory and activation energy 5.3 Unit 1 Assignment A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 172–173, 310–311 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 410–412 Reading Syllabus Distribution of molecular speed Effect of temperature change on molecular speeds Collision theory and activation energy Arrhenius equation III. Collision theory and activation energy A. Collision theory 1. Distribution of molecular speed (Maxwell-Boltzmann distribution) Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 140–142, 266–367, 369 Notes For a sample of gas at a specific temp, it is found that most particles possess speeds about average speed. Only very few have very high or very low speed (extreme speeds). This gives a Maxwell-Boltzmann distribution. Maxwell-Boltzmann distribution is different from normal distribution in the way that it is not symmetrical. Origin of Maxwell-Boltzmann distribution Gas particles are in constant random motion. During collisions, some particles gain energy from the others and some particles loss energy to the others. After a series of collisions, different particles will have different speed. 2. Effect of temperature change on molecular speeds Maxwell-Boltzmann distribution As temperature increases, collisions among the particles become more vigorous. This causes a wider spread in particle speed. The kinetic energy of the gas increases with increasing temperature, so the peak of most probable speed shifts to the right. The area under the curve represents the total no. of particle and should remain constant and independent of the temperature. This makes the height of the peak drops at higher temperature. III. Collision theory and activation energy 3. Unit 1 Page 2 Collision theory and activation energy To simplify the discussion, a sample of ideal gas is considered first. Except in self decomposition, all reacting particles must get in contact with other for reaction to take place. The collision frequency (Z) is proportional to the velocity of the particle (c). Z∝c Conclusion 1 – Reaction takes place as a result of collision. The simple collision theory is contradictory to the experimental finding. For most reactions, every 10 K increase in temperature will double the reaction rate. Meanwhile, an increase in 10 K at room temperature will only cause 1.6 % increase in particles velocity. Reaction is something more than just collision. Experimental finding finds that reaction rate increases exponentially with increasing temperature. Refering to the Maxwell-Boltzmann distribution of molecular speed, by mathematical approximation, the fraction of the shaded area is represented by e EA is called the activation energy. − EA RT , where = = shaded area total area no. of particles having energy greater than EA total no. of particles e − EA RT The value e temperature. − EA RT increases exponentially with In order to explain the expontential increase of reaction rate with temperature, it is assumed that a reaction will only take place if the collision has an energy exceeds the activation energy EA. Different reaction has different value of EA. Rate ∝ Z and therefore Rate ∝ e − EA RT − EA RT Rate ∝ Ze Z : collision frequency R : (universal) gas constant = 8.314 JK-1mol-1 EA : activation energy T : temperature in Kelvin Conclusion 2 – Reaction only occurs when the collision exceeds certain threshold energy, activation energy. III. Collision theory and activation energy Unit 1 Page 3 Since the rate of reaction is directly proportionally to the rate constant, k. Rate ∝ k k ∝ Ze − EA RT Z : collision frequency P : steric factor / orientation factor (a proportionality constant) where A = PZ A is called Arrhenius factor k = PZe k = Ae − − EA RT EA RT Through experiment, P is found to be a factor always smaller than 1. It is a factor reflects the steric factor and the fraction of collisions with effective orientation. Conclusion 3 – Reaction only occurs when the colliding particles are correctly oriented to one another. According to collision theory, a reaction will take place only if the following 3 conditions are satisfied and the collision is then called an effective collision. i. Collision ii. Exceed activation energy iii. Correct orientation of collision a) Arrhenius equation The equation k = Ae − EA RT is called Arrhenius equation. k: A: EA : e − EA RT This equation relates the temperature and activation energy with rate constant, thus rate of reaction. rate constant (a constant depends on temperature only) Arrhenius constant activation energy : activation state factor (fraction of particle has energy greater than EA at temperature T) Rate of reaction = k [Reactants]n = Ae − EA RT [Reactants]n From the above equation, it can be seen that the rate of a reaction is depending on 1. Temperature 2. Nature of the reaction (Activation energy, i.e. Steric factor & Orientation factor) 3. Concentration of reactants Glossary Maxwell-Boltzmann distribution Gas constant (R) threshold collision theory activation energy Arrhenius constant (factor) collision frequency (Z) III. Collision theory and activation energy Unit 1 Page 4 Past Paper Question 96 2A 1 b i ii 98 2A 3 b i ii 96 2A 1 b i ii 1b i Draw the Maxwell-Boltzmann curves for the distribution of molecular speeds at two different temperatures for an ideal gas. Plot of Boltzmann distribution curve at two temperatures T1 and T2, where T2 > T1 2 ii (½ marks for correct shape of the curves; ½ marks for indicating T2 > T1) Use your curves in (i) to explain why, for a fixed mass of an ideal gas at constant volume, the pressure increases as the temperature is raised. As temperature increases, fraction of molecules with high average speed / kinetic energy increases. 1 mark Hence, force exerted by the collision of molecules on the container wall / the chance in momentum of molecules upon collision / frequency of collision increases. 1 mark 2 98 2A 3 b i ii 3b The exothermic reaction E(g) → E'(g) (1) is a single stage reaction. i Sketch curves to show the distribution of molecular kinetic energy of the reactant, E(g), at two different temperatures. ii With reference to your answer in (i), explain why the rate of reaction (1) increases with temperature. III. Collision theory and activation energy Unit 2 Page 1 Topic Reference Reading III. Collision theory and activation energy 5.4 A-Level Chemistry Syllabus for Secondary School (1995), Curriculum Development Council, 199–200 Unit 2 A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 312 Reading Assignment Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 413–414 Advanced Practical Chemistry, John Murray (Publisher) Ltd., 100–101 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 365, 367–368 Determination of activation energy B. Determination of activation energy By rearranging the Arrhenius equation, the activation energy of a particular reaction can be determined experimentally. k = Ae − EA RT Syllabus Notes ln k = ln Ae − EA RT ln k = ln A + ln e − EA RT EA ln k = ln A - RT ln e EA ln k = ln A - RT where ln e = 1 1. By measuring the rate of reaction at different temperature Rate ∝ k Rate1 k1 Rate2 = k2 By measuring the rate of reaction at 2 different temperature, 2 set of data – Rate1, T1 and Rate2, T2 can be obtained. EA ln k1 = ln A - RT EA ln k2 = ln A - RT Eq. 1 - Eq. 2 Eq. 1 Eq. 2 EA EA ln k1 - ln k2 = (ln A - RT ) - (ln A - RT ) 1 2 1 2 EA 1 1 k1 ln k = - R ( T - T ) 2 1 2 EA 1 1 Rate1 ln Rate = - R ( T - T ) 2 1 2 By putting the Rates and temperature determined, R = 8.314 J-1K-1mol-1. The activation energy, EA, can be determined. III. Collision theory and activation energy 2. By plotting graph of ln k versus 1 T Unit 2 Page 2 1 To be more accurate, A and EA can then be determined by plotting a graph with ln k versus T k at different temperature can be determined by measuring the rates of a reaction at different temperatures according to the integral form of rate equation. (Refer to order of reaction and rate constant). 3. By plotting graph of ln rate versus 1 T 1 Actually, it is very time consuming to use plotting of ln k versus T to determine the activation energy. In order to determined a value of k at a certain temperature, a set of values of rate / concentration have to be determined first. (Refer to order of reaction and rate constant). To determine the Ea, many sets of values of rate / concentration at different temperature have to be measured. Since Rate ∝ k Rate ∝ Ae − and EA RT k = Ae − EA RT Rate = rAe Rate = qe − EA RT r : a proportionality constant q : another proportionality constant = rA − EA RT E −A RT ln rate = ln qe ln rate = ln q + ln e − EA RT EA ln rate = ln q - RT ln e EA ln rate = ln q - RT EA 1 ln rate = ln q - R T EA 1 By plotting a graph ln rate versus T , the slope of the graph will equal to - R . This provides a more convenient way to determine the activation energy if the determination of the Arrhenius constant is not required. where ln e = 1 III. Collision theory and activation energy Unit 2 Page 3 Glossary Past Paper Question Arrhenius equation 91 2A 3 a i ii 92 1A 2 e 93 2A 2 a 94 2A 1 c iii 96 1A 1 b iii 97 2A 3 d 98 2A 3 a i 91 2A 3 a i ii 3a For the reaction 2XY(g) → X2(g) + Y2(g), the rate constant is 3.91 × 10-4 mol-1dm3s-1 at 370ºC and 4.05 × 10-2 mol1 dm3s-1 at 470ºC. Generally the rate constant of a reaction is related to the temperature by k = A exp(-Ea/RT). Calculate i the activation energy, E ln k = ln A − a RT kT Ea 1 1 ln = (−) 1 mark kT R T1 T2 T1 = 273 + 370 = 643 K ½ mark T2 = 273 + 470 = 743 K −2 E 4.05 × 10 1 1 ) ln ½ mark = a( + 3.91 × 10 −4 8.31 643 743 Ea = 184.2 kJmol-1 (wrong unit -½) (not 3-4 sig.fig. -½) 2 marks ii the rate constant at 450ºC (Gas constant, R = 8.31 JK-1mol-1) 450 ºC ≡ 723 K k 723 184.2 × 103 1 1 ( ) ln( )= − −4 8.31 643 723 3.91 × 10 (wrong unit -½) 2 marks k723 = 1.77 × 10-2 mol-1dm3s-1 Ci Some candidates were not familiar with calculations involving the use of natural logarithm. ii Some weaker candidates failed to realize that the absolute temperature should be used in the calculation. 2 1 4 2 92 1A 2 e 2e The diagram below gives the Maxwell-Boltzmann energy distribution of a system of molecules at two temperatures: 2 C What do the shaded areas of the curves represent and why are they different at different temperatures? These curves plot the distribution of molecular kinetic energies at different temperatures. It is intended to reflect ½ mark that the most sensitive parameter which affects the reaction rate is the activation energy, Ea. The shaded area under either one of these curves corresponds to E ≥ Ea and is approximately equal to the fraction of molecules that collide with kinetic energy ≥ Ea. ½ mark The increase of the reaction rate with temperature corresponds closely to the ratio of the corresponding areas ½ mark where Ea represents the minimum collision energy necessary for the reaction to occur. Therefore, as the temperature changes, the area under the distribution curve for E ≥ Ea changes. ½ mark The poor performance to the first part of this question was disappointing. This is a standard question which has appeared almost on a yearly basis, yet candidates continued to fail to clearly explain the meaning of the various features of the Maxwell-Boltzmann distribution curve. III. Collision theory and activation energy Unit 2 93 2A 2 a 2a Discuss, in terms of the Arrhenius equation, the effect of temperature on the rate of a reaction. The rate of reaction is normally increased by raising the temperature. Ea Arrhenius Equation: ln k = -RT + C or k = Ae − Ea RT Page 4 2 1 mark 1 Ea Increase in T makes T smaller and -RT larger, hence ln k (k) increases. 1 mark 94 2A 1 c iii 1c The table below lists the concentration of the reactant C as a function of time at 298 K for the following reaction. C→D Time / s 0 60 120 180 240 300 -2 -3 [C] /10 mol dm 20.0 16.1 13.5 11.6 10.2 9.1 (Gas constant R = 8.31 JK-1mol-1) iii The rate constant of the above reaction is found to be doubled when the temperature is raised from 298 K to 306 K. Determine the activation energy of the reaction. Arrhenius Equation Ea ln k = C - RT Ea 1 (1) ln k = C - R ( 298 ) Ea 1 (2) ln k + ln 2 = C - R ( 306 ) Ea 1 1 (2) - (1) ln 2 = R ( 298 - 306 ) 1 mark 1 mark Ea = 65.7 kJ (no unit -½ mark) 96 1A 1 b iii 1b Carbon-14, 14C, is radioactive, emitting β-particles. iii All radioactive decay has zero activation energy. Comment on the effect of temperature upon the rate of decay. According to Arrhenius Equation, rate of decay is independent of temperature. 1 mark 97 2A 3 d 3 The reaction of iodine with propanone in acidic solutions can be represented by the following equation I2(aq) + CH3COCH3(aq) → CH2ICOCH3(aq) + H+(aq) + I-(aq) 3d For a given set of initial concentrations the initial rate doubles when temperature is increased from 298 K to 308 K. Calculate the activation energy of the reaction. 98 2A 3 a i 3a The table below lists the rate constants, k, at different temperatures, T, for the first order decomposition of a dicarboxylic acid, CO(CH2CO2H)2, in aqueous solution : CO(CH2CO2H)2(aq) → CH3COCH3(aq) + 2CO2(g) T/K 273 293 313 333 353 k / s-1 2.46 × 10-5 4.75 × 10-4 5.76 × 10-3 5.48 × 10-2 ? i Determine the activation energy for the reaction by plotting an appropriate graph. 2 1 2 9 IV. Energy profile of reaction Unit 1 Page 1 Topic Reference Reading IV. Energy profile of reaction 5.5 Unit 1 A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 313–315 Reading Assignment Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 417–420 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 373–375 Energy profile Transition state Reaction mechanism Rate determining step Consider the reaction between bromide and bromate(V) ion 5Br-(aq) + BrO3-(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l) Kinetic studies show that this is an overall fourth order reaction. Rate = k[Br-][BrO3-][H+]2 Syllabus Notes The theory of collision and activation energy cannot account for the order of the reaction determined. IV. Energy profile of reaction During a reaction, bonds are in the process of making and breaking. The idea of transition state (activated complex) comes from an assumption that in any reaction, the reacting molecules/ions, if acquired sufficient energy to overcome the energy barrier, would first form a transition state. The reaction would then continue to proceed as the transition state, which is energetically unstable, immediately decompose to form the desired products. Consider a simple reaction of hydrolysis of 1-bromobutane in aqueous alkaline solution, as it proceeds via the following mechanism: A. Transition state Transition state is an activated complex at the top of the potential curve, which is so unstable and can never be isolated. In the hydrolysis of 1-bromobutane, the transition state is a pentavalent trigonal bipyramidal complex. The transition state is highly unstable having 10 valence electrons. Activation energy (Ea) – the energy difference between the reactant and the transition state (the threshold energy level leading to a reaction) The energy profile for the reaction can be depicted as follows. IV. Energy profile of reaction 1. Unit 1 Page 2 Order of reaction leads to interpretation of reaction mechanism at molecular level Hydrolysis of 1-bromobutane is a second order reaction, the rate of reaction is directly proportional to the concentration of OH-(aq) ion and concentration of 1-bromobutane. Rate ∝ [OH-][CH3CH2CH2CH2Br] Rate = k[OH-][CH3CH2CH2CH2Br] It is proposed that the transition state is formed upon the collision between a OH- particle and a CH3CH2CH2CH2Br. Since the formation involves only 1 OH- particle and 1 CH3CH2CH2CH2Br molecule, the rate of formation of the transition state is directly proportional to [OH-] and [CH3CH2CH2CH2Br]. Hydrolysis of 1-bromobutane is called a bimolecular nucleophilic substitution reaction (SN2 reaction), since the formation of the transition state involves 2 molecules. The molecularity of the reaction is said to be 2. N.B. Molecularity is not the same as order the reaction. Molecularity is the number of the molecule involved in the formation of the transition state in a step of the reaction. While, order of reaction is the sum of the power of the concentration terms in the rate equation. They may or may not have the same value but their meaning are indeed different. a) Multisteps reaction (1) Alkaline hydrolysis of 2-chloro-2-methylpropane Alkaline hydrolysis of 2-chloro-2-methylpropane is found to be a first order reaction. CH3 CH3 C Cl + OH(aq) CH3 CH3 CH3 C OH + Cl-(aq) CH3 Rate = k[(CH3)3CCl] It is very clear that the reaction is not a SN2 reaction and OH- ion is not involved in the formation of the transition state. Indeed, it is a multisteps reaction consists of 2 steps Step 1 (slow) CH3 CH3 C Cl CH3 CH3 CH3 C + + ClCH3 Step 2 (fast) CH3 CH3 C+ CH3 -OH CH3 CH3 C OH CH3 As formation of the transition state for step 1 involves a higher activation energy Ea’, the step 1 would be slower and known as the rate determining step. The reaction is called unimolecular nucleophilic substitution (SN1 reaction) because only 1 molecule is involved in the formation of the transition state of the slowest step. The reaction, therefore, is only first order with respect to (CH3)3CCl. CH3 CH3 C+ CH3 is called a carbocation or carbonium ion. It is an intermediate of this reaction. Because it is at the bottom of a potential well, unlike the highly unstable transition state, it is possible to isolate the intermediate from the reaction mixture. IV. Energy profile of reaction Unit 1 Page 3 (2) Reaction between bromide and bromate(V) in acidic medium For the reaction, 5Br-(aq) + BrO3-(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l) Rate = k[Br-][BrO3-][H+]2 If the transition state is formed by collision between Br-, BrO3- and 2 H+ particles, the reaction rate would be extremely slow. The probability of having four particles collide at the same time is extremely low. b) Rate determining step For multisteps reaction, the rate of reaction is only controlled by the step with the slowest rate. The step with the slowest rate is called the rate determining step. Since the rate of the reaction is independent of the other steps. It is proposed that the reaction between bromide and bromate(V) in acidic medium involves 5 steps. Step 1 Step 2 Step 3 Step 4 Step 5 H+ + Br- → HBr H+ + BrO3- → HBrO3 HBr + HBrO3 → HBrO + HBrO2 HBrO2 + HBr → 2HBrO HBrO + HBr → H2O + Br2 Fast Fast Slow Fast Fast Since Step 3 is the slowest step, it would be the rate determining step. Rate of reaction ∝ [HBr][HBrO3] [HBr] ∝ rate of formation of HBr ∝ [H+][Br-] [HBrO3] ∝ rate of formation of HBrO3 ∝ [H+][BrO3-] Therefore rate of reaction ∝ [HBr][HBrO3] ∝ ([H+][Br-])([H+][BrO3-]) ∝ [Br-][BrO3-][H+]2 Rate = k[Br-][BrO3-][H+]2 N.B. For the concentration of HBr, the rate of disappearing of HBr in Step 4 and 5 need not to be considered. Since Step 4 and 5 is following the rate determining step. The rate of Step 4 and 5 would only be the same as Step 3. IV. Energy profile of reaction 2. Unit 1 Page 4 Energetic stability and kinetic stability The reaction N2(g) + 3H2(g) → 2NH3(g) is exothermic. N2(g) + 3H2(g) → 2NH3(g) ∆H = -92 kJmol-1. However, N2(g) does not react with H2(g) at room temperature. This is because the reaction involves a very high activation energy, EA = 668 kJmol-1. A mixture of N2(g) and H2(g) is said to be energetically unstable but kinetically stable. The high activation energy is assoicated with the high bond energies of N≡N and H–H bond. E(N≡N) E(H–H) 945.4 kJmol-1 435.9 kJmol-1 Glossary energy profile transition state (activated complex) reaction mechanism molecularity unimolecular nucleophilic substitution (SN1) bimolecular nucleophilic substitution (SN2) carbocation / carbonium ion intermediate rate determining step energetic stability kinetic stability 91 1A 2 b iii 93 1A 1 c i ii 94 2A 2 a iii 95 2B 4 c ii 99 2A 3 a i ii iii Past Paper Question 91 1A 2 b iii 2b The energy profile of the reaction A(g) + B(g) d C(g) under two different catalysis X and Y are represented below. C iii Compare the effect of increasing temperature on the rate of reaction in the system. ↑ temp. will increase the rate of reaction of both systems, the % increase in rate will be large for system X Many candidates discussed only one system when the question asked for the effect on each system. 2 1 mark 1 mark IV. Energy profile of reaction Unit 1 93 1A 1 c i ii 1c Consider the energy diagram shown below for a certain reaction which takes place with three steps. Page 5 i ii Which is the rate-determining step? Explain. Step 3. It is the intermediate step with highest activation energy, ∴ it is the slowest step. Is the reaction endothermic or exothermic? Explain. Exothermic since the potential energy of the products is lower than that of the reactant. i.e. ∆H is negative. 2 2 marks 2 1 mark 1 mark 94 2A 2 a iii 2a Given the following thermochemical data. Reaction ∆Ho98 / kJ mol-1 2 C (graphite) + 2H2(g) → CH4(g) - 75.0 C (graphite) + O2(g) → CO2(g) - 393.5 H2(g) + ½O2(g) →H2O(l) - 285.9 iii The enthalpy change of combustion of diamond at 298K is -395.4 kJ mol-1. Which allotrope of carbon, diamond or graphite, is energetically more stable? Explain why carbon does not convert from the less stable allotrope to the more stable one at room temperature. ∆Hocomb.[C(diamond)] = - 395.4 kJmol-1 ∆Hocomb.[C(graphite)] = - 393.5 kJmol-1 ∴ Graphite is the relatively more stable allotrope. 1 mark Conversion from diamond to graphite involves the rearrangement of atoms in a giant covalent network would require a high activation energy. ∴ Conversion does not occur at room temperature. 1 mark C Many candidates erroneously said that diamond was the more stable allotrope of carbon. This indicated that their concept of 'energetic stability' was quite confused and/or that they could not relate the sign of enthalpy terms to the stability of compounds. 2 95 2B 4 c ii 4c Consider the data given below for the hydrogen halides and answer the questions that follow. Bond dissociation energy / kJmol-1 Standard enthalpy change of formation / kJmol-1 H–F -269.4 +562 H–Cl -92.8 +430 H–Br -36.8 +367 H–I +26.1 +298 ii At temperatures above 400 K, hydrogen iodide decomposes to produce violet fumes, but, hydrogen chloride and 4 hydrogen bromide do not decompose. Briefly explain this difference and write a balanced equation for the decomposition of HI. HI has the smallest bond dissociation energy, the activation energy for its decomposition is lowest. 2 marks And therefore it is the most easily decomposed HX. However, HBr and HCl do not decompose at temperature > 400 K. From the ∆Hf values, HI is the most unstable hydrogen halide with respect to decomposition to its elements. 1 mark 1 mark 2HI(g) → H2(g) + I2(g) C ii Based on the given information candidates should have been able to point out that HI decomposes readily at moderately low temperatures because it has a small bond dissociation energy corresponding to a low activation energy for decomposition. The positive and therefore the largest enthalpy change of formation amongst the hydrogen halides means that HI is the least stable hydrogen halide with respect to decomposition into its elements. 99 2A 3 a i ii iii 3a Consider the following data for the reaction A + B → products Initial concentration / mol dm-3 IV. Energy profile of reaction [A] 4.0 × 10-2 8.0 × 10-2 4.0 × 10-2 [B] 4.0 × 10-2 4.0 × 10-2 8.0 × 10-2 Unit 1 Initial rate / mol dm-3s-1 6.4 × 10-5 12.8 × 10-5 6.4 × 10-5 Page 6 For this reaction, i deduce its rate equation, ii calculate the rate constant, and iii sketch a possible energy profile. (6 marks) IV. Energy profile of reaction Unit 2 Page 1 Topic Reference Reading IV. Energy profile of reaction 5.6 Unit 2 A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 296, 315–317, 446 Reading Assignment Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 414–416 Advanced Practical Chemistry, John Murray (Publisher) Ltd., 102 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 370–373 Effect of catalyst Application of catalysts B. Effect of catalyst According to collision theory, only the collision with an energy exceeding the activation energy may lead to a reaction. There are 2 ways to make more particles having an energy higher than activation energy, thus increase the reaction rate. 1. 2. Increase the temperature of the particle ⇒ more particles possess an energy greater than Ea. Lower the activation energy ⇒ more particles at the original temperature posses an energy greater than the new Ea'. This can be done by using of catalyst. Syllabus Notes 1. Characteristics of catalyst Catalyst – a regenerated reagent which modify the rate of reaction. In general, a catalyst is chemically involved in a reaction. It is consumed in one reaction step and regenerated in a subsequent step. It can thus be used repeatedly without undergoing any permanent changes. The following examples can be used to illustrate this: Using of catalyst provides an alternative multisteps reaction pathway. A lower activation energy is involved in the rate determining step of the catalyzed reaction. Thus, catalytic reaction has a faster rate of reaction. IV. Energy profile of reaction Unit 2 Page 2 Catalyst is usually highly specific, i.e. a particular catalyst usually can catalyze only one reaction. Since catalyst is involved in the formation of reaction intermediate, it may be physically changed after the reaction. e.g. from granule form to powder form. As ∆H remains unchanged, the equilibrium constant K of the reaction is not changed. The catalyst does not change the equilibrium position of a reaction. It speeds up both forward and reverse reaction for the same extent. Note : It can be proved that ln K = constant ∆H RT Catalyst speeds up a reaction but has no effect on the yield of the reaction. 2. Theory of catalyst According to the physical state, catalysts can be classified into 2 categories. 1. Homogeneous catalyst – the catalyst has the same physical state as the reaction mixture 2. Heterogeneous catalyst – the catalyst has different physical state from the reaction mixture a) Homogeneous catalysis Example 1 Acid-catalysed esterification (will be further elaborated in organic chemistry) Esterification O carboxylic acid R' C O H Or.d.s. R' C O H R+ O H charge separation involved O- H R' C O+ H RO O R' C O ester R + H2O water ROH alcohol In esterification, the nucleophilic oxygen atom on alcohol is added to the carbonyl carbon. It can be seen that an intermediate with charge separation in the structure is involved. This makes the reaction have a high activation energy, thus, a rather slow rate. Then, the proton possessed by the alcohol is shifted to the hydroxyl group to convert it to a better leaving group, water H2O. In contrast , esterification in the presence of any aqueous acid is found to be catalyzed. Acid Catalyzed Esterification H O R C OH H O H O R' H O H R C OH +O H R' no charge separation involved f H+ +O R C OH f r.d.s. f RCOH + O R' fR O C OR' + H2O + H+ In the acid catalyzed pathway, the carbonyl oxygen is first protonated. This activates the carbonyl group by increasing the polarity of the C=O bond and make the carbonyl carbon more positive and more vulnerable to the attack of the nucleophile, alcohol. Furthermore, no reaction intermediate with charge separation is involved in the acid catalyzed mechanism and a much lower activation energy would result. IV. Energy profile of reaction Unit 2 Page 3 Example 2 Reaction between aqueous solutions of iodide and persulphate ions: 2I-(aq) + S2O82-(aq) → I2(aq) + 2SO42-(aq) The reaction rate of the above redox reaction is rather low. This is probably because the reaction involves collision among similarly charged particles. Addition of Fe3+(aq) or Fe2+(aq) into the reaction mixture speed up the reaction considerably. Oversimplified mechanism of the catalyzed reaction – Fe3+(aq) or Fe2+(aq) are served as a medium of e- transfer. 2Fe3+(aq) + 2I-(aq) → 2Fe2+(aq) + I2(aq) 2Fe2+(aq) + S2O82-(aq) → 2Fe3+(aq) + 2SO42-(aq) Fe3+(aq) is consumed Fe3+(aq) is regenerated The catalyzed reaction has a lower activation energy since the alternative pathway involves only collision between oppositely charged particle. b) Heterogeneous catalysis Example 1 Decomposition of hydrogen peroxide in the presence of manganese(IV) oxide (MnO2(s)) catalyst Without the presence of catalyst, hydrogen peroxide will only decompose slowly into water and oxygen. 2H2O2(aq) → 2H2O(l) + O2(g) The rate of reaction can be traced by monitoring the volume of oxygen evolved. In the presence of catalyst, MnO2(s), the rate of reaction will be increased tremendously. This reaction is usually used in the laboratory in the preparation of oxygen. N.B. Please note that the presence of the catalyst only affect the rate of the reaction but it has no effect on the amount of oxygen obtained eventually. However, in the presence of acid, the reaction will be slowed down. The acid is said to be a negative catalyst. Beside the volume of the oxygen, the reaction can also be traced by quenching the reaction using excess acid. The amount of H2O2(aq) can then be determined by back titrating with potassium iodide and sodium thiosulphate or directly with KMnO4(aq). H2O2(aq) + 2I-(aq) + 2H+(aq) → 2H2O(l) + I2(aq) or 5H2O2(aq) + 2MnO4-(aq) + 6H+(aq) → 2Mn2+(aq) + 8H2O(l) + 5O2(g) I2(aq) + 2S2O32-(aq) → 2I-(aq) + 2S4O62-(aq) Although the MnO2(s) granules used are chemically unchanged after the reaction, it will be physically changed into powder form. IV. Energy profile of reaction 3. Applications of catalysts Unit 2 Page 4 a) Use of catalyst in Haber process and hydrogenation Haber process Iron is used as the catalyst in the Haber process to speed up the formation of ammonia. With the Fe catalyst, the new activation energy is much lower than in the uncatalyzed reaction. This makes the direct combination of N2(g) and H2(g) viable. N2(g) + H2(g) d 2NH3(g) Hydrogenation of unsaturated oil All lipids are triesters of glycerol (propan-1,2,3-triol) and fatty acid (long chain carboxylic acid). The difference between oil and fat are only the degree of unsaturation of the fatty acids in the molecule. Oil contains more unsaturated fatty acid that fat does. Saturated carbon chain is larger than the unsaturated one with the same length. Therefore, the van der Waals' forces among the saturated carbon chain is stronger. That's why fat is a solid and oil is a liquid at room temperature. O CH2 CH CH2 O O O O O from propan-1,2,3-triol from different fatty acids Oil can be converted to fat by hydrogenation of the unsaturated carbon chain. This is done by bubbling H2(g) through oil under heat and pressure with the presence of Ni(s) catalyst. The hydrogenated oil formed is called margarine. b) Catalytic converter All the cars imported into HK after 1 January 1993 are installed with catalytic converter. In the catalytic converter, some pollutants from car exhaust (e.g. carbon monoxide, nitrogen monoxide and unburnt hydrocarbons) are converted into relatively harmless substances (e.g. carbon dioxide, nitrogen and water) with the use of metal catalysts such as platinum (or palladium) and rhodium. 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) The catalysts are coated on a honeycomb support to increase the surface area for better action. IV. Energy profile of reaction c) Enzymes Unit 2 Page 5 Enzymes are proteins that catalyse specific biochemical reactions. They are often called biological catalysts. Enzymes obtained from yeast have long been used in the production of alcohol by fermentation. The fermentation of glucose to form ethanol can be represented by the following equation : C6H12O6 → 2C2H5OH + 2CO2 enzyme Nowadays, enzymes are used in various aspects such as in the manufacture of biological washing powders. These washing powders contain enzymes which can break down stains caused by blood, egg, sweat as well as fats. They have the advantage of removing stains even at normal temperature. Glossary catalyst regenerated reagent Haber process hydrogenation 91 1A 2 b iv 95 1A 1 e i 98 2A 3 b iii homogeneous catalysis catalytic converter heterogeneous catalysis palladium rhodium adsorption enzyme Past Paper Question 91 1A 2 b iv 2b The energy profile of the reaction A(g) + B(g) d C(g) under two different catalysis X and Y are represented below. iv C Why could the use of a different catalyst change the order of the reaction? different reaction paths or different reaction mechanisms is involved with the catalyst added. In part (iv) some candidates related the order of reaction to activation energy. 2 2 marks 95 1A 1 e i 1e The reaction between ethanoic acid and ethanol can be represented by the following equation: CH3COOH(l) + C2H5OH(l) d CH3COOC2H5(l) + H2O(l) 12.01 g of ethanoic acid are treated with 4.61 g of ethanol in the presence of a catalyst. When the reaction reaches equilibrium at 298 K, 5.04 g of ethanoic acid are found to have reacted. i Name a suitable catalyst for this reaction in the forward direction. Concentrated sulphuric(VI) acid / hydrochloric acid / hydrogen chloride gas 1 mark 1 IV. Energy profile of reaction Unit 2 98 2A 3 b iii 3b The exothermic reaction E(g) → E'(g) (1) is a single stage reaction. iii In the presence of a catalyst, C, reaction (1) will proceed at a faster rate via the following mechanism : E(g) + C(g) → EC(g) EC(g) → C(g) + E'(g) (EC is the reaction intermediate.) Sketch labelled energy profiles for the conversion of E(g) to E'(g), with and without the catalyst. Explain why reaction (1) proceeds faster in the presence of the catalyst. Page 6 V. Factors influencing the rate of reaction Page 1 Topic Reference Reading V. Factors influencing the rate of reaction 5.7 Assignment A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 293–296 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 399–401 Reading Laboratory experiments for general, organic & biochemistry (2nd ed.), Saunders College Publishing, 187–196, 522 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 354–356 Factors influencing the rate of reaction V. Factors influencing the rate of reaction A. Concentration The collision frequency among the particles increases with increasing concentration. B. Temperature Increase in temperature causes increase in energy of the particles. A larger fraction of particles will have energy greater than EA. Furthermore, the collision frequency also increase. C. Pressure For gaseous reactant, pressure is equivalent to concentration for solution. A gas under pressure means that there are more gaseous particle in a given volume. D. Surface area For solid reactant, a finer particles means a higher surface area to volume ratio. A larger surface area would lead to more collisions per unit mass of the solid reactant and higher reaction rate. E. Catalyst Use of catalyst lower the activation energy EA, this makes a larger fraction of particles having an energy greater than EA. Syllabus Notes F. Light Beside heat, light is also another source providing energy for the reaction. Glossary Past Paper Question 90 2A 1 c 90 2A 1 c 1c Give, with explanations, two factors that would increase the rate of a reaction. 4 Temperature: An increase in temperature increases the number of particles having sufficient energy to produce fruitful collision, and also that the average kinetic energy will be increased. This will result in higher velocities and hence more collision per unit time. 2 marks Concentration: An increase in concentration will give more particles per unit volume and hence more collision per unit time. 2 marks Catalyst: A catalyst that increases a reaction rate does so by providing a new mechanism. If the activation energy of the limiting step of the new mechanism is lower than the activation energy of the limiting step of the uncatalyzed reaction, then the rate is increased. 2 marks Maximium of 4 marks for any 2 correct answers. C Many candidates failed to point out explicitly that only in gas-phase reactions would an increase in pressure lead to an increase in reaction rate. Chemical Equilibria Page 1 Chemical Equilibria I. Nature of equilibrium A. B. Dynamic nature Examples of equilibrium 1. Bromine water in acidic and alkaline medium 2. Potassium dichromate in acidic and alkaline medium 3. Hydrolysis of bismuth chloride Equilibrium law 1. Introduction to different equilibrium constant (K) – Kc, Kp, Kw, Ka, Kb, KIn, Kd 2. Equilibrium constant (Kc and Kp) 3. Degree of dissociation (α) 4. Concentration of solid 5. Determination of equilibrium constant (Kc) a) Esterification b) Fe3+(aq) + NCS-(aq) d FeNCS2+(aq) 6. Relationship between rate constant and equilibrium constant Effect of change in concentration, pressure and temperature on equilibria (Le Chaterlier's principle) 1. Concentration 2. Pressure 3. Temperature ∆H a) Equation : ln K = constant T 4. Effect of catalyst on equilibria 5. Examples of calculation C. D. Chemical Equilibria Page 2 II. Acid-base Equilibria A. Acid-base Theory 1. Arrhenius definition 2. Bronsted-Lowry definition 3. Lewis definition Strength of acid 1. Leveling effect Dissociation of water 1. Ionic product of water pH and its measurement 1. Definition of pH 2. Temperature dependence of pH 3. Measurement of pH a) Use of pH meter (1) Calibration of pH meter Colour of indicator B. C. D. b) E. Use of indicator (1) Strong and weak acids/bases 1. Measuring of pH or conductivity of acid / base 2. Dissociation constant (Ka and Kb) a) Dissociation of polybasic acid (1) Charge effect 3. F. G. H. 4. Buffer 1. Principle of buffer action a) pH of an acidic buffer b) pH of an alkaline buffer 2. Calculation of buffer solution Theory of Indicator Acid-base titration 1. Difference between equivalence point and end point 2. Titration using pH meter 3. Titration using indicator a) Choosing of indicator 4. Thermometric titration 5. Conductimetric titration a) Strong acid vs Strong base b) Weak acid vs Strong base c) Weak acid vs Weak base Calculation involving pH, Ka and Kb a) Relationship between Ka and Kb (pKa and pKb) b) Relationship between pH, pOH and pKw c) Some basic assumptions applied Experimental determination of Ka Chemical Equilibria Page 3 III. Redox Equilibria A. Redox reaction 1. Balancing of redox reaction a) By combining balanced half-ionic equation (1) (2) Steps of writing balanced half-ionic equation Combining half-ionic equations B. C. b) By the change in oxidation no. 2. Faraday and mole 3. Calculation of mass liberated in electrolysis Electrochemical cells 1. Measurement of e.m.f. a) Potentiometer b) High impedance voltmeter / Digital multimeter 2. Use of salt bridge a) Requirement of a salt bridge b) An electrochemical cell does not need salt bridge 3. Cell diagrams (IUPAC conventions) Electrode potentials 1. Standard hydrogen electrode 2. Relative electrode potential (Standard reduction potential) a) Meaning of sign and value 3. Electrochemical series 4. Use of electrode potential a) Comparing the oxidizing and reducing power b) Calculation of e.m.f. of a cell c) Prediction of the feasibility of redox reaction (1) Disproportionation reaction D. E. Secondary cell and fuel cell 1. Lead-acid accumulator 2. Hydrogen-oxygen fuel cell Corrosion of iron and its prevention 1. Electrochemical process of rusting 2. Prevention of rusting a) Coating b) Sacrificial protection c) Alloying I. Nature of equilibrium Part 1 Page 1 Topic Reference Reading I. Nature of Equilibrium 6.0 – 6.1.0 Practical Chemistry (3rd ed.), Heinemann, 11–12 Part 1 Assignment A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 83–86, 228–229, 231 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 325–329 Reading Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 289–296 Nature of equilibrium Dynamic nature I. Nature of equilibrium Syllabus Notes A. Dynamic nature Equilibrium means a state of balance. There are two kinds of equilibrium : Static equilibrium and Dynamic equilibrium. Static equilibrium Dynamic equilibrium The positions of the two boys remain unchanged because they are not in motion. They are in a state of static equilibrium. A boy can also maintain a fixed position on a moving escalator. This can be done by running at the same speed of the escalator but at a different direction. The position of the boy remains unchanged but both the escalator and the boy are in motion. They are in a state of dynamic equilibrium. All equilibria studied in chemistry are dynamic in nature. In A-Level, we are going to study chemical equilibrium, acid-base equilibrium, redox equilibrium and phase equilibrium. Chemical equilibrium includes all equilibria involving any chemical reaction. Acid-equilibrium and redox equilibrium are 2 examples of chemical equilibrium. Phase equilibrium is the study of equilibrium involving interconversion of phases. For examples 1. Solid phase, liquid phase and gaseous phase. 2. Aqueous phase and organic phase. 3. Moving phase and stationary phase Static equilibrium E quilibrium Dynamic equilibrium Chemical equilibrium Phase equilibrium A cid-base equilibrium Redox equilibrium I. Nature of equilibrium Part 1 Page 2 All chemical equilibria are dynamic in nature. Consider the following reversible reaction, Reactant forward reaction backward reaction Product Rate of forward reaction = k1 [Reactant] Rate of backward reaction = k-1 [Product] Concentrations of reactant and product Concentration / mol dm -3 Concentration of reactant Concentration of product 0 Time At the beginning, there is only reactant but no product. Once the reaction proceeds, the concentration of the reactant drops and the concentration of the product rises. Since the rate of reaction increases with concentrations, the rate of the forward reaction will decrease and the rate of the backward reaction will increase. Rate of reaction Rate / mol dm s -3 -1 Rate of forward reaction Rate of backward reaction 0 Time At equilibrium, forward rate = backward rate When the increasing backward rate reaches the decreasing forward rate, there will be no net conversion between product and reactant. The amounts of reactant and product will remain constant and an equilibrium is established. However, the forward reaction and backward reaction do not stop after the equilibrium has established, there is always a constant conversion between the reactant and the product. Therefore, the equilibrium is said to be dynamic in nature. I. Nature of equilibrium Consider another example Part 1 Page 3 If H2O(l) is heated in a sealed pressure cooker at a constant temperature, H2O(l) vaporizes to steam H2O(g). At the same time, steam also condenses back to water. Eventually, the amounts of water and steam will become constant. This system doesn’t involve any chemical change but only interconversion between liquid phase (water) and gaseous phase (steam). This is an example of phase equilibrium. All equilibria only exist in close systems. In the above example, no equilibrium will exit if the cover of the pressure cooker is removed. I. Nature of equilibrium B. Examples of chemical equilibrium 1. Part 1 Page 4 Bromine water in acidic and alkaline medium Br2(aq) + orange H2O(l) d colourless HBr(aq) + colourless HBrO(aq) colourless Since HBr(aq) is a strong acid while HBrO(aq) is a weak acid, the equation may be rewritten as follow : Br2(aq) + orange H2O(l) d colourless H+(aq) + colourless Br-(aq) + colourless HBrO(aq) colourless In this equilibrium, Br2(aq) is orange in colour and all the other species are colourless. Once the equilibrium is established, the forward rate of reaction will be the same as the backward rate of reaction. And the concentrations of all species will remain constant. The equilibrium can be disturbed by either addition of acid or alkaline. Addition of acid Br2(aq) + orange H2O(l) d colourless H+(aq) + colourless Br-(aq) + colourless HBrO(aq) colourless Upon addition of acid, the concentration of H+(aq) increases suddenly accompanying with a sudden increase in backward reaction rate. More product is converted back to reactant. The concentration of product drops and the concentration of the reactant rises. A new equilibrium position will be established when the concentrations of product and reactant reaches a level that the forward rate and the backward rate become the same again. At the new equilibrium position, the concentration of product will be lower that the original concentration while the concentration of reactant will become higher. The position of the equilibrium is said to be shifted to the left by the addition of acid. And the colour of the equilibrium mixture will become darker. I. Nature of equilibrium Addition of alkali Br2(aq) + orange H2O(l) d colourless Part 1 Page 5 H+(aq) + colourless Br-(aq) + colourless HBrO(aq) colourless Conversely, the equilibrium position can also be shifted to the right by addition of alkaline. Upon addition of alkaline, the hydrogen ions are neutralized. H+(aq) + OH-(aq) → H2O(l) The decrease in concentration of H+(aq) slows down the backward reaction. Eventually, the equilibrium will adjust to a new position. The colour intensity of the equilibrium mixture will become lighter as the concentration of Br2(aq) decreases. Questions : What if NaBr(s) is added ? 2. Potassium dichromate in acidic and alkaline medium 2CrO42-(aq) yellow O O - + 2H+(aq) colourless O - d Cr2O72-(aq) orange O + H2O(l) colourless O O - - Cr O O + O Cr O O + H+ + H+ O Cr O Cr O O + H2O chromate(VI) ion dichromate(VI) ion Upon addition of acid, the equilibrium position can be shifted to the right and becomes orange. While upon addition of alkali, the equilibrium position can be shifted to the left and becomes yellow. I. Nature of equilibrium 3. Hydrolysis of bismuth chloride BiCl3(aq) colourless + H2O(l) colourless d Part 1 Page 6 BiOCl(s) white ppt. + 2HCl(aq) colourless When solid bismuth chloride (BiCl3) is mixed with water, a white suspension of bismuth oxychloride (BiOCl) is formed. Upon addition of drops of HCl(aq), the white disappears and the white ppt. of bismuth oxychloride reappear upon addition of drops of water. Although not all reactions are reversible, it would be beneficial to consider them all reversible. And consider an irreversible reaction only an equilibrium lying completely onto one side. Glossary equilibrium / equilibria static equilibrium dynamic equilibrium chemical equilibrium acid-base equilibrium redox equilibrium phase equilibrium reversible reaction forward rate backward rate close system shifting in equilibrium position chromate(VI) ion dichromate(VI) ion bismuth chloride bismuth oxychloride 92 2B 4 b i ii 98 2B 8 b i Past Paper Question 92 2B 4 b i ii 4b i Draw the structures of the chromate(VI) and dichromate(VI) ions. 2 ii C i ii 1 mark each Write an equation showing the action of aqueous acid on the chromate(VI) ion. 2CrO42- + 2H+ d Cr2O72- + H2O 1 mark Candidates should take care in providing a 3-D diagram, which shows the correct bond orders and charge(s). Some candidates stated that the shape of the chromate(VI) ion is square-planar. the chromate(VI) - dichromate(VI) equilibrium is not a redox reaction, as stated by some candidates. 1 98 2B 8 b i 8b For each of the following, state the expected observation and write the relevant balanced equation(s). i Dilute sulphuric(VI) acid is added dropwise to a solution of potassium chromate(VI). 6 I. Nature of equilibrium Part 2 Page 1 Topic Reference Reading I. Nature of Equilibrium 6.1.1 A-Level Chemistry Syllabus for Secondary School (1995), Curriculum Development Council, 201–202 Chemistry – Experimental foundation, Prentice Hall Inc., 51–55 Part 2 Assignment Advanced Practical Chemistry, John Murray (Publisher) Ltd., 53–56 Calculations for A-Level Chemistry, Stanley Thornes (Publishers) Ltd, 230–238 Reading rd A-Level Chemistry (3 ed.), Stanley Thornes (Publisher) Ltd., 86–87, 229–231, 232–233, 236–238 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 333–340 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 296–303 Equilibrium law Equilibrium constant (Kc and Kp) C. Equilibrium law 1. Introduction to different equilibrium constant (K) – Kc, Kp, Kw, Ka, Kb, KIn, Kd Considering the following reversible reaction aA+bBdcC+dD [C]c[D]d At equilibrium, it is found that the expression [A]a [B]b is a constant at a given temperature. The expression is called equilibrium constant, denoted by the symbol K or Keq. The unit of K is depending on the values of stoichiometric coefficients. [C]c[D]d K or Keq = [A]a [B]b This equation describes the relationship between the concentrations of reactant and product. The equation is also known as Law of equilibrium or Equilibrium law. a) Relationship between rate constant and equilibrium constant At equilibrium, the rate of forward reaction equals the rate of backward equation. Moreover, because the whole reaction is in equilibrium, all intermediate steps of the reaction are in equilibrium. Therefore, the rate of reaction is no longer depending on the rate determining step only. Instead, the rate of reaction is depending on the concentration and stoichiometric coefficients. Forward rate at equilibrium = k1[A]a[B]b Backward rate at equilibrium = k-1[C]c[D]d Forward rate at equilibrium k1[A]a[B]b k1 [C]c[D]d k-1 = [A]a[B]b = K k1 ∴ K=k N.B. = = where k1 is the rate constant of forward reaction k-1 is the rate constant of backward reaction Syllabus Notes Backward rate at equilibrium k-1[C]c[D]d -1 This is applicable only if the reaction is at equilibrium. If the reaction is not at equilibrium, the rate equation must be determined experimentally. I. Nature of equilibrium Part 2 Page 2 There are many different kinds of expression of equilibrium constant depending on the type of equilibrium concerned. e.g. Kc, Kp, Kw, Ka, Kb, KIn, Kd. Type Chemical Chemical Meaning Equilibrium constant in term of concentration, Kc Equilibrium constant in term of partial pressure, Kp Ionic product of water, Kw Acidity constant, Ka Bascity constant, Kb Indicator constant, KIn Partition coefficient / distribution coefficient, Kd Example 2CrO42-(aq) + 2H+(aq) d Cr2O72-(aq) + H2O(l) N2(g) + 3H2(g) d 2NH3(g) H2O(l) d H+(aq) + OH-(aq) CH3COOH(aq) + H2O(l) d H3O+(aq) + CH3COO-(aq) NH3(aq) + H2O(l) d NH4+(aq) + OH-(aq) HIn(aq) + H2O(l) d H3O+(aq) + In-(aq) I2(aq) d I2(hexane) Equilibrium Law [Cr2O72-(aq)] Kc = [CrO 2- ]2[H+ ]2 (aq) 4 (aq) pNH32 Kp = p · p 3 N2 H2 Kw = [H+(aq)][OH-(aq)] [H3O+(aq)][CH3COO-(aq)] [CH3COOH(aq)] [NH4+(aq)][OH-(aq)] Kb = [NH3(aq)] [H3O+(aq)][In-(aq)] KIn = [HIn(aq)] [I2(hexane)] Kd = [I ] 2(aq) Ka = Chemical Acid-base Acid-base Acid-base Phase 2. Equilibrium constant (Kc and Kp) The equilibrium constant K or Keq describes the relationship between the concentrations of reactants and products. However, there are many different ways to express concentration. In general, for a solution, concentration is expressed in term of molarity. for a gas, concentration is expressed in term of partial pressure (which is related to mole fraction). a) Concentration of pure solid and liquid Molarity is only one way of expressing concentration. Indeed, concentration is a very general concept concerning the relationship between the amount of a substance and the volume occupied by that substance. For a pure solid / liquid, the amount is usually measured in term of mass. Therefore, the concentration of a pure solid / liquid is the equivalent to its density which is a constant at a given temperature. Concentration of pure solid / liquid = amount of solid / liquid mass of solid / liquid = volume occupied = density volume occupied Example : Calculation of the concentration of water in term of molarity at 25ºC. Given : density of water at 25ºC is 1.00 gcm-3 = 1000.00 gdm-3. Concentration of pure water in molarity = = = no. of mole of water volume occupied volume × density / molar mass volume occupied 1000.00 gdm-3 18.0 gmol-1 = 55.6 moldm-3 = = mass / molar mass volume occupied density molar mass I. Nature of equilibrium Part 2 Page 3 b) Equilibrium constant in term of concentration, Kc Consider the following equilibrium 2CrO42-(aq) + 2H+(aq) d Cr2O72-(aq) + H2O(l) [Cr2O72-(aq)][H2O(l)] K = [CrO 2- ]2[H+ ]2 (aq) 4 (aq) Besides being a product, water is also the solvent in this equilibrium. Because it is in large excess comparing with the solutes, it can be considered as a pure liquid in here. Since the concentration of water is constant, it may be incorporated into the equilibrium constant and the expression will become [Cr2O72-(aq)] K = [CrO 2- ]2[H+ ]2 = Kc [H2O(l)] (aq) 4 (aq) The expression excludes the concentration of pure solid and liquid is known as Kc, equilibrium constant in term of concentration. In the writing of any equilibrium constant, the concentration of a pure solid or liquid in the equilibrium law should be excluded. The exclusion of concentration of pure solid / liquid in an equilibrium law is not only doctrinal. It has a significant physical meaning. Consider a saturated sugar solution, the concentration of the solution is not depending on the amount of sugar remains undissolved. sugar(s) + aq d sugar(aq) A saturated sugar solution A saturated solution with more undissolved sugar crystal Consider another equilibrium, Fe2+(aq) + Ag+(aq) d Fe3+(aq) + Ag(s) Equilibrium mixture with less Ag(s) suspension Equilibrium mixture with more Ag(s) suspension It is found that the addition of extra Ag(s) into the mixture has no effect on the concentrations of Fe2+(aq), Ag+(aq) and Fe3+(aq). Therefore, the equilibrium position is not depending on the amount of solid silver whose concentration is constant. This fact should be reflected in the equilibrium law. [Fe3+(aq)] Kc = [Fe2+ ][Ag+ (aq) (aq)] I. Nature of equilibrium c) Part 2 Page 4 Equilibrium constant in term of partial pressure, Kp Consider the thermal decomposition of CaCO3(s), CaCO3(s) d CaO(s) + CO2(g) K= [CaO(s)][CO2(g)] [CaCO3(s)] [CaCO3(s)]K [CaO(s)] = [CO2(g)] = Kc Or if the concentration of CO2(g) is expressed in term of partial pressure, the equilibrium constant is known as Kp. Kp = pCO2(g) Consider the Haber process, N2(g) + 3H2(g) d 2NH3(g) [NH3(g)]2 Kc = [N ][H ]3 2(g) 2(g) or pNH3(g)2 Kp = p 3 N2(g) pH2(g) I. Nature of equilibrium 3. Degree of dissociation (α) Part 2 Page 5 Sometimes, the equilibrium positions of dissociation reactions can be described by degree of dissociation. Degree of dissociation is the percentage of molecules that undergoes dissociation and represented by α. Consider the dissociation of carboxylic acid dimer in gaseous state, H HC H C OH O O HO H CCH H H 2H C H C OH O (CH3COOH)2(g) d 2CH3COOH(g) Consider 1 mole of dimer, the mole no. would be numerically the same as no. in percentage (i.e. mole fraction). Initial no. of mole no. of mole at equilibrium Partial pressure at equilibrium 1 1-α p(CH3COOH)2(g) 0 2α pCH3COOH(g) The equilibrium law in term of partial pressure can be expressed as pCH3COOH(g)2 Kp = p Let P be the total pressure, P = p(CH3COOH)2(g) + pCH3COOH(g) ni pi Since mole fraction is related to partial pressure by the equation, mole fraction = n = P ni pi = n P = mole fraction × P 1-α 1-α P= P (1 - α) + 2α 1 +α 2α 2α P= P pCH3COOH(g) = mole fraction of CH3COOH(g) × P = (1 - α) + 2α 1+α p(CH3COOH)2(g) = mole fraction of (CH3COOH)2(g) × P = 2α 2 2 pCH3COOH(g)2 (1 + α ) P 4α2 Kp = p = = P 1 - α2 1-α (CH3COOH)2(g) P 1+α In this equilibrium, unlike equilibrium constant Kp, dissociation constant (α) is also depending on the total pressure of the gaseous mixture on top of temperature. For example, T/K Kp / Nm-2 298 72.9 303 121.1 308 182.7 313 302 (CH3COOH)2(g) At 303 K and pressure of 1400 Nm-2 4α2 × 1400 Nm-2 121.1 Nm-2 = 1 - α2 At 303 K and pressure of 200 Nm-2 4α2 × 200 Nm-2 121.1 Nm-2 = 1 - α2 α = 0.145 α = 0.363 Thus, more carboxylic acid dimers will dissociate at a lower pressure. I. Nature of equilibrium 4. Part 2 Page 6 Determination of equilibrium constant (Kc) The equilibrium constant Kc can be determined by measuring the concentrations of different species in the equilibrium mixture. Indeed, sometimes it is not necessary to measure the concentrations of all the species. By knowing the initial concentration of the reactant and the equilibrium concentration of the reactant or product, the concentration of the other species can be determined according to the law of conservation of mass. Consider an equilibrium mixture with initial concentrations of 1 mol dm-3 Fe3+(aq) and 2 mol dm-3 of NCS-(aq) respectively. + Fe3+(aq) iron(III) ion pale yellow e.g. Initially At equilibrium NCS-(aq) thiocyanate ion colour d FeNCS2+(aq) thiocyanato iron(III) ion dark red complex + NCS-(aq) 2 mol dm-3 d FeNCS2+(aq) 0 mol dm-3 FeNCS2+(aq) x mol dm-3 Fe3+(aq) 1 mol dm-3 + Fe3+(aq) (1 - x) mol dm-3 NCS-(aq) d (2 - x) mol dm-3 By knowing the equilibrium concentration of either Fe3+(aq), NCS-(aq) or FeNCS2+(aq), the equilibrium constant can be determined. [FeNCS2+(aq)] x Kc = [Fe3+ ][NCS- ] = (1 - x) (2 - x) dm3 mol-1 (aq) (aq) I. Nature of equilibrium a) Fe3+(aq) + NCS-(aq) d FeNCS2+(aq) Part 2 Page 7 First of all, equilibrium constant is a value affected by temperature. It is better to keep the temperature constant by immersing the solution to be studied in a water bath. For the equilibrium, Fe3+(aq) + NCS-(aq) d FeNCS2+(aq), the initial concentrations of Fe3+(aq) and NCS-(aq) can be determined by prepared corresponding standard solution of Fe(NO3)3(aq) and KNCS(aq). The problem is how to determine the equilibrium concentration of either Fe3+(aq), NCS-(aq) or FeNCS2+(aq). If titration is used to determine the concentration, the reaction must be quenched first. Otherwise, when the chemical is being titrated, the equilibrium position will also be shifted. Since FeNCS2+(aq) is deep red in colour, its concentration can be determined by colorimetric measurement without quenching. Firstly, a calibration curve which relates the concentration of FeNCS2+(aq) and the colour intensity must be constructed. This can either be done by using a calorimeter or visual comparison. In the construction of calibration curve, samples of standard FeNCS2+(aq) with various concentrations are prepared. These are prepared by adding excess NCS-(aq) into standard Fe3+(aq). It is assumed that all Fe3+(aq) ions are converted to FeNCS2+(aq). Therefore, the concentration of FeNCS2+(aq) can be determined from concentration of Fe3+(aq). Equilibrium mixture with known initial Fe3+(aq) and NCS-(aq) concentrations are prepared. The equilibrium concentration of FeNCS2+(aq) in the mixture is determined by a colorimeter or by visual comparison of the [FeNCS2+(aq)] intensity of the samples prepared before. Hence, the equilibrium constant Kc = [Fe3+ ][NCS- ] can be (aq) (aq) determined. Several trials with different concentrations should be done to obtain an average value. I. Nature of equilibrium b) Esterification CH3COOH(aq) + CH3CH2OH(aq) Part 2 Page 8 d CH3COOCH2CH3(aq) + H2O(l) Both esterification and hydrolysis of ester are very slow, even in the presence of acid catalyst, it takes 48 hours for the system to reach a state of equilibrium. Hence, the equilibrium concentration of CH3COOH(aq) can be determined by titration even without quenching. In this example, water is not the solvent and is not in large excess. It cannot be considered as a pure substance and its concentration must be reflected in the equilibrium law. [CH3COOCH2CH3(aq)][H2O(l)] Kc = [CH COOH ][CH CH OH ] 3 (aq) 3 2 (aq) Since aqueous acid is required to catalyze the reaction, it would be more convenient to approach the problem by treating it as a hydrolysis problem. This reduces the number of measurement of from 3 liquids (ethanoic acid, alkanol and aqueous acid) to 2 liquids at the preparatory stage. CH3COOCH2CH3(aq) + H2O(l) d CH3COOH(aq) + CH3CH2OH(aq) [CH3COOH(aq)][CH3CH2OH(aq)] 1 Kc = Kc' = [CH3COOCH2CH3(aq)][H2O(l)] Procedure An accurately known volume of standard HCl(aq) is pipetted into a test tube containing known mass of CH3COOCH2CH3(l). The solution is shaken occasionally for 48 hours to ensure equilibrium is reached. The initial amount of water is determined by subtracting the mass of HCl from the mass of HCl(aq). The initial amount of CH3COOCH2CH3(l) is determined by weighing. The total no. of mole of HCl(aq) and CH3COOH(aq) at equilibrium are determined by titrating with standard NaOH(aq). Experimental result Mass of the empty test tube = 12.01 g Mass of the empty test tube with CH3COOCH2CH3(l) = 16.52 g Mass of the empty test tube with HCl(aq) and CH3COOCH2CH3(l) = 5.20 g Volume of HCl(aq) added = 5.00 cm3 Concentration of HCl(aq) = 1.953 moldm-3 Calculation I. Nature of equilibrium Part 2 Page 9 Glossary equilibrium law / law of equilibrium equilibrium constant (K/Keq) equilibrium constant in term of partial pressure (Kp) equilibrium constant in term of concentration (Kc) degree of dissociation (α) dimer thiocyanate ion thiocyanato iron(III) ion calibration curve 90 2A 1 a i ii 94 2A 1 b 95 1A 1 e ii 96 1A 1 f i ii 97 1A 2 a i 98 1A 2 c 99 2A 4 a i ii Past Paper Question 90 2A 1 a i ii 1a Consider the following equilibrium at constant pressure: A2(g) + 3B2(g) d 2AB3(g) A mixture of 4.0 mol of A2(g) and 12.0 mol of B2(g) was placed in a vessel of volume 20.0 dm3, and heated to 565K. When the system had reached equilibrium, it was found that 4.0 mol AB3(g) was present. i Calculate the concentration of each species at equilibrium. A2(g) + 3B2(g) d 2AB3(g) Since 4 moles of AB3 will consume 2 moles of A2 and 6 moles of B2 ∴ at equilibrium: 12 - 6 6 [B2] = 20 = 20 moldm-3 ½ mark 4-2 2 [A2] = 20 = 20 moldm-3 ½ mark 4 [AB3] = 20 moldm-3 ½ mark ii Calculate the equilibrium constant of this reaction at 565K. [AB3]2 4 20 20 K = [A ][B ]3 = (20)2 ⋅ ( 2 ) ⋅ ( 6 )3 = 14.82 dm6mol-2 (wrong unit -½) 2½ mark 2 2 Ci Some candidates mistakenly used the initially given concentration rather than the equilibrium concentration in their calculation of equilibrium constant. ii Some weaker candidates failed to give the correct unit for the equilibrium constant. 1½ 2½ I. Nature of equilibrium Part 2 94 2A 1 b 1b A colorimetric method can be used to provide data for the determination of the equilibrium constant of the following reaction. Fe3+(aq) + NCS-(aq) d FeNCS2+(aq) Outline such a method for the determination of the equilibrium constant Kc of the above equilibrium. FeNCS2+(aq) ions absorb visible light in the electromagnetic spectrum / hence [FeNCS2+(aq)] can be determined by measuring the absorbance of the solution. 1 mark Calibrate a colorimeter by measuring the absorbance of several solutions which contained known concentrations of FeNCS2+(aq). (In these solutions, the NCS- ions are in excess, so that all Fe3+ ions present can be considered as 1 mark existing in the complex form, FeNCS2+) By mixing comparable volumes of standard Fe3+(aq) and NCS-(aq) solutions and measuring the [FeNCS2+] formed with the colorimeter, [Fe3+(aq)], [NCS-(aq)] and [FeNCS2+(aq)] at equilibrim can be found. 1 mark [FeNCS2+(aq)] Hence, the equilibrim constant, Kc = [Fe3+ ][NCS- ] , can be determined. 1 mark (aq) (aq) C Poorly-answered. This indicated that most candidates were not familiar with the use of a colorimetric method in the determination of equilibrium constants. A large number of candidates did not mention the calibration of the colorimeter. 95 1A 1 e ii 1e The reaction between ethanoic acid and ethanol can be represented by the following equation: CH3COOH(l) + C2H5OH(l) d CH3COOC2H5(l) + H2O(l) 12.01 g of ethanoic acid are treated with 4.61 g of ethanol in the presence of a catalyst. When the reaction reaches equilibrium at 298 K, 5.04 g of ethanoic acid are found to have reacted. ii Calculate the equilibrium constant, Kc , for the reaction at 298 K. CH3COOH(l) + C2H5OH(l) d CH3COOC2H5(l) + H2O(l) initial no. of moles 12.01 60.052 = 0.2 12.01 - 5.04 60.052 = 0.116 0.116 V 4.61 46.068 = 0.1 0.1 - 0.084 = 0.016 0.016 V 0.084 V 0.084 V – – Page 10 4 3 no. of mole at equilibrium 0.084 0.084 2 marks concentration at equilibrium C 0.084 0.084 V×V 1 marks Kc = 0.116 0.016 = 3.80 (no unit) ×V V A common mistake in the answers was the omission of the concentration of water from the equilibrium expression. I. Nature of equilibrium Part 2 96 1A 1 f i ii 1f SO2(g) and O2(g) were mixed in the mole ratio of 3 : 1 at 1000 K in the presence of a catalyst. When equilibrium was attained at 373 kPa pressure, one half of the SO2(g) had been converted into SO3(g). i Write an expression for Kp for the reaction of SO2(g) and O2(g) to give SO3(g). (PSO3)2 (PSO3) Kp = (P )2(P ) or (P )(P )½ 1 mark SO2 O2 SO2 O2 ii Calculate Kp for the above reaction under the given conditions. 2SO2 + O2 d 2SO3 initial 3 1 0 at equilibrium 1.5 0.25 1.5 total = 3.25 1.5 PSO3 = PSO2 (= 3.25 × 373 kPa) 0.25 PO2 = 3.25 × 373 kPa (PSO3)2 1 3.25 Kp = (P )2(P ) = P = 0.25 × 373 kPa-1 = 0.035 kPa-1 SO2 O2 O2 or, Kp = 1 = P O2 3.25 -½ -½ 0.25 × 373 (kPa) = 0.19 (kPa) Page 11 1 2½ ½ mark ½ mark ½ mark 1 mark (1 mark) (½ mark for numerical answer, ½ mark for unit) (The answer and the expression (part (i)) for Kp should be consistent with each other.) 97 1A 2 a i 2a Consider the following dissociation reaction: PCl5(g) d PCl3(g) + Cl2(g) At 400 K and 101 kPa pressure, the percentage dissociation of PCl5(g) is 86%. i Calculate Kp for the reaction at 400 K. 4 98 1A 2 c 3 2c At 4200 K, the equilibrium constant for the following reaction is 1.2 × 10-2. N2(g) + O2(g) d 2NO(g) 1.0 mol of O2(g) and 2.0 mol of N2(g) are allowed to react in a 2.0 dm3 closed container. Calculate the concentration of N2(g), in moldm-3. in the equilibrium mixture at 4200 K. 99 2A 4 a i ii 4a In the Haber process, ammonia is synthesized by the exothermic reaction of nitrogen and hydrogen at around 723K. N2(g) + 3H2(g) d 2NH3(g) In a simulation of the process, a mixture of nitrogen and hydrogen was placed in a closed container. The initial concentrations of nitrogen and hydrogen were 0.50 mol dm-3 and 1.50 mol dm-3 respectively. When the equilibrium was attained at 723 K, 25.0% of the original nitrogen was consumed. i Calculate the respective concentrations of nitrogen, hydrogen and ammonia in the equilibrium mixture. ii Calculate Kc for the reaction at 723 K. I. Nature of equilibrium Part 3 Page 1 Topic Reference Reading I. Nature of Equilibrium 6.1.2.0–6.1.2.4 AS-Level Chemistry Syllabus for Secondary School (1991), Curriculum Development Council, 160 Part 3 Assignment A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 87, 231–232 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 343–354 Reading Syllabus Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 305–309 Effect of change in concentration, pressure and temperature on equilibria (Le Chaterlier's principle) Effect of catalyst on equilibria Le Chaterlier's principle D. Effect of change in concentration, pressure and temperature on equilibria (Le Chaterlier's principle) 1. Concentration Br2(aq) + orange H2O(l) d colourless HBr(aq) + colourless HBrO(aq) colourless Notes Upon addition of extra Br2(aq) into the equilibrium system, some Br2(aq) added will be converted to HBr(aq) and HBrO(aq). Therefore, the net change in concentration of Br2(aq) is always less than the actual amount added. Because more products are formed upon addition of the reactant, the position of the equilibrium is said to be shifted to the right upon the addition of Br2(aq). Kc = [HBr(aq)][HBrO(aq)] [Br2(aq)] [HBr(aq)][HBrO(aq)] smaller than the original Kc. [Br2(aq)] [HBr(aq)][HBrO(aq)] Therefore, the equilibrium will adjust in a way to restore the expression back to the original [Br2(aq)] K c. An increase in [Br2(aq)] will make the value of the expression + Fe3+(aq) iron(III) ion pale yellow NCS-(aq) thiocyanate ion colour d FeNCS2+(aq) thiocyanato iron(III) ion dark red complex [FeNCS2+(aq)] Kc = [Fe3+ ][NCS2+ ] (aq) (aq) Similarly, upon addition of Fe3+(aq) or NCS-(aq), the solution will turn darker. It shift the equilibrium position to the right. However, upon addition of Na3PO4(aq) or NaHPO4(aq), the colour will get paler. PO43-(aq) ions form complex with free Fe3+(aq) and lower its concentration. This shifts the equilibrium position to the left. I. Nature of equilibrium 2. Pressure Part 3 Page 2 Since only gas is compressible, pressure has only effect on the equilibrium involving gaseous species. 2NO2(g) brown d N2O4(g) colourless Once the plunger of the gas syringe is pressed, the colour of the gas gets darker because of an immediate increase of the concentration of brown NO2(g). However, the gas turns paler gradually. At equilibrium, forward rate = backward rate and forward rate = k1[NO2(g)]2 backward rate = k-1[N2O4(g)] The forward reaction is a second order reaction but the backward reaction is only a first order reaction. The increase in pressure increases the forward rate more than the backward rate. Hence a new equilibrium position is established eventually. I. Nature of equilibrium 3. Temperature 2NO2(g) brown N2O4(g) colourless Part 3 Page 3 d ∆H = -58.0 kJmol-1 When the tube containing the mixture of NO2(g) and N2O4(g) is immersed in hot water, the colour of the gas gets darker. Conversely, when it is put inside icy water, the colour of the gas gets paler. This is an exothermic reaction. From the energy profile of the reaction, it can be seen that the forward reaction and backward reaction involve different activation energy. The forward reaction has a lower activation energy Ea1. i.e. Ea1 < Ea-1 Once the temperature is increased, the rate of both forward reaction and backward reaction are increased. Moreover, from the diagram of distribution of molecular speed, the reaction with higher activation energy will have a higher percentage increase in rate. ∴ An increase in temperature will increase the values of k1 and k-1 for different extent, thus the equilibrium k1 constant and equilibrium position. The expression K = k will become smaller. Therefore, a higher -1 temperature will shift the equilibrium position of an exothermic reaction to the left. I. Nature of equilibrium a) Equation : ln K = constant ∆H RT Part 3 Page 4 Or the effect of the temperature can be interpreted from another point of view. It can be proved that the equilibrium constant is related to the ∆H according to the equation ∆G = ∆H - T∆S = - RTln K where ∆G is Free energy change, which indicates the total free energy change of the system. T is the temperature in Kelvin. ∆H is enthalpy change of the system. It is a constant for a specific reaction. ∆S is entropy change of the system, which is a measure of the change in degree of disorderness. It is a constant for a specific reaction. N.B. Free energy change and entropy change are not required in A-Level. ∆H + constant RT ∆G = ∆H - T∆S = - RTln K - RTln K = ∆H - T∆S ∆H T∆S ln K = - RT + RT ∆H ∆S ln K = - RT + R ∆H ln K = - RT + constant (1) Derivation of ln K = - By rearranging the equation N.B. The actual derivation is not required in A-Level but you are expected to use the equation to explain the effect of temperature on an equilibrium. ∆H For an exothermic reaction, ∆H is negative, therefore, the expression - RT is positive. An increase in ∆H temperature will make the expression - RT less positive, hence K will become smaller. This shifts the equilibrium position to the reactant side. Conversely, the equilibrium position of an exothermic reaction will be shifted to the product side if the temperature is lowered. I. Nature of equilibrium 4. Effect of catalyst on equilibria Part 3 Page 5 Addition of catalyst has no effect on concentrations nor equilibrium constant. Therefore, it has no effect on the position of an equilibrium. It only increases the forward rate and backward rate for the same extent and shortens the time required to attain equilibrium. 5. Le Chaterlier's principle The effect of a change in condition on the position of an equilibrium can be generalized as a principle, called Le Chaterlier's principle. Le Chaterlier's principle states that when a change is imposed on an equilibrium system, the system will response in a way to minimize the effect of the change. Effect of concentration + NCS-(aq) Fe3+(aq) pale yellow colourless FeNCS2+(aq) dark red d Upon addition of FeCl3(s), some Fe3+(aq) added with combine with NCS-(aq) to form FeNCS2+(aq). As a result the net increase of concentration of Fe3+(aq) will be reduced and the equilibrium position is shifted to the right. If PO43-(aq) is added to reduce the concentration of Fe3+(aq), some FeNCS2+(aq) will dissociate to give more Fe3+(aq). Consequently, the net decrease of concentration of Fe3+(aq) will be reduced and the equilibrium position is shifted to the left. Effect of pressure 2NO2(g) d N2O4(g) brown colourless Any increase in pressure can be reduced by converting 2 NO2(g) molecules to 1 N2O4(g) molecule Therefore, increase in pressure will shift the equilibrium to the right. If the pressure is decreased, the decrease in pressure can be reduced by converting 1 N2O4(g) molecule to 2 NO2(g) molecules. Effect of temperature 2NO2(g) d N2O4(g) brown colourless ∆H = -58.0 kJmol-1 The effect of decrease in temperature can be reduced by converting some NO2(g) to N2O4(g) because the reaction is exothermic. Therefore, a decrease in temperature will shift the equilibrium position of an exothermic reaction to the right. While a increase in temperature will shift the equilibrium position to the left. Le Chaterlier's principle is only a rule which helps to predict the outcome of the change. In order to explain the change, the changes in forward rate and backward rate must be considered. Glossary equilibrium position shifting of equilibrium position disorderness Le Chaterlier's principle free energy change entropy change I. Nature of equilibrium Part 3 Page 6 Past Paper Question 91 1A 2 b i ii 92 2C 7 c 94 2A 2 b ii 95 1A 1 e iv 96 1A 1 f iii 97 1A 2 a ii 95 2B 4 a iii 97 2B 8 a iii 97 2B 8 b i 91 1A 2 b i ii 2b The energy profile of the reaction A(g) + B(g) d C(g) under two different catalysis X and Y are represented below. i ii C What is the effect of increasing temperature on the equilibrium of each system? ↑ temp. will shift the equilibrium to the left for system X and system Y. 1½ mark What is the effect of decreasing pressure on the equilibrium of each system? ↓ pressure will shift the equilibrium to the left for system X and system Y. 1½ mark Many candidates discussed only one system when the question asked for the effect on each system. Many candidates did not distinguish kinetics from chemical equilibria. 1½ 1½ 92 2C 7 c 7 A carboxylic acid P, with a relative molecular mass less than 100, contains C, 55.8%; H, 7.0%; and O, 37.2% by mass . An attempt to convert P to its methyl ester Q by prolonged refluxing of P with methanol in the presence of aqueous H2SO4 gave the desired ester Q but with much of the starting material P unchanged. (Relative atomic masses : H 1.0; C 12.0; O 16.0) 7c Suggest, with explanations, two ways which would make the esterification go towards completion. 4 The reaction is reversible 1 mark 1 mark RCOOH + CH3OH d RCOOCH3 + H2O 1 mark Use large excess of CH3OH; mass action to shift equilibrium 1 mark Remove H2O as it is formed to shift equilibrium to the right ½ mark Use conc. H2SO4 or other drying agent C Some candidates misunderstood the question and gave alternative synthetic routes from acid to ester. A number of candidates failed to point out that esterification is reversible and the yield of product depends on the equilibrium position of the reaction. Some candidates confused equilibrium and rate of reaction thinking that by changing reaction conditions to increase the rate of reaction would make the reversible reaction go to completion. 94 2A 2 b ii 2b Account for each of the following: ii The melting point of ice decreases with an increase in pressure. In ice, the intermolecular attraction is H-bond, each H2O molecule is tetrahedrally surrounded by 4 other H2O 1 mark molecules. C.N. = 4, i.e. an open structure. In liquid water, the molecules are packed closer together ∴ water has a higher density than water. 1 mark For the reaction, H2O(s) d H2O(l), increase in pressure, shifts the equilibrium to the right ∴ m.p. of ice decreases with increase in pressure. 1 mark 3 I. Nature of equilibrium Part 3 95 1A 1 e i iv 1e The reaction between ethanoic acid and ethanol can be represented by the following equation: CH3COOH(l) + C2H5OH(l) d CH3COOC2H5(l) + H2O(l) 12.01 g of ethanoic acid are treated with 4.61 g of ethanol in the presence of a catalyst. When the reaction reaches equilibrium at 298 K, 5.04 g of ethanoic acid are found to have reacted. i Name a suitable catalyst for this reaction in the forward direction. Concentrated sulphuric(VI) acid / hydrochloric acid / hydrogen chloride gas 1 mark iv Would the addition of more of the same catalyst affect the value of Kc? Explain. No, catalyst can affect only the rate of (forward and backward reaction. 1 mark OR Catalyst does not affect position of an equilibrium, yield does not change. 1 mark C A common mistake in the answers was the omission of the concentration of water from the equilibrium expression. 95 2B 4 a iii 4a Explain the following facts: iii Iodine is more soluble in aqueous potassium iodide solution than in water. I2 forms soluble complex I3- with KI in solution. Therefore, I2 appears more soluble. I2(s) + KI(aq) d KI3(aq) or I2(s) + I-(aq) d I3-(aq) Page 7 1 1 2 1 mark 1 mark 96 1A 1 f iii 1f SO2(g) and O2(g) were mixed in the mole ratio of 3 : 1 at 1000 K in the presence of a catalyst. When equilibrium was attained at 373 kPa pressure, one half of the SO2(g) had been converted into SO3(g). iii If the above reaction takes place in the absence of the catalyst, but other conditions remain unchanged, will the value of Kp increase, decrease or remain the same ? remains the same ½ mark 97 1A 2 a ii 2a Consider the following dissociation reaction: PCl5(g) d PCl3(g) + Cl2(g) At 400 K and 101 kPa pressure, the percentage dissociation of PCl5(g) is 86%. ii State the effect of an increase in pressure (I) on Kp, and (II) on the percentage dissociation of PCl5(g). 97 2B 8 a iii 8a Suggest how the following nitrogen oxides can be prepared in the laboratory. In each case, state the reactant(s) used and the reaction conditions, and write balanced equation(s) for the reaction(s) involved. iii dinitrogen tetraoxide, N2O4 97 2B 8 b i 8b The synthesis of ammonia using the Haber Process involves the following: N2(g) + 3H2(g) d 2NH3(g) ∆Ho = -92 kJ mol-1 i State the effect of a change in temperature on the reaction at equilibrium. ½ 4 6 7 I. Nature of equilibrium Part 4 Page 1 Topic Reference Reading I. Nature of equilibrium 6.1.2.5 Part 4 Assignment A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 232–238 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 335–338 Reading Syllabus Notes 5. Examples of calculation Since the equilibrium law describes the relationship of the concentrations of different species in an n equilibrium system, all calculations must be based on the concentration ( ) instead of the absolute V amount (n). N.B. 1. 2. 3. Write an equilibrium law(s) to describe the system. Figure out the relationship among the equilibrium concentrations of different species, from initial concentrations, stoichiometric coefficients and law of conservation of mass. Solve the equation(s). Example 1 H2(g) + I2(g) d 2HI(g) When 46.0 g of I2(g) and 1.00 g of H2(g) are heated to equilibrium at 470ºC, the equilibrium mixture contains 1.9 g I 2. (a) How may moles of each gas are present in the equilibrium mixture? (b) Compute the equilibrium constant. (Given : Relative atomic masses: I, 126.9; H, 1.00) [HI(g)]2 Kc = [H ][I ] = 2(g) 2(g) = 2. no. of mole of HI(g) 2 ) V no. of mole of H2(g) no. of mole of I2(g) × V V no. of mole of HI(g)2 no. of mole of H2(g) × no. of mole of I2(g) ( (a) 1. 46.0 g initial no. of mole of I2(g) = 126.9 gmol-1 × 2 = 0.181 mol 1.00 g initial no. of mole of H2(g) = 1.00 gmol-1 × 2 = 0.500 mol 1.90 g no. of mole of I2(g) at equilibrium = 126.9 gmol-1 × 2 = 0.0075 mol H2(g) + I2(g) d 2HI(g) initial 0.181 mol 0.500 mol 0 produced used up 0.181 - 0.0075 = 0.174 mol 0.174 mol at equilibrium 0.0075 mol 0.500 - 0.174 = 0.326 mol 0 + 0.348 = 0.348 mol I2(g) H2(g) HI(g) (b) 3 0.174 × 2 = 0.348 mol no. of mole of HI(g)2 (0.348 mol)2 Kc = no. of mole of H × no. of mole of I = 0.326 mol × 0.0075 mol = 50 2(g) 2(g) I. Nature of equilibrium Example 2 Part 4 H2(g) + I2(g) d 2HI(g) Page 2 If 1.00 mole each of H2(g) and I2(g) are heated in a 30.0 dm3 evacuated chamber to 470ºC. Using the value of Kc from the Example 1, determine (a) how many mole of I2(g) remain unreacted when equilibrium is established, (b) the total pressure in the chamber. (c) Now if one additional mole of H2(g) is introduced into this equilibrium system, how many mole of the original iodine will remain unreacted? (Given : Universal gas constant (R) = 0.0821 atm mol-1K-1) (a) 1. 2. no. of mole of HI(g)2 Kc = no. of mole of H × no. of mole of I 2(g) 2(g) Let x mole be the no. of mole of I2(g) consumed. initial 1.00 mol 1.00 mol 0 produced used up x mol x mol at equilibrium (1.00 - x) mol (1.00 - x) mol 2x mol I2(g) H2(g) HI(g) 3. x mol × 2 (2x)2 Kc = (1.00 - x)(1.00 - x) = 50 4x2 1.00 - 2.00x + x2 = 50 x = 0.78 or 1.4 (rejected) no. of mole I2(g) remaining = (1.00 - 0.78) mol = 0.22 mol 4x2 = 50 - 100x + 50x2 46x2 - 100x + 50 = 0 (b) The number of moles of gas does not change as the reaction proceeds at 470ºC; hence 2.00 mol of gas remains at equilibrium. PV = nRT nRT 2.00 mol × 0.0821 atm mol-1K-1 × 743 K = 4.07 atm P= V = 30.0 dm3 (c) 1. 2. no. of mole of HI(g)2 Kc = no. of mole of H × no. of mole of I 2(g) 2(g) Since the final equilibrium position is not depending on when the additional H2(g) is introduced, it can be assumed that it is introduced with the original H2(g). Let x mole be the no. of mole of I2(g) consumed. initial produced I2(g) 1.00 mol H2(g) (1.00 + 1.00) mol = 2.00 mol HI(g) 0 z mol × 2 3. (2z)2 Kc = (2.00 - z)(1.00 - z) = 50 4z2 2.00 - 3.00z + z2 = 50 z = 0.93 or 2.3 (rejected) no. of mole I2(g) remaining = (1.00 - 0.93) mol = 0.07 mol 4z2 = 100 – 150z + 50z2 46z2 – 150z + 100 = 0 used up z mol z mol at equilibrium (1.00 – z) mol (2.00 – z) mol 2z mol I. Nature of equilibrium Part 4 stoichiometric coefficients law of conservation of mass Page 3 Glossary Past Paper Question equilibrium concentration 90 2A 1 a i ii iii 95 1A 1 e ii iii 90 2A 1 a i ii iii 1a Consider the following equilibrium at constant pressure: A2(g) + 3B2(g) d 2AB3(g) A mixture of 4.0 mol of A2(g) and 12.0 mol of B2(g) was placed in a vessel of volume 20.0 dm3, and heated to 565K. When the system had reached equilibrium, it was found that 4.0 mol AB3(g) was present. i Calculate the concentration of each species at equilibrium. A2(g) + 3B2(g) d 2AB3(g) Since 4 moles of AB3 will consume 2 moles of A2 and 6 moles of B2 ∴ at equilibrium: 12 - 6 6 [B2] = 20 = 20 moldm-3 ½ mark 4-2 2 [A2] = 20 = 20 moldm-3 ½ mark 4 [AB3] = 20 moldm-3 ½ mark ii Calculate the equilibrium constant of this reaction at 565K. [AB3]2 4 20 20 K = [A ][B ]3 = (20)2 ⋅ ( 2 ) ⋅ ( 6 )3 = 14.82 dm6mol-2 (wrong unit -½) 2½ mark 2 2 iii When the volume of the vessel was increased and the system allowed to come to a new equilibrium at the same temperature, 9.0 mol of B2(g) was found to be present. Calculate the new volume. At new equilibrium: 9 moles of B2 is present. Let V be the new volume 9 ½ mark [B2] = V 4-1 3 [A2] = V = V ½ mark 2 [AB3] = V ½ mark 2 VV K = 14.82 = (V)2 ⋅ ( 3 ) ⋅ ( 9 )3 ½ mark V= C i ii 14.82 × 93 × 3 = 90 dm3 1 mark 4 Some candidates mistakenly used the initially given concentration rather than the equilibrium concentration in their calculation of equilibrium constant. Some weaker candidates failed to give the correct unit for the equilibrium constant. 1½ 2½ 3 95 1A 1 e ii iii 1e The reaction between ethanoic acid and ethanol can be represented by the following equation: CH3COOH(l) + C2H5OH(l) d CH3COOC2H5(l) + H2O(l) 12.01 g of ethanoic acid are treated with 4.61 g of ethanol in the presence of a catalyst. When the reaction reaches equilibrium at 298 K, 5.04 g of ethanoic acid are found to have reacted. ii Calculate the equilibrium constant, Kc , for the reaction at 298 K. CH3COOH(l) + C2H5OH(l) d CH3COOC2H5(l) + H2O(l) initial no. of moles 4.61 g 12.01 g 46.068 gmol-1 60.052 gmol-1 = 0.2000 mol= 0.100 mol 12.01 - 5.04 60.052 = 0.116 0.116 V 0.100 - 0.0840.084 = 0.016 0.016 V 0.084 V 0.084 V – – 3 no. of mole at equilibrium 0.084 2 marks concentration at equilibrium I. Nature of equilibrium Part 4 0.084 0.084 V×V 1 marks Kc = 0.116 0.016 = 3.80 (no unit) ×V V iii What additional mass of ethanol would be required in order to use up a further 0.60 g of ethanoic acid ? 0.60 g of ethanoic acid ≡ 0.010 mole CH3COOH(l) + C2H5OH(l) d CH3COOC2H5(l) + H2O(l) no. of mole at equilibrium 0.0106 x 0.094 0.094 1 mark 0.0942 3.80 = 0.106 x ∴ x = 0.0219 mole ½ mark Need to add 0.016 mole of ethanol = 0.74 g ½ mark iii A common mistake in the answers was the omission of the concentration of water from the equilibrium expression. Few candidates were able to complete part (iii). Page 4 2 C II. Acid-base Equilibria Part 1 Page 1 Topic Reference Reading II. Acid-base Equilibria 6.2.1–6.2.2 Part 1 Assignment A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 255–258 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 219–222, 365–367, 369–370 Reading Chemistry in Context (5th ed.), Thomas Nelson and Sons Ltd., 198–201, 326 Acid-base Equilibria Acid-base Theory Dissociation of water II. Acid-base Equilibria A. Acid-base Theory At the early stage of development of acid-base theory, acid was only described as a sour substance. And base is only a substance which neutralizes the sour taste of an acid. Nevertheless, scientists find that acids and bases bear some other properties. According to these properties, they developed a series of definitions to describe acid and base. They are called Arrhenius definition, Brønsted Lowry definition and Lewis definition. The coverage of the latter definition is boarder than the former one. 1. Arrhenius definition Syllabus Notes B rφ n A rr h s ef wis d initio Le Lowry d n e d te nius defi f e Common Sense n it io in tion ni Arrhenius definition is based on ionization. In 1884, Swedish chemist Svante Arrhenius proposed that Acid – a hydrogen-containing compound that, when dissolved in water, produces hydrogen ions, H+(aq). Base – a substance which reacts with an acid to give salt and water only. Examples of Arrhenius acid e.g. H2SO4(l) + H2O(l) → H3O+(aq) + HSO4-(aq) HSO4-(aq) + H2O(l) → H3O+(aq) + SO42-(aq) Examples of Arrhenius base e.g. NaOH(aq) + HCl(aq) → Na+Cl-(aq) + H2O(l) CuO(s) + HCl(aq) → Cu2+Cl-2(aq) + H2O(l) This is the definition used in certificate level. Arrhenius theory is criticized that 1. Acid is restricted to hydrogen-containing species and base is restricted to oxide or hydroxide. 2. The theory is only applicable to aqueous medium where a lot of acid-base reaction takes place in the absence of water. II. Acid-base Equilibria 2. Brønsted-Lowry definition Part 1 Page 2 In 1923, Danish chemist Johannes Brønsted and British chemist Thomas Lowry proposed a boarder definition, in which Acid – a proton donor Base – a proton acceptor e.g. NH3(aq) + proton acceptor (base) e.g. HCl(aq) + proton donor (acid) H2O(l) d proton donor (acid) H2O(l) d proton acceptor (base) NH4+(aq) + proton donor (conjugate acid of NH3(aq)) H3O+(aq) + proton donor (conjugate acid of H2O(l)) OH-(l) proton acceptor (conjugate base of H2O(l)) Cl-(aq) proton acceptor (conjugate base of HCl(aq)) When a Brønsted-Lowry acid donates a proton, it becomes a potential proton acceptor and is called conjugate base of the acid. Stronger a proton donor, weaker will be the conjugate base. Similarly, when a Brønsted-Lowry base accepts a proton, it becomes a potential proton donor and is called conjugate acid of the base. Stronger a proton acceptor, weaker will be the conjugate acid. Water is a very special example, it can behave as both proton acceptor and proton donor. Therefore, it is a Brønsted-Lowry acid and a Brønsted-Lowry base. e.g. H2O(l) + proton acceptor (base) H2O(l) d proton donor (acid) H3O+(aq) + proton donor (conjugate acid of H2O(l)) OH-(aq) proton acceptor (conjugate base of H2O(l)) The above reaction is also known as self-dissociation / self-ionization of water. All Arrhenius acids and bases can be classified into Brønsted-Lowry acid and base accordingly. 3. Lewis definition The American chemist Gilbert N. Lewis (1923) proposed an even broader definition for acid and base. He proposed that Acid – an electron acceptor Base – an electron donor This definition offers many advantages, including i. The acids are not limited to compounds containing hydrogen. ii. It works with solvents other than water. iii. It does not require formation of a salt or an acid-base conjugate pair. Since all chemical species are potential electron acceptor or electron donor, virtually, all chemical species are either Lewis acid or Lewis base. e.g. H3N:(g) + BF3(g) → H3N→BF3(s) electron electron donor acceptor (Lewis base) (Lewis acid) Most of the chemical reaction is caused by redistribution of electrons which leads to rearrangement of atoms and formation of a new substance. Therefore, all kinds of reaction can be considered as Lewis acid-base reaction. Concept of Lewis acid-base is very useful in describing reaction mechanisms. II. Acid-base Equilibria N.B. Part 1 Page 3 In the description of an acid-base equilibrium in A-Level, the Brønsted-Lowry definition is used. However, in the description of reaction mechanism, Lewis definition is used. B. Strength of acid Strength of an acid can be measured by an equilibrium constant called acidity constant or acid dissociation constant, Ka. For an acid, HA(aq) in water HA(aq) + H2O(l) d H3O+(aq) + A-(aq) Acidity constant, Ka is defined as Ka = [H3O+(aq)][A-(aq)] = Keq[H2O(l)] [HA(aq)] where [H2O(l)] is a constant since water is the solvent in large excess. [H3O+(aq)][A-(aq)] Equilibrium constant, Keq = [HA ][H O ] (aq) 2 (l) For a stronger acid, more HA(aq) molecules will dissociate into H3O+(aq) ion and A-(aq) ion and give a larger value for K a. Similar to pH, Ka can also be expressed in a negative log scale. pH = - log [H+] pKa = - log Ka A low pH means a high concentration of H+ ion. A low pKa means a larger value for Ka and a stronger acid. Relative strength of selected acids and their conjugate bases Stronger an acid, weaker will be its conjugate base. Stronger a base, weaker will be its conjugate acid. N.B. Redox reaction can be considered as a competition for electron. oxidizing agent + e- d reducing agent In redox reaction, only strong oxidizing agent reacts with strong reducing agent. Similarly, acid base reaction can be considered as a competition for proton. conjugate base + H+ d conjugate acid In acid base reaction, only strong acid react with strong base. The strength of an acid or a base is also an indicator of their stability. A strong acid or base is less stable than a weak acid or base. A strong acid tends to react with a strong base to form a weak conjugate base and weak conjugate acid. e.g. NH2- + HC≡CH → NH3 + HC≡Cbut not NH3 + HC≡C- → NH2- + HC≡CH II. Acid-base Equilibria 1. Leveling effect Part 1 Page 4 Acids stronger than hydroxonium ion H3O+(aq), do not show any difference in acidity in water. The water will convert all those acid molecules into H3O+(aq) ions. HCl(aq) + strong acid HBr(aq) + strong acid H2O(aq) → H3O+(aq) + strong weak base acid H2O(aq) → H3O+(aq) + strong weak base acid Cl-(aq) weak base Br-(aq) weak base This is why we consider all HCl(aq), H2SO4(aq) and HNO3(aq) are strong acid though they don't have the same strength. Similarly, a base stronger than hydroxide ion does not exist in aqueous medium. The water will convert all those base molecules into OH-(aq) ions. For example, NaNH2(s) is a very strong base, a non-aqueous solvent (e.g. liquid NH3(l)) must be used. NaNH2(s) + H2O(l) → Na+(aq) + OH-(aq) + NH3(aq) This is known as the leveling effect of solvent. C. Dissociation of water Pure water is found to be a poor conductor of electricity but not an insulator. Therefore, it must contain a very small amount of ions. Indeed, the ions are from self-dissociation / self-ionization of water molecule. H2O(l) + H2O(l) d H3O+(aq) + OH-(aq) K= 1. [H+(aq)][OH-(aq)] [H2O(l)] Ionic product of water or simply H2O(l) d H+(aq) + OH-(aq) H2O(l) d H+(aq) + OH-(aq) K= [H+(aq)][OH-(aq)] [H2O(l)] In aqueous medium, H2O(l) is in large excess, thus [H2O(l)] is a constant and can be incorporated in the equilibrium constant. K [H2O(l)] = [H+(aq)][OH-(aq)] = Kw Kw is known as ionic product of water. Like other equilibrium constant, it is a quantity depending on temperature and has the value of 1.00 × 10-14 mol2dm-6 at 25ºC. Pure water is neutral because the concentration of H+(aq) is the same as the concentration of OH-(aq). i.e. [H+(aq)] = [OH-(aq)] Acidic Neutral Alkaline II. Acid-base Equilibria For pure water at 25ºC Part 1 Page 5 Kw = [H+(aq)][OH-(aq)] = [H+(aq)]2 = 1.00 × 10-14 mol2dm-6 [H+(aq)] = 1.00 × 10-7 moldm-3 Recall that pure water has the concentration of 55.6 moldm-3. [H+(aq)] 1.00 × 10-7 moldm-3 -9 55.6 moldm-3 = 1.80 × 10 [H2O(l)] = 1 This means that only 1 out of every 556,000,000 ( 1.80 × 10-9 ) water molecule will dissociate into a pair of H+(aq) and OH-(aq). This explains the extremely low conductivity of pure water but it is not an insulator. acid-base equilibria Arrhenius acid-base Brønsted Lowry acid-base Lewis acid-base conjugate acid-base proton donor / acceptor electron acceptor / donor pH pKa leveling effect acidity constant / acid dissociation constant (Ka) self- dissociation / self-ionization of water ionic product of water (Kw) 94 1A 1 f i 97 2A 4 c i 99 1A 4 a 94 1A 2 c i ii Glossary Past Paper Question 94 1A 1 f i 1f i Write an equation to show that HPO42-(aq) can act as a Brφnsted base in water. HPO42-(aq) + H2O(l) d H2PO4-(aq) + OH-(aq) 1 1 mark 94 1A 2 c i ii 2c i What is a “Brφnsted acid”? “Brφnsted acid” is a proton donor 1 mark ii Write equations to show that nitric(V) acid is an acid in water, but a base in liquid hydrogen fluoride. HNO3 + H2O d H3O + NO31 mark + 1 mark HNO3 + HF d H2NO3 + F Ci Many candidates thought that an aqueous medium was required for the proton donation. ii The equation for nitric(V) acid acting as an acid was given correctly, but few candidates gave the correct equation for it acting as a base. 97 2A 4 c i 4c A solution is formed by mixing equal volumes of 0.20 M CH3CO2H(aq) and 0.20 M CH3CO2Na(aq). i Identify all Brφnsted acids and all Brφnsted bases in the solution. 99 1A 4 a 4a Write all the Bronsted acids present in aqueous ammonia. 1 2 II. Acid-base Equilibria Part 2 Page 1 Topic Reference Reading II. Acid-base Equilibria 6.2.3 Part 2 Assignment A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 258–259, 264, 268 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 367–369 Reading Chemistry in Context (5th ed.), Thomas Nelson and Sons Ltd., 325–329 pH and its measurement Temperature dependence of pH D. pH and its measurement 1. Definition of pH pH 0 1 2 3 4 [H+(aq)] 1 mol dm-3 0.1 mol dm-3 0.01 mol dm-3 0.001 mol dm-3 0.0001 mol dm-3 Syllabus Notes pH is defined as the negative log of the concentration of H+(aq) in molarity. pH = - log [H+(aq)] Because it is a log scale, every 1 unit increase in pH means ten folds decrease in the concentration of H+(aq). For pure water at 25ºC, Kw = [H+(aq)][OH-(aq)] = [H+(aq)]2 = 1.00 × 10-14 mol2dm-6 [H+(aq)] = 1.00 × 10-7 moldm-3 pH = - log[H+(aq)] = - log 1.00 × 10-7 = 7.00 Therefore, pure water has pH 7 at 25ºC. Find the pH of a 0.1 M HCl(aq) solution. Find the pH of a 10-8 M HCl(aq) solution. 2. Temperature dependence of pH Pure water has pH 7 only at 25ºC because Kw is temperature dependent. The degree of dissociation of water, thus the Kw, increases with increasing temperature. For example, at 50ºC, Kw = 5.47 × 10-14 mol2dm-6 Kw = [H+(aq)][OH-(aq)] = [H+(aq)]2 = 5.47 × 10-14 mol2dm-6 [H+(aq)] = 5.47 × 10-14 mol2dm-6 = 2.34 × 10-7 moldm-3 pH = - log[H+(aq)] = - log 2.34 × 10-7 = 6.63 Therefore, water at 50 ºC has a pH 6.63. N.B. Temperature /ºC 0 10 20 25 50 100 Kw / mol2 dm-6 0.11 × 10-14 0.30 × 10-14 0.68 × 10-14 1.00 × 10-14 5.47 × 10-14 51.3 × 10-14 A solution is said to be neutral because the concentration of H+(aq) is the same as the concentration of OH-(aq), not because of the pH. A neutral solution has the pH 7 only at 25 ºC. Acidic Neutral Alkaline II. Acid-base Equilibria 3. Measurement of pH Part 2 Page 2 pH meter, indicator and pH paper are commonly used to determine the pH of a solution. a) Use of pH meter pH meter is an electronic instrument which gives the pH value of a solution directly. However, pH meter needs to be calibrated before use. (1) Calibration of pH meter pH meter is calibrated by immersing the electrode into a buffer solution with known pH and adjusting the knob on the pH meter. Buffer solution is a solution whose pH is insensitive to the addition of small amount of acid or base. For example, blood is a buffer solution which has a constant pH of about 7.4 which is independent of the acidity of our diet. b) Use of indicator Indicator is a chemical which has different colour at different pH. Usually, it is a weak acid or a weak base whose conjugate acid and conjugate base have different colours. (1) Colour of indicator Indicator litmus methyl orange phenolphthalein 0 | | | 1 red 2 red 3 4 5 | pH value 678 purple | 10 11 12 13 14 blue | yellow | |pale pink| red | 9 |orange| colourless Universal indicator / pH indicator is a mixture of several indicators. As a result, the universal indicator gives a series of colour change at different pH instead of just 2. pH paper is a filter paper is soaked in universal indicator. Like other testing paper, it should be dipped into the solution to be tested and removed immediately. If the paper is immersed in the solution, the dye will diffuse into the solution and the colour observed would be paler than the standard. Glossary Past Paper Question pH scale indicator. 94 2A 3 b i 99 1A 4 d pH meter calibration of pH meter buffer solution indicator universal 94 2A 3 b i 3b Account for each of the following: i At 323K, the pH of pure water is less than 7.0. 2 + 1 mark The dissociation of water, 2H2O(l) d H3O (aq) + OH (aq), is an endothermic process. Increase in temperature, the equilibrium shifts to the right and hence [H3O+(aq)] increases. In pure water, [H3O(aq)+] at 323 K is higher than that at 298 K, ∴ pH of pure water at 323 K is less than 7. 1 mark Ci Many candidates were not aware of the fact that the dissociation of water is an endothermic process. Some candidates erroneously pointed out that the low pH of water at 323 K was caused by the dissolution of some acidic gases. 99 1A 4 d 4d Constant boiling hydrochloric acid contains 20.2% by mass of HCl. Calculate the mass of constant boiling II. Acid-base Equilibria Part 2 hydrochloric acid required to prepare 1.00 dm3 of HCl(aq) of pH 2.0 at 298 K. Page 3 II. Acid-base Equilibria Part 3 Page 1 Topic Reference Reading II. Acid-base Equilibria 6.2.4 Advanced Practical Chemistry, John Murray (Publisher) Ltd., 62–64 Modern Physical Chemistry, Bell and Hyman, 283 Part 3 rd Assignment A-Level Chemistry (3 ed.), Stanley Thornes (Publisher) Ltd., 258–262, 273 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 369 Reading Chemistry in Context (5th ed.), Thomas Nelson and Sons Ltd., 329–330 Syllabus Strong and weak acids/bases Measuring of pH or conductivity of acid / base Dissociation constant (Ka and Kb) Calculation involving pH, Ka and Kb D. Strong and weak acids/bases An acid or a base is said to be strong if the degree of dissociation is high. Usually, this is measure in term of pH, the conductivity of the solution or the dissociation constant. 1. Measuring of pH or conductivity of acid / base Obviously, with the same concentration, a strong acid (HCl(aq)) has lower pH than a weak acid (CH3COOH(aq)). And a strong alkali (e.g. NaOH(aq)) has higher pH than a weak base (e.g. NH3(aq)). Owing to the higher ionic concentration, the conductivity of a strong acid / base is also higher than that of a weak acid / base. 2. Dissociation constant (Ka and Kb) For the equilibrium, HA(aq) + H2O(l) d H3O+(aq) + A-(aq) [H3O+(aq)][A-(aq)] K = [HA ][H O ] (aq) 2 (l) Since H2O(l) is in excess, its concentration can be incorporated into the equilibrium constant and the new dissociation constant is called acidity constant Ka. HA(aq) + H2O(l) d H3O+(aq) + A-(aq) [H3O+(aq)][A-(aq)] K = [HA ][H O ] (aq) 2 (l) K[H2O(l)] = [H3O+(aq)][A-(aq)] = Ka [HA(aq)] Solution 1 M HCl(aq) 0.1 M HCl(aq) 0.1 M CH3COOH(aq) pure water 0.1 M NH3(aq) 0.1 M NaOH(aq) 1 M NaOH(aq) pH 0 1 3 7 11 13 14 Notes e.g. CH3COOH(aq) + H2O(l) d CH3COO-(aq) + H3O+(aq) Ka = [CH3COO-(aq)][H3O+(aq)] = 1.7 × 10-5 mol dm-3 [CH3COOH(aq)] II. Acid-base Equilibria Part 3 Page 2 Similarly, the dissociation constant of a base (basicity constant Kb) can also be defined in a similar way. B(aq) + H2O(aq) d HB+(aq) + OH-(aq) [HB+(aq)][OH-(aq)] K = [B ][H O ] (aq) 2 (l) K[H2O(l)] = [HB+(aq)][OH-(aq)] = Kb [B(aq)] e.g. NH3(aq) + H2O(l) d NH4+(aq) + OH-(aq) Kb = [NH4+(aq)][OH-(aq)] = 1.8 × 10-5 mol dm-3 [NH3(aq)] a) Dissociation of polybasic acid Consider the dissociation of tribasic acid H3PO4(aq) in water, it dissociates in water through 3 consecutive steps. Acid H3PO4(aq) H2PO4-(aq) HPO42-(aq) Ka(298K)/mol dm-3 7.9 × 10-3 6.2 × 10-8 4.4 × 10-13 Equilibrium H3PO4(aq) d H+(aq) + H2PO4-(aq) H2PO4-(aq) d H+(aq) + HPO42-(aq) HPO42-(aq) d H+(aq) + PO43-(aq) [H+(aq)][ H2PO4-(aq)] [ H3PO4(aq)] [H+(aq)][ HPO42-(aq)] K a2 = [ H2PO4-(aq)] + [H (aq)][ PO43-(aq)] Ka3 = [ HPO 2- ] 4 (aq) K a1 = Phosphoric acid is a tribasic acid which is not very strong. Only 25% of the molecules dissociate in water. The first dissociation constant Ka1 is low. Moreover, the second dissociation constant Ka2 is lower than the first one and the third dissociation constant Ka3 is even lower. Therefore, in a solution of H3PO4(aq), it consists of mainly water and H3PO4(aq) molecules. Relative abundance of different species H2O(l) >> H3PO4(aq) > H+(aq) > H2PO4-(aq) > HPO42-(aq) > PO43-> OH-(aq) Furthermore, the dissociation constants of a tribasic acid can be expressed as : H3PO4(aq) d 3H+(aq) + PO43-(aq) Ka(overall) = K a1 × K a2 × K a3 [H+(aq)][ H2PO4-(aq)] [H+(aq)][ HPO42-(aq)] [H+(aq)][ PO43-(aq)] = × × [ HPO 2- ] [ H3PO4(aq)] [ H2PO4-(aq)] 4 (aq) + 3 3[H (aq)] [ PO4 (aq)] = [ H3PO4(aq)] (1) Charge effect For a polybasic acid, the first dissociation constant Ka1 is always larger than the second dissociation constant Ka2. This is because when the first proton is removed from the acid molecule, it is removed from an electrically neutral molecule. When the second proton is removed from the acid molecule, it is removed from a negatively charged species which is energetically less favorable. The effect of the charges on the dissociation of acid molecule is known as charge effect. II. Acid-base Equilibria 3. Part 3 Page 3 Calculation involving pH, Ka and Kb a) Relationship between Ka and Kb (pKa and pKb) Sometimes the value of Ka and Kb can also be expressed in negative log scale known as pKa and pKb. pKa = -log[Ka] pKb = - log[Kb] For the hydrolysis of a conjugate acid-base pair in water, HA(aq) + H2O(l) d H3O+(aq) + A-(aq) Ka = [H3O+(aq)][A-(aq)] [HA(aq)] A-(aq) + H2O(l) d HA(aq) + OH-(aq) Kb = [HA(aq)][OH-(aq)] [A-(aq)] [H3O+(aq)][A-(aq)] [HA(aq)][OH-(aq)] × = [H3O+(aq)][OH-(aq)] = Kw [HA(aq)] [A-(aq)] K a × Kb = Ka × Kb = Kw = 1 × 10-14 mol dm-3 OR -log (Ka × Kb) = - log Kw -log Ka - log Kb = - log Kw pKa + pKb = pKw = 14 b) Relationship between pH, pOH and pKw Kw = [H3O+(aq)][OH-(aq)] = 1 × 10-14 mol dm-3 -log Kw = -log ([H3O+(aq)][OH-(aq)]) -log Kw = -log [H3O+(aq)] - log [OH-(aq)] pKw = pH + pOH = 14 II. Acid-base Equilibria c) Some basic assumptions Part 3 Page 4 Depending on whether the small quantity is omitted, there may be two possible answers for a single question. Calculate the pH of 0.100 M CH3COOH(aq). Ka of CH3COOH(aq) is 1.70 × 10-5 mol dm-3. 1. The equation is solved accurately concentration at equilibrium (M) Ka = CH3COOH(aq) 0.100 - x d CH3COO-(aq) + H+(aq) x x [CH3COO-(aq)][H3O+(aq)] x·x = 0.100 - x mol dm-3 = 1.70 × 10-5 mol dm-3 [CH3COOH(aq)] or -1.31 × 10-3 (rejected) x = 1.29 × 10-3 pH = - log x = - log (1.29 × 10-3) = 2.89 2. The equation is solved with the small quantity omitted. concentration at equilibrium (M) Ka = CH3COOH(aq) 0.100 - x d CH3COO-(aq) + H+(aq) x x [CH3COO-(aq)][H3O+(aq)] x·x = 0.100 - x mol dm-3 = 1.70 × 10-5 mol dm-3 [CH3COOH(aq)] Since CH3COOH(aq) is a weak acid, x must be very small comparing with the original concentration (0.1M). i.e. 0.100 - x ≈ 0.100. x·x -5 0.100 = 1.70 × 10 x = 1.30 × 10-3 pH = - log x = - log (1.30 × 10-3) = 2.89 It can be seen that no matter the small quantity is omitted or not, they give similar results. This is because the degree of dissociation of ethanoic acid is only about 1%. The dissociation constant of any weak acid with Ka less than 1 × 10-4 mol dm-3 can be neglected in the expression : (original concentration – x) M. This is applicable to most weak acids. II. Acid-base Equilibria 3. 1. Basic assumptions Part 3 Page 5 Attention must be taken that the small quantity can be omitted only if it is added/subtracted to another number, but not multiplied/divided to another number. If x is very small, 0.1 - x ≈ 0.1 but 0.1 · x ≠ 0.1 2. All salts are strong electrolyte and completely ionize in water, only very small percentage is hydrolyzed. e.g. When NH4Cl(s) is dissolved in water NH4Cl(s) → NH4+(aq) + Cl-(aq) water NH4+(aq) + H2O(l) d NH3(aq) + H3O+(aq) [NH4+(aq)] >> [NH3(aq)] The amount of NH4+(aq) hydrolyzed is considered negligible comparing with the amount of NH4+(aq) from the ionization of salt (a strong electrolyte). 3. In alkaline medium, weak acid will be neutralized. CH3COOH(aq) + OH-(aq) → CH3COO-(aq) + H2O(l) And because the solution is alkaline, hydrolysis of CH3COO-(aq) is very minimal. CH3COO-(aq) + H2O(l) d CH3COOH(aq) + OH-(aq) [CH3COO-(aq)] >> [CH3COOH(aq)] Similarly, in acidic medium, weak alkali will also be neutralized. H3O+(aq) + NH3(aq) → NH4+(aq) + H2O(l) Hydrolysis of the salt is very minimal in acidic medium NH4+(aq) + H2O(l) d NH3(aq) + H3O+(aq) [NH4+(aq)] >> [NH3(aq)] II. Acid-base Equilibria 4. Example Part 3 Page 6 Some Useful Relationships Kw = [H+(aq)][OH-(aq)] = Ka × Kb = 1 × 10-14 mol2 dm-6 pKw = pH + pOH = pKa + pKb = 14 Calculate the pH of 0.100 M CH3COO-Na+(aq) solution. Ka of CH3COOH(aq) is 1.70 × 10-5 mol dm-3 and Kw of water is 1.00 × 10-14 mol2dm-6. Answer 1 CH3COO-Na+(aq) is a strong electrolyte which completely ionizes in water. CH3COO-Na+(aq) → CH3COO-(aq) + Na+(aq) A very small percentage of CH3COO-(aq) hydrolyzes in water and makes the solution alkaline. CH3COO-(aq) + Inital concentration (mol dm-3) Equilibrium concentration (mol dm-3) [CH3COOH(aq)][OH-(aq)] [CH3COO-(aq)] 0.100 0.100 - x x·x 0.100 -x = [CH3COOH(aq)] + [CH3COO-(aq)][H+(aq)] × [H (aq)][OH (aq)] = 1 K a × Kw H2O(l) d CH3COOH(aq) + OH-(aq) 0 x 0 x Kc = = = [CH3COOH(aq)][OH-(aq)] [H+(aq)] × [H+ ] [CH3COO-(aq)] (aq) Kw x·x 0.100 - x = Ka Since x << 0.100, therefore 0.100 - x ≈ 0.100 1.00 × 10-14 x2 = 1.70 × 10-5 0.100 x = 7.67 × 10-6 pOH = - log [OH-(aq)] = - log 7.67 × 10-6 = 5.12 pH = 14 - pOH = 14 - 5.12 = 8.88 II. Acid-base Equilibria Anwser 2 Part 3 Page 7 Alternatively, the problem can be interpreted as hydrolysis of the base CH3COO-(aq) which is the conjugate base of CH3COOH(aq). CH3COO-(aq) + H2O(l) d CH3COOH(aq) + OH-(aq) Inital concentration (mol dm-3) Equilibrium concentration (mol dm-3) 0.100 0.100 - x 0 x 0 x Since the dissociation constant of the conjugate acid-base pair is related by the relationship Kw = Ka × Kb. Kw 1.00 × 10-14 Kb = K = 1.70 × 10-5 = 5.88 × 10-10 a 5.88 × 10-10 = [CH3COOH(aq)][OH-(aq)] x·x = 0.100 - x [CH3COO-(aq)] or - 7.67 × 10-6 (rejected) If the equation is solved exactly, x = 7.67 × 10-6 pOH = - log [OH-(aq)] = - log 7.67 × 10-6 = 5.12 pH = 14 - pOH = 14 - 5.12 = 8.88 4. Experimental determination of Ka By half-neutralization By other methods Glossary Past Paper Question conductivity charge effect 91 1A 3 b ii 92 2A 3 c ii 93 1A 1 e 94 1A 1 f ii 95 2A 2 c i ii iii 96 2A 1 c i 97 1A 2 b dissociation constant acidity constant Ka pK b Kw pH pOH pKa basicity constant Kb polybasic acid pK w hydrolysis of salt 94 2A 3 c i 91 1A 3 b ii 3b Write equation(s) to describe the reaction of ii NaHCO3(s) with water. NaHCO3(s) → Na+(aq) + HCO3-(aq) HCO3-(aq) + H2O(l) d H2CO3(aq) + OH-(aq) C ii Most knew the hydrolysis of HCO3-(aq) to form H2CO3(aq) and OH-(aq). 1 1 mark II. Acid-base Equilibria Part 3 92 2A 3 c ii 3c ii 40.0 cm3 of an aqueous solution of a weak acid, HA(aq), was titrated with a strong base, MOH(aq), at 298K. The initial pH, before the addition of base, was 2.70. At the equivalence point of the titration, the pH was 8.90. Calculate (I) the initial concentration of the acid. (II) the volume of the base added to reach the equivalence point. (III) the concentration of the base. (The dissociation constant of the weak acid and the ionic product of water at 298K are respectively Kw = 1.0 × 10-14 mol2 dm-6 ) Ka = 1.8 × 10-5 mol dm-3 (I) pH = 2.70 ⇒ -log[H+] = 2.70 ⇒ [H+] = 1.995 × 10-3 Let a be the concentration of the weak acid HA d H+ + Aa-x x x [ H + ][ A − ] x ⋅ x . Ka = = = 18 × 10 −5 [ HA] a−x Since HA is a weak acid a - x ≈ a ∴ Concentration of weak acid = a = 0.2212M 2 marks (II) Let V be the volume (in cm3) of MOH added. The total volume of the solution at the equivalence point = 40.0 + V (cm3) 40 × 0.2212 = 0.008847 mole Total no. of mole of HA present = 1000 At equivalence point, the amount of acid is the same as the amount of base. Thus, a pure salt solution is formed. It was because of hydrolysis, the solution is basic. A- + H2O d HA + OHi.e. the [HA] = [OH-] at equivalence point Since pH = 8.90 at equivalence point ⇒ [H+] = 1.2589 × 10-9 10 −14 10 -14 and [HA] = [OH-] = = = 7.943 × 10-6 [H + ] . 12589 × 10 −9 Also the amount of A- at equivalence point can be assumed to be equal to the amount of HA in the original solution because HA is a weak acid and the solution is alkaline. + ATherefore, HA d H+ 0.008847 1.2589 × 10-9 at equivalence point, 7.943 × 10-6 (40.0 + V) / 1000 0.008847 1.2589 × 10 −9 × [H + ][A - ] (40.0 + V) / 1000 -5 Ka = = 1.8 × 10 = [HA] 7.943 × 10 −6 0.008847 1.2589 × 10 −9 × × 1000 −6 (40.0 + V) 7.943 × 10 (40.0 + V) = 77.89 cm3 2 marks or V = 37.89 cm3 0.008847 × 1000 = 0.2335 M 1 mark (III) Concentration of MOH = 37.89 (Comment : Wrong no. of significant figures in the calculation) Few candidates obtained all three answers correctly, but marks were awarded for all intermediate steps. Page 8 5 1.8 × 10-5 = C ii II. Acid-base Equilibria Part 3 93 1A 1 e 1e A solution is prepared by dissolving potassium hydrogencarbonate in water at 298 K. Write chemical equations for four equilibrium reactions, each involving the hydrogencarbonate ions, that occur in the solution and calculate the value of the equilibrium constant for each reaction. Give that at 298 K: K1 = 4.3 × 10-7 moldm-3; K2 = 4.8 × 10-11 moldm-3 for H2CO3 Kw = 1.0 × 10-14 mol2dm-6 for H2O K2 = 4.8 × 10-11 moldm-3 HCO3-(aq) d H+(aq) + CO32-(aq) Kw HCO3-(aq) + H2O(l) d H2CO3(aq) + OH-(aq) K = K =2.33 × 10-8 moldm-3 1 1 + 6 -1 3 HCO3 (aq) + H (aq) d H2CO3(aq) K1 = 2.33 × 10 mol dm K2 HCO3-(aq) + HCO3-(aq) d CO32-(aq) + H2CO3(aq) K = K = 1.12 × 10-4 1 K2 2HCO3 (aq) + OH (aq) d CO3 (aq) + H2O(l) K = K = 4.8 × 103 mol-1dm3 w 4 marks(½ for equation, ½ for value) for any 4 of the above equilibria 1 mark for correct dimension of equilibrium constants C Most candidates could get the two equilibria involving HCO3- and H+, but failed to realize the two equilibria involving HCO3- and OH- indicating that the candidates were not familiar with the interplay between Ka and Kb of the respective conjugate acid and conjugate base. 94 1A 1 f ii 1f ii For the following equilibrium at 298 K, the equilibrium constant Kc = 2.84 × 10-22 (mol dm-3)5. Ag3PO4(s) + H2O(l) d 3Ag+(aq) + HPO42-(aq) + OH-(aq) Calculate the solubility, in mol dm-3, of silver phosphate(V) in 0.10 M disodium hydrogenphosphate(V) solution at pH 10. Given : Ionic product of water at 298 K = 1.0 × 10-14 (mol dm-3)2 (Assume the extent of dissociation of HPO42- ions at pH 10 is negligible. At pH = 10 [OH-(aq)] = 10-4 moldm-3 [HPO42-(aq)] = 10-1 moldm-3 Kc = [Ag+(aq)]3[HPO42-(aq)][OH-(aq)] = 2.84 × 10-22 ∴ [Ag+(aq)]3 = 2.84 × 10-17 [Ag+(aq)] = 3.05 × 10-6 moldm-3 Q 1 mole of Ag3PO4 produces 3 moles of Ag+(aq) ions. ∴ Solubility of Ag3PO4 = 1.017 × 10-6 moldm-3 3 marks 94 2A 3 c i 3c Given: Ka for CH3(CH2)2COOH = 1.5 × 10-5 moldm-3 at 298K Calculate the pH of i an aqueous solution of 0.10M CH3(CH2)2COOH; [H+][PrCOO-] Ka = [PrCOOH] Assumming [H+] = [PrCOO-], [PrCOO-] = 0.1 [H+]2 1.5 × 10-5 = 0.1 [H+] = 1.225 × 10-3 M pH = 2.91 Page 9 5 3 2 1 mark 1 mark II. Acid-base Equilibria Part 3 95 2A 2 c i ii iii 2c The following reversible reaction occurs in an aqueous solution of ammonium ethanoate : (I) CH3COO-(aq) + NH4+(aq) d CH3COOH(aq) + NH3(aq) i Write an expression for the dissociation constant, Ka , of ammonium ion. NH4+ d NH3 + H+ [NH3][H+] Ka = [NH +] 1 mark 4 ii Calculate the equilibrium constant of reaction (I) at 298 K. (At 298 K, the dissociation constants of ethanoic acid and ammonium ion are 1.76 × 10-5 mol dm-3 and 5.59 × 1010 mol dm-3 respectively.) NH4+ + CH3COO- d NH3 + CH3COOH [NH3][CH3COOH] K = [NH +][CH COO-] 1 mark 4 3 Ka(NH4+) 5.59 × 10-10 = K (CH COOH) = 1.76 × 10-5 = 3.18 × 10-5 1 mark a 3 iii For a 0.10 M solution of ammonium ethanoate at 298 K, calculate (1) the concentration of ammonia, and (2) the pH of the solution. (1) Let x = [NH3] = [CH3COOH] 1 mark [NH4+] = [CH3COO-] = 0.100 - x x2 -5 K = (0.100 - x)2 = 3.18 × 10 1 mark x = 5.60 × 10-4 M (2) Let y = [H+] Ka(CH3COOH) = y (100 - x) x 1 mark y = 9.96 × 10-8 M 1 mark pH = -log [H+] = 7.00 OR y (x) Ka(NH4+) = (0.100 - 5.60 × 10-4) y (5.60 × 10-4) 5.59 × 10-10 = (0.100 - 5.60 × 10-4) 1 mark y = 9.93 × 10-8 M pH = 7.00 1 mark The performance of candidates in this question was poor, indicating that they were not able to apply what they had learnt to a slightly varied situation. Page 10 1 2 4 C 96 2A 1 c i 1c i At 298 K the pH of 0.050 M CH3CH2COOH is 3.10. Calculate the Ka of CH3CH2COOH at 298 K. CH3CH2COOH d CH3CH2COO- + H+ [CH3CH2COO-][H+] Ka = [CH CH COOH] 3 2 Assuming [CH3CH2COO-] = [H+] (10-3.10)2 Ka = (0.05 - 10-3.10) = 1.26 × 10-5 mol dm-3 (1.25 – 1.30 × 10-5 mol dm-3) (Deduct ½ mark for no / wrong unit) 97 1A 2 b 2b Account for the difference in Ka values given in the following equilibrium reactions: H2SO4(aq) + H2O(l) d H3O+(aq) + HSO4-(aq) Ka = 7.94 × 102 mol dm-3 HSO4-(aq) + H2O(l) d H3O+(aq) + SO42-(aq) Ka = 0.10 mol dm-3 2 ½ mark ½ mark 1 mark II. Acid-base Equilibria Part 4 Page 1 Topic Reference Reading II. Acid-base Equilibria 6.2.5 Part 4 Assignment Advanced Practical Chemistry, John Murray (Publisher) Ltd., 65–67 A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 271–272 Reading Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 374–377 Chemistry in Context (5th ed.), Thomas Nelson and Sons Ltd., 334–337 Buffer E. Buffer Buffer is a solution which resists to pH change upon addition of small amount of acid or alkali. Blood is an example of buffer solution, which has a constant pH of about 7.4. 1. Principle of buffer action Since buffer is capable to absorbs both H+(aq) and OH-(aq), it must be a mixture of an acid and a base. However, strong acid and strong base cannot coexist, they neutralize each other. Buffer must be a mixture of a conjugate acid-base pair of a weak acid or weak base. e.g. A mixture of CH3COOH(aq) (acid) and CH3COO-(aq) (conjugate base), or A mixture of NH3(aq) (base) and NH4+(aq) (conjugate acid) Syllabus Notes For the buffer consists of CH3COOH(aq) and CH3COO-Na+(aq). This involves the equilibrium of CH3COOH(aq) d CH3COO-(aq) + H+(aq) Upon addition of H+(aq), H+(aq) will be removed by the base CH3COO-(aq). CH3COO-(aq) + H+(aq) → CH3COOH(aq) Upon addition of OH-(aq), OH-(aq) will be removed by the acid CH3COOH(aq). CH3COOH(aq) + OH-(aq) → CH3COO-(aq) + H2O(l) Therefore, the pH of the solution can be maintained. II. Acid-base Equilibria a) pH of an acidic buffer Part 4 Page 2 For a buffer consists of an acid and its conjugate base (e.g. CH3COOH(aq) and CH3COO-Na+(aq)), CH3COOH(aq) d CH3COO-(aq) + H+(aq) Ka = [CH3COO-(aq)][H+(aq)] [CH3COOH(aq)] Ka [CH3COOH(aq)] [CH3COO-(aq)] [H+(aq)] = pH = [CH3COOH(aq)] -log Ka - log [CH COO- ] (aq) 3 pKa - log [acid] [salt] pH = In a buffer solution, acid and its salt share the same volume ∴ pH = pKa - log ( no. of mole of acid / V no. of mole of acid ) = pKa - log no. of mole of salt / V no. of mole of salt The pH is depending on the nature of the acid i.e. pKa and the ratio of concentrations of the acid and its salt. Furthermore, the more concentrate the acid and its salt, the higher will be the buffering capacity of the solution because it will be able to absorb more base or acid. b) pH of an alkaline buffer For an buffer solution involving a weak base and its conjugate acid (e.g. NH3(aq) and NH4+(aq)) NH3(aq) + H2O(l) d NH4+(aq) + OH-(aq) [NH4+(aq)][OH-(aq)] [NH3(aq)] Kb = Kb[NH3(aq)] [OH-(aq)] = [NH + ] 4 (aq) [NH3(aq)] pOH = pKb - log [NH + ] 4 (aq) pOH = pKb - log [base] [salt] The base and salt share the same solution, therefore, pOH = pKb - log no. of mole of base no. of mole of salt Since pKw = pH + pOH = 14, pH = pKw - pOH [base] [base] pH = pKw - ( pKb - log [salt] ) = 14 - pKb + log [salt] ) II. Acid-base Equilibria 2. Calculation of buffer solution Part 4 Page 3 Important relationships pH = pKa - log [acid] [salt] OR pH = pKa - log no. of mole of acid no. of mole of salt no. of mole of base no. of mole of salt pOH = pKb - log [base] [salt] OR pOH = pKb - log pKw = pH + pOH = 14 a) pH of buffer solution Calculate the pH of the buffer solution consists of 0.100 M NH3(aq) and 0.0500 M NH4+(aq) respectively. Kb of NH3(aq) 1.58 × 10-5 mol dm-3. Initial concentration (mol dm ) Equilibrium concentration (mol dm-3) -3 NH3(aq) + 0.100 0.100 - x H2O(l) d NH4+(aq) 0.0500 0.0500 + x + OH-(aq) 0 x Like the example mentioned in the last unit, the equation can be solved accurately or approximately. A. The equation is solved accurately. Kb = [NH4+(aq)][OH-(aq)] (0.0500 + x) x = 0.100 - x = 1.58 × 10-5 [NH3(aq)] x2 + 0.0500 x - 1.58 × 10-6 = 0 x = 3.16 × 10-5 or - 0.0500 (rejected) pOH = - log x = - log (3.16 × 10-5) = 4.50 pH = pKw - pOH = 14 - 4.50 = 9.50 B. The equation is solved with the approximation applied. It can be assumed that the 1. hydrolysis of ammonia is very minimal in the presence of NH4+. ∴ [NH3(aq)] = (0.100 - x) M ≈ 0.100 M 2. hydrolysis of ammonium ion is also very minimal in the presence of NH3(aq). [NH4+(aq)] = (0.0500 + x) M ≈ 0.0500 M [NH4+(aq)][OH-(aq)] 0.0500 · x = 0.100 = 1.58 × 10-5 [NH3(aq)] Kb = x = 3.16 × 10-5 pOH = - log x = - log (3.16 × 10-5) = 4.50 pH = pKw - pOH = 14 - 4.50 = 9.50 II. Acid-base Equilibria Part 4 Page 4 C. Solving by applying the equation with approximations applied. [NH3(aq)] pOH = pKb - log [NH + ] 4 (aq) [NH3(aq)] = (0.100 - x) M ≈ 0.100 M [NH4+(aq)] = (0.0500 + x) M ≈ 0.0500 M 0.100 pOH = -log (1.58 × 10-5) - log 0.0500 = 4.50 pH = pKw - pOH = 14 - 4.50 = 9.50 [NH3(aq)] Practically, the equation pOH = pKb - log [NH + ] can only be used with the approximations applied. This is 4 (aq) because a quadratic equation cannot be set up without considering the expression of Kb. No matter which method is used, they all give the same result (pH = 9.50). This means that the approximations are valid. N.B. In the following examples, approximation is used whenever applicable. b) Effect of addition of acid / alkali into a non-buffered solution 1. Calculate the pH of a 0.200 M CH3COOH(aq) solution. Ka of CH3COOH(aq) is 1.70 × 10-5 mol dm-3. At equilibrium d CH3COO-(aq) + H+(aq) CH3COOH(aq) 0.200 - x ≈ 0.200 x x x = 1.84 × 10-3 x·x -5 0.200 = 1.70 × 10 pH = - log [H+(aq)] = - log (1.84 × 10-3) = 2.74 2. Calculate the pH of the solution if 0.0500 mole of solid NaOH(s) is added into 600 cm3 of 0.200 M CH3COOH(aq). No. of mole of CH3COOH(aq) in 600 cm3 of 0.200 M = 0.200 mol dm-3 × 0.600 dm3 = 0.120 mol NaOH is a strong alkali which will neutralize 0.0500 mole of CH3COOH(aq). In the equilibrium mixture, no. of mole of CH3COOH(aq) = (0.120 - 0.0500) mol = 0.070 mol no. of mole of CH3COO-(aq) = no. of mole of CH3COOH(aq) neutralized = 0.0500 mol no. of mole of acid 0.070 pH = pKa - log no. of mole of salt = - log (1.70 × 10-5) - log 0.050 = 4.6 3. Calculate the pH of the solution if 0.0500 mole of HCl(g) is dissolved in 600 cm3 of 0.200 M CH3COOH(aq). Since HCl(aq) is a strong acid, the solution will become very acidic. In acidic medium, the ionization of CH3COOH(aq) can be neglected comparing with the ionization from HCl(aq). The effect of change in volume can be neglected when HCl(g) is being dissolved. 0.0500 mol [H+(aq)] = [HCl(aq)] = 0.600 dm-3 = 0.0830 mol dm-3 pH = - log 0.0830 = 1.08 II. Acid-base Equilibria c) 1. Part 4 Page 5 Effect of addition of acid / alkali into a buffer solution Calculate the pH of a buffer solution with 0.300 M of CH2ClCO2H(aq) and 0.200 M of CH2ClCO2-Na+(aq) respectively. Ka of CH2ClCO2H(aq) is 1.21 × 10-3 mol dm-3. [acid] 0.300 pH = pKa - log [salt] = - log (1.21 × 10-3) - log 0.200 = 2.74 2. Calculate the pH of the solution if 0.0500 mole of solid NaOH(s) is added into 600 cm3 of the buffer. no. of mole of CH2ClCOOH(aq) in 600 cm3 of the buffer = 0.300 M × 0.600 dm3 = 0.180 mol no. of mole of CH2COO-(aq) in 600 cm3 of the buffer = 0.200 M × 0.600 dm3 = 0.120 mol When 0.0500 mole of NaOH(s) is added, 0.0500 mole of CH2ClCO2H(aq) will be neutralized and 0.0500 mole of CH2ClCOO-(aq) will be produced. In the equilibrium mixture, no. of mole of CH2ClCOOH(aq) = (0.180 - 0.0500) mol = 0.130 mol no. of mole of CH2COO-(aq) = (0.120 + 0.0500) mol = 0.170 mol no. of mole of acid 0.130 pH = pKa - log no. of mole of salt = - log (1.21 × 10-3) - log 0.170 = 3.03 3. Calculate the pH of the solution if 0.0500 mole of HCl(g) is dissolved in 600 cm3 of the buffer. no. of mole of CH2ClCOOH(aq) in 600 cm3 of the buffer = 0.300 M × 0.600 dm3 = 0.180 mol no. of mole of CH2ClCOO-(aq) in 600 cm3 of the buffer = 0.200 M × 0.600 dm3 = 0.120 mol CH2ClCOOH(aq) d CH2ClCOO-(aq) + H+(aq) When 0.0500 mole of HCl(g) is dissolved, 0.0500 mole of CH2ClCOOH(aq) will be produced and 0.0500 mole of CH2ClCOO-(aq) will be consumed. In the equilibrium mixture, no. of mole of CH2ClCOOH(aq) = (0.180 + 0.0500) mol = 0.230 mol no. of mole of CH2COO-(aq) = (0.120 - 0.0500) mol = 0.070 mol no. of mole of acid 0.230 pH = pKa - log no. of mole of salt = - log (1.21 × 10-3) - log 0.070 = 2.40 d) Comparison of the action of acid and alkali Non-buffered Acid Buffer solution Original pH 2.74 2.74 Alkali added 4.6 3.03 Acid added 1.08 2.40 Comparatively, the pH of a buffer solution is more resistant to addition of alkali or acid. For a non-buffered acid, it contains no base to absorb the extra H+(aq) ions if acid is added. Furthermore, with [acid] the acid alone, addition of alkali will cause a big change in [salt] . This causes a relatively big change in pH. [acid] i.e. pH = pKa - log [salt] II. Acid-base Equilibria 3. Preparation of buffer solution Part 4 Page 6 Usually, a buffer solution is prepared by mixing a weak acid / base with its salt. Or it can be prepared by halfway titration. For example, if 1dm3 of 0.5 M CH3COOH(aq) requires 0.5 mole of NaOH(aq) for complete neutralization. 0.25 mole of NaOH(aq) can be added so that only half of the CH3COOH(aq) will be neutralized. Initial amount Final amount d CH3COOH(aq) + OH-(aq) ≈ 0.5 mole ≈ 0.25 mole ≈ 0.25 mole ≈ 0 mole CH3COO-(aq) + H2O(l) ≈ 0 mole ≈ 0.25 mole no. of mole of acid Since pH = pKa - log no. of mole of salt = pKa - log 1 = pKa This method is used to prepared a buffer solution with the pH equals pKa of the acid. For any pH not the same as pKa, the ratio of no. of mole of acid to no. of mole of salt can be adjusted to prepare a solution of desired pH. Glossary Past Paper Question buffer hydrolysis half-way titration 91 2A 2 c i ii 94 2A 3 c ii iii 96 2A 1 c i ii iii 97 2A 4 c ii 98 2A 1 a i ii 99 2A 4 c ii 91 2A 2 c i ii 2c A weak base MOH has an ionization constant Kb = 2.0 × 10-5 mol dm-3, where Kb = [M+][OH-]/[MOH]. Solution S is made up of 0.10 mol of MOH in 1.0 dm3, and a second solution T has 0.10 mol of MOH and 0.50 mol of MCl in 1.0 dm3. i Calculate the pH of solutions S and T. MOH d M+ + OHSolution S: Since MOH d M+ + OHinitial 0.1 0 0 at eqm. 0.1 - x x x x2 -5 ∴ Kb = 2 × 10 = 0.1 - x , assuming (0.1 -x) ≈ 0.1 1 mark ∴ x = 2 × 10-6 = 1.414 × 10-3 M pOH = -log(1.44 × 10-3) = 2.85 pH = 14 - 2.85 = 11.15 1 mark Solution T: Since MOH d M+ + OHinitial 0.1 0.5 0 at eqm. 0.1 - x 0.5 + x x x × (0.5 + x) -5 with 0.5 + x ≈ 0.5 and 0.1 - x ≈ 0.1 0.1 - x = 2 × 10 0.1 = 4 × 10-6 M 1 mark x = 2 × 10-5 × 0.5 -log (4 × 10-6) = 5.40 pH = 14 - 5.40 = 8.60 1 mark 4 II. Acid-base Equilibria ii Part 4 Page 7 4 C i 0.01 mol of a strong acid HX is added separately to the solutions S and T. The new pH value for solution S is 10.26; calculate that for solution T. What conclusion can you draw from this result? (Kw = 1.0 × 10-14 mol2dm-6) Solution T: + OHSince MOH d M+ Before addition of acid and hydrolysis 0.1 0.5 Upon addition of acid 0.1 - 0.01 0.5 + 0.01 0 Upon hydrolysis at eqm. 0.09 - x 0.51 + x x 1 mark x × (0.51 + x) -5 with 0.51 + x ≈ 0.51 and 0.09 - x ≈ 0.09 0.09 - x = 2 × 10 9 × 10-5 = 3.53 × 10-6 1 mark x=2× 51 -log (3.53 × 10-4) = 5.45 pH = 8.55 1 mark Solution T can resist the change in pH to a greater extent than solution S, it is generally known as a buffer. 1 mark Most candidates were not able to work out the final pH of solution T. 94 2A 3 c ii iii 3c Given: Ka for CH3(CH2)2COOH = 1.5 × 10-5 moldm-3 at 298K Calculate the pH of ii an aqueous solution of 0.050M CH3(CH2)2COONa and 0.050M CH3(CH2)2COOH; and Assuming [PrCOOH] = 0.050 M [PrCOO-] = 0.050 M 0.50 1.5 × 10-5 = [H+] × 0.50 [H+] = 1.5 × 10-5 M pH = 4.82 iii 1.0 dm3 of the solution in (ii) after the addition of 1.0 × 10-3 mol of solid NaOH. After addition of 1.0 × 10-3 moles solid NaOH [PrCOOH] = 0.050 - 1.0 × 10-3 = 0.049 M [PrCOO-] = 0.050 + 1.0 × 10-3 = 0.051 M [H+] (0.051) 1.5 × 10-5 = (0.049) [H+] = 1.44 × 10-5 M pH = 4.34 2 1 mark 1 mark 3 1 mark 1 mark 1 mark 96 2A 1 c i ii iii 1c i At 298 K the pH of 0.050 M CH3CH2COOH is 3.10. Calculate the Ka of CH3CH2COOH at 298 K. CH3CH2COOH d CH3CH2COO- + H+ [CH3CH2COO-][H+] Ka = [CH CH COOH] ½ mark 3 2 + ½ mark Assuming [CH3CH2COO ] = [H ] (10-3.10)2 Ka = (0.05 - 10-3.10) 1 mark = 1.26 × 10-5 mol dm-3 (1.25 – 1.30 × 10-5 mol dm-3) (Deduct ½ mark for no / wrong unit) ii Calculate the pH at 298 K of a solution which is 0.050 M with respect to CH3CH2COOH and to CH3CH2COONa. In the solution, [CH3CH2COO-] = [CH3CH2COOH] = 0.05 M K a = [ H +] ∴ pH = pKa 1 mark = 4.90 (4.88 – 4.90) 1 mark iii 5.0 × 10-4 mol of solid NaOH were added separately to (1) 100 cm3 of 0.050 M CH3CH2COOH and (2) 100 cm3 of the solution in (ii) above. Assuming that the change in volume upon the addition of NaOH is negligible calculate the pH at 298 K of the solution in each case. Comment on the difference in the pH change of the two solutions. (1) For 100 cm3 of 0.05 M CH3CH2COOH ½ mark [CH3CH2COO-] = 5.0 × 10-3 M 2 2 5 II. Acid-base Equilibria Part 4 [CH3CH2COOH] = 4.5 × 10-2 M [H+] (5.0 × 10-3) Ka = 4.5 × 10-2 + [H ] = 1.13 × 10-4 mol dm-3 (1.15 × 10-4 mol dm-3) pH = 3.95 (3.93 – 3.96) (2) For 100 cm3 of the solution in (ii), [CH3CH2COO-] = 5.5 × 10-2 M [CH3CH2COOH] = 4.5 × 10-2 M [H+] (5.5 × 10-2) Ka = 4.5 × 10-2 + [H ] = 1.03 × 10-5 mol dm-3 (1.05 × 10-5 mol dm-3) pH = 4.99 (4.97 – 5.00) 0.05 M CH3CH2COOH is not a buffer ∴ pH changes significantly or, the solution in (ii) is a buffer ∴ pH change is very small 97 2A 4 c ii 4c A solution is formed by mixing equal volumes of 0.20 M CH3CO2H(aq) and 0.20 M CH3CO2Na(aq). ii Calculate the concentration of each chemical species, excluding H2O, present in the solution at 298 K. (Ka of CH3CO2H = 1.76 × 10-5 mol dm-3 at 298 K) Page 8 ½ mark 1 mark ½ mark ½ mark 1 mark 1 mark (1 mark) 7 98 2A 1 a i ii 1a Solution A is 0.15 M lactic acid, and solution B is a mixture of equal volumes of 0.30 M lactic acid and 0.10 M aqueous sodium hydroxide solution. Note : (1) Lactic acid is a monprotic acid and its Ka at 298 K is 1.38 × 10-4 moldm-3. (2) For (ii) and (iii), you may assume that the volume changes are negligible. i Calculate the pH of A and of B at 298 K. ii A few drops of dilute hydrochloric acid are added to 50.0 cm3 of A and B. Compare the effect of such action on the pH of the two solutions. Explain your answer. 99 2A 4 c ii 4c ii Given the following materials and apparatus, describe a method to determine the dissociation constant, Ka, of HCO2H. approximately 0.1 M aqueous HCO2H, approximately 0.1 M aqueous NaOH, phenolphthalein indicator, titration apparatus, and a calibrated pH meter 9 II. Acid-base Equilibria Part 5 Page 1 Topic Reference Reading II. Acid-base Equilibria 6.2.6–6.2.7 Part 5 Assignment Advanced Practical Chemistry, John Murray (Publisher) Ltd., 68–69 Experiment – Acid-base titration using method of double indicator Reading A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 264–270 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 371–374 Chemistry in Context (5th ed.), Thomas Nelson and Sons Ltd., 331–334 Indicators Acid-base titration F. Theory of Indicator Acid-base indicator is a weak acid which has different colour at different pH. Actually, the conjugate acidbase pair of the acid have different colour. HIn(aq) + Colour I H2O(l) d H3O+(aq) + In-(aq) Colour II Kin = [H3O(aq)+][In-(aq)] [HIn(aq)] Syllabus Notes For example, methyl orange is a weak acid. The conjugate acid of methyl orange is red in colour while the conjugate base of methyl orange is yellow. Methyl orange (pKin of methyl orange = 3.7) H3C H3C Conjugate acid (red) O N NN S O OH + H2O H3C H3C Conjugate base (yellow) O N NN S O O + H3O+ - In acidic medium, the concentration of H+(aq) is high and the equilibrium position is lying on the left. Therefore, it will be red in colour in acidic medium. For a similar reason, it is yellow in colour in alkaline medium. Phenolphthalein is also a weak acid. Its conjugate acid is colourless while the conjugate base is red. Because methyl orange and phenolphthalein have different Kin, they change colour at different pH. Phenolphthalein (pKin of phenolphthalein = 9.3) OH O- HO C O C O + 3H2O HO C OH CO + + 2H3O O Conjugate acid (colourless) Conjugate base (red) II. Acid-base Equilibria Part 5 Page 2 However, a naked eye cannot distinguish colours accurately. It cannot notice any difference unless the intensity of one colour is ten times of another. For methyl orange, pKin = 3.7 HMe(aq) + H2O(l) d H3O+(aq) + Me-(aq) Kin = [H3O(aq)+][Me-(aq)] [HMe(aq)] [acid] Therefore, it will become red at pH = , i.e. [salt] = 10 [acid] pH = pKin - log [salt] = pKin - log 10 = 3.7 - 1 = 2.7 [acid] 1 The solution will become yellow at pH = , i.e. [salt] = 10 [acid] 1 pH = pKa - log [salt] = pKin - log 10 = 3.7 + 1 = 4.7 The pH range from 2.7 to 4.7 is called the working range of methyl orange. N.B. Working range of a pH indicator = pKin ± 1 Phenolphthalein works in the same way as methyl orange. It has pKin 9.3, therefore, the working range of phenolphthalein is from pH 8.3 to 10.3. Because indicator is actually an acid, only minimal amount of indicator should be added during a titration. So that the pH of the solution will not be affected. G. Acid-base titration Titration is a process for determining the volume of one solution required to react quantitatively with a given volume of another. One solution is added to the other, a small amount at a time until just sufficient has been added to complete the reaction. The equivalence point is determined as an end point. Titrations may be carried out by hand from a burette or automatically. 1. Difference between equivalence point and end point Equivalence point – the stage in a titration when the reactants and products are present in equivalent amounts according to the stoichiometry of the reaction. End point – the stage in a titration when a change (e.g. colour, pH, temp., conductivity changes) is observed indicating the end of the titration. The accuracy of a titration is depending on how close the end point (experimental value) is to the equivalence point (theoretical value). II. Acid-base Equilibria 2. Titration using pH meter Part 5 Page 3 About the equivalence point of most acid-base titration, the solution shows a great pH change upon addition of small amount of acid or base. By plotting a graph with the pH of the solution against the volume of titrant added, the end point can be determined. The graphy is usually called titration curve. For the titration between a weak acid and a weak base, plotting of a pH graph is not applicable because there is no abrupt pH change throughout the course. It is hard to decide where is the end point. II. Acid-base Equilibria 3. Titration using indicator Part 5 Page 4 a) Choosing of indicator Choosing of indicator has a great impact on the accuracy of titration. A wrongly chosen indicator will cause a large error. The selection is depending on the shape of the pH curve of titration. The indicator is chosen so that the abrupt change on the pH curve will fall across the working range of the indicator. The value of pKin should be close to the value of the pH of the solution at the end-point, so that the colour change occurs as closely as possible to the equivalence point. In the titration of strong acid versus strong base, the equivalence point will be at pH 7 because the salt formed will not hydrolyse in water. Although neither methyl orange nor phenolphthalein changes colour at pH 7, the colour will change with a volume very close to the equivalence point. Therefore, both methyl orange and phenolphthalein can be used in the titration between a strong acid and a strong base. Conversely, in the titration of weak acid versus strong base, the equivalence point will be higher than pH 7, because of hydrolysis of the salt. A-(aq) + H2O(aq) d HA(aq) + OH-(aq) Furthermore, the vertical portion of the pH curve will also be shifted to the higher pH range. If phenolphthalein is used, the end point will be very close to the equivalence point. However, if methyl orange is used, the colour will change far before the equivalence point is reached. And a large error will be caused. Therefore, only phenolphthalein can be used in the titration between a weak acid and a strong base. II. Acid-base Equilibria Part 5 Page 5 Litmus is usually not used in titration because the colour change is not as obvious as methyl orange and phenolphthalein. It is only used in the form of litmus paper where a white background is available. In a titration of a strong acid and weak base, methyl orange should be used. In a titration of a weak acid and a strong base, phenolphthalein should be used. In a titration of a strong acid and a strong base, either methyl orange or phenolphthalein can be used. pH value Indicator 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 methyl orange | red |orange| yellow | phenolphthalein | colourless |pale pink| red | For the titration between a weak acid and a weak base, no indicator is appropriate because there is no abrupt pH change throughout the course. The end point of a titration between a weak acid and a weak base can neither be detected by any indicator nor a pH meter. The end points are usually determined by thermometric titration or conductimetric titration. 4. Thermometric titration Since neutralization is exothermic, the temperature of the solution will increase until the equivalence point is reached. However, the addition of any cold excess titrant into the warm solution will cause a drop in temperature. According to this principle, by plotting a graph with the temperature versus the volume of the titrant added, the end point can be determined. The graph is composed of 2 straight lines. One representing the exothermic reaction and another represents the cooling. The end point is where the two straight lines intersect each other. N.B. According to the theory behind thermometric titration, the points on the graph should not be joined together by a smooth curve. II. Acid-base Equilibria 5. Conductimetric titration Part 5 Page 6 Depending on the nature of the acid, base and the salt formed, the conductivity / conductance of the solution will also change throughout the titration process. Conductivity of a solution depends on i) ionic concentration ii) mobility of the ions a) Strong acid vs Strong base If a strong acid is titrated against a strong base, the product will be water and the salt. For example, in the titration between NaOH(aq) and HCl(aq), NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq) At the beginning, the solution contains only H+(aq) and Cl-(aq). Upon neutralization, a pair of H+(aq) and OH-(aq) ions is converted to a H2O(l) molecule. Consequently H+(aq) in the acid is replaced by Na+(aq) which has lower conductivity. Therefore, the conductivity will drop before the end point. Beyond the end point, addition of a strong base (a strong electrolyte) increases the ionic concentration, thus the conductivity. Furthermore, the excess OH-(aq) ions are more conductive. The end point is determined by extrapolating the two sessions of the graph with 2 straight lines. b) Weak acid vs Strong base At the beginning, the acid solution contains mainly undissociated acid molecules and the conductance is minimal. The addition of a strong base will convert the molecular acid molecules to ionic salt. Thus, the conductivity will increase. Beyond the end point, the conductivity will increase even further upon the addition of excess strong base. c) Weak acid vs Weak base However, if the titration is between a weak acid and a weak base HA(aq) + BOH(aq) → H2O(l) + A-(aq) + B-(aq) Two covalent molecule reacts to form 1 water molecule and 2 ions. The increases in ionic concentration will cause an increase in conductivity, Beyond the end point, the addition of the weak acid will only dilute the solution and the conductivity will decreases. N.B. Sometimes, it is difficult to predict the actual shape of the graph but a sudden change in slope can be treated as the existence of end point. Glossary indicator methyl orange phenolphthalein Kin naked eye working range titration equivalence point end point titration curve (pH curve) indicator thermometric titration conductimetric titration conductivity / conductance ionic concentration extrapolating II. Acid-base Equilibria Part 5 Page 7 Past Paper Question 90 1A 3 a 92 2A 3 c i 94 2A 3 b ii 95 1B 4 e i ii 98 2A 1 a iii 99 1A 4 b i ii 99 1A 4 c 90 1A 3 a 3a Explain why phenolphthalein turns pink in a solution of sodium carbonate, but remains colourless in a solution of sodium hydrogencarbonate. CO32- and HCO3- hydrolyse to a different extent. ½ mark ½ mark In a solution of Na2CO3, CO32- + H2O d HCO3- + OHresulting in a pH sufficiently high ( and [H+] sufficiently low) 1 mark such that the equilibrium, In(red) + H+ d InH+(colourless) lies to the left and a pink colour is exhibited. ½ mark In a solution of NaHCO3, whose pH is also above 7, [H+] is sufficiently high for the equilibrium to shift to the right. C Few candidates were able to answer that a solution of Na2CO3 has a higher pH value than that of a solution of NaHCO3 because the hydrolysis of CO32- and HCO3- takes place to a different extent; and that at the higher pH, phenolphthalein exists in a form which has a resonance structure and the solution becomes pink. 92 2A 3 c i 3c i Explain the difference between the equivalence point and the end point of a titration. Equivalence point is the point in the titration where moles of acid equals moles of base, corresponding to a salt solution. The end point is the point in a titration where a particular indicator changes colour which occurs at pH 2 marks close to the value of pKa for the indicator. 94 2A 3 b ii 3b Account for each of the following: ii At 298K, in a solution of pH 7.0, the indicator methyl orange shows its alkaline colour (yellow), while phenolphthalein shows its acidic colour (colourless). Acid-base indicators are weak acids / bases. The dissociation of which can be represented by 1 mark HIn(aq) + H2O(l) d H3O+(aq) + In-(aq) The colour of an indicator depends on the relative concentrations of the acidic form, HIn and the alkaline form In½ mark (aq) which are of different colours. The dissociation constant Ki of different acid-base indicators are different, thus they change colour over different pH range. The pH range of methyl orange is below 7, while that of phenolphthalein is above 7. ∴ at pH 7, methyl orange shows its alkaline colour, while phenolphthalein shows its acidic colour. 1½ mark C ii Very few candidates mentioned the fact that an acid-base indicator is a weak acid or weak base. They also failed to relate the colour of the indicator to its pH range. 95 1B 4 e i ii 4e For each of the volumetric analyses (i) to (iii), state whether an indicator is required. If an indicator is required, select the appropriate one from those given below: litmus; methyl orange; phenolphthalein; potassium dichromate(VI) solution; starch solution i ethanedioic acid titrated with sodium hydroxide phenolphthalein 1 mark ii sulphuric(VI) acid titrated with aqueous ammonia methyl orange 1 mark 98 2A 1 a iii 1a Solution A is 0.15 M lactic acid, and solution B is a mixture of equal volumes of 0.30 M lactic acid and 0.10 M aqueous sodium hydroxide solution. Note : (1) Lactic acid is a monprotic acid and its Ka at 298 K is 1.38 × 10-4 moldm-3. (2) For (ii) and (iii), you may assume that the volume changes are negligible. 3 iii 50.0 cm of A and of B are titrated with a strong base. Compare the pH of the two solutions at the respective equivalence points of the titrations. Explain your answers. 99 1A 4 b i ii c 4b The graph below shows the variation of pH when 25.0 cm3 of 0. 10 M HCl(aq) is titrated against 0. 10 M NaOH(aq). 2½ 2 3 1 1 II. Acid-base Equilibria Part 5 Page 8 i ii On the above graph, sketch a curve to represent the variation of pH when 25.0 cm3 of 0. 10 M CH3CO2H(aq) is titrated against 0. 10 M NaOH(aq). From the table below, choose an appropriate indicator for the titration in (i). Explain your choice. Indicator pH range of colour change bromocresol green 3.8 - 5.4 bromothymol blue 6.0 - 7.6 thymolphthalein 8.3 - 10.6 99 1A 4 c 4c Given an aqueous solution of Na2CO3 and NaHCO3, suggest how to determine the concentrations of the two substances by titrimetric method using indicators. III. Redox Equilibria Part 1 Page 1 Topic Reference Reading III. Redox Equilibria 6.3.1 A-Level Chemistry Syllabus for Secondary School (1995), Curriculum Development Council, 188–189 Part 1 Assignment A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 74–80 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 21–29 Reading Syllabus Notes Redox Equilibria Faraday and mole III. Redox Equilibria Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 186–187 Redox equilibrium is similar to Acid-base equilibrium. An acid-base equilibrium can be considered as a competition of donation and acceptance of proton. e.g. Conjugate base NH3(aq) + + proton H+(aq) d d Conjugate acid NH4+(aq) Upon donation of proton, a conjugated acid will be converted to a conjugate base. Redox equilibrium can also be considered as a competition of donation and acceptance of electron. e.g. Oxidizing agent + + Mg2+(aq) electron 2e-(aq) d d Reducing agent Mg(s) Upon donation of electron, a reducing agent will be converted to an oxidizing agent. A. Redox reaction A complete redox reaction consists of two systems of redox equilibria, thus two half ionic equation. One half ionic equation represents the reduction and another represents oxidation. Reduction (acceptance of electron) Oxidation (donation of electron) Overall redox reaction 1. Balancing of redox reaction In a balanced equation, the no. of atoms and the no. of charges must be balanced at the two sides of the equation. Furthermore, in a redox reaction, the no. of electrons accepted by oxidizing agent must be the same as the no. of electrons donated by reducing agent. Therefore, a balanced redox reaction can be constructed in two ways a) By combining balanced half-ionic equation b) By the change in oxidation no. Ag+(aq) + e-(aq) d Ag(s) Cu(s) d Cu2+(aq) + 2e2Ag+(aq) + Cu(s) d 2Ag(s) + Cu2+(aq) III. Redox Equilibria Part 1 Page 2 a) By combining balanced half-ionic equation (1) Steps of writing Balanced Half ionic equation 1. 2. Write down the reactant and the product Cr2O72-(aq) → Cr3+(aq) Mass balance – by adding something already existed. Since the reaction is usually conducted in water medium, equation may be balanced by i. Adding H2O(l) for lack of O ii. Adding H+(aq) for lack of H (from dissociation of water, H2O d H+ + OH-) 2Cr2O7 (aq) → 2Cr3+(aq) (Balance the no. of Cr) 2Cr2O7 (aq) → 2Cr3+(aq) + 7H2O(l) (Add H2O(l) for lack of O) 14H+(aq) + Cr2O72-(aq) → 2Cr3+(aq) + 7H2O(l) (Add H+(aq) for lack of H) 3. Charge balance – Add e- to balance the charge → 2Cr3+(aq) + 7H2O(l) 14H+(aq) + Cr2O72-(aq) (14 × +1) + (-2) = +12 (2 × +3) + (0) = +6 14H+(aq) + Cr2O72-(aq) + 6e→ 2Cr3+(aq) + 7H2O(l) (14 × +1) + (-2) + (6 × -1) = +6 (2 × +3) + (0) = +6 4. Check once again → 2Cr3+(aq) + 7H2O(l) 14H+(aq) + Cr2O72-(aq) + 6eMass: 14H + 2Cr + 7O 2Cr + 14H + 7O Charge: (14 × +1) + (-2) + (6 × -1) = +6 (2 × +3) + (0) = +6 Balancing of half-ionic equation can only be started with correct reactant and product, which can only be memorized. Note: (2) Combining half-ionic equations In a redox reaction, an oxidizing agent must react with a reducing agent. i.e. Oxidizing agent does not react with another oxidizing agent. By combining an oxidation half equation and a reduction half equation, the ionic equation can be reconstructed. (First of all, the half ionic equations must be balanced.) For example, Reduction half ionic equation Ag+(aq) + e- → Ag(s) (Oxidizing agent: Ag+(aq)) Oxidation half ionic equation Cu(s) → Cu2+(aq) + 2e- (Reducing agent: Cu(s)) Since the no. of e- gained by Ag+(aq) must equal the no. of e- lost by Cu(s), ∴ 2 Ag+ are needed for each Cu. Reduction half ionic equation Oxidation half ionic equation → 2Ag(s) Cu(s) → Cu2+(aq) + 2e–––––––––––––––––––––––––––––––––––––––– 2Ag+(aq) + 2e- + Cu(s) → 2Ag(s) + Cu2+(aq) + 2e(left and right must have the same no. of e-) 2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq) 2Ag+(aq) + 2e- Overall ionic equation III. Redox Equilibria b) By the change in oxidation no. Part 1 Page 3 Since O.N. is the difference between the no. of electron and no. of proton associated with an atom, an increase in O.N. by 1 is equivalent to losing of 1 e-. Similarly, a decrease in O.N. by 1 is equivalent to gaining of 1 e-. Oxidizing agent MnO4-(aq) +7 → Mn2+(aq) +2 Each Mn gains 5 electrons Each Fe loses 1 electron Reducing agent Fe2+(aq) +2 → Fe3+(aq) +3 By the fact that the no. of electron gained by the oxidizing agent must be the same as the no. of electron lost by the reducing agent, each MnO4-(aq) must reacts with 5 Fe2+(aq). Therefore, the equation will become MnO4-(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) H2O(l). MnO4-(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) H+(aq). MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) again mass 1 Mn 4O 5 Fe 8H charge (-1) + 5×(+2) + 8×(+1) = +17 mass 1 Mn 4O 5 Fe 8H charge (+2) + 5×(+3) = +17 No. of O is not balanced, so add some No. of H is not balanced, so add some Check the no. of atoms and charges The equation, MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l), is balanced. This method is particularly useful in answering M.C. question where the completely balanced equation is usually not required. 2. Faraday and mole In counting the no. of atom, mole is used as the counting unit and method of counting by weighing is used. In counting the no. of electron, method of counting charges is used. The amount of charges carried by 1 mole of electrons is called 1 Faraday of charges (1 F). 1 Faraday of charges is found to be 96500 Coulombs (96500 C). 6.02 × 1023 electron charges ≡ 1 F of charges ≡ 96500 C 96500 1 1 electron charge ≡ 6.02 × 1023 F ≡ 6.02 × 1023 C 1 electron charge ≡ 1.66 × 10-24 F ≡ 1.60 × 10-19 C By measuring the size of the current and the time taken, the amount of charges flowing through a circuit can be measured. Charges (Q) = Current (I) × Time (t) If the current is measured in ampere (A) and the time is measured in second (s), the unit of charge will be coulomb (C). Charges (Q) = 1C = 1C = Current (I) × Time (t) 1A×1s 1 As III. Redox Equilibria 3. Part 1 Calculation of mass liberated in electrolysis In the electrolysis of CuSO4(aq) using graphite electrodes, Cu(s) will deposit on the cathode and O2(g) will evolve at the anode. By measuring the size of current and the time taken of electrolysis, the amount of Cu(s) and O2(g) can be calculated. In the circuit, rheostat and ammeter must be installed to adjust and measure the current. For example, if a beaker of CuSO4(aq) is electrolyzed at 200 mA for 20 minutes, what will be the mass of Cu(s) deposited on the cathode and the volume of oxygen evolved at the anode. All volume is measured at standard temperature and pressure. (Given : Relative atomic mass Cu : 63.55; O : 16.00 ) Molar volume of gas at s.t.p. is 22.4 dm3 At cathode At anode Cu2+(aq) + 2e- → Cu(s) 4OH-(aq) → O2(g) + 2H2O(l) + 4e- Page 4 By the relationship Q = It, No. of coulombs flowing through the circuit = Current in ampere × time in second 200 = 1000 A × (20 × 60) s = 240 C 240 charges in coulomb No. of mole of electron flowing through the current = Faraday constant = 96500 C = 2.49 × 10-3 mol Formation of each mole of Cu(s) requires 2 moles of electron, therefore, no. of mole of Cu(s) formed = no. of mole of electron 2.49 × 10-3 mol = = 1.25 × 10-3 mol 2 2 Mass of Cu(s) deposited = no. of mole of Cu(s) × molar mass of Cu = 1.25 × 10-3 mol × 63.55 g mol-1 = 0.0794 g Similarly, formation of each mole of O2(g) requires losing of 4 mole of electron. no. of mole of O2(g) formed = Volume of O2(g) at s.t.p. no. of mole of electron 2.49 × 10-3 mol = = 6.23 × 10-4 mol 4 4 = no. of mole of oxygen × molar volume of gas at s.t.p. = 6.23 × 10-4 mol × 22.4 dm3mol-1 = 0.0140 dm3 = 14.0 cm3 reducing agent conservation of mass and charge Glossary redox equilibrium oxidizing agent faraday (F) coulomb (C) III. Redox Equilibria Part 1 Page 5 Past Paper Question 91 2A 2 b i ii 92 2A 2 a i 93 1A 1 d i ii 93 2B 6 b 94 1A 1 d e i ii 95 1A 2 b i 97 1A 1 b 99 2A 4 b i ii 93 2A 3 a i ii 94 2B 4 e i ii 95 2B 4 b i ii 91 2A 2 b i ii 2b i CrO42- ions react with S2O32- ions to form Cr(OH)4- and SO42- ions in basic medium. Write balanced equations for 3 each half reaction and for the overall reaction. 1 mark CrO42- + 4H2O + 3e- → Cr(OH)4- + 4OHS2O32- + 10OH- → 2SO42- + 5H2O + 8e1 mark overall equation : 1 mark 8CrO42- + 3S2O32- + 17H2O → 8Cr(OH)4- + 6SO42- + 2OHii What volume of 0.50M Na2CrO4 solution is needed to react completely with 40.0 cm3 of 0.20M Na2S2O3 solution 2 in basic medium? No. of mole of S2O32- = 0.2 × 40 × 10-3 Since 3 moles of S2O32- ≡ 8 moles of CrO421 mark 8 -3 2∴ no. of moles of CrO4 = 0.2 × 40 × × 10 3 8 0.2 × 40 × × 10 −3 3 Volume of Na2CrO4 = = 0.04267 dm3 = 42.67 cm3 1 mark 0.50 (no unit -½) (more than 2 decimal places -½) Ci Some candidates failed to recognize that the given reaction takes place in basic medium. ii Most candidates were not able to write balanced equations for the half-reactions which take place in basic medium. 92 2A 2 a i 2a In acid solution, chlorate(V) ions, ClO3-, slowly oxidize chloride ions to chlorine. The following kinetic data are obtained at 25°C. [Cl-]/mol dm-3 [H+]/mol dm-3 Initial rate/mol dm-3 [ClO3-]/mol dm-3 0.08 0.15 0.20 1.0 × 10-5 0.08 0.15 0.40 4.0 × 10-5 0.16 0.15 0.40 8.0 × 10-5 0.08 0.30 0.20 2.0 × 10-5 i Write the balanced equation for this reaction. The balanced equation for this reaction ClO3- + 5Cl- + 6H+ → 3Cl2 + 3H2O 1 mark Ci Weaker candidates were not able to give the balanced equation. 93 1A 1 d i ii 1d Write balanced equations for: i the reaction between MnO2(s) and PbO2(s) in acidic solution to give MnO4-(aq) and Pb2+(aq). 2MnO2(s) + 3PbO2(s) + 4H+(aq) d 2MnO4-(aq) + 3Pb2+(aq) + 2H2O(l) 1½ mark ii the reaction between H2O2(aq) and Cr(OH)4-(aq) in alkaline solution to give CrO42-(aq). 3H2O2(aq) + 2Cr(OH)4-(aq) + 2OH-(aq) d 2CrO42-(aq) + 8H2O(l) 1½ mark C A surprisingly large number of candidates could not do the balancing of equation for redox reactions which should be basic work for A-Level candidates. 1 1½ 1½ III. Redox Equilibria Part 1 93 2A 3 a i ii 3a i Write balanced half equations for the redox reaction of Crn+ and MnO4- in acidic solution. In terms of n, how many moles of electrons per mole of Crn+, are involved in the oxidation reaction? Write the balanced overall chemical equation for the above reaction. The half reactions 7H2O + 2Crn+ → Cr2O72- + 14H+ + (12 - 2n)e2 marks MnO4- + 8H+ + 5e- → Mn2+ + 4H2O 1 mark 1 mark No. of moles of e- per Crn+ is (6 - n) Overall equation: 35H2O + 10Crn+ + (12 - 2n)MnO4- + (12 - 2n)8H+ → 5Cr2O72- + 7OH- + (12 - 2n)Mn2+ + (12 -2n)4H2O or 1 mark 10Crn+ + (12 - 2n)MnO4- + (26 - 16n)H+ → 5Cr2O72- + (12 - 2n)Mn2+ + (13 - 8n)H2O ii An acidified solution containing 1.50 × 10-3 mol of Crn+ is titrated with 0.0250 M KMnO4 solution. The equivalence point of the reaction is reached after 48.00 cm3 of KMnO4 has been added. Calculate the value of n. No. of moles of Crn+ is 0.00150 48.00 × 0.0250 = 0.00120 1 mark No. of moles of MnO4- is 1000 1 mark 5 moles of Crn+ ≡ 4 moles of MnO410 5 1 mark 12 - 2n = 4 n=2 1 mark C This part was unsatisfactorily answered. Many candidates started with wrong equations so that they were not able to obtain the correct final answer. The balancing of equations for the redox reactions should be given more attention. 93 2B 6 b 6b In aqueous solution, 1 mole of iodine reacts with 2 moles of thiosulphate ions, whereas 4 moles of bromine reacts with 1 mole of thiosulphate ions. Write balanced ionic equations for the two reactions. Predict the reaction between chlorine and thiosulphate ions in aqueous solution. Explain your prediction. 1 mark Equation for the reaction between I2 and S2O32- is I2 + 2S2O32- → 2I- + S4O62for the reaction between 4Br2 and S2O32-, 4Br2 + S2O32- + 5H2O → 8Br- + 2SO42- + 10H+ 2 mark Chlorine will react similarly as bromine because Cl2 is a stronger oxidising agent when compared with Br2 and +6 (in SO42-) is the maximium oxidation state for S. 2 marks C Few candidates were able to construct the equation for the reaction of bromine and thiosulphate ions. Some candidates suggested that chlorine could oxidize thiosulphate ions to peroxodisulphate(VI); or that chlorine was a stronger reducing agent. 94 1A 1 d e i ii 1d An aqueous solution of titanium (Ti) salt was electrolysed by passing a current of 5.00 A for 2.50 hours. As a result, 5.60 g of metallic Ti were deposited at the cathode. Deduce the charge on the Ti ion in the solution. Give : 1 faraday = 96500 Cmol-1 Tin+ + ne- → Ti i.e. 1 mole of Ti produced requires n moles of e5.60 no. of moles of Ti liberated = 47.90 = 1.169 × 10-1 1 mark 5.00 × 2.50 × 60 × 60 no. of moles of e- passed = = 4.663 × 10-1 1 mark 96550 -1 4.663 × 10 ∴ n = 1.169 × 10-1 = 3.989 ≈ 4 1 mark 1e Write a balanced equation for each of the following reactions. i The reaction between BrO3-(aq) and Br-(aq) in acidic solutions to give Br2(aq). 1½ mark BrO3-(aq) + 5Br-(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l) ½ mark for completing the equation 1 mark for balancing the equation ii The reaction between MnO4-(aq) and SO2(g) in alkaline solution to give MnO2(s) and SO42-(aq). 1½ mark 2MnO4-(aq) + 3SO2(g) + 4OH-(aq) → 2MnO2(s) + 3SO42-(aq) + 2H2O(l) ½ mark for completing the equation 1 mark for balancing the equation C Generally well-answered, except that some candidates still put electrons in the overall equation. Page 6 5 4 5 3 1½ 1½ III. Redox Equilibria Part 1 Page 7 94 2B 4 e i ii 4e In concentrated HCl, 1 mole of hydrazine H2NNH2 reacts with 1 mole of potassium iodate(V) to give 1 mole of iodine monochloride ICl with the evolution of a colourless gas. i Deduce, the oxidation state of nitrogen in the reaction product, and hence suggest what the colourless gas might 2 be. Q NH2NH2 : IO3- = 1 : 1 In the reduction of IO3-, O.S. of I decreases from +5(IO3-) to +1(ICl) ∴ In the oxidation of NH2NH2, 4 moles of e- should be released from each mole of NH2NH2. 1 mark In 1 mole of NH2NH2, there are 2 moles of N atoms. ∴ O.S. of N in the reaction product is 0. ½ mark ½ mark The colourless gas in probably N2. ii Write balanced half-equations and a balanced overall equation for the reaction. 3 1 mark IO3- + 5H+ + HCl + 4e- → ICl + 3H2O NH2NH2 → N2 + 4H+ + 4e1 mark IO3- + H+ + HCl + NH2NH2 → ICl + N2 + 3H2O 1 mark OR 1 mark IO3- + 5H+ + HCl + 4e- → ICl + 3H2O NH3NH32+ → N2 + 6H+ + 4e1 mark IO3- + HCl + NH3NH32+ → ICl + N2 + 3H2O + H+ 1 mark C This question demanded application of knowledge to an unfamiliar situation and was poorly-answered. Only a few candidates worked out the identity of the colourless gas from the change in oxidation states of nitrogen. Some candidates gave I+, instead of ICl, as the product in the reduction of IO3-. 95 1A 2 b i 2b i What is the essential feature of a "redox reaction" ? “Redox reaction” : a reaction involving the transfer of electron(s) from one reactant to another 1 mark OR A reaction involving a change in oxidation state / oxidation no. 1 mark Ci Generally well answered, except that some candidates did not point out the essential feature of a redox reaction. They just wrote that a redox reaction involves reduction and oxidation. 95 2B 4 b i ii 4b Give the oxidation states of iodine in the two equations below. Balance the equation in each case. i S2O32- (aq) + I2(aq) → S4O62- (aq) + I-(aq) Oxidation state of I in I2 = 0 Oxidation state of I in I- = -1 2S2O32- (aq) + I2(aq) → S4O62- (aq) + 2I-(aq) ii I-(aq) + IO3-(aq) + H+(aq) → I2(aq) + H2O(l) Oxidation state of I in I- = -1 Oxidation state of I in IO3- = +5 / V Oxidation state of I in I2 = 0 5I-(aq) + IO3-(aq) + 6H+(aq) → 3I2(aq) + 3H2O(l) C ii Many candidates could not give the correct oxidation state of iodine in IO3-. 1 2 1 mark 1 mark 2 1 mark 1 mark 97 1A 1 b 1b In a nickel-plating experiment, after passing a current of 5.0 A through an electroplating bath containing a nickel compound for 5.5 minutes, 0.50 g of metallic nickel was deposited at the cathode. Assuming that the current efficiency is 100%, deduce the oxidation state of nickel in the compound. 99 2A 4 b i ii 4b In an electrolysis experiment, inert electrodes were used and a constant current of 1.00 A was passed for 1800 s through two electrolytic cells connected in series. i One of the cells contained an aqueous solution of silver(I) nitrate(V) and 2.02 g of silver was deposited on the cathode. Deduce the Faraday constant and the charge carried by an electron. ii The other cell contained an aqueous solution of a gold salt and 1.23 g of gold was deposited on the cathode. Deduce the oxidation state of gold in the gold salt. (Assume 100% current efficiency in this experiment.) 3 III. Redox Equilibria Part 2 Page 1 Topic Reference Reading III. Redox Equilibria 6.3.2 Advanced Practical Chemistry, John Murray (Publisher) Ltd., 108–109 Physical Chemistry (3rd ed.), P.W. Atkins, 259–263 Part 2 Assignment A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 280–281 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 209 Reading Syllabus Electrochemical cells Measurement of e.m.f. Use of salt bridge Cell diagrams (IUPAC conventions) B. Electrochemical cells Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 188, 278–280 Notes If a reducing agent is mixed with an oxidizing agent, the electrons from the reducing agent flow to the oxidizing agent directly. The chemical energy of the two reagents will be converted to heat energy. e.g. Zn(s) strong reducing agent + Cu2+(aq) strong oxidizing agent → Zn2+(aq) weak oxidizing agent + Cu(s) weak reducing agent However, if a reducing agent is connected to an oxidizing agent through an external circuit and completed with a salt bridge, the electrons from the reducing agent will flow through the external circuit. In this situation, the chemical energy is converted to electrical energy. And an electrochemical cell is constructed. III. Redox Equilibria 1. Measurement of e.m.f. Part 2 Page 2 e.m.f. of a cell is defined as the potential difference across a cell when the circuit is open. That means the current is zero. If the current is not zero, owing to the internal resistance of a cell, the potential difference measured will be lower than the e.m.f. a) Potentiometer Potentiometer is the best device to measure the e.m.f. of a cell because the current flowing through the cell to be measured would be exactly zero. A standard accumulator is used as the reference in the potentiometer and the circuit is connected as the diagram shown. The sliding contacted is slided along the potentiometer wire until zero reading is attained on the galvanometer. The e.m.f. of the cell will be proportional to the potentiometer wire and given by the equation XB e.m.f. of the cell = e.m.f. of the accumulator × XY However, the setup and operation of a potentiometer is not very convenient. A voltmeter or multimeter may be used instead. b) High impedance voltmeter / Digital multimeter High impedance voltmeter or digital multimeter is not as good as potentiometer because the current flowing through the cell would not be zero. Although voltmeter and multimeter are not the ideal instruments for e.m.f. measurement, the operations of them are very simple. Moreover, both of them have very high resistance and the current flowing through them will be very minimal. The error will be within an acceptable level. 2. Use of salt bridge Salt bridge is made by a thread of cotton soaked in saturated KNO3(aq) or NH4Cl(aq) solution. It is used to complete the circuit when the two half cells are connected together. a) Requirement of a salt bridge Not all kind of strong electrolytes can be used as a salt bridge. Criteria of a salt bridge 1. The electrolyte must have no reaction with the chemicals in the cell. For example, NH4Cl(aq) salt bridge should not be used with Ag(s) | Ag+(aq) system, otherwise precipitation will occur. Ag+(aq) + Cl-(aq) → AgCl(s) 2. The cation and anion in the salt bridge should have the same migration speeds. If they have different speeds, e.m.f. of the cell will be affected by the choice of salt bridge. For this reason, KNO3(aq) and NH4Cl(aq) are the only two commonly used salt bridges. III. Redox Equilibria Part 2 Page 3 b) An electrochemical cell does not need salt bridge Salt bridge can be replaced by a porous partition if accurate supply of e.m.f. is not required. The tiny holes allows the ions to diffuse slowly but prevent the solutions from mixing. Although the setup is more convenient than the one with the salt bridge, it is not as good as salt bridge for two reasons. 1. 2. The difference in diffusion speed of the ions in the two partitions affects the e.m.f. of the cell. After a long period of time, the two electrolytes may mix together and an undesirable reaction may take place. Daniel cell invented in 1837 Moreover, some electrochemical cells even do not require the used of a porous partition. For the redox reaction 2AgCl(s) + H2(g) d 2Ag(s) + 2HCl(aq) Reduction Oxidation AgCl(s) + e- d Ag(s) + Cl-(aq) H2(g) d 2H+(aq) + 2e- Overall reaction 2AgCl(s) + H2(g) d 2Ag(s) + 2Cl-(aq) + 2H+(aq) In this example the two half cells share the same electrolyte, HCl(aq), therefore, neither salt bridge nor porous pot is required. 3. Cell diagrams (IUPAC conventions) Instead of drawing the actual diagram of an electrochemical cell, IUPAC has an agreed convention for writing an notation, called cell diagram, to represent a cell. In the notation, | || or MM M , + N.B. phase boundary (boundary between solid, liquid or gaseous phase) salt bridge porous partition used to separate different species in the same phase used to enclose different species in the same chemical system used between different species in the same chemical system The rule of writing of cell diagram is not very strict, it may be possible to construct more than 1 diagram for 1 electrochemical cell. III. Redox Equilibria Example 1 Part 2 Example 2 Example 3 Page 4 Zn(s) | Zn2+(aq) M Cu2+(aq) | Cu(s) Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) Pt(s) | H2(g) | HCl(aq) | AgCl(s) | Ag(s) Example 1 For the Daniel cell with the porous partition Zn(s) | Zn2+(aq) M Cu2+(aq) | Cu(s) e.m.f. of the cell = + 1.10 V By convention, the sign of e.m.f. of a cell is given by the polarity of the electrode at the right in a cell diagram. Therefore, if the cell diagram is written as Cu(s) | Cu2+(aq) M Zn2+(aq) | Zn(s), the e.m.f of the cell will become - 1.10 V. Example 2 For the Daniel cell with salt bridge Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) e.m.f. of the cell = + 1.10 V III. Redox Equilibria Example 3 Pt(s) | H2(g) | HCl(aq) | AgCl(s) | Ag(s) Part 2 Page 5 e.m.f of the cell = + 0.22 V Sometimes " | " and " , " can be used interchangeably depending on how do you define a phase. AgCl(s) and Ag(s) can be considered as two different phases because they are different substances. Therefore it is written as above. Example 4 Electrochemical cell is not limited to metal | metal ion system. For those system doesn't involve metal, an inert electrode is required. Graphite and platinum are the two commonly used inert electrodes. For the system doesn't involve metal, the cell diagram is written in this way (Pt) | RedL, OxL || RedR, OxR | (Pt) Red : Reduced form Ox : Oxidized form The two half-ionic equations involved in the cell on the right are Reduction half equation Oxidation half equation MnO4-(aq) + 8H+(aq) + 5e- d Mn2+(aq) + 4H2O(l) Fe2+(aq) d Fe3+(aq) + e- The cell diagram is written as C(graphite) | Fe2+(aq) , Fe3+(aq) || [Mn2+(aq) + 4H2O(l)] , [MnO4-(aq) + 8H+(aq)] | C(graphite) RedL Fe(II) N.B. Ox L Fe(III) RedR Mn(II) Ox R Mn(VII) " , " is used to used to separate different species in the same phase " [ ] " is used to enclose used to enclose different species in the same chemical system Glossary electrochemical cell e.m.f potential difference potentiometer potentiometer wire accumulator galvanometer impedance multimeter salt bridge migration speed porous partition (pot) Daniel cell IUPAC cell diagram convention phase boundary inert electrode half-cell half-ionic equation 90 2A 1 b i ii iv 95 2A 3 c i ii iii 96 2A 3 a ii 97 2A 4 a ii iii Past Paper Question III. Redox Equilibria Part 2 Page 6 90 2A 1 b i ii iv 1b A cell based on the reaction: 6Fe2+(aq) + Cr2O72-(aq) + 14H+(aq) → 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) consists of one platinum electrode dipped into a beaker containing an acidified solution of K2Cr2O7(aq), and another platinum electrode dipped into a separate beaker containing FeSO4(aq), with the two solutions being connected by a salt bridge. With reference to this cell: i write balanced equations for the reactions occurring at the cathode and the anode; 2 Reduction occurs at the cathode Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l) 1 mark Oxidation occurs at the anode 1 mark Fe2+(aq) → Fe3+(aq) + eii state the direction in which electrons move when the electrodes are connected externally; 1 From anode to cathode 1 mark iv represent the cell using the I.U.P.A.C. convention 2 Pt|Fe2+(aq),Fe3+(aq)||Cr3+(aq),[Cr2O72-(aq),H+(aq)]|Pt 2 mark Ci Many candidates did not give state symbols in the half equation. 95 2A 3 c i ii iii 3c The electromotive force of a new zinc-carbon dry cell is 1.5 V. When it is producing an electric current, the following changes occur at the two electrodes: Anode : Zn(s) reacts to give Zn2+(aq). Cathode: MnO2(s) and NH4Cl(aq) react to give Mn2O3(s) and NH3(g). i Write half equations for the reactions at the anode and at the cathode, and the equation for the overall reaction that 3 occurs in the dry cell. 1 mark At anode Zn(s) → Zn2+(aq) + 2e+ At cathode 2MnO2(s) + 2NH4 (aq) + 2e → Mn2O3(s) + 2NH3(g) + H2O(l) 1 mark or 2MnO2(s) + NH4+(aq) + 2e- → Mn2O3(s) + OH-(aq) + NH3(g) Overall equation 1 mark Zn(s) + 2MnO2(s) + 2NH4+(aq) → Zn2+(aq) + Mn2O3(s) + 2NH3(g) + H2O(l) or Zn(s) + 2MnO2(s) + NH4+(aq) → Zn2+(aq) + Mn2O3(s) + NH3(g) + OH-(aq) ii Write the cell diagram for the dry cell, using the IUPAC convention. 2 Zn(s) | Zn2+(aq) | MnO2(s) | Mn2O3(s) | C(s) 2 marks or Zn(s) | Zn2+(aq) M NH4+(aq) | MnO2(s) | Mn2O3(s) | C(s) or Zn(s) | Zn2+(aq) | [MnO2(s) + NH4+(aq)], [Mn2O3(s) + OH-(aq)] | C(s) or Zn(s) | Zn2+(aq) | [MnO2(s) + 2NH4+(aq)], [Mn2O3(s) + 2NH3(aq)] | C(s) (1 mark for the correct species at cathode and anode ½ mark for using vertical solid lines to indicate phase difference, ½ mark for using vertical double line (dotted line) to represent salt bridge or using vertical single line to represent porous partition) iii Explain why the electromotive force of the dry cell drops, 3 (1) after it has been used for some time; (2) after it has been stored for a long time without being used. (1) If a current is drawn for some time, NH3(g) will accumulate at the cathode, the equilibrium will shift to the left, 1½ mark leading to a drop in electrode potential. OR If a current is drawn for some time, Zn2+(aq) will accumulate at the anode, the equilibrium will shift to the left, making the electrode potential of Zn | Zn2+ less negative. (2) ½ mark If the cell is allowed to stand for some time, NH4+ which is an acid will react with Zn Zn(s) + 2NH4+(aq) → Zn2+(aq) + 2NH3(aq) + H2(g) ½ mark + + 2+ or Zn(s) + 4NH4 (aq) → Zn(NH3)4 (aq) + 4H (aq) + 2e ½ mark will decrease in [NH4+(aq)], the electrode potential will also drop. C As in previous years, this type of question on electrochemistry was poorly answered. The basic concept of the half cell was not clearly understood by most candidates, hence they were not able to write the correct half equations and the equation for the overall reaction. Many candidates could not write the IUPAC cell diagram for the zinccarbon dry cell. 96 2A 3 a ii 3a ii Consider the following electrochemical cell : 6 III. Redox Equilibria Part 2 Page 7 (1) Identify, with explanation, the negative terminal of the electrochemical cell. (2) Comment on the change in concentration of Cu2+(aq) ions in beakers D and E as electrochemical reaction occurs. (3) Using the IUPAC convention, write the cell diagram for the electrochemical cell. (1) The left hand electrode / Cu electrode in beaker D is the negative terminal ½ mark 1 mark [Cu2+(aq)] on LHS is lower than that on RHS / 1.0 M Oxidation occurs at left hand electrode / electrons flow from left hand electrode to right hand electrode ½ mark so that [Cu2+(aq)] in the beaker E decreases / the difference in [Cu2+(aq)] of the two beakers becomes smaller. (2) In beaker E, [Cu2+(aq)] decreases because Cu2+ is reduced to Cu or Cu2+(aq) +2e-→ Cu(s) ½ + 1 mark ½ mark In beaker D, [Cu2+(aq)] remains almost constant because Cu2+(aq) ions produced in the electrochemical reaction combine with OH-(aq) ions in the buffer to form Cu(OH)2(s) 1 mark (3) Cu(s) | Cu(OH)2(s) | OH-(aq) || Cu2+(aq) | Cu(s) or Cu(s) | Cu(OH)2(s) | OH-(10-4 M) || Cu2+(aq) | Cu(s) or Cu(s) | Cu2+(aq) (2.0 × 10-11 M) || Cu2+(aq) (1.0 M) | Cu(s) (Deduct ½ marks for each minor mistake) 1 mark (1 mark) (1 mark) 97 2A 4 a ii iii 4a The diagram below shows an electrochemical cell connected to a digital voltmeter. An electromotive force of 0.83 6 V was recorded at 298K. ii Write the overall equation for the electrochemical reaction. iii Write the cell diagram for the electrochemical cell, using the IUPAC convention. III. Redox Equilibria Part 3 Page 1 Topic Reference Reading III. Redox Equilibria 6.3.3 Part 3 Assignment Advanced Practical Chemistry, John Murray (Publisher) Ltd., 110–112, 176–177 A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 281–285 Reading Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 209–217 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 189–195 Electrode potentials Standard hydrogen electrode Relative electrode potential (Standard reduction potential) Electrochemical series Use of electrode potential C. Electrode potentials 1. Standard hydrogen electrode Since the actual e.m.f. of a cell is depending on both electrodes, the concentrations of electrolytes and temperature. It would be more convenient if a standard condition is agreed and a reference electrode is chosen for comparison. Standard condition is defined as 1 atm, 298 K with all solution having molar concentration. Hydrogen electrode is chosen by the scientist as the standard of reference. It consists of an inert platinum electrode coated with platinum black (platinum powder) which increases the area of contact. The electrode is enclosed in a hydrogen jacket at 1 atm and immersed in a 1M H+(aq) solution. Eventually, a redox equilibrium is set up and can be described by a half ionic equation. 2H+(aq) + 2e- d H2(g) If electron is supplied to the electrode, H+(aq) will be converted to H2(g). Or, if electron is taken from the electrode, H2(g) will be converted back to H+(aq). Cell diagram of hydrogen electrode : Pt(s)H2(g) | H+(aq) Syllabus Notes III. Redox Equilibria 2. Part 3 Relative electrode potential (Standard reduction potential) Page 2 In order to compare the ease of losing or gaining electron of different chemical systems (different half cells), hydrogen electrode is chosen arbitrarily as the standard of comparison. Standard hydrogen electrode is assigned an (relative) electrode potential of 0 V. The values are also called reduction potentials because the half equations are always expressed in reduction form when listed in the electrochemical series. Reduction half equation 2H+(aq) oxidizing agent + 2e- d H2(g) reducing agent Eo = 0 V At standard condition, the electrode potential measured is also called standard reduction potential. Determination of relative electrode potential (standard reduction potential) In the determination of the standard reduction potential of an unknown half cell, a cell with the hydrogen electrode as the left-hand electrode and the unknown electrode as the right-hand electrode is set up. The e.m.f. of the cell is then determined. For example when the half-cell Pt|Cl2(g)|Cl-(aq), is used as the right-hand cell and the hydrogen electrode is used as the left-hand cell, the e.m.f. is determined to be 1.36 V. However, a sign should be assigned to the e.m.f. value to denote the direction of flow of electrode (i.e. the relative affinity to electron). By definition, the polarity of the right-hand electrode is assigned to the e.m.f. value to do the job. Therefore, in this example, the e.m.f. of the cell is +1.36 V Cell diagram Pt | H2(g) | H+(aq) || Cl-(aq) | Cl2(g) | Pt Eo = + 1.36 V Furthermore, the relationship between the overall e.m.f. of a cell and the reduction potential of the 2 half cells is given by Eocell = Eoright - Eoleft When hydrogen electrode is chosen as the left-hand electrode, the value of Eoleft will be zero. Eocell = Eoright - Eoleft = Eoright Therefore, by definition, electrode potential of an electrode is given by the e.m.f. of the cell if the LEFT electrode in the cell diagram is a hydrogen electrode. Cl2(aq) + 2e- d 2Cl-(aq) Eo = + 1.36V III. Redox Equilibria Part 3 Consider another example where the positions of the cell diagram and the physical setup are different. Cell diagram Pt | H2(g) | H+(aq) || Sn2+(aq) | Sn(s) Sn2+(aq) + 2e- d Sn(s) N.B. Eo = - 0.14 V Page 3 Eo = - 0.14 V When we are discussing the position of the electrode, we are referring to the cell diagram instead of the position of the actual set up. a) Meaning of sign and value A positive electrode potential means that the electrode accepts electrons more readily than a hydrogen electrode. As the sign and value of the electrode potential are depending on the readiness of a system to accept / donate electrons, electrode potential serves as a very good indicator of the oxidizing and reducing power of different species. When the reduction equations are listed according to the electrode potential, the list is called electrochemical series. III. Redox Equilibria 3. Electrochemical series Part 3 Page 4 Reduction equation Weak oxidizing agent K+(aq) + ePbSO4(s) + 2eSn2+(aq) + 2e2H+(aq) + 2e4H+(aq) + SO42-(aq) + 2eAgCl(s) + ePbO2(s) + H2O(l) + 2eO2(g) + 2H2O(l) + 2e2H+(aq) + O2(g) + 2eAg+(aq) + e2NO3-(aq) + 4H+(aq) + 2eMnO2(s) + 4H+(aq) + 2eCr2O72-(aq) + 14H+(aq) + 6eCl2(aq) + 2ePbO2(s) + 4H+(aq) + 2eMnO4-(aq) + 8H+(aq) + 5ePbO2(s) + SO42-(aq) + 4H+(aq) + 2eMnO4-(aq) + 4H+(aq) + 3eStrong oxidizing agent N.B. Strong reducing agent K(s) Pb(s) + SO42-(aq) Sn(s) H2(g) H2SO3(aq) + H2O(l) Ag(s) + Cl-(aq) PbO(s) + 2OH-(aq) 4OH-(aq) H2O2(aq) Ag(s) N2O4(g) + 2H2O(l) Mn2+(aq) + 2H2O(l) 2Cr3+(aq) + 7H2O(l) 2Cl-(aq) Pb2+(aq) + 4H2O(l) Mn2+(aq) + 4H2O(l) PbSO4(s) + 2H2O(l) MnO2(s) + 2H2O(l) Weak reducing agent Eo / V d d d d d d d d d d d d d d d d d d - 2.92 - 0.36 - 0.14 0 + 0.17 + 0.17 + 0.28 + 0.40 + 0.68 + 0.80 + 0.80 + 1.23 + 1.33 + 1.36 + 1.46 + 1.51 + 1.69 + 1.70 The values are only correct at standard conditions. The electrode potentials are measured at standard condition. If the condition is not standard, another value would be expected. For example, concentrated sulphuric acid has a concentration of about 18 M instead of 1 M. Reduction potential of 1M H2SO4(aq) 4H+(aq) + SO42-(aq) + 2e- d H2SO3(aq) + H2O(l) Eo = +0.17 V Concentrated sulphuric acid has a much more positive electrode potential and can be considered as a strong oxidizing agent, while dilute sulphuric acid is not. 4. Use of electrode potential a) Comparing the oxidizing and reducing power On the electrochemical series, the chemical species are arranged according to their electrode potentials. Those oxidizing agents low on the list accept electron readily and are strong oxidizing agents. e.g. MnO4/H+(aq). And those reducing agents high on the list donate electron readily and are strong reducing agents. e.g. K(s). III. Redox Equilibria b) Calculation of e.m.f. of a cell Part 3 Page 5 The e.m.f of a cell can be calculated from the values of the electrode potentials. Consider an overall cell reaction, Sn(s) + 2Ag+(aq) → Sn2+(aq) + 2Ag(s) It is comprised two half cells presented by 2 half equations Sn2+(aq) + 2e- d Sn(s) Ag+(aq) + e- d Ag(s) Eo = - 0.14 V Eo = + 0.80 V The electrode potential is reversed when the direction of reaction is reversed. Sn(s) d Sn2+(aq) + 2eAg+(aq) + e- d Ag(s) Eo = + 0.14 V Eo = + 0.80 V The overall electrode potential can then calculated by adding the 2 values together. Sn(s) + 2Ag+(aq) → Sn2+(aq) + 2Ag(s) Or consider a cell diagram, Sn(s) | Sn2+(aq) || Ag+(aq) | Ag(s) The e.m.f. of the cell can be calculated by using the equation, Eocell = Eoright - Eoleft Sn2+(aq) + 2e- d Sn(s) Ag+(aq) + e- d Ag(s) Eo = - 0.14 V Eo = + 0.80 V Eo = (+ 0.14 V) + (+ 0.80 V) = + 0.94 V Eocell = Eoright - Eoleft = (+ 0.80 V) - (- 0.14 V) = + 0.94 V Since the sign of the e.m.f is depending on the direction of reaction of a cell, when a cell diagram is converted to an equation or vice versa, the following convention should be observed. RedL + OxR → RedR + OxL Sn(s) + Ag+(aq) → 2Ag(s) + Sn2+(aq) Eo = + 0.94 V Furthermore, e.m.f. of a cell is depending on the nature of the two electrodes only. It is not depending on the current (or no. of electrons) flowing through the circuit. In the calculation of e.m.f. of cell, the electrode potentials are only simply added together and are not multiplied by the no. of electrons involved. In contrast, in the balancing of a redox reaction, the half equations are multiplied whenever necessary to make the no. of electrons conserve. In the above example Oxidation half ionic equation Reduction half ionic equation Overall ionic equation Sn(s) d Sn2+(aq) + 2e×2 Ag+(aq) + e- d Ag(s) ------------------------------------------Sn(s) + 2Ag+(aq) → Sn2+(aq) + 2Ag(s) III. Redox Equilibria c) Part 3 Prediction of the feasibility of redox reaction In general, oxidizing agent reacts with reducing agent. Indeed, only strong oxidizing agent reacts with strong reducing agent. For example, both Cu(s) and Ag(s) are reducing agents while Cu2+(aq) and Ag+(aq) are oxidizing agents. But only Cu(s) reacts with Ag+(aq) but Ag(s) doesn't reacts with Cu2+(aq). Cu2+(aq) + 2e- d Cu(aq) Ag+(aq) + e- d Ag(s) Eo = + 0.34 V Eo = + 0.80 V Eocell = (- 0.34 V) + (+ 0.80 V) = + 0.46 V Eocell = (+ 0.34 V) + (- 0.80 V) = - 0.46 V Page 6 Cu(s) + 2Ag+(s) → Cu2+(aq) + 2Ag(s) Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+(s) (energetically feasible) (energetically infeasible) The reaction will only be energetically feasible if the sum of electron potentials is positive. N.B. Relationship between ∆G and Eo is not required in A-Level) When 1 e- jumps across a potential difference of 1V, the energy acquired by the electron is called 1 eV (electron volt). 1 eV = 1.60 × 10-19 C × 1 V = 1.60 × 10-19 J The e.m.f. of a cell is related to the ∆G by the expression, ∆Go = -nFEocell where n is the no. of mole of e- involved, F is the Faraday constant Eocell is the e.m.f. of the cell ∆Go is the total free energy change of the reaction. Its value is depending on the total no. of mole of electron passing through the cell. In contrast, Eocell is a reflection of the electron pressure whose value is only depending on the nature of the two electrode. According to the expression ∆Go = -nFEocell , a positive Eocell means a negative ∆Go. This means an energetically feasible reaction. Note : ∆G = ∆H - T∆S = - RT ln K = - nFE III. Redox Equilibria (1) Disproportionation reaction Part 3 Page 7 Disproportionation – A process in which the same element undergoes oxidation and reduction simultaneously. e.g. 2Cu+(aq) → Cu2+(aq) + Cu(s) A chemical species will disproportionate if the reduction potential to the next lower oxidation state is more positive than that from the next higher state. i.e. E2 > E1. Eo = + 0.52 V Cu+(aq) + e- → Cu(s) Cu+(aq) → Cu2+(aq) + eEo = - 0.15 V ---------------------------------------------------------------------------------------→ Cu(s) + Cu2+(aq) Eo = + 0.37 V (energetically feasible) 2Cu+(aq) Furthermore, it must be noticed that the overall reduction potential of Cu2+(aq) to Cu(s) is not the sum of the reduction potentials of the two steps. i.e. E3 ≠ E1 + E2. Two half-cell potentials cannot be added to give a reduction potential for a different half-cell. Glossary standard hydrogen electrode e.m.f. reference electrode standard condition platinum electrode relative electrode potential (standard reduction potential) electrochemical series feasibility disproportionation 90 1A 2 a 92 2A 1 c i 95 2B 6 c iii 96 2B 6 d i ii e 97 1A 2 c i ii 99 2A 1 b i ii 90 2A 1 b iii 90 2B 6 a ii Past Paper Question 97 2A 4 a i 90 1A 2 a 2a A convenient laboratory method for determining electrode potentials makes use of a standard hydrogen electrode. 6 Draw a labelled diagram of a hydrogen electrode. What conditions should be maintained when a hydrogen electrode is used as a standard hydrogen electrode? State how the conditions can be met in the laboratory. C 3 marks for correctly labelled diagram. Deduct 1 mark for each wrong/missing label Deduct ½ mark if ‘Platinized’ omitted Conditions: Maintaining the stated conditions: 298K/25ºC thermostat/water-bath 1 mark pressure gauge 1 mark H2 gas at 1 atm pressure the use of standard acid solution 1 mark [H+] = 1 M Diagrams of a hydrogen electrode were generally not well drawn. Only about a third of the candidates mentioned the use of platinized platinum or platinum black, and even fewer considered the depth of its immersion in the solution containing hydrogen ions. III. Redox Equilibria Part 3 90 2A 1 b iii 1b A cell based on the reaction: 6Fe2+(aq) + Cr2O72-(aq) + 14H+(aq) → 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) consists of one platinum electrode dipped into a beaker containing an acidified solution of K2Cr2O7(aq), and another platinum electrode dipped into a separate beaker containing FeSO4(aq), with the two solutions being connected by a salt bridge. With reference to this cell: iii calculate the e.m.f. of the cell, given that the standard reduction potentials of Fe3+/Fe2+ and acidified Cr2O72-/Cr3+ are respectively +0.77V and +1.33V; and e.m.f. = +1.33 - (+0.77) = 0.56V 1 mark 90 2B 6 a ii 6a The following species are either impossible to prepare or very unstable. Explain, in each case, why this is so. ii SI6 S(VI) has a high redox potential and would oxidize I- to I2. SI6 is therefore very unstable with respect to disproportionation. 1½ mark 92 2A 1 c i 1c Given the following standard reduction potentials, Eo : 1.33V (Cr2O72-(aq)+14H+(aq)),(2Cr3+(aq)+7H2O(l))|Pt(s) -0.41V Cr3+(aq),Cr2+(aq)|Pt(s) -0.76V Zn2+(aq)|Zn(s) i Predict the products of the reaction of zinc with dichromate(VI) solution. Explain your prediction. In aqueous acid solution the overall reaction: Zn(s) → Zn2+(aq) + 2eCr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l) ------------------------------------------------------------------------------Eo = -(-0.76) + 1.33 = 2.09 V Cr2O72-(aq) + 14H+(aq) + 3Zn(s) → 3Zn2+(aq) + 2Cr3+(aq) + 7H2O(l) is favourable because of the large positive Eo value. The overall reaction: → Zn2+(aq) + 2eZn(s) 3+ 2e + 2Cr (aq) → 2Cr2+(aq) ------------------------------------------------Eo = - (-0.76) + (-0.41) = 0.35 V 2Cr3+(aq) + Zn(s) → Zn2+(aq) + 2Cr2+(aq) is also favourable because of the positive Eo value. 2 marks 1 mark Therefore, the products are Zn2+(aq) and Cr2+(aq). Ci Candidates were not able to use all the given electrode potential data to predict the products : Zn2+ and Cr2+. Page 8 1 1½ 3 95 2B 6 c iii 6c In aqueous solutions, TiO2+ is colourless. It can be reduced to give a violet solution containing [Ti(H2O)6]3+. iii The violet solution formed should be kept in a sealed vessel or handled in an inert atmosphere. Using the 4 following data, predict, giving a balanced equation, what will happen to the violet solution if it is not kept in a sealed vessel or not handled in an inert atmosphere. Eo / V 3+ 2+ [Ti(H2O)6] (aq) + e → [Ti(H2O)6] -0.37 TiO2+(aq) + 5H2O(l) + 2H+(aq) + e- → [Ti(H2O)6]3+(aq) +0.11 O2(g) + 2H2O(l) + 4e- → 4OH-(aq) +0.40 O2(g) + 4H+(aq) + 4e- → 2H2O(l) +1.23 From the Eo values, O2 is a stronger [O] agent than TiO2+. 1 mark At standard conditions, the Eo of the following reaction is +1.23 - (+ 0.11) V = +1.12 V 1 mark 4Ti3+ + O2 + 2H2O → 4TiO2+ + 4H+ 1 mark Therefore, in the presence of oxygen, [Ti(H2O)6]3+ will be oxidized to TiO2+(aq) / the violet solution will turn colourless. 1 mark C iii Candidates were expected to predict that if [Ti(H2O)6]3+ is not kept in a sealed container, the reaction 4[Ti(H2O)6]3+ + O2 → 4TiO2+ + 4H3O+ + 18H2O takes place because the overall Eo for the reaction is +1.12 V. However, most of them failed to do so. 96 2B 6 d i ii e 6d You are provided with the following standard reduction potentials : Eo/V + 0.15 Cu2+(aq) + e- d Cu+(aq) + 0.34 Cu2+(aq) + 2e- d Cu(s) III. Redox Equilibria Cu+(aq) + e- d Cu(s) I2(s) + 2e- d 2I-(aq) Cu2+(aq) + I-(aq) + e- d CuI(s) i Part 3 + 0.52 + 0.54 + 0.92 Page 9 ii 6e Using the above information explain why copper(I) compounds are unstable in aqueous solutions; and 2 In aqueous solutions, Cu+ disproportionates to give Cu(s) and Cu2+(aq) ½ + ½ mark 2Cu+(aq) → Cu(s) + Cu2+(aq) (1 mark) 1 mark because the above reaction has a +ve Eº of + 0.52 - (+0.15) = + 0.37 V predict what will be observed when a potassium iodide solution is added to a copper(II) sulphate(VI) solution. 3 Explain your prediction and write a balanced equation for the reaction involved. A brown solution and a (white) precipitate are formed because the reaction ½ + ½ mark 2Cu2+ + 4I- → 2CuI + I2 1 mark has a Eº value of (0.92 - 0.54) = + 0.38 V 1 mark In acidic conditions, VO2+(aq) can be reduced by sulphur dioxide to give VO2+(aq). Write a balanced equation for the 2 reaction. ½ mark VO2+(aq) + 2H+(aq) + e- → VO2+(aq) + H2O(l) +) SO2(g) + 2H2O(l) → SO42-(aq) + 4H+(aq) + 2e½ mark --------------------------------------------------------------------------------------------- 2VO2+(aq) +SO2(g) → 2VO2+(aq) + SO42-(aq) (Award FULL marks if only the overall equation is given.) 97 1A 2 c i ii 2c Consider the standard reduction potentials listed below: Eº = + 0.40 V [FeF6]3-(aq) + e- d [FeF6]4-(aq) Eº = + 0.54 V I2(aq) + 2e- d 2I-(aq) Eº = + 0.76 V Fe3+(aq) + e- d Fe2+(aq) Explain the following observations, giving a balanced equation in each case. i When Fe(NO3)3(aq) is added to KI(aq), a brown solution is formed. ii When concentrated KF(aq) is added to the resulting solution in (i) above, the brown colour fades. 1 mark 4 97 2A 4 a i 4a The diagram below shows an electrochemical cell connected to a digital voltmeter. An electromotive force of 0.83 6 V was recorded at 298K. i Write half-equations for the reaction at the anode and at the cathode, and give their corresponding standard electrode potentials. 99 2A 1 b i ii 1b i What is 'standard reduction potential' ? ii With reference to the standard reduction potentials listed below, explain why hydrochloric acid can be used to acidify potassium dichromate(VI) solution but cannot be used to acidify potassium manganate(VII) solution. Give a balanced equation for any reaction that takes place. Eo/ V +1.33 Cr2O72-(aq) + 14H+(aq) + 6e- d 2Cr3+(aq) + 7H2O(l) +1.36 Cl2(g) + 2e- d 2Cl-(aq) +1.51 MnO4-(aq) + 8H+(aq) + 5e- d Mn2+(aq) + 4H2O(l) III. Redox Equilibria Part 4 Page 1 Topic Reference Reading III. Redox Equilibria 6.3.4 Part 4 Assignment A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 288–290 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 217–219 Reading Syllabus D. Secondary cell and fuel cell 1. Lead-acid accumulator Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 196–198 Lead-acid accumulator is commonly used in vehicle to start the engine. Comparing with the other secondary cell, it produces a much larger current. Notes The two electrodes are lead alloy grids filled with spongy PbO2(s) (cathode) and spongy Pb(s) (anode) immersed in the electrolyte of dilute H2SO4(aq). Upon discharge, thin layers of PbSO4(s) will be coated on both the cathode and anode. However, the reaction is reversible. When the battery is charged up, PbSO4(s) will be converted back to PbO2(s) and Pb(s). Question : What would happen if a Daniel cell is recharged ? PbO2 cathode (Reduction) PbO2(s) + 4H+(aq) + SO42-(aq) + 2e- → PbSO4(s) + 2H2O(l) Pb anode (Oxidation) Pb(s) + SO42-(aq) → PbSO4(s) + 2eOverall reaction Eo = + 1.69 Eo = + 0.36 Pb(s) + PbO2(s) + 2H2SO4(aq) d 2PbSO4(s) + 2H2O(l) Eocell = + 2.05 V In a fully charged lead-acid accumulator, the 36% H2SO4(aq) has the density 1.27 gcm-3. Upon discharging, H2SO4(aq) is consumed and H2O(l) is produced, the density of the electrolyte will decrease. The condition of charge in the cell can be determined by measuring the density of the acid with a hydrometer. Cell diagram Pb(s) | PbSO4(s) | H2SO4(aq) | PbSO4(s) | PbO2(s) | Pb(s) Ox L RedR Ox R RedL Pb(0) Pb(II) Pb(II) Pb(IV) No salt bridge or porous partition is required because the two electrodes share the same electrolyte, H2SO4(aq). III. Redox Equilibria 2. Hydrogen-oxygen fuel cell Part 4 Page 2 Fuel cell is the battery used in spaceship to produce electricity. H2 Anode (Oxidation) O2 Cathode (Reduction) Overall Reaction H2(g) → 2H+(aq) + 2eO2(g) + 2H2O(l) + 4e- → 4OH-(aq) 2H2(g) + O2(g) d 2H2O(l) Eo = 0 V Eo = + 0.40 V Eo = + 0.40 V On top of electrical energy, the fuel cell also provide the warmth and water for the astronaut in the spaceship. Glossary Past Paper Question Secondary cell lead-acid accumulator hydrogen-oxygen fuel cell 91 1A 2 a i ii iii 93 2A 1 c i ii iii 98 2A 4 a i ii iii alloy grid spongy hydrometer 91 1A 2 a i ii iii 2a An electrochemical cell containing an oxygen cathode and a hydrogen anode is shown below : The pistons above the gas chambers are frictionless. i Write balanced equations for the half reactions and for the overall reaction in the cell. cathode: 2H+ + ½O2 + 2e- d H2O anode: H2 d 2H+ + 2eoverall: H2 + ½O2 → H2O wrong coefficient -½ reverse labelling of cathode and anode reaction -½ wrong species in each reaction -½ wrong direction of reaction (e.g. if a single arrow) -½ 2 2 marks III. Redox Equilibria ii Part 4 Page 3 C 1 How does the concentration of H2SO4 affect the equilibria of the half reactions ? ↑[H+] will shift equilibrium of 2H+ + ½O2 + 2e- d H2O to right, and ½ mark each shift equilibrium of H2 d 2H+ + 2e- to left. iii If weights are added to the pistons of both chambers, how would the reading of the voltmeter change ? Explain 2 your answer. voltage reading increase 1 mark ½ mark pressure in O2 and H2 chamber increase or [O2] and [H2] increase equilibrium of H2 + ½O2 → H2O is shifted to right and favours the discharge of battery ½ mark Surprisingly, many candidates could not apply knowledge of chemical equilibria to an electrochemical reaction. A fuel cell and an electrolytic cell could be the same problem if viewed from the equilibrium principles. Over half of the candidates gave OH- as the product in one or both of the half reactions. They also confused the anode and the cathode and the direction of overall reaction. 93 2A 1 c i ii iii 1c i Draw a labelled diagram of a cell that makes use of the reaction: 2H2SO4(aq) + PbO2(s) + Pb(s) → 2PbSO4(s) + 2H2O(l) Indicate the direction of electron flow on your diagram. 2 C Labelled diagram 1 mark Direction of electron flow 1 mark ii Write a balanced equation for the reaction at each electrode. At PbO2 (cathode) PbO2(s) + HSO4-(aq) + 3H+(aq) + 2e- → PbSO4(s) + 2H2O(l) or PbO2(s) + 4H+(aq) + SO42-(aq) + 2e- → PbSO4(s) + 2H2O(l) 1 mark At Pb (anode) Pb(s) + HSO4-(aq) → PbSO4(s) + 2e- + H+ or Pb(s) + SO42-(aq) → PbSO4(s) + 2e1 mark iii Using the IUPAC convention, write the cell diagram for the above cell. General format: (Pt)|RedL,OxL||RedR,OxR|(Pt) Pb(s)|PbSO4(s)|H2SO4(aq)|PbSO4(s)|PbO2(s)|Pt(s) or Pb(s)|PbSO4(s)|H2SO4(aq)|PbO2(s)|PbSO4(s)|Pt (without physical state -1) 2 marks or PbSO4(s)|PbO2(s)|H2SO4(aq)|PbSO4(s)|Pb(s) ii Many candidates had difficulties in writing balanced equations. iii Few candidates gave the right cell diagram. Many candidates were not aware that in writing cell diagrams, a vertical line " | ", instead of a comma ",", should be used to denote a phase boundary. 2 2 98 2A 4 a i ii iii 4a A lead-acid rechargeable cell is formed by dipping a lead plate coated with PbO2(s) and another lead plate coated with PbSO4(s) in H2SO4(aq). i You are provided with the following standard reduction potentials at 298 K. 2H2O(l) + 2e- d H2(g) + 2OH-(aq) PbSO4(s) + 2e- d Pb(s) + SO42-(aq) Pb2+(aq) + 2e- d Pb(s) O2(g) + 4H+(aq) + 4e- d 2H2O(l) PbO2(s) + SO42-(aq) + 4H+(aq) + 2e- d PbSO4(s) + 2H2O(l) Eo/V - 0.83 - 0.36 - 0.13 + 1.23 + 1.69 8 (I) Write half-equations for the reaction at the anode and at the cathode of the cell. (II) Write the overall equation for the electrochemical reaction and hence determine the standard electromotive force (e.m.f.) of the cell. III. Redox Equilibria Part 4 ii Write the cell diagram in accordance with the IUPAC convention. iii Explain why (I) the voltage of the cell drops upon discharge: (II) the cell is rechargeable. Page 4 III. Redox Equilibria Part 5 Page 1 Topic Reference Reading III. Redox Equilibria 6.3.5 Part 5 Assignment A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 516–518 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 309–310 Reading Syllabus Notes Corrosion of iron and its prevention E. Corrosion of iron and its prevention 1. Electrochemical process of rusting Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 271–272 Formation of rust is a complicate process. Rusting only occurs when iron gets in contact with dissolved oxygen in water. Rusting can be considered as an electrochemical process. Initially, Fe(s) atom loses 2 electrons and becomes Fe2+(aq) ion. Anode (oxidation) Fe(s) → Fe2+(aq) + 2eEo = + 0.44 V The dissolved oxygen in water accepts the electrons and becomes hydroxide ion. Cathode (reduction) O2(aq) + 2H2O(l) + 4e- → 4OH-(aq) Eo = + 0.40 V The overall e.m.f. of the cell = (+ 0.44 V) + (+ 0.40 V) = + 0.84 V (energetically feasible) Fe2+(aq) precipitates with OH-(aq) ion in water to form solid iron(II) hydroxide.. Fe2+(aq) + 2OH-(aq) → Fe(OH)2(s) Iron(II) hydroxide is then further oxidized by the dissolved oxygen to iron(III) hydroxide. 4Fe(OH)2(s) + O2(aq) + 2H2O(l) → 4Fe(OH)3(s) Upon standing, the iron(III) hydroxide changes to a reddish brown solid – rust, Fe2O3·nH2O(s). Since Fe2O3·nH2O(s) has a very loose structure, it flakes off from the iron surface and exposes the inside of the iron block to further corrosion. Because rusting involves flowing of charges, rusting would be faster in the presence of an electrolyte. i.e. iron rusts much faster in sea water (sodium chloride solution) than in distilled water. Acidic condition also accelerates rusting by reacting with iron metal. III. Redox Equilibria Part 5 Rust indicator - Potassium hexacyanoferrate(III) Since Fe2+(aq) ion is the one of the initial product in rusting, formation of rust can be monitored by detecting the presence of Fe2+(aq) ion. Potassium hexacyanoferrate(III), K3Fe(CN)6(aq), is a yellow chemical which turns deep blue in the presence of Fe2+(aq) ion. It serves as a very good indicator of rust. [Fe(CN)6]3-(aq) + Fe2+(aq) yellow → [FeFe(CN)6]-(aq) deep blue Page 2 When a drop of solution containing phenolphthalein and potassium hexacyanoferrate(III) is dropped onto an iron surface, the rim of the drop will turn pink and the center of the drop will turn blue gradually. This shows that the rim is the cathodic area where the concentration of dissolved oxygen is higher. And, the center is the anodic area. After a long time, most rust will form somewhere between the cathodic area and anodic area, where Fe2+ and OH- ion meet.. 2. Prevention of rusting a) Coating Since rust is only formed in the presence of oxygen and water, iron can be protected from rusting by putting a coating on the surface. A. Paint or grease Painting is the most cost effective method for large iron object but if the painted surface is scratched, iron will be vulnerable. Grease or oil is used in the moving part of machinery where the paint will be scratched. B. Galvanising (zinc dipping) Iron is dipped into a tank of molten zinc or molten zinc is spray onto iron so that the surface of the iron is coated with a layer of zinc. The process is called galvanising or galvanization. The iron is called galvanised iron. Although zinc also reacted with oxygen in the air to form zinc oxide, zinc oxide has a very tight structure and protects the zinc underneath from further corrosion. The body of most automobiles are galvanized before the paint is applied. This make it more resistant to rusting. C. Tin-plating Since zinc is toxic, it is not used in protecting the food can from rusting. Instead, a layer of tin is coated on the surface of iron can to exclude the air and water. However, if the tin-plating is scratched and the iron underneath is exposed, the tin-plated can will rust even faster than an ordinary iron can. This is because iron is more reactive than tin and iron will lose electrons through tin quickly. III. Redox Equilibria Part 5 Page 3 D. Electroplating of other metal Chromium is another metal which is coated on iron to prevent rusting. Chromium is frequently used because of its silvery appearance. e.g. Most water taps are electroplated with chromium. b) Sacrificial Protection Iron is first changed to iron(II) ion in the rusting process. If the formation of iron(II) ion can be suppressed, iron can be protected from rusting. Fe(s) d Fe2+(aq) + 2eA reactive metal loses electron more readily than a less reactive metal. By connecting iron to a more reactive metal, formation of Fe2+(aq) can be retarded. Once a Fe2+(aq) ion is formed, it will acquire the electrons from the more reactive metal and changes back to Fe. Consequently, iron is protected from corrosion but the more reactive metal is sacrificed instead. This method is preferable for the iron object which is hard to be repainted e.g. iron pipe under the ground, hull of ship. Sacrificial protection by galvanization Besides isolating iron from air and water, Zn(s) also protects iron by sacrificial protection because Zn(s) is more reactive than Fe(s). For the same reason, the body of a car is always connected to the negative terminal of the car battery. This make the car body negatively charged and less vulnerable to rusting. c) Alloying Pure iron rust easily in moist air. But if appropriate amounts of carbon, chromium and nickel are added, iron will become stainless steel which is more resistant to rusting. Glossary rusting (corrosion of iron) rust (hydrated iron(III) oxide) anodic area rust indicator (potassium hexacyanoferrate(III)) coating galvanising sacrificial protection tin-plating electroplating alloying 97 2A 4 b i ii 99 2A 1 b iii cathodic area zinc dipping Past Paper Question 97 2A 4 b i ii 4b Discuss and write relevant equations for the electrochemical processes involved when broken surfaces of the following are exposed to moist air. i galvanized iron ii tin-plated iron 99 2A 1 b iii 1b iii Explain why cathodic protection can be used to prevent the corrosion of iron. (7 marks) Phase Equilibria I. One component system A. Water 1. Phase diagram of water 2. Isotherm of water 3. PVT surface of water Carbon dioxide 1. Phase diagram of carbon dioxide 2. Isotherm of carbon dioxide Ideal system 1. Raoult's law 2. Composition of the liquid mixture comparing with composition of the vapour 3. Boiling point-composition diagram a) Definition of boiling point b) Boiling point-composition diagram c) Fractional distillation Non-ideal solution 1. Deviation from Raoult's law a) Positive deviation b) Negative deviation c) Enthalpy change of mixing 2. Elevation of boiling point and depression of freezing point by an involatile solute 3. Boiling point of two immiscible liquid a) Steam distillation Partition of a solute between two phases 1. Solvent extraction a) Calculation 2. Paper chromatography a) Rf value b) Separation of amino acids B. II. Two components system A. B. III. Three components system A. Phase Equilibria Unit 1 Page 1 Topic Reference Reading Assignment Reading Syllabus Phase Equilibria 7.1 Unit 1 A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 239240 One component system PVT surface Phase diagram Isotherm I. One component system Notes One component system is a system consists of one substance. At different condition, the same substance may have different physical state or phase. There are 3 physical states : solid, liquid and gas. Moreover, the same substance with the same physical state may not be exactly the same. e.g. graphite and diamond are two allotropes of carbon, the two different phases of solid carbon. Diatomic oxygen and ozone is another pair of example. By measuring the physical properties of a substance at different pressure, volume and temperature, a PVT surface can be constructed. Different regions of the surface represents different phases of a substance. PVT surface is solely a summary of experimental findings and can only be obtained through experiment. Phase Equilibria Unit 1 Page 2 Depending on the properties, gaseous phase can be further divided into vapour and gaseous phases. Vapour is defined as a gas which can be liquefied by compression alone. Conversely, a gas cannot be compressed into liquid without cooling. However, it is rather difficult to study pressure, volume and temperature altogether. It would be easier to investigate a 2-dimensional projection of the 3-dimensional PVT surface along the temperature or volume axes. The projection along the volume axis is called phase diagram. The projection along the temperature axis is called isotherm. A. Water 1. Phase diagram of water A solid line on the diagram represents the condition at which two phases coexist. The temperature at which solid and liquid coexist is called melting point. The temperature at which liquid and gas coexist is called boiling point. The temperature at which solid and gas coexist is called sublimation point. The point where the 3 solid lines joined together is called triple point. It is the only condition at which solid, liquid and gaseous phases can coexist in equilibrium. It is an unique condition for a specific substance. For example, 6.03 × 10-3 atm and 273.16 K for water 5.11 atm and 216.6 K for carbon dioxide. Phase diagram of water BD – melting/freezing curve AB – sublimation curve BC – vaporization/condensation curve EB – supercooling curve Water is a very special substance which expands upon freezing. Its freezing curve (BD) has a negative slope. Comparatively, carbon dioxide has a freezing curve with positive slope. The freezing curve (BD) represents the conditions at which water and ice are at equilibrium. i.e. melting point of ice at different pressure. From the negative slope of the freezing curve, it can be seen that the melting point decreases with increasing pressure. If pressure is exerted onto a cube of ice, it will melt because the freezing point of water is lowered. But, how is this related to the volume of ice ? This can be explained by Le Chatelier's principle. Water (liquid) d higher density (smaller volume) Ice (solid) lower density (larger volume) When pressure is exerted onto ice, the system will response in a way so that the effect of increased pressure can be reduced. This is done by converting ice to water which has lower volume. Once the pressure is removed, water will freeze again. At molecular level, when a high pressure is exerted onto ice, the open structure will be broken and ice will be converted to water. Like all kind of change, freezing process also involves activation energy, though it is very small comparing with other chemical process. If a beaker of water is cooled calmly, it may reach a temperature below 0ºC without freezing. The liquid water which has a temperature lower than 0ºC is called supercooled liquid. If the supercooled water is disturbed (i.e. activation energy is provided), the water will freeze immediately. The curve EB represents the supercooling curve of water. The existence of water extends into the ice region.. Supercooled water is energetically less stable than ice. Supercooled water is a metastable state, which means highly unstable. Phase Equilibria 2. Isotherm of water Unit 1 Page 3 Isotherm is the curve representing the relationship between the pressure and volume of a substance at the same temperature. The isotherm has the shape of a hyperbola only at high temperature. This means that the pressure is inversely 1 proportional to volume (i.e. P α V which is stated by Boyles' Law). At high temperature, the gaseous molecules are moving at great speed. The intermolecular attractions among particles become insignificant comparing with thermal motions. This makes a gas behave as an ideal gas. At a temperature higher than the critical temperature, a gas cannot be liquefied by compression alone. This has important implication in the manufacturing of liquid air. If the temperature is too higher, the air can never be compressed into liquid no matter how high the pressure is. At a temperature lower than critical temperature, vapour can be compressed into liquid. The vapour pressure, at which the vapour and liquid coexist, is called saturated vapour pressure. Saturated vapour pressure is a quantity independent of volume. If the volume of the system is decreased, more vapour will condense into liquid and the saturated vapour pressure will remain unchanged. It is only depending on temperature. Water vapour d Liquid water 3. PVT surface of water The curve abcdef represents the change of water when it is warmed at 1 atm pressure. Along ab, ice expands a little bit. Along bc, ice starts to melt and contract. Ice and water coexist at melting point. Along cd, water is heated up and expands. Along de, water boils and turns to water vapour. Water and water vapour coexist at boiling point. Along ef, the temperature and volume of water vapour increase. At a temperature beyond critical temperature, it becomes a gas. The curve ghjk represents the compression of water vapour to liquid water at room temperature. The curve nopqrs represents the compression of water vapour at a temperature lower than triple point temperature. Along no, the vapour is compressed. Along op, the vapour sublimes directly into ice. Along pq, ice is compressed and the volume gets a little bit small. Along qr, ice melts because of the increasing pressure. Along rs, liquid water is compressed and the volume change is very minimal. PVT surface of water Phase Equilibria B. Carbon dioxide Unit 1 Page 4 Liquid carbon dioxide doesn't exist at atmospheric pressure. From the phase diagram of carbon dioxide, the triple point pressure (5.1 atm) is higher than atmospheric pressure. When dry ice is warmed, it sublimes directly to carbon dioxide gas. Carbon dioxide behaves like many other substances. It contracts upon freezing and has a freezing curve with a positive slope. 1. Phase diagram of carbon dioxide 2. Isotherm of carbon dioxide Glossary one component system sublimation phase allotrope PVT surface vapour gas projection phase diagram isotherm triple point freezing curve sublimation curve vaporization curve supercooled liquid supercooling curve metastable state hyperbola ideal gas critical point vapour pressure (saturated) vapour pressure 90 1A 2 e i ii iii 92 1A 2 g i ii iii iv 98 1A 2 a i ii iii Past Paper Question 90 1A 2 e i ii iii 2e i Each of the graphs A to D in the above diagram shows the variation of pressure with volume for a particular substance at different temperatures T1, T2, T3 and T4. Name the point P and comment on its significance. Critical point ½ mark At points on isotherm above P, the gaseous state of the substance cannot be liquefied by the increased in pressure alone. 1 mark 1½ Phase Equilibria ii Unit 1 Page 5 C What is the significance of the horizontal portion of graph D? 1 Along the horizontal portion, the gaseous and liquid states of the substance can co-exist. 1 mark Or The horizontal portion signifies the liquefaction (vaporization) of the gaseous (liquid) state of the substance. iii Which graph provides experimental evidence that PV = nRT? Explain your answer. 1½ Graph A ½ mark 1 mark Q it has the shape of a hyperbola or Q it represents the relationship that P is inversely proportional to 1/V. iii Candidates were expected to know that Graph A provides experimental evidence that PV = nRT because the graph gives the relationship in the form of a hyperbola in which P is inversely proportional to V (at a given temperature). 92 1A 2 g i ii iii iv 2g For the phase diagram below: i Identify the point X and Y. Point X - critical temperature (substance can no longer exist in the liquid state beyond this temperature) ½ mark Point Y - triple point (solid, liquid and vapour are all at equilibrium) ½ mark ii Which two points in the diagram define the sublimation curve? Y, Z iii Mark on the diagram, using the symbol B, the normal boiling point. 1 1 1 iv Point B is the boiling point. Suggest the meaning of the curve YW. Metastable state or Supercooling occurs. 1 mark 1 1 mark 98 1A 2 a i ii iii 2a The phase diagram for carbon dioxide is shown below. Phase Equilibria Unit 1 Page 6 i Identify points B and C. ii What is the physical meaning of point C ? iii Starting with a sample of CO2(g) at room temperature and atmospheric pressure, suggest how the following can be obtained. (I) solid CO2 (dry ice) (II) liquid CO2 1 1 2 Phase Equilibria Unit 2 Page 1 Topic Reference Reading Phase Equilibria 7.2.1 Modern Physical Chemistry ELBS pg. 372–377 Unit 2 Assignment A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 181–186 Reading Syllabus Two components system Ideal system Raoult's law II. Two components system Two components system is a system consists of 2 different substances. If a system is comprised of one more substance, the physical behavior will be more complicate. On top of pressure, volume and temperature, the physical properties of a mixture is also depending on the relative abundance (mole fraction) of individual component. A 4-dimensional space will be required to represent the experimental findings. It would be easier if the investigation is limited to a given temperature. Therefore, the effect of temperature can be excluded. The situation can further be simplified by considering the liquid and vapour phases only. The saturated vapour pressure of a liquid is also commonly known as the vapour pressure of the liquid. In the presence of a liquid, the saturated nature of the vapour is understood. At a given temperature in an one component system, the (saturated) vapour pressure is a constant independent of volume. If the volume of the mixture is decreased, some vapour will be compressed into liquid and the (saturated) vapour pressure will remain a constant. Liquid d Saturated vapour In general, the more volatile a liquid, the higher will be the (saturated) vapour pressure at a given temperature. However, if two volatile liquids are mixed, the (saturated) vapour pressure will no longer depend on temperature only. It will also be depending on the abundance of individual component. PAº = vapour pressure of pure A (more volatile) PBº = vapour pressure of pure B (less volatile) Notes Phase Equilibria By Dalton's law of partial pressure, Total vapour pressure PT A. Ideal system = = Unit 2 Page 2 partial pressure of vapour A PA + + partial pressure of vapour B PB It is found that if the intermolecular forces among all molecules in a mixture have similar magnitude, i.e. the intermolecular forces ALA, ALB and BLB are all similar, the (saturated) vapour pressure will be jointly proportional to the mole fractions of the two components. The ideal behavior of the solution is further confirmed by the nearly zero enthaply change of mixing of the two components. 1. Rauolt's Law The system in which the total (saturated) vapour pressure is jointly proportional to the mole fractions of the two components is called an ideal system. The relationship PA = XA × PAº is called Raoult's law. Or in words, the partial vapour pressure of any volatile component of an ideal solution is equal to the vapour pressure of the pure component multiplied by the mole fraction of that component in the solution. N.B. Meaning of superscript º : pure substance o : standard state Vapour pressure-composition diagram By Dalton's law of partial pressure, Total vapour pressure PT = = = PT = partial pressure of vapour A PA mole fraction of component A × + + vapour pressure of + pure A + partial pressure of vapour B PB mole fraction of × component B XB × PBº vapour pressure of pure B XA × PAº A vapour pressure-composition curve can also be considered as a kind of phase diagram. At an external pressure higher than the equilibrium vapour pressure, the vapour will tend to condense to form more liquid. And if the external pressure is lower than the equilibrium vapour pressure, the liquid will tend to vaporize to form more vapour. Phase Equilibria 2. Unit 2 Composition of the liquid mixture comparing with composition of the vapour Will the composition of the vapour be the same as the composition of the liquid ? Obviously, the more volatile component vaporizes more easily and will have a greater contribution in the total vapour pressure (which is related to concentration) at a certain temperature. Therefore, when an equal molar solution of 2 liquids evaporates, the vapour will be enriched by the vapour of the more volatile component. Quantitatively, the composition of the vapour of an ideal solution at a given temperature can be calculated according to Rauolt's law. This requires the liquid composition of the mixture and the (saturated) vapour pressure of individual component at the given temperature. Mole fraction of A in the liquid = XA Mole fraction of B in the liquid = XB = (1- XA) Partial vapour pressure of A = PA = XA PAº Partial vapour pressure of B = PB = XB PBº Vapour pressure of the liquid mixture (according to Rauolt's law) = PT = XA PAº + XB PBº = XA PAº + (1- XA) PBº Mole fraction of A in the vapour (according to Dalton's law of vapour pressure) X A P Aº X A P Aº PA = = = X A' = PA + PB XA PAº + XB PBº XA PAº + (1 - XA )PBº Data (Saturated) vapour pressure of A at 300 K = 1.0 × 104 Pa (Saturated) vapour pressure of B at 300 K = 3.0 × 103 Pa Calculation Determine the composition of the vapour of the equimolar mixture of A and B at 300 K. Vapour pressure-composition diagram Page 3 By repeating the calculation for the mixtures with different compositions, a vapour composition curve can be constructed on the same vapour pressure-composition diagram. Phase Equilibria 3. Boiling point composition diagram Unit 2 Page 4 a) Definition of boiling point A liquid is said to be boiling when the (saturated) vapour pressure reaches the atmospheric pressure. A more volatile liquid has a higher (saturated) vapour pressure at a given temperature, thus it has a comparatively lower boiling point. b) Boiling point-composition diagram By measuring the boiling pointsof mixtures with different composition and the composition of the vapour, a boiling composition diagram can be constructed. A boiling point composition diagram also consists of 2 lines. The first one is the boiling point of the mixture with different composition. The second one is the composition of the vapour when the mixture boils. The diagram on the right is the boiling pointcomposition diagram of an ideal system consisting of propan-1-ol and 2-methylpropan-1-ol. A B C D E F G H I J K Mole fraction of propan-1-ol Boiling point of the mixture Boiling point of the distillate Mole fraction of propan-1-ol in the liquid / ºC / ºC in the vapour (prepared experimentally) (measured) (measured) (read from the graph) 0 109.0 109.0 0 0.10 105.2 99.1 0.24 0.20 100.9 94.2 0.41 0.30 97.4 91.3 0.53 0.40 94.3 88.6 0.65 0.50 92.0 86.8 0.74 0.60 89.5 85.3 0.82 0.70 87.3 84.4 0.90 0.80 85.6 84.0 0.94 0.90 84.2 83.7 0.98 1 83.3 83.3 1 Actually, it is quite difficult to determine the composition of the vapour chemically. It would be more convenient to determine the boiling point of the distillate (condensed from the vapour) and determine the composition of the vapour using the liquid composition curve as the calibration curve. (The boiling point composition curve cannot be constructed from the Rauolt's law, Rauolt's law does not provide any relationship between the vapour pressure and the temperature. It only provides the relationship between vapour pressure and mole fraction of the liquid.) Phase Equilibria c) Fractional distillation The boiling point composition curve can be used to predict the purity of the distillate. For example, when a mixture of 0.2 mole propan-1-ol and 0.8 mole 2-methylpropan-1-ol is boiled, from the curve, the vapour will have 0.37 mole of propan-1-ol and 0.63 mole of 2-methylpropan-1-ol. The distillate will have the same composition as the vapour. It should be noted that boiling is a constant temperature process, in which the liquid mixture and the vapour will have the same temperature (Line CD). On the other hand, condensation of the vapour is accompanied with a decrease in temperature. (Line DE) If the 0.37 mole propan-1-ol and 0.63 mole 2-methylpropan-1-ol mixture is distilled again, from the curve, the distillate obtained will have 0.58 mole propan-1-ol and 0.42 mole 2-methylpropan-1-ol. If the distillation is continued, the final distillate will be 100% propan-1-ol. Unit 2 Page 5 While more propan-1-ol is removed, the percentage of the less volatile 2-methylpropan-1-ol remain in the flask will increase. As a result, the boiling point of the liquid will increase gradually. Instead of separate distillations, successive distillations can be accomplished by using a fractionating column, thus fractional distillation. Upon heating, the liquid mixture vaporizes and condenses on the glass beads when it moves further from the heat source. The thin liquid film condensed on the glass bead will evaporate again while more hot vapour is coming up and the distillation will continue. In general, in the fractionating column, the composition will change when a liquid mixture is vaporized but the composition will not change when a vapour mixture is condensed. Glossary two components system 4-dimensional space (saturated) vapour pressure Rauolt's law ideal system/solution vapour pressure-composition diagram vapour composition curve boiling point composition diagram boiling point fractionating column glass bead boiling process condensation process 91 2A 3 b i ii 93 2A 1 a i ii 94 2A 2 c i 95 2A 1 c i 96 2A 2 c i 99 2A 2 a i ii jointly proportional liquid composition curve fractional distillation liquid film Past Paper Question 91 2A 3 b i ii Phase Equilibria Unit 2 3b i A liquid is added as a solute to a solvent forming an ideal solution. Comment on the vapour pressure of the solution as compared with that of the pure solvent. If a volatile solute is added to a solvent, the total vapour pressure will be the sum of the partial pressures of the solute and solvent. If the vapour pressure of the solute is higher than the vapour pressure of the solvent then the total vapour pressure is higher and vice versa. 3 marks ii The vapour pressures of pure methanol and ethanol at 25ºC are 89.2 mm Hg and 44.6 mm Hg respectively., A mixture containing 32.0 g of methanol and 184.0 g of ethanol behaves as an ideal solution. Calculate the partial vapour pressure of each component, the total vapour pressure of the mixture and the composition of the vapour. (Relative atomic masses : H, 1.0; C, 12.0; O, 16.0) CH3OH mol. wt. = 32 mol. wt. = 46 C2H5OH 32 =1 No. of moles of CH3OH = 32 184 No. of moles of C2H5OH = =4 46 1 Mole fraction of CH3OH = = 0.2 5 4 Mole fraction of C2H5OH = = 0.8 1 mark 5 By Raoult’s Law PA = x PAº 1 mark 1 Partial pressure of CH3OH = × 89.2 = 17.84 mmHg 5 4 Partial pressure of C2H5OH = × 44.6 = 35.68 mmHg 1 mark 5 Total vapour pressure = 17.84 + 35.68 = 53.52 mmHg ½ mark 17.84 1 = The composition of CH3OH in the vapour = 53.53 3 12 The composition of C2H5OH in the vapour = 1 - = ½ mark 33 i.e. ratio of CH3OH : C2H5OH is 1 : 2 Ci Most candidates were not able to point out that the vapour pressure of the solution could be higher or lower than that of the pure solvent. ii Careless mistakes in the calculation of mole fractions were common. 93 2A 1 a i ii 1a Liquid A is miscible with Liquid B over the whole composition range, forming ideal solutions. The vapour pressures of pure A and pure B, measured at 350 K, are 24.0 kPa and 12.0 kPa respectively. i Calculate the vapour pressure of a liquid mixture consisting of 60 mole % A and 40 mole % B at 350 K. PA = PAºXA 1 mark PB = PBºXB Partial pressure of A = 24 × 0.6 = 14.4 kPa Partial pressure of B = 12 × 0.4 = 4.8 kPa 1 mark Total vapour pressure PT = 14.4 + 4.8 = 19.2 kPa (wrong unit -½) 1 mark ii The mixture in (i) is distilled at 350 K and a small amount of the distillate is collected. If this distillate is distilled for a second time at 350 K, calculate the composition of the first droplet of the second distillate. Composition of distillate (1st time) PA 14.4 XA = P = 19.2 = 0.75 T 1 mark XB = 1 - 0.75 = 0.25 Total vapour pressure of second distillation = 0.75 × 24 + 0.25 × 12 = 21 kPa 1 mark Composition of second distillate 0.75 × 24 = 0.86 1 mark XA = 21 1 mark XB = 1 - 0.86 = 0.14 C ii Many candidates were not able to get the final answers correct because they did not realise that the ratio of partial pressures of A to B in the first distillate was the same as the mole ratio of A to B in the condensed phase. This indicated that the underlying principles of fractional distillation were not well understood by these candidates. Page 6 3 4 3 4 Phase Equilibria Unit 2 94 2A 2 c i 2c Two miscible liquids E and F form ideal solutions on mixing. At 298 K, the vapour pressure E is p0 while that of pure F is 2p0. i Sketch a graph of vapour pressure versus mole fraction for the above solutions at 298K. For a mixture of E and F with mole ration E to F = 1 : 4, determine its vapour pressure, in terms of p0, at 298K. Page 7 3 4 4 Mole fraction of F = 4 + 1 = 5 or 0.8 From the graph, vapour pressure of the mixture = 1.8 p0 OR BY CALCULATION PF = 0.8 (2p0) = 1.6p0 PE = 0.2p0 ∴ vapour pressure of mixture = PE + PF = 0.2p0 + 1.6p0 = 1.8p0 95 2A 1 c i 1c Two miscible liquids A and B form ideal solutions upon mixing. The boiling point of A is higher than that of B. i Draw the boiling point-composition graph at 1 atm for mixtures of A and B and label the graph in detail. Boiling point-composition curve 3 C 1 mark for labelling the x- and y-axis1 mark for the liquid curve 1 mark for the vapour curve (deduct 1 mark if TB > TA) This question was well answered except that some candidates mistook the boiling point-composition diagram as the vapour pressure-composition diagram 96 2A 2 c i 2c i Two miscible liquids A and B form ideal solutions when mixed. At 298 K, the vapour pressures of pure A and pure B are 32 kPa and 16 kPa respectively. For a mixture of 1 mol of A and 3 mol of B at 298 K, calculate (1) the vapour pressure of the mixture; (2) the mole fraction of A in the vapour which is in equilibrium with the mixture. 1 (1) Vapour Pressure = 4 × 32 + 4 × 16 1 + 1 marks = 20 kPa 1 mark (Deduct ½ mark for wrong / no unit) 82 (2) Mole fraction of A = 20 = 5 (Deduct ½ mark for including a unit in the answer.) 99 2A 2 a i ii 1 mark 4 Phase Equilibria Unit 2 2a A liquid mixture consists of 70.0 mol % of methanol and 30.0 mol% of ethanol. Suppose that the mixture is an ideal solution. Calculate, at 298 K, (At 298 K, the vapour pressures of methanol and ethanol are 12.5 kPa and 9.9 kPa respectively.) i the vapour pressure of the mixture, and ii the concentration of methanol, in mol%, in the vapour phase at equilibrium. (3 marks) Page 8 Phase Equilibria Unit 3 Page 1 Topic Reference Reading Phase Equilibria 7.2.2–7.2.3 A-Level Chemistry Syllabus for Secondary School (1995), Curriculum Development Council, 203–204 Unit 3 Assignment Advanced Practical Chemistry, John Murray (Publisher) Ltd., 78–80 A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 186–187 Reading Syllabus Non-ideal system Deviation from Raoult's law Enthalpy change of mixing B. Non-ideal system However, most solutions do not behave ideally. Some of them will give a (saturated) vapour pressure higher than the expected and some of them will give a lower value. 1. Deviation from Raoult's law This can be interpreted as the change in the strength of intermolecular forces. Positive deviation If the intermolecular forces among the two kinds of particles (ALB) is weaker the original forces in pure components (ALA & BLB), it will be easier for the particles to escape from the liquid mixture and give a (saturated) vapour pressure higher than the expected at a given temperature. This is called positive deviation from Raoult's law. For an ideal solution, the liquid composition curve is linear in shape. If the solution is slightly positively deviated, the curve will curve up a little bit. If the solution is highly deviated, a curve with a maximium point will result. This means that the mixture with that particular composition is even more volatile than either component. Slightly deviated Highly deviated Notes Negative deviation Similarly if the intermolecular forces among the two kinds of particles (ALB) is stronger, it will give a (saturated) vapour pressure lower than the expected at a given temperature. This is called negative deviation from Raoult's law. The minimium vapour pressure in the liquid composition curve represents a mixture which is less volatile than either component. Slightly deviated Highly deviated Phase Equilibria a) Positive deviation Unit 3 Page 2 A mixture of ethanol and benzene is a positive deviation from Raoult's law. It is a largely deviated system. The mixture has a minimium boiling point at 0.55 mole fraction of ethanol. At this point, both liquid and vapour have the same composition. This mixture is known as azeotropic mixture or azeotrope. Azeotropic mixture is a constant boiling mixture and cannot be separated by fractional distillation. If a 0.20 mole fraction of ethanol is distilled, the first distillate will contain 0.39 mole fraction of ethanol. If the distillate is distilled continuously, eventually, only a 0.55 mole fraction of ethanol distillate will be obtained. Boiling point-composition curve for ethanol / benzene Ethanol and water is another example of highly positive deviation. Fractional distillation can only purify the mixture up to 0.94 mole fraction ethanol or 97% by mass of ethanol. The remaining 3% water can only be removed by drying agent e.g. calcium oxide followed by distillation. This explains why absolute ethanol is much more expensive than 97% ethanol. Boiling point-composition curve for ethanol / water Phase Equilibria b) Negative deviation Unit 3 Page 3 Trichloromethane and propane is a system of highly negative deviation. The mixture has a maximium boiling point at 0.65 mole fraction of trichloromethane. If a 0.30 mole fraction of trichloromethane is distilled, the vapour will contain less trichloromethane (0.20 mole fraction) and more propanone. And the liquid remain in the flask will contain more trichloromethane. Eventually, a constant boiling mixture of 0.65 mole fraction of trichloromethane will be left in the flask and cannot be separated into trichloromethane and propanone. The distillate obtained will be pure propanone. Boiling point-composition curve for trichloromethane / propanone Similar if a 0.80 mole fraction of trichloromethane is distilled, the solution remain in the flask will be 0.65 mole fraction of trichloromethane. But the distillate obtained will be pure trichloromethane. Hydrochloric acid is also a highly negative deviation from Rauolt's law. Its azeotropic mixture has a composition of 0.11 mole fraction of HCl. Boiling point-composition curve for hydrogen chloride / water Phase Equilibria c. Enthalpy change of mixing Unit 3 Page 4 The change in strength of intermolecular forces can further be confirmed by the enthalpy change of mixing of components. When the components of a ideal system is mixed, the temperature change is almost zero. This means no special intermolecular attractions are formed or broken. When the components of a system showing positive deviation is mixed, the temperature decreases. This means that some intermolecular attractions are weakened. (i.e. bond breaking is an endothermic process.) When the components of a system showing negative deviation is mixed, the temperature increases. This means that some intermolecular attractions are strengthened. (i.e. bond formation is an exothermic process.) Glossary Past Paper Question non-ideal system / solution positive deviation from Raoult's law negative deviation from Raoult's law azeotropic mixture / azeotrope / constant boiling mixture enthalpy change of mixing 92 1A 2 f i ii iii 93 2A 1 b 94 2A 2 c i ii 95 2A 1 c i ii 96 2A 2 c ii 97 2A 1 c 99 2A 2 b i ii 92 1A 2 f i ii iii 2f The vapor pressure versus mole fraction diagram for a two component system (A and B) is shown below: i C State the reason(s) which give(s) rise to the shapes of the curves in the diagram. The attractive forces between A and B are weaker than those between molecules of the same kind. 1 mark As a result, there is an increased tendency to escape form solution and that the vapour pressure of each of the 1 mark components is greater than that predicted by Raoult’s law. ii What do points P and Q represent? P, Q are the vapour pressure of pure A and pure B respectively. 1 mark iii What assumption(s) would have to be met for the curves to be linear? Give an equation to represent this behaviour. The intermolecular attraction between molecule A and molecule B is similar to that between molecule A and molecule A and that between molecule B and molecule B. ½ mark Ideal behaviour (Obeys Raoult’s Law) OR PA=XAPAº ½ mark P=XAPAº+XBPBº iii Some candidates confused ideal gas law with Raoult's Law for ideal solution. 2 1 1 Phase Equilibria Unit 3 93 2A 1 b 1b Draw vapour pressure versus composition graphs of two types of non-ideal solutions formed by mixing two miscible liquids. Interpret these graphs in terms of intermolecular attractions. Vapour pressure – composition diagrams for negative and positive deviations from Rauolt’s Law are : Page 5 4 C (no dotted line -½) If the attraction between the molecules A & B is weaker (stronger) than the average intermolecular attraction in pure A & B, positive (negative) deviation results. Weaker (stronger) intermolecular attraction means molecules have greater (smaller) chance to escape, hence vapour pressure of the system is higher (lower) than that expected by Raoult’s Law. Many ca`ndidates were not aware of the relation between intermolecular interactions and non-ideal behaviour of real solutions. 94 2A 2 c ii 2c Two miscible liquids E and F form ideal solutions on mixing. At 298 K, the vapour pressure E is p0 while that of pure F is 2p0. i Sketch a graph of vapour pressure versus mole fraction for the above solutions at 298K. For a mixture of E and F with mole ration E to F = 1 : 4, determine its vapour pressure, in terms of p0, at 298K. 3 4 4 Mole fraction of F = 4 + 1 = 5 or 0.8 From the graph, vapour pressure of the mixture = 1.8 p0 OR BY CALCULATION PF = 0.8 (2p0) = 1.6p0 PE = 0.2p0 ∴ vapour pressure of mixture = PE + PF = 0.2p0 + 1.6p0 = 1.8p0 If E and F do not form ideal solution on mixing, in what possible ways could the vapour pressure of the above mixture deviate from the value determined in (i)? Account for the deviation(s). If deviation is negative, P < 1.8p0 ½ mark ½ mark If deviation is positive, P > 1.8p0 In case of negative (positive) deviation, the interaction betwee E and F is stronger (weaker) than the average of interaction between E and E and between F and F. The molecules has a smaller (greater) chance to escape, thus 2 marks resulting in a lower (higher) vapour pressure. [Max. 2 marks, if only one of the cases is discussed.] ii 3 Phase Equilibria Unit 3 95 2A 1 c i ii 1c Two miscible liquids A and B form ideal solutions upon mixing. The boiling point of A is higher than that of B. i Draw the boiling point-composition graph at 1 atm for mixtures of A and B and label the graph in detail. Boiling point-composition curve Page 6 3 ii C 1 mark for labelling the x- and y-axis 1 mark for the liquid curve 1 mark for the vapour curve (deduct 1 mark if TB > TA) Using your graph, explain how liquid B can be separated from a mixture containing A and B, by fractional 4 distillation. If a solution with composition X1 is heated to T1, the gaseous phase in equilibrium with it has the composition X2 which is richer in B. (1 marks for explanation; 1 mark for indication on the graph) This vapour is cooled to T2 and condenses to liquid with a composition of X2. Vaporization of this liquid leads to a new vapour with a composition X3, which is richer in B than that with a composition X2 and finally pure B is obtained. (1 mark for explanation; 1 mark for indication on the graph) This question was well answered except that some candidates mistook the boiling point-composition diagram as the vapour pressure-composition diagram 96 2A 2 c ii 2c ii Predict whether the following pairs of liquids when mixed, would give solutions showing positive deviation or negative deviation from Raoult's Law. Explain your prediction. CH3 4 (1) CH3CH2OH and (2) C2H5OC2H5 and CHCl3 (1) Mixtures of ethanol and methylbenzene will show positive deviation from Raoult's Law. ½ mark ½ mark In pure EtOH, the intermolecular attraction is H-bond. In the mixture EtOH molecules are surrounded by methylbenzene, the attraction between the molecules is mainly van der Waals' force. ½ mark Weakening of the intermolecular force causes an increase in vapour pressure / escaping tendency of molecule. ½ mark ½ mark (2) Mixtures of C2H5OC2H5 and CHCl3 will show negative deviation from Raoult's Law. ½ mark In the pure, liquids the intermolecular attraction is dipole-dipole interaction. In the mixture C2H5OC2H5 forms H-bond with CHCl3 Thus, the mixture will have a lower vapour pressure than the pure liquids. ½ mark 97 2A 1 c 1c Three types of boiling point - composition curves for two miscible liquids A and B at one atmosphere pressure are 7 shown below. Phase Equilibria Unit 3 Page 7 Account for the characteristics of these curves. 99 2A 2 b i ii 2b i At 1 atm pressure, the boiling points of pure nitric(V) acid and water are 359 K and 373 K respectively. At this pressure, nitric(V) acid and water form an azeotrope which boils at 394 K. The azeotrope consists of 35.3 mol %of HNO3, Sketch a labelled phase diagram for the nitric(V) acid-water system at 1 atm pressure showing (I) the boiling point - liquid composition relationship, and (II) the boiling point - vapour composition relationship. ii Describe all changes in the fractional distillation of an aqueous solution of 20 mol % of HNO3. (5 marks) Phase Equilibria Unit 4 Page 1 Topic Reference Reading Phase Equilibria 7.2.4 Chemistry (Structure and Dynamics), McGraw-Hill, 331–334 Unit 4 Assignment A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd.,187–189 Reading Syllabus Elevation of boiling point and depression of freezing point by an involatile solute Boiling point of two immiscible liquids Steam distillation 2. Elevation of boiling point by an involatile solute A volatile solvent (A) has weak intermolecular forces among the solvent particles (ALA). A solute (B) is involatile only if the intermolecular forces among the particles (BLB) are strong. If an involatile solute (B) is soluble in the volatile solvent (A), the forces among the solute and solvent particles (ALB) must be stronger than the solvent-solvent interaction (ALA) and solute-solute interaction (BLB). Otherwise, the solute will not be soluble. Notes Since the solute-solvent interaction is the strongest, when an involatile solute is dissolved in a volatile solvent, this must be a negative deviation from Rauolt's law. The vapour pressure of the mixture will be lower than the pure solvent at any temperature. Recall the phase diagram of pure water, pure water boils at 100 ºC and freezes at 0 ºC. If salt is dissolved in water, a higher temperature will be required to rise the vapour pressure of the mixture to atmospheric pressure. Therefore, the presence of involatile solute increases the boiling point of the solvent. For example, curve AB shows the vapour pressure of pure water at different temperature. Normally, its vapour pressure will reach atmospheric pressure at 100 C. Curve CD is the vapour pressure of the aqueous salt solution. At any particular temperature, the vapour pressure is always lower than that of pure water. And a higher temperature is required to make the solution boil. Phase Equilibria Unit 4 If the lowering of vapour pressure is extended to the lower temperature range, the vaporization curve will meet the sublimation curve at a lower temperature. This will lead to a lower triple point temperature as well as a lower melting point. For this result, salt is sprayed onto the icy road to melt the ice in winter. The degree of elevation of boiling point and depression of freezing point depends on the nature of the solvent and the concentration of the solute. (Surprisingly, it is not depending on the nature of the solute.) Page 2 3. Boiling point of a mixture of two immiscible liquids For a two components system, if the intermolecular forces between the two components (ALB) are very small comparing with the original forces (ALA & BLB), the two liquids will be immiscible with each other. This is an extremely case of positive deviation from Rauolt's law. Under this circumstance, the two components will vaporize independently, the vapour pressure will be the sum of the vapour pressures of the two pure components and independent of the mole fraction of individual component. PT = PAº + PBº Comparing with an ideal solution : PT = XA PAº + XB PBº Since the total vapour pressure will be higher than the vapour pressure of either component at any temperature, the boiling point of the mixture will also be lower than either component. Nitrobenzene and water are two immiscible liquids. Nitrobenzene and water have boiling points 211 ºC and 100 ºC respectively. When they are mixed together, independent of the relative abundance, the mixture boils at 98 ºC. Phase Equilibria a) Steam distillation Unit 4 Page 3 According to the above principle, a heat sensitive organic compound can be separated by steam distillation. Sometimes, ordinary distillation cannot be used to purify heat sensitive organic compound. This is because some organic compounds decompose well before they boil. In order to separate them, the boiling point must be lowered. Most organic compounds are immiscible with water. Thus, water can be added to lower the boiling point of the compound. This is done by passing steam through the organic mixture to be separated or heat the mixture in the presence of large amount of water. Steam provides both the heat and water for the distillation. The method is called steam distillation. Steam distillation Since water and the organic compound are immiscible, water in the distillate can be separated by a separating funnel. Glossary elevation of boiling point depression of freezing point separating funnel vaporization curve immiscible liquids sublimation curve steam distillation triple point temperature steam generator safety tube Phase Equilibria Unit 4 92 2A 2c ii Page 4 Past Paper Question 92 2A 2 c ii 2c ii The total vapour pressure exhibited by a mixture of immiscible liquids is equal to the sum of their individual 4 saturated vapour pressures. (I) Explain why steam distillation enables some heat-sensitive organic compounds to be distilled at a lower temperature than their normal boiling points. (II) A mixture of two immiscible liquids F and G boils at 98°C. At this temperature, the vapour pressures of F and G are 733 mmHg and 27 mmHg respectively. Calculate the percentage of G by mass in the vapour when the mixture boils. (The relative molecular masses of F and G are 18.0 and 123.0 respectively.) (I) A liquid boils when its vapour pressure reaches the value exerted by external pressure. Thus the total vapour pressure exerted by an immiscible mixture of liquids will reach atmospheric pressure at a temperature below the boiling-point of the heat sensitive / most volatile constituent. 1 mark (II) PF = vapour pressure of F = 733 torr PG = vapour pressure of G = 27 torr Since the no. of mole of F is proportional to PF. 1 mark Similarly, the no. of mole of E is proportional to PG as PV = nRT. Let mF and mG be the masses of F and G in the system mF PF nF MF PG = nG = mG MG mf PFMF 733 × 18.0 or m = P M = 27 × 123.0 = 3.97 G GG ⇒ mf = 3.97 mG 1 mark C ii mG mG mG 1 mark Percentage by mass of G = m + m = 3.97 m + m = 4.97 m = 0.201 or 20.1% f G G G G To answer the questions, candidates had to realize that in the vapour phase, the number of moles of F and G are proportional to their vapour pressures and then the candidates needed to set up the relationship PG / PE = nG / nE = (m/M)G / (m/M)E Phase Equilibria Unit 5 Page 1 Topic Reference Reading Phase Equilibria 7.3 Unit 5 Assignment A-Level Chemistry Syllabus for Secondary School (1995), Curriculum Development Council, 205–206 A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd.,189–193 Reading Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 330–333 Three components system Partition of a solute between two phase Solvent extraction Paper chromatography III. Three components system The three components systems that we are going to study consist of 1 solute and 2 solvents. A. Partition of a solute between two phases When a solute is added into a mixture of two immiscible solvents, it dissolves in both of them. At equilibrium, the speed of diffusion from one solvent to another is the same as the reverse speed. Therefore, the concentrations of the solute in both solvents will remain constant. Solute in lower layer d Solute in upper layer And the equilibrium constant of the system is known as partition coefficient or distribution coefficient, Kd. By convention, the concentration of solute in the less dense solvent (upper layer) will be used as the numerator in the equilibrium law. N.B. CCl4 and CHCl3 are the only two common organic solvents denser than water. concentration in the upper layer [Solute(upper layer)] Kd = concentration in the lower layer or [Solute (lower layer)] Kd can also be expressed as [Solute(upper layer)] molar mass of the solute concn in gram per dm3 in the upper layer [Solute(lower layer)] × molar mass of the solute = concn in gram per dm3 in the lower layer Since the numerator and denominator always have the same dimension, Kd always has no unit. The concentration can be expressed in molarity, gcm-3 or gdm-3, the choice of unit has no effect on the value of Kd. Syllabus Notes Phase Equilibria 1. Solvent extraction Unit 5 Page 2 Solvent extraction is an useful technique in organic chemistry. It allows partial removal of solute from one solvent into another immiscible one. This is done by shaking the sample to be purified with an immiscible solvent in a separating funnel. Ether is a slightly polar organic solvent lighter than water. It is commonly used in solvent extraction because it is immiscible with water and volatile. Once the solute is extracted, ether can be evaporated at reduced pressure to obtain the pure sample. N.B. Ether can never be evaporated by a naked flame because it is highly immflammable. High temperature is also undesirable because ether forms explosive peroxide with oxygen at high temperature. a) Calculation Example 1 At 291 K, the partition coefficient of butanoic acid, CH3CH2CH2COOH between ether and water is 3.5. Calculate the mass of butanoic acid extracted by shaking 100 cm3 of water containing 10 g of butanoic acid with 100 cm3 of ether. [Solute(ethereal phase)] Kd = [Solute ] = 3.5 (aqueous phase) Let x g be the mass of acid extracted by 100 cm3 of ether. x / 100 (10 - x) / 100 = 3.5 x = 7.78 Therefore, 7.78 g of butanoic acid will be extracted. Example 2 For the same aqueous solution of butanoic acid, calculate the mass of butanoic acid extracted if two portions of 50 cm3 ether are used. Let x1 g and x2 g be the masses of acid extracted in the two portions. In the first extraction, x1 / 50 (10 - x1) / 100 = 3.5 In the second extraction, = = The mass of acid remaining in the aqueous phase original mass - mass extracted in the first extraction (10 - 6.36) g = 3.64 g x2 = 2.32 x1 = 6.36 x2 / 50 (3.64 - x2) / 100 = 3.5 The total mass extracted by the two 50 cm3 portions of ether = 6.36 g + 2.32 g = 8.68 g Although the same total amount of ether is used, the mass of acid extracted in two portions is larger. For this reason, solvent extraction is usually done with a small amount of solvent at a time. Phase Equilibria 2. Paper chromatography Unit 5 Page 3 Partition of a solute between two phases is not limited to 2 immiscible liquids only. Paper chromatography is also a kind of distribution equilibrium. The solute partitions between the stationary phase and the mobile phase. [Solute(moving phase)] Kd = [Solute ] (stationary phase) In paper chromatography, stationary phase is the layer of water adsorbed on the natural cellulose fibres of filter paper. And, moving phase is the solvent used to carry the solute along the paper. In a mixture of dyes, different dyes have different partition coefficients between the moving phase and stationary phase. The dye with larger partition coefficient will move faster on the paper when the solvent is soaking up. Because of these differences, different dyes can be separated. a) Rf value The colour pattern obtained from paper chromatography is called paper chromatogram. Each spot on the chromatogram is identified by a value called Rf value. It is defined as : distance moved by the coloured spot Rf = distance moved by the solvent front Rf is for a particular compound depends upon the solvent used and the temperature. It always has a value smaller than 1. It is possible to characterize a particular compound separated from a mixture by its Rf value. b) Separation of amino acids A mixture of unknown amino acids can be separated and identified by means of paper chromatography. However, all amino acid are colourless. They must be developed first. The positions of the amino acids in the chromatogram can be detected by spraying with ninhydrin (a developer), which reacts with amino acids to yield highly coloured products. Glossary three components system partition partition coefficient / distribution coefficient separating funnel explosive peroxide paper chromatography stationary phase chromatogram characterize ninhydrin developer adsorbed Rf value solvent extraction mobile phase Phase Equilibria Unit 5 92 2A 2 c i 96 1A 1 e i ii 98 2A 4 b i ii 99 1A 2 c i ii iii Page 4 Past Paper Question 92 2A 2 c i 2c i The distribution coefficient of a compound X between two immiscible liquids D and E is 12. X is more soluble in 3 D. What mass of X can be extracted form 50cm3 of a solution of E which initially contains 4g of X when shaken with 100cm3 of D? Let α g be the mass of X in 1 cm3 of D and β g be the mass of X in 1 cm3 of E Since distribution coefficient is 12. ∴ α = 12 β and 100α + 50β = 4 50α =4 100α + 12 48 α= = 0.0384 1250 The mass of X in liquid D = 100α = 3.84 g 3 marks Ci Many candidates gave the mass of X remaining in E instead of the mass of X extracted from E. They probably did not read the question carefully. 96 1A 1 e i ii 1e At 298 K, 50 cm3 of a solution of I2 in CCl4 were mixed and shaken with 200 cm3 of distilled water in a separating funnel until equilibrium was attained. The two layers were then separated. 20.0 cm3 of the CCl4 layer required 12.15 cm3 of 0.105 M S2O32-(aq) for titration, whereas 50.0 cm3 of the aqueous layer required 8.25 cm3 of 0.0050 M S2O32-(aq). i Calculate the distribution coefficient of I2 between water and CCl4 at 298 K. 2 8.25 × 0.0050 50 water ratio of concentration of I2 : CCl = 12.15 × 0.105 1 mark 4 20 = 0.012934 = 0.013 1 mark (Accept distribution coefficient expressed as [I2(CCl4)]/[I2(aq)], in which case the value calculated is 77.3 (77)) (Award ½ marks for a correct expression of Kd ) ii If the 200 cm3 of distilled water contain some dissolved KI, will this affect the value of the distribution 1 coefficient ? Briefly explain. No change ½ mark because distribution coefficient is a function of temperature only ½ mark (Also accept K will change because there is a change in ionic strength/activity (1/0)) 98 2A 4 b i ii 4b Even though the partition coefficient of ethanoic acid between water and 2-methylpropan-1-ol is 3.05 at 298 K, 2methylpropan-1-ol is still used to extract ethanoic acid from aqueous solutions. Calculate the efficiency of ethanoic acid extraction (in terms of %) at 298 K by shaking 100 cm3 of a 0.50 M aqueous solution of ethanoic acid with i 200 cm3 of 2-methylpropan-1-ol; ii two successive portions of 100 cm3 of 2-methylpropan-1-ol. Decide whether (i) or (ii) is the better extraction method. Explain your answer. 99 1A 2 c i iii 2c A mixture of amino acids, A, B and C, was separated by paper chromatography using an appropriate solvent. The chromatogram obtained is shown below: Phase Equilibria Unit 5 Page 5 (× is the starting point of the mixture) i Briefly describe the principle underlying the separation of A, B and C by paper chromatography. ii The amino acid spots are invisible to naked eyes. Suggest how to make them visible. iii Calculate the Rf value for A. ...
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