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1422-notes-full-2010 - Honors: General Chemistry CHEM 1422...

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Unformatted text preview: Honors: General Chemistry CHEM 1422 Section 1 Spring, 2010 9:10 - 10:30 T-Th E137 Howe Russell Prof. George G. Stanley 614 Choppin Hall 578-3471 E-mail: gstanley@lsu.edu Chapter 15 (not much on E) Thermodynamics: Enthalpy, Entropy & Gibbs Free Energy Thermo 2 Thermodynamics: thermo = heat (energy) dynamics = movement, motion Some thermodynamic terms chemists use: System: the portion of the universe that we are considering open system: energy & matter can transfer closed system: energy transfers only isolated system: no transfers Surroundings: everything else besides the system Isothermal: a system that is kept at a constant temperature by adding or subtracting heat from the surroundings. Heat Capacity: the amount of heat energy required to raise the temperature of a certain amount of material by 1°C (or 1 K). Specific Heat Capacity: 1 g by 1°C Molar Heat Capacity: 1 mole by 1°C Thermo 3 Calorie: the amount of heat required to raise the temperature of 1g of water by 1°C. specific heat of water = 1 cal/g °C 1 calorie = 4.18 joules Specific Heats and Molar Heat Capacities Substance Specific Heat (J/°Cg) Molar Heat (J/°Cmol) Al Cu Fe CaCO3 0.90 0.38 0.45 0.84 24.3 24.4 25.1 83.8 Ethanol 2.43 112.0 Water 4.18 75.3 Air 1.00 ~ 29 important to: engineers chemists EXAMPLE: How many joules of energy are needed to raise the temperature of an iron nail (7.0 g) from 25°C to 125°C? The specific heat of iron is 0.45 J/°Cg. Heat energy = (specific heat)(mass)(T) Heat energy = (0.45 J/°Cg)(7.0 g)(100°C) = 315 J Note that T can be ºC or K, but NOT ºF. When just T is being used in a scientific formula it will usually be kelvin (K). Thermo 4 Problem: How much energy does it take to raise the body temperature 2.5ºC (a fever of just over 103ºF) for someone who weighs 110 pounds (50 kg). Assume an average body specific heat capacity of 3 J/ºC.g. Problem: What would be more effective at melting a frozen pipe – hot water or a hair dryer (hot air gun). Why? Thermo 5 State Functions System properties, such as pressure (P), volume (V), and temperature (T) are called state functions. The value of a state function depends only on the state of the system and not on the way in which the system came to be in that state. A change in a state function describes a difference between the two states. It is independent of the process or pathway by which the change occurs. For example, if we heat a block of iron from room temperature to 100°C, it is not important exactly how we did it. Just on the initial state and the final state conditions. For example, we could heat it to 150°C, then cool it to 100°C. The path we take is unimportant, so long as the final temperature is 100°C. Miles per gallon for a car, is NOT a state function. It depends highly on the path: acceleration, speed, wind, tire inflation, hills, etc. Most of the thermodynamic values we will discuss in this chapter are state functions. Thermo 6 Energy: "The capacity to do work and/or transfer heat" Forms of Energy: Kinetic (Ekinetic = ½mv 2) Heat Light (& Electromagnetic) Electricity Sound Potential Gravitational Chemical Nuclear - Matter (E = mc 2) WORK Thermo 7 First Law of Thermodynamics: The total amount of energy (and mass) in the universe is constant. In any process energy can be changed from one form to another; but it can never be created nor destroyed. “You can't get something for nothing” Thermo 8 Enthalpy (Heats) of Reaction The amount of heat released or absorbed by a chemical reaction at constant pressure (as one would do in a laboratory) is called the enthalpy or heat or reaction. We use the symbol H to indicate enthalpy. Sign notation (EXTREMELY IMPORTANT!!): +H indicates that heat is being absorbed in the reaction (it gets cold) endothermic H indicates that heat is being given off in the exothermic reaction (it gets hot) Standard Enthalpy = H° (° is called a “not”) Occurring under Standard Conditions: Pressure Concentration 1 atm (760 torr) 1.0 M Temperature is not defined or part of Standard Conditions, but is often measured at 298 K (25°C). Thermo 9 Standard Enthalpy of Formation -- H f The amount of heat absorbed (endothermic) or released (exothermic) in a reaction in which one mole of a substance is formed from its elements in their standard states, usually at 298 K (25°C). Also called heat of formation. H = 0 for any element in its standard state (the f natural elemental form at 1 atm or 1 M) at 298 K. EXAMPLES: C(graphite, s) + O 2 (g) H° = 0 kJ/mol rxn CO2 (g) 0 kJ/mol elements in their standard states negative sign heat released -- exothermic rxn 393.5 kJ/mol product (one mole) H° (CO2 ) = f 393.5 kJ/mol Thermo 10 2H2 (g) + O 2 (g) H° = 0 kJ/mol rxn 0 kJ/mol 2H2 O (g) 483.6 kJ/ 2 mol elements in their standard states product (two moles) divide by 2 to put on per mole basis!! negative sign heat released -- exothermic rxn H° (H2 O) = f 241.8 kJ/mol Note that I usually will not have you calculate Hfº on homeworks or tests – so you generally don’t have to worry about normalizing your answer to a per mole basis. Hess's Law -- Adding Reactions The overall heat of reaction (Hrxn) is equal to the sum of the Hf (products) minus the sum of the Hf (reactants): H° = rxn (# eqiv) H° (products) (# eqiv) H° (reactants) f f Therefore, by knowing Hf of the reactants and products, we can determine the Hrxn for any reaction that involves these reactants and products. Thermo 11 EXAMPLE: CO2 is used in certain kinds of fire extinguishers to put out simple fires. It works by smothering the fire with "heavier" CO2 that replaces oxygen needed to maintain a fire. CO2 is not good, however, for more exotic electrical and chemical fires. 2Mg(s) H° = 0 kJ/mol f + CO2 (g) 2MgO(s) - 393 kJ/mol - 602 kJ/mol REACTANTS H° = rxn H° = rxn + C(s) 0 kJ/mol PRODUCTS (# eqiv) H° (products) (# eqiv) H° (reactants) f f (2 eqiv)(-602 kJ/mol) + (1 eqiv)(0 kJ/mol) (2 eqiv)(0 kJ/mol) + (1 eqiv)(-393 kJ/mol) H° = (-1204 kJ/mol) rxn H° = (-1204 kJ/mol) rxn H° = rxn 811 kJ/mol (-393 kJ/mol) + 393 kJ/mol } highly exothermic rxn !! Therefore, Mg will "burn" CO2 ! Thermo 12 You can also add two reactions together to get the Hrxn for another new reaction: Calculate H°xn for the following reaction: r C2 H4 (g) + H 2O(l) H° = ?? rxn C 2 H 5OH(l) Given these two reactions and thermodynamic data: a) C 2 H 5OH(l) + 3O 2(g) 2CO2 (g) + 3H 2 O(l) H° = -1367 kJ/mol rxn b) 2CO2 (g) + 2H 2 O(l) H° = -1411 kJ/mol rxn C2 H4 (g) + 3O 2(g) How to solve: 1) C 2 H 5OH is on the product side of the first reaction -- so we want to switch equation a) around to get C 2 H 5OH also on the product side: 2CO2 (g) + 3H 2 O(l) C 2 H 5OH(l) + 3O 2(g) H° = +1367 kJ/mol rxn * note that when we reverse the reaction, H° rxn changes sign!!! 2) Now we can add the two reactions together to give us the desired net reaction: 1 2CO2 (g) + 3H 2 O(l) C 2 H 5OH(l) + 3O 2(g) H° = +1367 kJ/mol rxn + C2 H 4 (g) + 3O 2(g) 2CO2 (g) + 2H 2 O(l) C2 H 4 (g) + H 2O(l) C 2 H 5OH(l) H° = -1411 kJ/mol rxn H° = -44 kJ/mol rxn If we have to multiply one (or more) of the reactions by some constant to get them to add correctly, then we also would have to multiply Hrxn for that reaction by the same amount. Thermo 13 Chemists use bomb calorimeters to measure Enthalpies of formation or reaction. Bomb Calorimeter Stirrer Electrical contacts to initate sample combustion Thermocouple to measure temperature Sample placed inside inner container Water Highly insulated outside container Thick-walled inner container (bomb) to contain combustion of sample (pressurized with O2) In order to use this effectively one must know the heat capacity of the bomb (inner part) and water bath. By measuring the temperature increase of the water one can calculate the amount of heat given off during the combustion process. Thermo 14 Problem: Calculate Hrxn for the following reactions given the following H values: f H (SO2, g) = 297 kJ/mol f H (SO3, g) = 396 kJ/mol f H (H2SO4, l) = 814 kJ/mol f H (H2SO4, aq) = 908 kJ/mol f H (H2O, l) = 286 kJ/mol f H (H2S, g) = 20 kJ/mol f a) S(s) + O2(g) b) 2SO2(g) + O2(g) SO2(g) 2SO3(g) c) SO3(g) + H2O(l) H2SO4(l) d) 2H2S(g) + 3O2(g) 2SO2(g) + 2H2O(l) Thermo 15 Internal Energy -- E The internal energy, E, represents all the energy contained within a material. It includes kinetic energy (heat), intra- and intermolecular forces (bond energies, electrostatic forces, van der Waals), and any other forms of energy present. As with enthalpy, H, the absolute value can’t (or is extremely difficult) to define. What we can track is the change in E: E = Efinal – Einital = Eproducts – Ereactants A key relationship is: E = q + w Where q = heat and w = work performed on or by the system. Sign notations: q = positive = heat added to system (adds energy) q = negative = heat removed (removes energy) w = positive = work done on system (adds energy) w = negative = work done by system (removes energy) Thermo 16 The most common type of work involves pressure/volume changes: e.g., explosion of gasoline vapors in an internal combustion engine. The explosion creates a dramatic pressure and volume increase that pushes the piston and creates work. If one has a constant volume situation, then no pressure/volume work will be done and w = 0. So under constant volume conditions: E = q The change in internal energy is, therefore, equal to the amount of heat added or removed from the material (system). Thermo 17 Relationship Between E & H E = q (at constant volume and temperature) H = q (at constant pressure and temperature) The difference is that volume changes occur for H and that typically involves work of some type. Remember that significant volume changes only occur when gasses are involved. So we only need to be concerned about volume work when there is a change in the amount of gas produced in a chemical reaction. The relationship between H and E is defined, therefore, as: H = E + (n)RT Where R = gas constant; T = temperature in kelvin, and: n = equivalents (moles) of product gas – equivalents (moles) of reactant gas If n = 0, then H = E. But even when n ≠ 0, the PV work component is usually small. See example in textbook (pages 574-575). Thermo 18 Entropy The final state of a system is more energetically favorable if: 1. Energy can be dispersed over a greater number and variety of molecules. 2. The particles of the system can be more dispersed (more disordered). The dispersal of energy and matter is described by the thermodynamic state function entropy, S. The greater the dispersal of energy or matter in a system, the higher is its entropy. The greater the disorder (dispersal of energy and matter, both in space and in variety) the higher the entropy. Adding heat to a material increases the disorder. Ice - well ordered structure water vapor - most disordered water - more disordered Thermo 19 Unlike H, entropy can be defined exactly because of the Third Law of Thermodynamics: Third Law of Thermodynamics: Any pure crystalline substance at a temperature of absolute zero (0.0 K) has an entropy of zero (S = 0.0 J/K mol). Sign notation (EXTREMELY IMPORTANT!!): +S indicates that entropy is increasing in the reaction or transformation (it's getting more disordered -- mother nature likes) S indicates that entropy is decreasing in the reaction or transformation (it's getting less disordered {more ordered} -- mother nature doesn't like, but it does happen) Thermo 20 Qualitative "Rules" About Entropy: 1) Entropy increases as one goes from a solid to a liquid, or more dramatically, a liquid to a gas. 250 Entropy (J/mol) Solid 200 Gas Liquid 150 phase transitions 100 50 0 Temperature (K) 2) Entropy increases if a solid or liquid is dissolved in a solvent. 3) Entropy increases as the number of particles (molecules) in a system increases: N2O4(g) S° = 304 J/K (1 mole) 2NO2(g) S° = 480 J/K (2 moles) These first 3 above are most important for evaluating Srxn. Thermo 21 The rules below are mainly for comparing the entropy of individual molecules or materials. 4) The Entropy of any material increases with increasing temperature 5) Entropy increases as the mass of a molecule increases S°(Cl2(g)) > S°(F2(g)) S° = 165 J/Kmol S° = 158 J/Kmol 6) Entropy is higher for weakly bonded compounds than for compounds with very strong covalent bonds S°(graphite) > S°(diamond) S° = 5.7 J/Kmol S° = 2.4 J/Kmol Note that for individual molecules (materials) the higher the entropy, the more likely the molecule will want to “fall apart” to produce a number of smaller molecules. Thermo 22 7) Entropy increases as the complexity (# of atoms, # of heavier atoms, etc.) of a molecule increases Entropy of a Series of Gaseous Hydrocarbons H H C Methane H H S° = 186 J/Kmol H H C C S° = 201 J/Kmol Acetylene H H C Ethylene C H H H H C C H H H S° = 230 J/Kmol H Ethane H H S° = 220 J/Kmol H H C C C H H H H Propane S° = 270 J/Kmol Thermo 23 What are the biggest factors for evaluating Srxn for a chemical rxn? 1) phase change 2) change in # of molecules Problem: For the following reactions, is the entropy of the reaction increasing or decreasing? a) Ag+(aq) + Cl-(aq) AgCl(s) b) H2CO3(aq) H2O + CO2(g) c) Ni(s) + 4CO(g) d) H2O(s) Ni(CO)4(l) H2O(l) e) graphite diamond 2Na+(aq) + 2OH-(aq) + H2(g) f) 2Na(s) + 2H2O g) H2S(g) + O2(g) h) 2H2O(l) H2O(l) + SO(g) 2H2(g) + O2(g) i) CO2(g) + CaO(s) CaCO3(s) j) CaCl2(s) + 6H2O(l) k) 2NO2(g) N2O4(g) CaCl26H2O(s) Thermo 24 Just as with enthalpies, one can calculate entropies of reaction. S ° = rxn (# eqiv) S°(products) (# eqiv) S°(reactants) EXAMPLE: 2Mg(s) + CO2 (g) 2MgO(s) Sf° = 32 J/K·mol 215 J/K·mol 27 J/K·mol REACTANTS S ° = rxn S ° = rxn + C(s) 6 J/K·mol PRODUCTS (# eqiv) S°(products) (# eqiv) S°(reactants) (2 eqiv)(27 J/K·mol) + (1 eqiv)(6 J/K·mol) (2 eqiv)(32 J/K·mol) + (1 eqiv)( 214 J/K·mol) S ° = (60 J/K.mol) rxn S ° = rxn 218 J/K.mol (278 J/K.mol) } entropy is decreasing (reaction is becoming more ordered) Thermo 25 Spontaneous Processes A process that takes place without the net input of energy from an external source is said to be spontaneous (not instantaneous). 1) Rxn of sodium metal with water: 2Na+(aq) + 2OH-(aq) + H2(g) 2Na(s) + 2H2O 2) Combustion rxns: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) 2H2(g) + O2(g) 2H2O(l) 3) Expansion of a gas into a vacuum xCO2(g) yCO2(s) + zCO2(g) (x = y + z) 4) A salt dissolving into solution: NH4NO3(s) + H2O(l) NH4+(aq) + NO3-(aq) Thermo 26 Second Law of Thermodynamics: In any spontaneous process the entropy of the universe increases Suniverse = Ssystem + Ssurroundings Second Law (variant): in trying to do work, you always lose energy to the surroundings. You can't even break even! Neither entropy (S) or enthalpy (H) alone can tell us whether a chemical reaction will be spontaneous or not. An obvious (?) conclusion is that one needs to use some combination of the two. Thermo 27 Gibbs Free Energy The combination of entropy, temperature and enthalpy explains whether a reaction is going to be spontaneous or not. The symbol G is used to define the Free Energy of a system. Since this was discovered by J. Willard Gibbs it is also called the Gibbs Free Energy. "Free" energy refers to the amount of energy available to do work once you have paid your price to entropy. Note that this is not given simply by H, the heat energy released in a reaction. Gº = Hº TSº When G is negative, it indicates that a reaction or process is spontaneous. A positive G indicates a non-spontaneous reaction. Thermo 28 G = H TS S + G = negative G = ?? spontaneous at all temperatures spontaneous at high temperatures - + 0 G = ?? G = positive spontaneous at low temperatures H non-spontaneous at all temperatures Spontaneous = exoergic (energy releasing) Non-spontaneous = endoergic (energy releasing) Thermo 29 Remember that entropies are given in units of J/Kmol while enthalpies and free energies are in kJ/mol. DON'T forget to convert all units to kJ or J when using both S and H in the same equation!! Thermo 30 Gº vs. G: Standard vs. Non-Standard Conditions Remember that the º (“not”) on Gº indicates that the numerical value of Gº is based on the reaction at standard conditions (1 M solution concentration, 1 atm gas pressure). Temperature is NOT part of standard conditions! As soon as one has a concentration different than 1 M or 1 atm pressure, the º “not” goes away and one has G. Consider the reaction: Initial: 1 atm 1 atm 2SO2(g) + O2(g) 1 atm 2SO3(g) Gºrxn = 142 kJ/mol The Gºrxn of 142 kJ/mol is for when each gas is present with a concentration of 1 atm. This indicates that the reaction under these conditions will proceed to make products (spontaneous). As the reactants start reacting, however, their concentrations decrease (SO2 twice as fast as O2) and Gº turns into G and becomes less negative. When G = 0 the reaction has reached equilibrium. Although for this rxn, SO2 is probably the limiting reagent (not enough present to complete the rxn). Thermo 31 Example: Calculate Gºf for CO2 at 298 K. Hf ° (CO2) = 393 KJ/mol, S° (O2) = 205 J/mol•K, S° (C) = 6 J/mol•K, S° (CO2) = 213 J/mol•K C(graphite) + O2(g) CO2(g) Gºf = Hºf TSºf Note change in Sºf = Sºprod Sºreact units – J to KJ Sºf = (213 J/mol•K) (205 + 6 J/mol•K) Sºf = 2 J/mol•K) Gºf = (393 KJ/mol) (298 K)(0.002 KJ/mol•K) Gºf = (393 KJ/mol) (1 KJ/mol) Gºf = 394 KJ/mol DANGER!! Common mistake!! Problem: Calculate Gºf for CO at 298 K. Hf ° (CO) = 110 KJ/mol, S°(O2) = 205 J/mol•K, S°(C) = 6 J/mol•K, S°(CO) = 198 J/mol•K 2C(graphite) + O2(g) 2CO(g) Thermo 32 Just as with enthalpies and entropies, one can calculate free energies of reaction. G° = rxn (# eqiv) G° (products) (# eqiv) G° (reactants) f f EXAMPLE: 2Mg(s) G° = 0 kJ/mol f + CO2 (g) 2MgO(s) - 394 kJ/mol - 570 kJ/mol REACTANTS G° = rxn G° = rxn + C(s) 0 kJ/mol PRODUCTS (# mol) G° (products) (# mol) G° (reactants) f f (2 mol)(-570 kJ/mol) + (1 mol)(0 kJ/mol) (2 mol)(0 kJ/mol) + (1 mol)(-394 kJ/mol) G° = (-1140 kJ) rxn (-394 kJ) G° = (-1140 kJ) + 394 kJ rxn G° = rxn 746 kJ } SPONTANEOUS rxn! highly exothermic rxn !! Compare to H ° which was -811 kJ for the same rxn. rxn The "missing" 65 kJ of energy went to ENTROPY! Thermo 33 Example: To make iron, a steel mill takes Fe2O3 (rust or iron ore) and reacts it with coke (a complex, impure form of carbon) to make iron and CO2. Based on the data below, this is a non-spontaneous reaction at room temperature, but it becomes spontaneous at higher temperatures. Assuming that H° and S° do not change much with temperature, calculate the temperature above which the reaction becomes spontaneous (i.e., G°rxn = 0). H°rxn = +465 kJ/mol S°rxn = +552 J/molK (or 0.552 kJ/molK) G°rxn = +301 kJ/mol (at 298 K) G°rxn = H°rxn TS°rxn as we raise the temperature, G° will eventually reach 0 and then go negative & spontaneous, so let G° = 0 and solve for T, the temperature at which this will happen: 0 = H°rxn TS°rxn rearranging to solve for T gives: T = (H°rxn) / (S°rxn) T = (465 kJ/mol) / (0.552 kJ/molK) T = 842 K (above this temperature G°rxn will be negative – we will have a spontaneous reaction) Thermo 34 Problem: Calculate Gºrxn for the following. Gºf (SO2, g) = 300 kJ/mol Gºf (SO3, g) = 371 kJ/mol Gºf (H2SO4, l) = 690 kJ/mol Gºf (H2SO4, aq) = 742 kJ/mol Gºf (H2O, l) = 237 kJ/mol Gºf (H2S, g) = 34 kJ/mol a) S(s) + O2(g) b) 2SO2(g) + O2(g) SO2(g) 2SO3(g) c) SO3(g) + H2O(l) H2SO4(l) d) 2H2S(g) + 3O2(g) 2SO2(g) + 2H2O(l) Thermo 35 Comparisons of Hºrxn and Gºrxn S(s) + O2(g) SO2(g) Hºrxn = 297 kJ/mol Gºrxn = 300 kJ/mol 2SO2(g) + O2(g) Sºrxn = 11 J/mol K 2SO3(g) Hºrxn = 198 kJ/mol Gºrxn = 142 kJ/mol SO3(g) + H2O(l) H2SO4(l) Hºrxn = 132 kJ/mol Gºrxn = 82 kJ/mol 2H2S(g) + 3O2(g) Hºrxn = 1126 kJ/mol Gºrxn = 1006 kJ/mol 2SO2(g) + 2H2O(l) Thermo 36 Entropy of Fusion and Vaporization While the entropy of a substance increases steadily with increasing temperature, there is a considerable jump in the entropy at a phase transition: 250 Entropy (J/mol) Solid 200 Gas Liquid 150 100 phase transitions 50 0 Temperature (K) This jump at the melting point is called the entropy of fusion, Sfusion, and as you might expect, it is related to the enthalpy (or heat) of fusion, Hfusion: Hfusion S fusion Tm Thermo 37 The jump in entropy at the boiling point is called the entropy of vaporization, Svaporization, and it is related to the enthalpy of vaporization, Hvaporization: Hvaporization S vaporization Tb EXAMPLE: What is the boiling point of bromine (Br2)? Svapor = 93.2 J/Kmol and Hvapor = 30.9 kJ/Kmol. Hvaporization S vaporization Tb but we want to solve for Tb, the boiling point temperature, so we need to rearrange the formula: Hvaporization Tb Svaporization 30.9 103 J / mol Tb 93.2 J / K mol Tb 331.5K 58.3 C Note how we have to convert kJ/mol to J/mol in order for the units on H to be the same as the units on S!! To convert kJ to J we need to multiply the kJ value by 1000. Thermo 38 Problem: Calculate the boiling point for ethanol (CH3CH2OH) from the data in the thermodynamic tables. (literature value = 78.5°C) Thermo 39 Thermo 40 th From “General Chemistry”, 7 Ed, by Whitten, Davis, Peck & Stanley. Thomson Brooks/Cole Publisher. Kinetics 1 Chapter 16 (no derivation of kinetic rate laws) Kinetics 2 Chemical Kinetics is the area of chemistry that is concerned with finding out how fast a chemical rxn will go and what is the mechanism by which it proceeds? The mechanism of a rxn is how the reactants react in a step-wise fashion to form the product molecules. Thermodynamics only tells us if the rxn is energetically possible (exothermic, endothermic, spontaneous) – nothing about the rate of the rxn or how it will proceed (i.e., the mechanism of the rxn). Kinetics provides us with the techniques for measuring the rate of a reaction and gaining information about how the reaction is proceeding. This, in turn, can give us valuable information about the mechanism of the rxn. Kinetics 3 Consider the very spontaneous and exothermic rxn of H2 + O2 to produce H2O: Note: most text books use E (potential energy) for the vertical axis. I will generally use ΔG The activation energy or activation barrier, when present and large enough, is what prevents a spontaneous reaction from reacting instantly! Kinetics 4 The activation energy arises due to kinetic effects, i.e., the rates at which molecules react together and the factors that affect those rates of reactions. Consider the rearrangement of methyl isocyanide into the more stable acetonitrile molecule: H3C N C methyl isocyanide N C CH3 acetonitrile This reaction involves the migration (movement) of the methyl group (H3C or CH3) from the nitrogen atom over to the terminal carbon atom. Kinetics 5 The initial movement of the methyl group involves partial breaking of chemical bonds between the methyl group and nitrogen atom and the C-N triple bond, both of which take energy. The topmost part of the activation energy barrier is called the transition state, which represents the highest energy part of the reaction pathway from reactants to products. The structure of the molecule at the transition state is something that one can never directly observe or isolate!! We can only speculate (or calculate from fancy quantum mechanical programs) what it may look like. But once we hit the transition state and the methyl group starts moving towards the terminal carbon atom to form a new and stronger bond, the same amount of energy used to move up to the top of the activation barrier from the reactant side is released. So there is NO NET CONSUMPTION OF ENERGY from the activation barrier – the total amount of energy released in the reaction is still just ΔG. Kinetics 6 But if the activation barrier is high enough, we may need to add some (or a lot of) heat (energy) to the reactants to give them enough energy to make it over the barrier. But we get that energy back when they go down the other side of the activation barrier to the products!! On the other hand, if a reaction has a very small activation barrier, the reactants will react extremely quickly (sometimes instantaneously) to make products! For endoergic reactions, the activation energy is defined as the entire height of the barrier: Kinetics 7 If the potential energy curve/activation barrier has a “dip” in it, the reaction is proceeding through an intermediate. An intermediate is a molecule that is stable enough to exist for some period of time. Unlike the transition state, which only exists for an instant in time. The deeper the “dip” or “well”, the more stable the intermediate species (molecule) will be. Sometimes the intermediate will be stable enough to isolate and study. Kinetics 8 Problem: Consider these potential energy diagrams: a) P ΔG b) c) R ΔG ΔG R R Rxn Coordinate e) P Rxn Coordinate P ΔG P g) ΔG R ΔG R Rxn Coordinate R P Rxn Coordinate f) ΔG d) P Rxn Coordinate Rxn Coordinate h) ΔG R R P Rxn Coordinate P Rxn Coordinate Which is the: • the most endoergic reaction? • the spontaneous reaction that will go the slowest? • he reaction that will go the fastest? • the reaction that will have the most stable intermediate species? • the lowest temperature dependence? • the highest temperature dependence? • the highest activation energy (a little tricky) • the most exoergic reaction? • the reaction that is going through the most reaction steps? Kinetics 9 A factor that helps “create” the activation barrier is that when two molecules collide to react together, they usually must: 1) collide with enough energy to react 2) collide with the proper orientation to react 3) react! Consider the following two molecules that are trying to react. CO CO OC Mo CO OC no reaction! CO CO OC Mo CO OC no reaction! CO CO CO CO OC Mo CO OC no reaction! CO CO OC Mo CO OC no reaction! CO CO CO OC Mo CO OC CO OC Mo CO OC CO CO Kinetics 10 Elementary or Fundamental Reaction Steps These are the simplest reactions that can occur and they pretty much fall into two categories: unimolecular and bimolecular reaction steps. A unimolecular reaction is one in which a single molecule does something by itself and converts into another molecule. It is not dependent on a collision between two molecules. For example: CO CO OC Mo CO OC Mo CO OC CO OC CO + CO CO A bimolecular reaction is one in which two molecules collide in the proper orientation to react together to form a new product(s). CO CO OC Mo CO OC CO + PMe3 OC Mo CO OC CO PMe3 Kinetics 11 Putting these two fundamental steps together allows us to propose a mechanism for the substitution of a carbonyl ligand on the Mo(CO)6 with a PMe3 ligand. Overall rxn: Proposed mechanism consistent with experimental data & kinetic measurements: Of these two fundamental rxn steps, #1 is the slow or rate determining step. Kinetics 12 Factors that affect the kinetics (rate) of a rxn: 1) The nature of the reactants and products (thermodynamics, ΔG) 2) Concentration – rxns usually go faster when the reactants are more concentrated 3) Temperature – rxns almost always go faster at higher temperatures 4) The presence of an effective Catalyst will speed up a chemical reaction 5) The surface area of a solid reactant or solid catalyst – the higher the surface area the faster the reaction (dissolving, catalysis) will occur For chemical reactions rates are usually expressed as: moles liter−1 sec−1 or M sec−1 Kinetics 13 The rate or speed of a reaction is defined as the change in concentration of reactants (which are consumed) or products (which are formed) over some period of time. Consider the reaction of A → B. Let’s start with a 1.0 M solution of A, with no B initially present. The following is a plot of the conversion of A into B versus time. Note that we can either monitor the production of product or the loss of reactant in this reaction. Kinetics 14 The rate for most reactions constantly changes (except at the end when it is zero) during the course of the reaction. The rate at any point in time is calculated by taking the slope of the tangent to the production or consumption curves. At very early times (for example, the first 10 seconds in the rxn above), the rate of formation is essentially linear and gives one the fastest rate of the reaction. This is called the initial rate of reaction. Kinetics 15 Once the rate of product formation starts to curve downwards, one needs to measure the rate graphically by the tangent method or via kinetic rate expressions (math formulas) that we will discuss below. If we look at the data that we plotted more carefully we can see a trend from which we can construct a mathematical relationship. A (M) rate (M sec-1) 1.00 14.0 x 10-3 0.50 7.00 x 10-3 0.25 3.50 x 10-3 0.125 1.75 x 10-3 We find that as the concentration of A doubles, the rate of reaction doubles (or if it is cut in half, the rate is cut in half). This leads us to the following mathematical relationship: rate of reaction = k[A] This is called a kinetic rate expression. Kinetics 16 k is a constant (it does change with temperature!) that we call the rate constant. Using the first line of data from the table when A = 1 M, we can easily solve for the value of k = 0.014 sec-1. From this we can calculate the rate of the reaction at any concentration of A. The bigger the rate constant, the faster the reaction (so long as there are some reactants present). Note that as the concentration of the reactant gets lower, the reaction goes slower. When it gets to 0, the reaction stops and is over. Since there is only one reactant (raised to the first power) in the kinetic rate expression, we refer to this reaction as a First Order Reaction. The units on the rate constant are set so as to ensure that the rate of reaction is always moles liter-1 sec-1 (or M sec-1). Kinetics 17 First Order Reactions are fairly common – for example, all radioactive atoms decay via first order kinetics. If we carefully examine the plot of first order data again we can see where the term HalfLife comes from and that it applies to a reaction with first order kinetic behavior. A half-life (often abbreviated t½ ) is the amount of time it takes for ½ of a reactant to be converted into product. Kinetics 18 For a First Order rxn, the half-life is independent of the reactant (or product) concentration and is defined as: t½ = 0.693 k Where k is the rate constant. The unit on t½ is typically seconds, the reciprocal of that on k (sec−1) While the kinetic rate expression for a first order reaction looks quite simple, it takes calculus to convert it into an equation that allows you to calculate the concentration of the reactant or product over time (as shown in the plots on the previous pages). One also needs to determine the rate constant k from experimental measurements. [ A t ] = [ A0 ]e - k t [At] is the concentration of reactant A at time t. [A0] is the initial or starting concentration of A at the beginning of the reaction. k is the rate constant, and t is the time in the same units as the rate constant. Kinetics 19 Bimolecular reactions (such as A + B → C) have Second Order Kinetics, for which we can write the following rate equation: rate = k[A][B] The total order of a reaction is defined as the sum of the powers on the reactant concentrations in the rate expression (equation). In this case [A] and [B] are both raised to the 1st power, so the total order = 1 + 1 = 2. So this is a Second Order Rxn. It is important to note that the order on each reactant concentration (the power to which it is raised) can be 0, fractional, or even negative!! THE ORDER IS NOT RELATED TO THE COEFFICENTS ON THE BALENCED CHEMICAL EQUATION, as with equilibrium constants!!! IT MUST BE DETERMINED EXPERIMENTALLY!!! Although a bimolecular rxn is second order and a unimolecular rxn is first order, the opposite is not necessarily true!! Kinetics 20 The orders of fundamental rxns, however, are based on their coefficients. The trick is knowing which fundamental rxn defines the rate determining step, and then to express the overall rate law in terms of reactants. Problem: Consider the following reaction: 3H2 + N2 2NH3 Which of the following correctly describes the kinetic rate expression for this reaction? 3 a) rate = k[H2][N2] 3 b) rate = k[H2] [N2] 2 3 c) rate = k[H2] [N2] d) rate = k[H2][N2] e) not enough data to determine Problem: Consider the following bimolecular reaction: Ni(CO)3 + CO → Ni(CO)4 Which of the following correctly describes the kinetic rate expression for this reaction? a) rate = k[Ni(CO)3][CO] b) rate = k[Ni(CO)3]3[CO] c) rate = k[Ni(CO)3] d) rate = k[CO] e) not enough data to determine Kinetics 21 Second order rate equations: 1 1 − =kt [ A t ] [ A0 ] or, rearranging to solve for [A]: [At ] = 1 ⎛ 1⎞ ⎜ k t + [A ] ⎟ ⎝ 0⎠ The half-life for a second order reaction is defined as: 1 t 1 /2 = k[A 0] Plotted below is a comparison of the first order and second order kinetic plots, both with rate constants set to 0.014. Kinetics 22 Problem: Qualitatively explain why the production of product (or consumption of reactants is significantly slower for the second order rxn relative to the first order rxn in the graph above? Kinetics 23 Example: Given the following experimental kinetic data, determine the kinetic rate expression and overall order of reaction. 2HgCl2 + C2O42− Hg2Cl2 + 2CO2 + 2Cl− Exp # [HgCl2] [C2O42-] Initial Rate 1 0.10 M 0.15 M 2 x 10-5 2 0.10 M 0.30 M 8 x 10-5 3 0.05 M 0.30 M 4 x 10-5 Our tentative rate expression for this reaction is: rate = k[HgCl2]x[C2O42−]y We need to determine what the exponents (orders) x and y are on the reactant concentrations. Then we can solve for the rate constant k. Step #1: find two experiments where the concentration of one of the reactants stays the same. Since the rate constant k is a constant and the concentration of one of the components is not changing (also a constant for this comparison), we only have to worry about the one reactant that is changing. Kinetics 24 We can then set up the following proportionality: y [Aexp#2] y [Aexp#2] Rate(exp #2) = y = [A [Aexp#1] Rate(exp #1) exp#1] 8 x 10 -5 2 x 10 -5 = 0.30 y 0.15 y (2) 4= If you can’t solve this by inspection, then take the log of each side of the equation: log(4) = y log(2) Rearrange and solve for y: log(4) y= log(2) = 0.602 0.301 =2 Kinetics 25 Step #2: Repeat for the other reactant concentration terms in the rate equation: [Aexp#2] x Rate(exp #2) = Rate(exp #3) [Aexp#3] 8 x 10 -5 4 x 10 -5 = x 2 = (2) 0.10 x 0.05 x=1 Now that we have solved for the orders of the kinetic rate expression, we can write out the rate equation: rate = k[HgCl2][C2O42−]2 The overall order of the rate expression is the sum of the individual orders (x + y): 1 + 2 = 3. So this is called a third order rxn or rate law. Note that the orders DO NOT correspond to the coefficients on the chemical equation. Many times they do, but just as many times they won’t. There is NO FORMAL CONNECTION (except for fundamental rxns)!! Common Mistake ! Kinetics 26 Step # 3: Calculate the rate constant (if asked for). We now just plug in the experimental data from any one of the experiments and rearrange and solve for k, the rate constant. For this example we will use the data from experiment # 1. rate = k[HgCl2 ] [C2 O42- ] 2 k= k= k= rate [HgCl2 ] [C2 O4 2- ] 2 2 x 10-5 M sec -1 (0.10 M) (0.15 M ) 2 2 x 10-5 M sec -1 2.25 x 10-3 M 3 -3 k = 8.88 x 10 M -2 sec -1 Kinetics 27 Example: Given the following experimental kinetic data, determine the kinetic rate expression and overall order of reaction. Mo(CO)6 + PMe3 Mo(PMe3)(CO)5 + CO Exp # [Mo(CO)6] [PMe3] Initial Rate 1 0.10 M 0.10 M 0.3 2 0.10 M 0.30 M 0.3 3 0.30 M 0.10 M 0.9 Our tentative rate expression for this reaction is: rate = k[Mo(CO)6]x[PMe3]y First, let’s solve for [PMe3]y: Rate(exp #2) Rate(exp #1) 0.3 0.3 = 1 = (3) 0.30 = [Bexp#2] y [Bexp#1] y 0.10 y y=0 So the order on PMe3 is zero! This means that it does NOT appear in the kinetic rate expression! Kinetics 28 Now, let’s solve for [Mo(CO)6]x: Rate(exp #3) = Rate(exp #1) 0.9 0.3 = x 3 = (3) [Aexp#3] x 0.30 [Aexp#1] x 0.10 x=1 So the final kinetic rate law (expression, equation) is: rate = k[Mo(CO)6] This, therefore, is a First Order Rxn. Kinetics 29 Problem: What is the kinetic rate law for the following reaction (figure out w, x, y and z below) H catalyst = OC Co CO H2C CH2 + C O + H2 ethylene Exp H2C=CH2 CO CO O H C H2 CH3 adehyde CO H2 Catalyst Initial Rate 1 2M 0.04 M 0.06 M 0.001 M 4 × 10−4 2 1M 0.04 M 0.06 M 0.001 M 2 × 10−4 3 1M 0.02 M 0.06 M 0.001 M 4 × 10−4 4 1M 0.02 M 0.03 M 0.001 M 2 × 10−4 5 1M 0.02 M 0.03 M 0.0005 M 1 × 10−4 rate = k[H2C=CH2]w[CO]x[H2]y[Catalyst]z Kinetics 30 Problem: Consider the following reaction and kinetic data. What is the kinetic rate expression? Ni(CO)4 + CNCH3 Ni(CO)3(CNCH3) + CO Exp # [Ni(CO)4] [CNCH3] Initial Rate (Msec−1) 1 2 3 4 0.04 0.04 0.08 0.08 0.06 0.12 0.06 0.12 0.2 × 10−5 0.2 × 10−5 0.4 × 10−5 0.4 × 10−5 Problem: What is the rate constant k for the following reaction: O C C OC C O O O C O C O C C Mn Mn C O O C CO + PMe3 OC C C O O O O C Mn Mn C O PMe3 C O CO C O Exp [Mn2] [PMe3] Initial Rate 1 0.2 M 0.5 M 1 M/sec 2 0.4 M 0.5 M 1.4 M/sec 3 0.4 M 1.0 M 2.8 M/sec + CO Kinetics 31 Problem: Consider the following reaction and kinetic data. What is the rate law and rate constant? CO(g) + H2O(g) CO2(g) + H2(g) Exp # [CO] [H2O] Initial Rate (M sec−1) 1 2 3 4 5 1.5 1.5 3.0 1.5 3.0 1.5 4.5 4.5 0.75 3.0 0.2 0.6 1.2 0.1 0.8 Kinetics 32 Temperature, Rates & the Arrhenius Equation One can calculate the rate constant for a reaction using the Arrhenius equation: k = Ae − Ea / RT A is a constant representing the fraction of collisions between molecules having the correct orientation to react when the reactants have a concentration of 1 M. It is generally not known. But A can be factored out if one ratios the equation at two temperatures. One typically uses this modified Arrhenius equation to either calculate Ea values from rate constants derived from experiments at different temperatures, or a rate constant at a different temperature if one has already determined Ea. 1⎞ ⎛ k 2 ⎞ Ea ⎛ 1 ln ⎜ ⎟ = ⎜T −T ⎟ ⎝ k1 ⎠ R ⎝ 1 2⎠ k2 R ln k1 Ea = 1⎞ ⎛1 ⎜T −T ⎟ ⎝1 2⎠ Kinetics 33 Example: Chemists have a “rule of thumb” that many reactions will double their rate when the temperature increases by 10ºC (or K). Calculate the Ea that fits this doubling of reaction rate. First step: rate and rate constant are different, but if one assumes that one is keeping concentrations of the reactants the same, the rate is directly related to the rate constant. So we want to use the Arrhenius equation with a k2/k1 ratio = 2. We will also assume a room temperature reaction with temperatures of 300K = T1 and 310K = T2. k2 R ln k1 8.314 J/molK × ln ( 2 ) 5.76 Ea = = = 1⎞ 1⎞ 1.07 × 10 − 4 ⎛1 ⎛1 ⎜T −T ⎟ ⎜ 300 − 310 ⎟ ⎝1 ⎝ ⎠ 2⎠ Ea = 53, 832 J = 53.8 kJ Warning: watch your units!! Kinetics 34 Problem: How much will the rate increase if a rxn has a Ea = 70 kJ and the temperature increases from 300K to 400K? Problem: A reaction has k = 1.6 × 10−5 s−1 at 600 K. When the temperature is increased to 700 K the new measured k = 6.36 × 10−3 s−1. Calculate the Ea value for this reaction. Kinetics 35 Catalysis [catalyst] A+B C A catalyst is a substance that increases the rate of rxn without itself being consumed in the reaction. After the rxn is finished you should be able to recover the catalyst from the rxn mixture unchanged. A catalyst speeds up the rate at which a chemical reaction reaches equilibrium. The overall thermodynamics of the rxn is NOT changed by the catalyst. Therefore, very endothermic (nonspontaneous) reactions are usually NOT suitable for catalytic applications. A catalyst provides a lower energy pathway for the production of products from reactants, thus allowing the rxn to proceed faster. It lowers the activation energy for a rxn (kinetics) – it does NOT change the thermodynamics of a rxn. Kinetics 36 Catalyzed rxn proceeding through an intermediate Ea Ea catalyzed ΔG Reactants ΔG Products Reaction Coordinate A catalyst provides an alternate mechanism (or pathway) for the reactants to be transformed into products. The catalyzed mechanism has an activation energy that is lower than the original uncatalyzed rxn. An excellent catalyst will lower the activation energy the most. Kinetics 37 Problem: Which of the following represents a correct catalyzed potential (free) energy diagram based on this uncatalyzed energy diagram. Why? Kinetics 38 There are two broad classes of catalysts: Heterogeneous Catalysis: Technically speaking this is when the catalyst is in a different phase than the reactants and products. Practically, it is usually when the catalyst is a solid and the reactants and products are liquids or gases. On a solid catalyst, only the surface of the catalyst is where the reaction will occur – so the more surface area available the more catalysis can occur. Homogeneous Catalysis: This is when the catalyst is in the same phase as the reactants & products. Practically, this is usually in the liquid or solution phase. Homogeneous catalysts are usually molecules dissolved in solution. Kinetics 39 Advantages/Disadvantages of Homogeneous Catalysts Relative to Heterogeneous Catalysts Good homogeneous catalysts are: good generally more selective for a single product far more active more easily studied far more easily optimized bad far more sensitive to deactivation for separating product & catalyst Heterogeneous catalysts dominate chemical and petrochemical industry – ~ 95% of all chemical processes use heterogeneous catalysts. Homogenous catalysts are used when selectivity is critical and product-catalyst separation problems can be solved. Kinetics 40 Catalysis Terminology Turnover (TO) – one loop through the catalyst cycle. Typically one equivalent of reactant is converted to one equivalent of product (per equivalent of catalyst). Turnover Frequency (TOF) or Turnover Rate – the number of passes through the catalytic cycle per unit time (typically sec, min or hrs). This number is usually determined by taking the # of moles of product produced, dividing that by the # of moles of catalyst used in the rxn, then dividing that by the time to produce the given amount of product. The units, therefore, are usually just time−1. Note that the rate of a batch catalytic reaction is fastest at the very beginning when the reactant concentration is the highest and generally slows down as the reaction proceeds – stopping when all the reactant is used up. Note the graph below for the production of aldehyde product from the homogeneously catalyzed reaction of vinyl acetate, H2, and CO. Kinetics 41 Vinyl Acetate Hydroformylation sampling from autoclave causes pressure glitches 0.3mM catalyst -- 85°C/90 psi H2 /CO 2,000 1,800 4.5 Uptake curve 1,600 4 1,400 Equiv Aldehyde 1,200 kobs = 0.0076 min 1,000 3.5 -1 3 Ln plot Prod 800 Ln (Δ P) 2.5 600 Initial TOF 8 TO/min 476 TO/hr 400 2 1.5 200 1 0 0 5 10 15 20 Time (hours) The TOF, therefore, will vary throughout the course of a batch reaction. The Initial TOF is defined as the initial part of a catalytic reaction where the rate is the fastest and essentially linear. A far better measure of rate, however, is the observed rate constant kobs, which allows one to reproduce the entire product production curve given a set of reactant & catalyst concentrations. In the above graph, the reaction is pseudo-first order in excess reactant alkene (vinyl acetate concentration ~ 0.6 M, catalyst 0.3 mM) and kobs is determined from a ln plot of the change in H2/CO pressure (the reactant concentration) versus time for this rxn. Kinetics 42 Turnover Number (TON) – the absolute number of passes through the catalytic cycle before the catalyst becomes deactivated. Academic chemists sometimes report only the turnover number when the catalyst is very slow (they don’t want to be embarrassed by reporting a very low TOF), or decomposes quite rapidly. Industrial chemists are interested in both TON and TOF. A large TON (e.g., 106 - 1010) indicates a stable, very long-lived catalyst. Kinetics 43 Heterogeneous Catalyst Examples Heterogeneous catalysts are often very small particles of inert support (alumina, silica, MgO) coated with a very thin layer of catalytically active metal or alloy (mixture of metals). Remember that the catalysis only occurs on the surface of a heterogeneous catalyst. Hydrogenation of CO to produce CH3OH: Cu/ZnO CO(g) + H2(g) CH3OH(g, l) This is one of the most selective heterogeneous catalytic reactions known since methanol is almost the only product formed. Methanol has many important uses in the chemical industry. It can also be used as a very clean fuel. Kinetics 44 Fischer-Tropsch Catalysis: Fe xCO(g) + yH2(g) CH4(g) + CH3(CH2)zCH3(g,l) + H2O(g) z = 0 to 20+ GASOLINE (z = 6-10)!! This very important catalytic process was developed in Germany during the 1930's and made WWII possible by allowing the Germans to convert coal into gasoline. New Zealand and South Africa both have Fisher-Tropsch plants to turn coal into hydrocarbon mixtures. Zeolites Currently there is a great deal of interest in a very important class of naturally occurring aluminosilicates called Zeolites, which have the general formula: Mx/nn+[AlxSiyO2x+2y]-x Kinetics 45 Zeolites have porous structures that have tunnels and chambers in which catalytic reactions can occur: Red = O Yellow = Si Blue = Al Purple = Na+ ZSM-5 Kinetics 46 Zeolites are very popular because they do a very good job at catalysis. Consider the following rxn: ZSM-5 CH3OH(l) "Gasoline" + H2O(l) Unlike the Fisher-Tropsch catalysis discussed earlier which gave a wide variety of products which had to be separated apart (expensive!), the zeolite ZSM-5 developed by Mobil produces a gasoline blend of hydrocarbons quite selectively and under mild conditions (sorta cheap). Why?? The catalytic rxns are going on inside the chambers in the zeolite which limits the size of the products that can be produced! Kinetics 47 Metal-Based Homogeneous Catalysis Hydroformylation (also called the “oxo” rxn) is a catalytic rxn that takes organic compounds with C=C double bonds (alkenes or olefins), reacts them with H2 and CO and turns them into new, more useful organic compounds called aldehydes: O O H2 C=CHR H 2 CO Rh or Co catalyst H-C-CH2 CH2R + linear C H H 3 C-CHR branched Aldehydes * Largest homogeneous catalytic process * > 12 billion pounds of aldehydes (alcohols) per year * Selectivity to linear (normal) or branched (iso) products is important Otto Roelen discovered hydroformylation in 1938. He was actually studying the heterogeneously catalyzed Fischer-Tropsch rxn using cobalt metal and found that the cobalt metal was dissolving into solution to produce a homogeneous catalyst that did hydroformylation. Kinetics 48 Roelen and co-workers figured out that HCo(CO)4 was the active catalytic species, but did not understand how it worked. In 1960 and 1961 Heck and Breslow proposed what is now accepted as the general mechanism for hydroformylation: Scheme 1. Heck/Breslow Hydroformylation Mechanism O C + H2 OC CO Co Co2 (CO) 8 H O C O H + CO - CO R + alkene + CO R Co2 (CO) 7 H OC Co OC + H2 HCo(CO) 4 Co R O C O - CO CO OC H H C O rate determining step + CO anti-Markovnikov alkene addition to M-H bond - CO proposed bimetallic pathway -- since shown to be of no importance under normal catalytic conditions O C O C + CO CO OC Co C O CO OC Co O 2.5 atm -- 1.6:1 linear to branched 90 atm -- 4.4:1 linear to branched R { C O R increasing CO partial pressure keeps back rxns from occuring -- this limits alkene isomerization and the corresponding opportunity for making branched aldehyde Kinetics 49 The kinetic rate expression for this rxn is: d ( aldehyde) = k [ alkene][ Co][ H2][ CO] − 1 dt The first hydroformylation plant in the USA was build in 1947 in Baton Rouge by what is now called Exxon Chemicals. It is located just north of the state capital and still uses the HCo(CO)4 technology. In the late 1960’s it was discovered that Rh was a far more active catalyst for hydroformylation and when combined with PPh3 (triphenylphosphine) a very selective catalyst for making linear aldehyde product was produced. The first Rh/PPh3 hydroformylation plant was build in the early 1970’s and currently about 70-75% of all hydroformylation plants use the Rh/PPh3 technology (or closely related variant). Kinetics 50 Problem: Consider the reaction: F + G and the following experimental data: Exp # [F] [G] Rate (Msec−1) 1 1.0 1.0 1 2 1.0 2.0 2 3 2.0 2.0 4 What is the kinetic rate expression for this reaction? a) rate = k[F] b) rate = k[G] c) rate = k[F][G] d) rate = k[F]2[G] e) rate = k[G]2 f) not enough information to determine H Chapters 17 & 20-1 to 20-3 Chemical Equilibria Equilibrium 2 Chemical Equilibrium: It is the condition of a chemical reaction in which the rate of formation of products (from reactants) equals the rate of formation of the reactants (from products). rate1 A+B C+D rate-1 Equilibrium occurs when rate1 = rate-1. Although chemists usually want reactions to go completely to products (and ideally only to a single product), many do not. Theoretically all reactions are in equilibrium. A reaction will not generally reach equilibrium if: 1) The rxn is very exothermic (exoergic) 2) One (or more) of the products (or reactants) is removed from the rxn 3) One of the products (or reactants) is insoluble Equilibrium 3 Consider the very important industrial reaction (the Haber process) of nitrogen and hydrogen to produce ammonia, which is used as a fertilizer: N2(g) + 3H2(g) 2NH3(g) This is a very difficult reaction (large activation barrier) that requires high temperatures, pressures and a catalyst. At the high temperatures required to make the reaction proceed at a reasonable rate, the thermodynamics favors the N2 + H2 reactants producing the following behavior: H2 P NH3 N2 Time Note that the rxn does not go to completion, rather the forward and backward rxns reach a state of chemical equilibrium. Equilibrium 4 Equilibrium is a dynamic process. This means that when a reaction has reached a state of equilibrium, the forward and backward reactions making up the overall reaction have not stopped!! The equilibrium definition states that equilibrium is reached when the forward and backward reaction rates become equal! For example consider a saturated solution of NaCl (no additional salt will dissolve): NaCl(s) Na+(aq) + Cl−(aq) If we add 5 g more NaCl(s) to this solution, the amount of solid NaCl in the container will increase by 5 g (that is, no additional solid NaCl will dissolve into solution). This does not mean, however that some of the new NaCl that we just added won't dissolve at all. Some of it will dissolve, while some Na+(aq) + Cl−(aq) in solution elsewhere will precipitate out! We could follow this by adding radioactive 24Na38Cl to the container: radioactive salt TIME Equilibrium 5 Demonstration: Equilibrium 6 Law of Mass Action One can set up a general mathematical expression to describe the following chemical equilibria: wA + x B yC + z D y [D]z [C] Keq = [A]w [B]x products reactants Keq is called the equilibrium constant The equilibrium expression for the Haber process reaction would be written as: N2 (g) + 3H2 (g) Keq = [NH3 ] 2 2NH 3 (g) products [N2 ] [H2 ]3 reactants Equilibrium 7 Consider, for example, the equilibrium between N2O4(g) and NO2(g): N2O4(g) 2NO2(g) [NO2]2 Keq = [N2O4 ] Listed below is experimental data giving initial concentrations for N2O4(g) and NO2(g). After some time the reaction reaches equilibrium and the concentrations listed. Initial @ Equilibrium N2O4 NO2 N2O4 NO2 Keq 0.00 0.02 0.0014 0.017 0.21 0.00 0.03 0.0028 0.024 0.21 0.00 0.04 0.0045 0.031 0.21 0.02 0.00 0.0045 0.031 0.21 Note how Keq is the same regardless of the initial concentrations. This is why it is called the equilibrium constant. Equilibrium 8 Some Features of Equilibrium Constants Keq usually depends on temperature If one reverses the way a reaction is written the new Keq is the inverse of the original value: H2 (g) + ½O2 (g) K eq = H2O(g) [H2 O] = 10 (@ high temp) [H2] [O2 ] ½ Reversing the above rxn we now write: H2O(g) * K eq = H2(g) + ½O2 (g) [H2] [O2 ] ½ [H2 O] 1 = 0.1 = K eq Multiplying a reaction by a constant factor results in raising Keq to that power: 2H2 (g) + O2 (g) # K eq = [H2O] 2 [H2 ]2[O2 ] 2H2O(g) = 100 (@ high temp) Equilibrium 9 Equilibrium constants have a number of very important functions: 1) whether a rxn will be spontaneous under a given set of conditions (equilibrium constants are directly related to ΔG – Gibbs Free Energy, see end of this chapter) 2) in which direction a reaction is going to proceed to reach equilibrium 3) allow us to calculate the concentrations of products and reactants at equilibrium Qualitatively, the magnitude of Keq should immediately tell you in what direction a reaction is going to proceed and how far it will go before reaching equilibrium. For example: Keq >> 1 reaction will go mainly to products Keq ~ 1 reaction will produce roughly equal amounts of product and reactant Keq << 1 reaction will go mainly to reactants Equilibrium 10 Problem: Will the following equilibria proceed mainly to products, reactants or produce an approximately equal amount of both? a) N2(g) + 3H2(g) b) 2SO3(g) 2NH3(g) 2SO2(g) + O2(g) c) 2HBr(g) + Cl2(g) Keq = 0.001 Keq = 2 2HCl(g) + Br2(g) Keq = 104 d) 2H2O(g) 2H2(g) + O2(g) Keq = 10-28 e) N2O4(g) 2NO2(g) Keq = 1 f) H2(g) + Cl2(g) g) PCl3(sol) + Cl2(sol) h) 2HCl(g) + Br2(g) 2HCl(g) Keq = 1044 PCl5(sol) Keq = 0.1 2HBr(g) + Cl2(g) Keq = 10-4 Equilibrium 11 Units on Keq There are typically no units on equilibrium constants. This is because one is formally supposed to use the activities of compounds instead of their concentrations. The activity of a compound in an ideal mixture is the ratio of its concentration (or partial pressure) to a standard concentration (or pressure). Since the activity is defined as a ratio, the units cancel out. Despite the fact that we should use dimensionless activities, most chemists still refer to equilibrium concentrations in terms of M or pressure units (atm). One does need to watch out when one is working with gases. An equilibrium constant calculated with gas pressures (atm is the standard unit for gases) will not have the same numerical value as one calculated using molarity values if there are different # of gas molecules on each side of the equilibrium. This is because of the relationship between pressure and molarity as defined by the ideal gas law. Chemists use various subscripts on the equilibrium constant K to indicate different types of equilibria: Kp = gases (pressure), Kc = solutions (molarity), Ka = acids, Kb = bases, Ksp = solubility product (slightly soluble solids). Equilibrium 12 Reaction Quotient If a reaction is at equilibrium, then the equilibrium relationship will hold true: wA + x B yC + z D y [D]z [C] Keq = [A] w [B]x But, what if one is NOT at equilibrium? Then we find that the equilibrium expression is redefined as the reaction quotient, Q: [C] y [D]z Keq = w [B]x = [A] this will not be equal to Keq if the reaction is not at equilibrium Q when this occurs, the equilibrium expression is defined as being equal to Q, the Reaction Quotient Equilibrium 13 By comparing Q to Keq, one can tell in which direction a reaction will go to reach a state of equilibrium: wA + x B Keq ? = Q > Keq yC + z D More products? y [D]z [C] = Q [A] w [B]x More reactants? reverse reaction will be spontaneous Q = Keq reaction @ equilibrium Q < Keq forward reaction will be spontaneous Equilibrium 14 EXAMPLES: CO2(g) + H2(g) CO(g) + H2O(g) [CO][H2O] = Keq = 4.4 [CO2][H2] If the initial concentrations of all species are 1 M, which way will the reaction proceed to reach equilibrium? [CO] [H2O] [CO2 ] [H2 ] (1) (1) = (1) (1) = 1=Q K eq = 4.4 Q=1 Q < Keq } therefore, the rxn will go FORWARD to reach equilibrium Equilibrium 15 What if we increase the [CO] concentration to 10 M? [CO] [H2O] [CO2 ] [H2 ] (10) (1) = (1) (1) = 10 = Q K eq = 4.4 Q = 10 Q > Keq } therefore, the rxn will go BACKWARD to reach equilibrium If you are ever given a problem where the product and reactant concentrations are all non-zero, you MUST calculate Q and compare it to Keq in order to figure out which way the reaction has to go to reach equilibrium. DANGER!! Common mistake!! Equilibrium 16 Problem: Are the following rxns @ equilibrium? If not, which way must they proceed to reach a state of equilibrium? a) CO(g) + 2H2(g) 5M 1M b) CO2(g) + H2(g) 1M 2M c) Br2(g) + I2(g) 0.2 M 0.2 M d) N2(g) + O2(g) 2M 0.0001 M CH3OH(g) 5M Keq = 1 H2O(g) + CO(g) 2M 3M Keq = 4 2BrI(g) 0.1 M Keq = 1 x 10−4 2NO(g) 2M Keq = 9 Equilibrium 17 Numerical Problems -- Using Keq EXAMPLE: Consider the following rxn. The initial concentrations are [I2] = [H2] = 2M, [HI] = 0M, and Keq = 16. What will be the various concentrations when the reaction reaches equilibrium? I2(g) + H2(g) 2HI(g) Step 1: Write out your initial and @ equilibrium conditions: Initial cond: [I2] = [H2] = 2M [HI] = 0M note that since [HI] = 0M, the reaction must proceed to the right to make more product. Thus, for this example, we will lose reactants and gain product. What we don't know is how much. Therefore, we will setup an algebraic expression to solve for x, the amount of product being produced and the amounts of reactant that we are losing. It is CRITICAL to remember to multiply x by the appropriate coefficient from the balanced chemical equation. @ Equilibrium: [I2] = [H2] = (initial conc.) − (coefficent)(x) = 2−x [HI] = (initial conc.) + (coefficent)(x) DANGER!! Common = 0 + 2x = 2x mistake!! Equilibrium 18 Step 2: Write out your equilibrium expression: [HI] 2 Keq = = 16 [H2] [I2] now substitute in the @equilibrium conditions: [HI] 2 [H2] [I2] (2x)2 = 16 = (2-x) (2-x) now solve for x: (2x)2 = (2-x) (2-x) (2x)2 (2-x) 2 = 16 DANGER!! Common mistake!! make sure that you don't miss common algebraic simplifications!! take the square root of each side: (2x)2 (2-x) 2 (2x) = 16 (2-x) 2x = (4) (2-x) 6x = 8 =4 2x = 8 - 4x x = 8/6 x = 1.33 Equilibrium 19 Step 3: Substitute the value for x that you solved back into the equilibrium conditions that you wrote out in Step 1 above: @ Equilbrium: [I2] = [H2] = 2 - x = 2 - 1.33 = 0.66 M [HI] = 2x = 2.66 M [I2] = [H2] = 0.66 M [HI] = 2.66 M Step 4: Substitute the equilibrium concentrations you just found back into the equilibrium expression to see if you calculate the correct value for Keq: [HI] 2 Keq = [H2] [I2] = 16 substitute in the calculated equilibrium concentrations and see if you get K eq (2.66) 2 [HI] 2 = 16 = [H2] [I2] (0.66) (0.66) Step 5: Carefully read the question and make sure that you are picking the right answer. Note that what you solve for x may not be the answer (make sure you do Step 3!)!! DANGER!! Common mistake!! Equilibrium 20 Problem: Starting with [CO2] = 2 M, [H2] = 2 M, [CO] = 0 M and [H2O] = 0 M, what will be the various concentrations @ equilibrium. Keq = 9. CO2(g) + H2(g) CO(g) + H2O(g) Equilibrium 21 More Difficult EXAMPLE: Calculate Keq for the following reaction. Initial concentrations are: [SO2] = 4 M, [O2] = 4 M, [SO3] = 6 M. At equilibrium [SO2] = 3 M. 2SO2(g) + O2(g) 2SO3(g) Solution: This is really just a stoichiometry problem. Given the initial concentrations and a single equilibrium concentration (along with some algebra) one can solve for the other equilibrium concentrations. Once you have obtained all the equilibrium concentrations, one can put them into the equilibrium expression to solve for Keq. First one must figure out which way the reaction is going to go in order to reach equilibrium. We can't use the Reaction Quotient, Q, because we don't know Keq. We do, however, have our initial and one final equilibrium condition to tell us which way the reaction will shift: initial [SO2] = 4 M, @ equilib [SO2] = 3 M. So we are losing [SO2], therefore, the reaction will go to make more product and to lose reactant. Now we can setup and solve for the other equilib values: Equilibrium 22 4 initial: 4 6 2SO2 (g) + O2 (g) 4-2x @ equilib: 2SO3(g) 4-x 6+2x Normally we would substitute our x values into the equilibrium expression and solve for x. Here, however, we actually know one of the equilibrium values: @ equilib: (given) 3 ? ? So we can let 4-2x = 3 and solve for x: 4-2x = 3 2x = 1 x = 0.5 Now, substitute x into our @ equilibrium formulas: 4-2x 4-x 6+2x 3M @ equilib: 3.5 M 7M Finally, substitute the final values into the equilibrium expression to solve for Keq : 2SO2 (g) + O2 (g) [SO3 ] 2 Keq = [SO2 ]2[O2 ] 2SO3(g) (7) 2 = (3)2(3.5) = 1.6 Equilibrium 23 Problem: Consider the following reaction: CO(g) + 2H2(g) CH3OH(g) Initially we start with [CO] = 10 M and [H2] = 11 M. When the reaction reaches equilibrium there is 5 M [CH3OH]. Calculate Keq for this reaction. Equilibrium 24 As Tough as We Get EXAMPLE: Calculate equilibrium concentrations for the following reaction. Initial values are: [CO2] = 1 M, [H2] = 2 M, [CO] = 6 M, and [H2O] = 6 M. Keq = 2. CO2(g) + H2(g) CO(g) + H2O(g) Solution: First we need to determine in which direction the reaction will shift to reach equilibrium because none of the concentrations are zero. To do this we use the Reaction Quotient, Q, and the initial concentrations: [CO] [H2O] (6) (6) = = 18 = Q [CO2] [H2] (1) (2) K eq = 2 Q = 18 Q > Keq } therefore, the rxn will go BACKWARDS to reach equilibrium Now we can write out our initial and (most importantly) @ equilibrium values using x's: Initial cond: [CO2] = 1 M [CO] = 6 M @ Equilibrium: [CO2] = 1 + x [CO] = 6 - x [H2] = 2 M [H2O] = 6 M [H2] = 2 + x [H2O] = 6 - x Equilibrium 25 K eq = (6-x) (6-x) [CO] [H2O] = [CO2] [H2] (1+x) (2+x) =2 (x2 - 12x + 36) =2 (x2 + 3x + 2) x 2 - 12x + 36 = 2x 2 + 6x + 4 x 2 + 18x - 32 = 0 } this is a quadratic equation solve by using quadratic formula: ax 2 ± bx ± c = 0 x= x= x= - 18 ± (18) 2 - 4(-32) 2 - 18 ± 2 452 = b 2 - 4ac 2a -b± = - 18 ± - 18 ± 21.3 2 x = 1.7 Note that x is NOT OUR ANSWER!!!! (324) + (128) 2 = 1.7 or -19.7 physically impossible DANGER!! Common mistake!! Equilibrium 26 @ Equilibrium: [CO2] = 1 + x [CO] = 6 - x [H2] = 2 + x [H2O] = 6 - x Substituting in x = 1.7 we can get the equilibrium values: @ Equilibrium: [CO2] = 2.7 M [H2] = 3.7 M [CO] = 4.3 M [H2O] = 4.3 M Double-check that these numbers are correct by recalculating Keq and comparing to the value given to you in the problem: [CO][H2O] (4.3)(4.3) = = 1.85 ≈ 2 [CO2][H2] (2.7)(3.7) You don’t get exactly 2.0 due to round-off error (I only carried one decimal point in my calculation) Equilibrium 27 A Simple EXAMPLE (but looks really hard if you don't think): If 2 moles of H2O are placed into a 5L container, what will be the equilibrium concentration of H2, O2 and H2O? 2H2O(g) 2H2(g) + O2(g) Kc = 6.0 × 10-28 [H2]2[O2] = 6.0 × 10 − 28 Kc = [H2O]2 initial cond: [H2O] = 2.0 moles/5 L = 0.40 M [H2] = [O2] = 0 M [H2] = 2x [O2 ] = x [H2O] = 0.40 - 2x substituting into our equilibrium expression we get: @ equilibrium: [2 x ]2[ x ] = 6.0 × 10 − 28 [0.4 - 2 x ]2 4 x 3 − 24.0 × 10 − 28 x 2 + 9.6 × 10 − 28 x − 0.96 × 10 − 28 = 0 But this is a cubic equation!!! Almost impossible for you to solve!!! OH MY GOD, WHAT DO I DO NOW!! What is this idiot Professor doing to me!! Equilibrium 28 HOWEVER, consider the physical reality of the situation. Kc = 6.0 × 10-28 is extremely small, this means that very little H2O will decompose to form H2 or O2!! That means that the amount of H2 or O2 forming will be very, very small. That means that x will be very, very small. Small enough that we can ignore it in the [H2O] = 0.40 − 2x expression. This considerably simplifies the math: [2 x ]2[ x ] = 6.0 × 10 − 28 [0.4]2 4x 3 = 6.0 × 10 − 28 0.16 x 3 = 0.24 × 10 − 28 x = 2.9 × 10 − 10 M Since x is indeed much, much smaller than 0.40, the approximation was a very good one. So our concentrations at equilibrium are: @ equilibrium: [H2] = 5.8 × 10-10 M [O2] = 2.9 × 10-10 M [H2O] = 0.40 M Actually, one didn't have to do any calculations for this problem! Because Kc is so very small, you should know that virtually no products will be produced. Therefore: @ equilibrium: [H2] = [O2] = 0 M [H2O] = 0.40 M For this course and this kind of problem, there isn't much difference between 5.8 × 10-10 M and 0 M!! Equilibrium 29 Heterogeneous Equilibria So far all the equilibrium examples we have used have involved gases or solutions. What happens if we have other states of matter present -- such as solids or pure liquids?? How do they affect the equilibrium?? It turns out that as long as some solid or liquid is present, the equilibrium will be independent of the amount of that solid or liquid that is present! EXAMPLE: Br2(l) Br2(g) Keq = [Br2(g)] [Br2(l)] What is the concentration of [Br2(l)]? M= moles of Br2(l) volume The density of Br2(l) is 3.12 g/mL, so the # of moles is: # moles = (3.12 g/mL) / (159.8 g/mol) # moles = 0.0195 Equilibrium 30 The M can now be calculated for liquid bromine: M= moles of Br2(l) volume = 0.0195 moles of Br2(l) 0.001 L M = 19.5 mol/L Now we can look at our equilibrium expression: Keq = [Br2(g)] [Br2(l)] = [Br2(g)] 19.5 M note that this is a constant concentration -- it is independent of the amount of liquid bromine present, as long as some is there!! Because the [Br2(l)] concentration is a constant value we can multiply the equilibrium expression by that amount and incorporate it into Keq : (Keq ) (19.5 M) = [Br2(g)] * K eq = [Br2(g)] DANGER!! VERY Common mistake!! Therefore, one should NOT include solids or pure liquids in equilibrium expressions!! A more technical, but simpler explaination is that we are actually using activities instead of concentrations (see section on units), and the activity of a solid or pure liquid is defined as being = 1. Thus it factors out of the equilibrium expression. Equilibrium 31 Problem: Write out equilibrium expressions for the following reactions: a) CaO(s) + CO2(g) CaCO3(s) b) Ag+(aq) + Cl−(aq) AgCl(s) c) Br2(l) + Ni(CO)4(l) d) H2CO3(aq) e) HCl(g) + H2O(l) NiBr2(s) + 4CO(g) H2O(l) + CO2(g) H+(aq) + Cl−(aq) Equilibrium 32 Numerical Example: What is the equilibrium concentration for [Ag+] in the following reaction: Ag2(SO4) (s) 2Ag+(aq) + SO42−(aq) Ksp = 4 × 10−9 Answer: Ksp refers to equilibria involving solubility products, that is, solids that are slightly soluble in water (or other solvents). Note that the reactant in this problem is a solid and, as such, will NOT appear in the final equilibrium expression. We also usually do NOT give the amount of the solid and assume that there is excess present, since only a little bit will dissolve in solution. Otherwise, set it up and solve just like a regular equilibrium problem: Initial: excess solid Ag2(SO4) (s) @Equilib: 0M 0M 2Ag+(aq) + SO42−(aq) less excess solid 2x x The equilibrium expression for this rxn is: Ksp = [Ag+]2[SO42−] (the solid Ag2SO4 doesn’t appear in the equilibrium expression because it is a solid!). Plug in the @equilb values and solve for x: [2x]2[x] = 4 × 10 −9 4x3 = 4 × 10−9 x3 = 1 × 10−9 x = 1 × 10−3 DANGER!! BUT, watch out, x is NOT our answer!! [Ag+] = 2x, so [Ag+] = 2 × 10−3 M. Common mistake!! Equilibrium 33 Problems: a) What is the equilibrium concentration for OH− in the following reaction: Ca(OH)2(s) Ca2+(aq) + 2OH−(aq) Ksp = 4 × 10−6 b) What is the equilibrium concentration for Ag+ for the following system: Ag3(AsO4)(s) 3Ag+(aq) + AsO43−(aq) Ksp = 1 × 10−22 Equilibrium 34 Kp – Kc Relationship When the number of equivalents of gas phase reactants and products is not equal the following relationship relates Kc (concentration in M) and Kp (concentration in pressures - atm). This is true even if we technically use dimensionless activities due to the relationship between molarity and pressure (even when units are factored out). Kp = Kc(RT)Δn Kc = Kp(RT)−Δn Kp Kc = -or(RT)∆n Δn = (ngas prod) – (ngas react) No gas molecules? Then Δn = 0. Equal number of gas molecules on reactant & product side? Then Δn = 0. Pressures must be expressed in atmospheres (atm). Equilibrium 35 Le Chatelier's Principle When a system in a state of equilibrium is acted upon by some outside stress, the system will, if possible, shift to a new equilibrium position to oppose the effect of the stress. What do we mean by "stress"? Stress means that we are disturbing the reaction by: adding or removing reactants or products; increasing or decreasing the temperature; and increasing or decreasing the pressure (if gases are involved). Once we do one of these things, the reaction will (usually) no longer be in equilibrium and will have to shift to make more reactants or products to reattain a state of chemical equilibrium. A+B original rxn in equilibrium C+D + C but now we have added the reaction has to shift backwards to consume some of the products and make more reactants! more product C and there is too much product Equilibrium 36 Concept Demonstration: Reactants Keq > 1 Δ G rxn < 0 Products Energy reaction coordinate Chemical Demonstration: 2 CoCl4 (sol) + xH2 O 2+ Co(H 2 O) 6 (sol) + 4Cl (sol) H2O H 2O Co H 2O O2H OH2 H2O Equilibrium 37 Listed below are how various "disturbances" affect equilibria: 1) Adding products (unless one of the products is a solid!) to a reaction will cause the equilibrium to shift back to produce more reactants. 2) Adding reactants (unless one of the reactants is a solid!) to a reaction will cause the equilibrium to shift forward to produce more products. 3) Removing reactants (unless one of the reactants is a solid and as long as there is some left) will cause the equilibrium to shift back to produce more reactants. 4) Removing products (unless one of the products is a solid and as long as there is some left) will cause the equilibrium to shift forward to produce more products. Equilibrium 38 5) The effect of temperature on a reaction is dependent on whether the reaction is exothermic (ΔHrxn = negative) or endothermic (ΔHrxn = positive): Exothermic rxn: A+B C + heat } heat is one of the products a) increasing the temperature (adding heat) will cause the equilibrium to shift back to make more reactants b) decreasing the temperature (removing heat) will cause the equilibrium to shift forward to make more products Endothermic rxn: heat is one of the reactants A + B + heat C a) increasing the temperature (adding heat) will cause the equilibrium to shift forward to make more products b) decreasing the temperature (removing heat) will cause the equilibrium to shift back to make more reactants Equilibrium 39 6) The effect of changing pressure depends on the number of gasous reactants and products present: a) if there are NO gas phase species present then pressure will have NO effect on the equilibrium (actually, there is an effect at very high pressures -- but we won't worry about this). b) if there are gas phase species present, but there are the same number of gaseous molecules on each side of the reaction, pressure will have NO effect on the equilibrium. c) if there are different numbers of gas phase species present on the reactant and product sides of the equilibrium, then: i) increasing the pressure will favor the side of the equilibrium with the smaller number of gas phase molecules. ii) decreasing the pressure will favor the side of the equilibrium with the larger number of gas phase molecules. Equilibrium 40 Qualitative EXAMPLE: Consider the following equilibrium: N2O4(g) ½ original pressure Original pressure Twice original pressure 2NO2(g) Note how the N2 O4 molecules have dissociated in order to produce more NO2 molecules. These more efficiently fill up the "empty" space present in this system. = N2O4 = NO 2 Note how the NO2 molecules have combined in order to produce fewer N2 O4 molecules. These more efficently save space, which is in short supply in this compressed system. Equilibrium 41 Mathematical EXAMPLE: Effect of doubling the pressure (halving the volume) on N2O4(g) original equilib { N2 O4 2NO2(g) 2NO2 Keq = 0.21 0.031 M 0.0045 M compress N2 O4 0.009 M ? 2NO2 0.062 M } is this in equilibrium? Calculate Q the reaction quotient to determine the direction of the reaction to reach equilibrium: [NO 2]2 (0.062)2 Q= = = 0.427 [N2O4] (0.009) since Q > Keq the rxn has to go backwards to reach equilibrium. That means that some of the NO2 has to disappear. @ equilibrium: [N2O4] = 0.009 + x } we are gaining N2O4 [NO2] = 0.062 − 2x } we are losing NO2 substituting [N2O4] = 0.009 + x and [NO2] = 0.062 − 2x into our equilibrium expression we can solve for x: (0.062 − 2 x )2 = 0.21 0.009 + x 4 x 2 − 0.458 x + 0.0019 = 0 Equilibrium 42 [solve using quadratic equation] x = 0.004, 0.106 x = 0.106 M is physically unreasonable (that would give us a negative concentration, which is impossible), we can forget it. So x = 0.004 M. Substituting this back into our equilibrium conditions we can find the final equilibrium concentrations: [N2O4] = 0.009 + x = 0.013 M [NO2] = 0.062 - 2x = 0.054 M So the N2O4 concentration has increased and the NO2 concentration has decreased: exactly what one would qualitatively predict from Le Chatilier's principle!! Equilibrium 43 Problem: What are the various things that one can do to the following reactions to shift the equilibria to a) favor the reactants; b) favor the products? (you can add or remove products and reactants; change the temperature; change the pressure) a) CaO(s) + CO2(g) CaCO3(s) ΔHrxn = - 179 kJ/mol b) N2(g) + 3H2(g) 2NH3(g) c) Ca+2(aq) + 2Cl−(aq) CaCl2(s) ΔHrxn = + 95 kJ/mol d) H2(g) + I2(g) e) Br2(l) + Ni(CO)4(l) f) H2CO3(aq) g) Ag+(aq) + Cl−(aq) 2HI(g) NiBr2(s) + 4CO(g) H2O(l) + CO2(g) AgCl(s) Equilibrium 44 Le Chatelier's Principle II: Common Ion Effect The common ion effect is Le Chatelier’s principle – just under a different name. You will see another variant of this at the end of the Acid/Base chapter when we discuss Buffer solutions. Consider the following equilibrium: AgCl(s) Ag+(aq) + Cl−(aq) Ksp = 1.8 × 10−10 What happens to the Ag+(aq) concentration if we add enough NaCl to raise the Cl−(aq) concentration to 0.1 M? Qualitatively, of course, from Le Chatelier’s principle, adding product (Cl−) to the solution will push the equilibrium backwards to produce more reactant (AgCl(s)). This will decrease the free Ag+(aq) concentration in solution. The Na+ cations will not have any effect, so we can pretty much ignore them (spectator ions). In this case the Cl− anion is the Common Ion between the NaCl and AgCl. Equilibrium 45 Let’s set this up and mathematically solve for the concentration of Ag+ after adding 0.1 M Cl−. Initial: solid AgCl(s) @ equilib: 0 0.1 Ag+(aq) + Cl−(aq) less solid (-x) x 0.1 + x Ksp = [Ag+][Cl−] = 1.8 × 10−10 (x)(x + 0.1) = 1.8 × 10−10 x2 + 0.1x – 1.8 × 10−10 = 0 solve via the quadratic equation to get: x = 1.8 x 10−9 -- or -- −0.1000000003 Physically x = [Ag+] = 1.8 × 10−9 M impossible!! Equilibrium 46 But, there is a simple approximation we can use to make our math a lot easier. Before we add any extra Cl−, let's show that the concentrations of [Ag+] and [Cl−] present are very small: Initial: solid AgCl(s) @ equilib: less solid (-x) 0 0 Ag+(aq) + Cl−(aq) x x Ksp = [Ag+][Cl−] = 1.8 × 10−10 (x)(x) = 1.8 × 10−10 x 2 = 1.8 × 10−10 x = [Ag+] = [Cl−] = 1.3 x 10−5 M Concentrations with no extra Cl- anion added to the solution! So the Ag+ and Cl− concentrations in solution from AgCl(s) are 1.3 x 10−5 M. Le Chatelier’s principle tells us that adding more Cl− will decrease the Ag+ and Cl− (x values) from the AgCl(s) dissociation even further. Small enough that we can make the approximation that 0.1 + x in the original problem is essentially 0.1. This will reduce our quadratic expression down to a very simple algebra problem: Equilibrium 47 Initial: solid AgCl(s) @ equilib: 0 Ag+(aq) + Cl−(aq) less solid (- x) x Ksp = [Ag+][Cl−] = 1.8 × 10−10 (x)(0.1) = 1.8 × 10−10 divide through by 0.1 0.1 Approximation! 0.1 + x Simplify to just 0.1 because x is very small x = [Ag+] = 1.8 × 10−9 M Note that this is the same as what we calculated from the quadratic equation ([Ag+] = 1.8 × 10−9 M). And it is a LOT quicker and easier to calculate! So using this approximation, when appropriate, will save you a lot of time. Typically it is OK to drop x in a (# + x) or (# − x) algebraic expression when x is going to be more than an order of magnitude smaller than the # it is being added or subtracted to AND it will simplify the algebra. You will see similar approximations a lot in Acids & Bases for weak acid and base equilibrium calculations. Equilibrium 48 Problem: What is the concentration of Ag+ in a 0.01 M K2SO4 solution to which excess Ag2(SO4) is added. Ksp (Ag2(SO4)) = 4 × 10−9 Initial: Ag2(SO4) (s) @Equilib: 2Ag+(aq) + SO42−(aq) Equilibrium 49 Problem: What is the concentration of H+ in a 0.1 M acetic acid (HOAc) solution to which 0.1 M Na+OAc− is added. Keq (HOAc) = 2 × 10−5 Initial: 0.1 HOAc(aq) @Equilib: 0.1 − x 0 0.1 H+(aq) + OAc−(aq) x 0.1 + x Note: This is called a Buffer Solution (see Acids/Bases) Equilibrium 50 ΔGº & the Equilibrium Constant As I’ve mentioned during the first part of this chapter, the equilibrium constant is directly related to the Gibbs Free Energy, ΔG. ΔGº = negative Keq > 1 (spontaneous) ΔGº = zero Keq = 1 ΔGº = positive Keq < 1 (non-spontaneous) (rare) The mathematical relationship for calculating ΔGº, given Keq is: ΔGº = − R T l n K eq R = 8.314 J/K (gas constant) T = Temp in ºK Given the value of ΔGº, we can rearrange the above equation to solve for Keq: Keq = e−(ΔGº/RT) Thus, given ΔGº (or ΔHº and ΔSº) we can calculate Keq at a given temperature. Similarly, given Keq, we can calculate ΔGº. Equilibrium 51 ΔGº vs. ΔG: Standard vs. Non-Standard Conditions Remember that the º (“not”) on ΔGº indicates that the numerical value of ΔGº is based on the reaction at standard conditions (1 M solution concentration, 1 atm gas pressure). Temperature is NOT part of standard conditions! As soon as one has a concentration different than 1 M or 1 atm pressure, the º “not” goes away and one has ΔG. Consider the reaction: Initial: 1 atm 1 atm 2SO2(g) + O2(g) ΔGºrxn = −142 kJ/mol 1 atm 2SO3(g) The ΔGºrxn of −142 kJ/mol is for when each gas is present with a concentration of 1 atm. This indicates that the reaction under these conditions will proceed to make products (spontaneous). As the reactants start reacting, however, their concentrations decrease (SO2 twice as fast as O2) and ΔGº turns into ΔG and becomes less negative. When ΔG = 0 the reaction has reached equilibrium. Equilibrium 52 Example: A reaction has a ΔGº value of –40 KJ/mol at 25ºC. What is the Keq for this reaction? First convert the temperature from ºC to K: Temp (K) = 25ºC + 273 = 298 K Now we can use the formula for calculating Keq: Keq = e−(ΔGº / RT) Don’t forget to carry along the sign on ΔGº Keq = e−((−40000 J/mol) / (8.314 J/mol K)(298K)) Important Note: R the gas constant has units of J/molK, while we usually express ΔGº in KJ/mol. The units must match!!! The easiest thing is to multiply the ΔGº value in KJ/mol by 1000 to give J/mol. Keq = e (16.14) Keq = 1.02 × 107 DANGER!! Common mistake!! The negative ΔGº represents a spontaneous reaction. See how this converts over to a large positive Keq value, indicating that the reaction goes mainly to products. Note also how there are no units when you calculate Keq this way. Equilibrium 53 Example: A reaction has a Keqvalue of 0.01 at 25ºC. What is ΔGº for this reaction? First convert the temperature from ºC to K: Temp (K) = 25º + 273 = 298 K Now we can use the formula for calculating ΔGº: ΔGº = − R T l n K eq ΔGº = − (8.314 J/mol K)(298K) ln(0.01) ΔGº = − (2477 J/mol)(−4.6) ΔGº = +11,394 J/mol -- or -- +11.4 KJ/mol Don’t forget to convert J/mol to KJ/mol for the ΔGº value!! DANGER!! Common mistake!! Equilibrium 54 Problem: A reaction has a Keq value of 10 at 25ºC. What is ΔGº for this reaction? Equilibrium 55 Catalysts Catalyst: a material that speeds up the RATE of a reaction without being consumed in the reaction. A catalyst will NOT change an equilibrium, only the speed (rate) at which equilibrium is reached! Remember that equilibrium is directly related to thermodynamics. Catalysts never affect the thermodynamics of a reaction. They only lower the energy of activation (kinetics) of a reaction. Equilibrium 56 Chemical Demonstration: Oscillating Iodine Reaction IO3- H2O2 Mn2+ to Mn3+ HOI I2 O O HO H2O2 O I3 O HO - OH I + H2O Mn3+ to Mn2+ H2O2 I- 2CO2 HCOOH OH Chapters 18 & 19 Acids & Bases H2O + HF(aq) H3O+(aq) + F−(aq) Acids & Bases 2 Acid and Base Definitions Arrhenius Acid increases H + concentration increases OH- concentration Base Brønsted-Lowry (1923) Acid donates a H + combines with or accepts H + Base Lewis Acid Base electron acceptor electron donor Acids & Bases 3 Water itself has some ionic character: H2O(l) + H2O(l) H3O+(aq) + OH−(aq) Shorthand: H2O(l) H+(aq) + OH−(aq) This is called self-ionization. Although most chemists simply write H+, it is important to realize that H+ by itself represents a naked proton. In water the H+ is hydrated, that is, it forms ionic-diople interactions with other water molecules. A common way of more accurately representing the fact that it is hydrated (interacting with waters) is to write H3O+. Acids & Bases 4 H2O(l) H+(aq) + OH−(aq) [H + ] [OH− ] Keq = [H2O] But H2O is a pure liquid and its concentration is constant ([H2O] = 55.6 M, activity = 1), so it is not included in the equilibrium expression: K w = [H+ ][OH−] K w = 1.0 × 10 −14 @ 20 C [H+] [OH−] = 1 × 10−14 [H+] = [OH−] = 1 × 10−7 M So the [H+] and [OH−] concentration in pure water is 1 × 10−7 M We use Kw to indicate the water self- or autoionization. This is still a Keq or Kc. Chemists use many subscripts on the equilibrium constant K to indicate specific types of equilibria: Kb = base equilibria Ka = acid equilibria Ksp = slightly soluble (solubility product) equilibria Because of the often very small nature of H+ concentrations, chemists (and others) have devised Acids & Bases 5 a logarithmic scale to simplify expressing these values: pH = −log [H+] The negative sign in front of the log makes sure that most small concentrations of acid are given by positive values. [H+] = 1 × 10−7 M pH = −log(1 × 10−7) = 7.0 The greater the [H+] concentration, the LOWER the pH value!!! pH < 7.0 pH = 7.0 pH > 7.0 Acidic Neutral Basic (Alkaline) Acids & Bases 6 Substance pH 10 M HCl −1.0 1 M HCl 0.0 Stomach Acid (HCl) 1.4 Lemon Juice 2.1 Orange Juice 2.8 Wine 3.5 Black Coffee 5.0 Urine 6.0 Pure Water 7.0 Blood 7.4 Baking Soda Solution 8.5 Ammonia Solution 11.9 1 M NaOH 14.0 10 M NaOH 15.0 Acids & Bases 7 We can also define: pOH = −log −] [OH Although most chemists mainly use pH, pOH can be useful in base equilibrium numerical problems that we will run into later. Another definition we use is: pKw = −log Kw = 14 So, for a given water solution: pOH + pH = pKw = 14 or: pOH = 14 − pH pH = 14 − pOH Acids & Bases 8 Example: The pH of wine is 3.5. What is the [H+]? What is the [OH−]? What is the pOH? pH = −log[H+] = 3.5 [H+] = antilog(−3.5) = 10−3.5 = 3.16 × 10−4 M 1 × 10 −14 Kw = 3.1 × 10 −11 [OH− ] = + = [H ] 3.16 × 10 − 4 pOH = −log[OH−] = −log(3.1 × 10−11) = 10.5 ---- or ---pOH = pKw − pH = 14 − 3.5 = 10.5 Problem: The pH in your stomach is around 1. What is the [H+]? What is the [OH−]? What is the pOH? Acids & Bases 9 Dissociation Equilibrium Constants An acid is a compound that will ionize in solution (usually water) to form a H+(aq) and a counteranion. This can be writen in a general fashion as: HA(aq) H+(aq) + A−(aq) The equilibrium expression for this reaction is: [H+ ][ A− ] Keq = [HA] The equilibrium constant, Keq, is often given a special name for acids: Ka or the acid dissociation constant. The larger the Ka value the more H+ is being produced (the lower the corresponding pH), therefore, the stronger the acid!!!! Just as with pH we can also define pKa as: pKa = −log Ka The smaller (and more negative) the pKa the stronger the acid!!!! This can be confusing, but is very important: pKa’s are commonly used in biology & chemistry! Acids & Bases 10 Acid Dissociation Constants and pKa Values Acid HA A- Ka pKa hydroiodic HI I− ∼1010 ∼ −10 hydrobromic HBr Br− ∼1010 ∼ −10 perchoric HClO4 ClO4− ∼108 ∼ −8 hydrochoric HCl Cl− ∼108 ∼ −8 sulfuric H2SO4 HSO4− ∼108 ∼ −8 nitric HNO3 NO3− ∼108 ∼ −8 trichloroacetic Cl3COOH Cl3COO− 2 × 10−2 0.7 oxalic HOOCCOOH HOOCCOO− 5.9 × 10−2 1.2 sulfurous SO32− 1.5 × 10−2 1.8 SO42− 1.2 × 10−2 1.9 phosphoric H2SO3 HSO4H3PO4 H2PO4− 7.5 × 10−3 2.1 nitrous HNO2 4.6 × 10−4 3.3 hydrofluoric HF NO2− F− 3.5 × 10−4 3.5 formic HCOOH HCOO− 1.8 × 10−4 3.8 benzoic C6H5COOH C6H5COO− 6.5 × 10−5 4.2 oxalic (2nd H+) HOOCCH2COO− −OOCCH2COO− 6.4 × 10−5 4.2 acetic CH3COOH carbonic hydrogen sulfide sulfuric (2nd H+) 1.7 × 10−5 4.7 H2CO3 CH3COO− HCO3− 4.3 × 10−7 6.4 H2S HS- 9.1 × 10−8 7.1 HPO42− 6.2 × 10−8 7.2 ammonium ion H2PO4− NH4+ 5.6 × 10−10 9.2 hydrocyanic HCN NH3 CN- 4.9 × 10−10 9.3 phosphoric (2nd) Acids & Bases 11 Strong Acids These are acids that are essentially completely dissociated in solution (usually water). H+(aq) + Cl−(aq) HCl(aq) [H+ ][Cl− ] ≈ 10 8 K= a [HCl] In general, Ka > 1 for a strong acid (although there is no firm dividing line!) The common strong acids that you are expected to know are: HCl, HBr, HI (the hydrohalic acids) H2SO4 (sulfuric acid) HNO3 (nitric acid) HClO4 (perchloric acid) HCl, H2SO4, and HNO3 are often referred to as mineral acids. A strong acid completely dissociates the first H+, so the H+ concentration is the same as the given acid concentration. ∴ the pH of a strong acid is just −log of the acid concentration (= −log[H+]). Acids & Bases 12 Problem: What are the pH's of the following solutions? a) 0.001 M HCl b) 0.1 M HNO3 c) 1 × 10−5 M H2SO4 d) 10 M HBr e) 1 M HI f) 0.1 M HF g) 0.01 M HCl h) 1 × 10−4 M HNO3 i) 1 M acetic acid j) 1 × 10−14 M HCl Acids & Bases 13 Explanation for j) on the last page: The actual proton concentration for any acid dissolved in water is more precisely defined as: [H+]total = [H+]acid + [H+]water Normally, [H+]water is 1 × 10−7 M and is much less than [H+]acid, so that we usually ignore it. BUT, in this example, [H+]acid turns out to be only 1 × 10−14 M, which is much, much less than the [H+]water. So in this example we can actually ignore the tiny amount of H+ contributed from the strong acid and only consider the H+ naturally present in water: This will work for acid concentrations of 1 × 10−8 M and lower. It gets complicated mathematically right around 1 × 10−7 M (not dealt with in this course). Acids & Bases 14 Lewis Dot Structures: Where’s the Proton? When we write the formula for HNO3, it does NOT mean that the proton (H+) is attached to the nitrogen atom. The proton always binds to, or is associated with a lone pair of electrons, usually on one of the outer negatively charged atoms of a polyatomic anion. Consider the Lewis Dot structures for NO3−, SO42−, and ClO4−: when N has 4 bonds, it is f or mally cationic O Cl can hav e mor e than 8 v e- S can hav e mor e than 8 ve- O O N O 2− S O O O Cl O O O sulfate dianion nitrate anion O perchlorate anion Resonance, of course, will spread out the negative charges and bonding over all the O atoms. H+ in each of these cases binds to the oxygen atom(s) that is (are) negatively charged: H N O O O O O S OH O Cl OH O OH nitric acid sulfuric acid O perchloric acid Acids & Bases 15 Problem: Draw the Lewis Dot structure that best minimizes the formal charges for the following acids (some strong, some weak). Cleary show where the H+ is coordinated. a) HBF4 (fluoroboric acid, strong) b) H2CO3 (carbonic acid, weak) c) H3PO4 (phosphoric acid, weak) d) CF3SO3H (triflic acid, strong) e) HCO2H (formic acid, weak) Acids & Bases 16 Strong & Weak vs. pH A common mistake is to confuse pH with strong or weak acids/bases. For example, if I tell you a solution has a pH of 5.0, most of you would incorrectly assume that it is a weak acid. Maybe, maybe not. A pH of 5.0 tells you that it is a weakly acidic solution. This solution might have been made from a medium to considerable amount of a weak acid, or a very small amount of a strong acid. Without knowing the concentration (or amount and nature) of the acid used (not just the resulting H+ concentration or pH), you can’t tell whether it is composed of a strong or weak acid. It is true that solutions with very low pH’s (for example, −1.0) can pretty much only be composed of strong acids. So you need to be very careful with your language in dealing with strong & weak acids and solutions that are strongly or weakly acidic. Acids & Bases 17 Polyprotic Acids Sulfuric Acid: − H2SO4(aq) H+(aq) + HSO4 (aq) HSO4 (aq) H+(aq) + SO42 (aq) − − Ka1 ≅ 108 Ka2 ≅ 10-2 Phosphoric Acid: H3PO4(aq) − H2PO4 (aq) − HPO42 (aq) − H+(aq) + H2PO4 (aq) Ka1 ≅ 10-3 − H+(aq) + HPO42 (aq) Ka2 ≅ 10-8 − H+(aq) + PO43 (aq) Ka3 ≅ 10-13 Carbonic Acid: − H2CO3(aq) H+(aq) + HCO3 (aq) HCO3 (aq) H+(aq) + CO32 (aq) − − Ka1 ≅ 10-7 Ka2 ≅ 10-11 Acids & Bases 18 Note that for polyprotic acids only the first dissociation (the first H+) is likely to occur. The second (or third) dissociation process is far less likely, so the contribution of these subsequent dissociations to the overall [H+]total . So, generally we only have to worry about the first proton and the first dissociation constant unless one is doing a titration or other acidbase reaction!! Acids & Bases 19 Strong Bases These are usually (and most commonly) alkali metal hydroxides that dissociate completely in solution: NaOH(aq) Kb = Na+(aq) + OH−(aq) [Na + ][OH− ] ≈ 10 8 [NaOH] The common strong bases that you are expected to know are: LiOH, NaOH, KOH, RbOH, & CsOH The alkaline earth hydroxides Ca(OH)2, Sr(OH)2, Ba(OH)2 are medium strong bases. Be(OH)2 & Mg(OH)2 are considered to be weak bases since they only partially dissociate in water. The smaller ionic radius of Be2+ and Mg2+ cations polarize coordinated H2O enough to promote hydrolysis (that is, loss of H + from H2O). Acids & Bases 20 It is very important to remember that there are almost always two steps in converting from [OH−] to pH: 1) convert [OH−] to [H+]: Kw 1 × 10 − 14 [H+ ] = = [OH− ] [OH− ] 2) then convert [H+] to pH. ---- or ---1) convert [OH−] to pOH 2) then convert pOH to pH: pH = 14 − pOH Acids & Bases 21 Problem: What are the pH's of the following solutions? What are the pOH's? a) 0.001 M NaOH b) 0.1 M CsOH c) 1 × 10−5 M KOH d) 10 M RbOH e) 1 M NaOH f) 0.1 M Be(OH)2 g) 0.01 M LiOH h) 1 × 10−4 M NaOH i) 1 M ammonia j) 1 × 10−14 M KOH Acids & Bases 22 Explaination for j) on the last page: The actual hydroxide concentration for any base dissolved in water is more precisely defined as: [OH−]total = [OH−]base + [OH−]water Normally, [OH−]water is 1 × 10−7 M and is much less than [OH−]base, thus we can usually ignore it. BUT, in this example, [OH−]base turns out to be only 1 × 10−14 M, much less than the [OH−]water. So in this example we can actually ignore the tiny amount of OH− contributed from the base and only consider the OH− naturally present in water: This will work for base concentrations of 1 × 10−8 M and lower. It gets complicated mathematically right around 1 × 10−7 M (not dealt with in this course). Acids & Bases 23 Conjugate Acid-Base Pairs Bronsted-Lowry Definition of a Base: a substance that combines or accepts a H+. Consider the dissociation of an acid: H2O + HF(aq) H3O+(aq) + F−(aq) Because this is an equilibrium, F−(aq) is back reacting with H+(aq) to produce undissociated acid HF(aq). Therefore, F−(aq) is acting like a base! Since it was originally part of the acid (HF), there’s a special name for it: conjugate base. Strong acids have weak conjugate bases. Weak acids have stronger conjugate bases (but usually not as strong as OH−). To indicate that an equilibrium favors one side of a rxn chemists sometimes use the double equilibrium arrows where one arrow is shorter. For a strong acid (weak conjugate base) one could write: HCl(aq) H+(aq) + Cl−(aq) Acids & Bases 24 Bronsted-Lowry Definition of a Acid: a substance that donates a H+. Consider the reaction of a weak base with water: NH3(aq) + H2O NH4+(aq) + OH−(aq) Because this is an equilibrium, NH4+(aq) donates a H+(aq) that reacts with OH−(aq) to produce the original NH3 (and water). Therefore, NH4+(aq) is acting like an acid! Since it started as a weak base (NH3), we call NH4+ a conjugate acid. Strong bases have weak conjugate acids. Weak bases have stronger conjugate acids. In the two reactions shown on this page and the previous, water is acting as either an acid or a base. Any chemical that can act as either an acid or base is called amphoteric. In the following reaction, Na+ is more accurately called a conjugate Lewis Acid (not a proton donor). NaOH(aq) Na+(aq) + OH−(aq) Acids & Bases 25 Weak Acids These are acids that only dissociate to produce a small amount of H+ in solution: HF(aq) H+(aq) + F−(aq) [H+ ][F− ] Ka = = 3.5 × 10 −4 [HF] Most of a weak acid is dissolved in solution in its undissociated, neutral form. An acid is weak because its conjugate base (counter-anion) is a good base and likes to bind to H+. A weak acid typically has Ka < 1 × 10−3 Most weak acids are organic acids based on the carboxylic group: Some of the more common weak acids that we run into on a daily basis include: Acids & Bases 26 Acetic Acid Carbonic Acid Phosphoric Acid O O O C C P H3C OH HO active ingredient in vinegar HO OH OH OH used in soda (Coke, Pepsi) formed when CO2 dissolves in water Acetylsalicylic Acid Ascorbic Acid O HO H3C O O C H C C H C C H O HO C C CH2 OH O H HO OH H H Aspirin Vitamin C Note that the “OH” groups in all these examples are not hydroxides!! The oxygen atom of the OH is strongly bonded to the atom they are attached to and will not fall off as OH −. Instead they dissociate H+ due to the O’s electronegativity and ability to stabilize negative charge(s). Acids & Bases 27 Example: What is the pH of a 0.1 M solution of acetic acid? Ka ≅ 1 × 10−5 Abbreviation for acetic acid: HOAc Initial Cond: 0.1 M HOAc(aq) @ Equilib: 0.1 - x 0M 0M H+(aq) + OAc−(aq) x x Substitute our x values into the equilibrium expression and solve for x: [H+ ][ OAc− ] K= = 1 × 10 − 5 a [HOAc] ( x )( x ) = 1 × 10 − 5 (0.1 − x ) But, this will be a quadratic expression and we'll have to use the quadratic formula to solve for x (ugh!). There is, however, a very good approximation we can make to dramatically simplify the algebra. Acids & Bases 28 Because Ka is fairly small (10−5) we know that acetic acid is a weak acid and that we are only going to make small quantities of H+. That means that x should be a rather small number as well. This in turn means that 0.1 − x should be ≈ 0.1. Our assumption, therefore, will be that x will be much less than 0.1 M (initial concentration of acid). This really simplifies the algebra: ( x )( x ) = 1 × 10 − 5 (0.1 − x ) assume that x << 0.1 and drop from deonominator ( x )( x ) = 1 × 10 − 5 (0.1) Important Approximation!! x 2 = 1 × 10 − 6 } take square root of each side x = [H+] = [OAc−] = 1 × 10−3 M pH = −log(1 × 10−3) = 3.0 Acids & Bases 29 When can I drop x?? When is it a good approximation?? When Keq is 1 × 10−3 or smaller and x works out to be at least an order of magnitude smaller than the initial concentration that one is subtracting x from. This also works when one is adding x to the initial concentration (e.g., common ion problems). Problem: What is the pH of a 10 M solution of hydrogen sulfide? Ka ≅ 1 × 10−7 H2S(aq) H+(aq) + HS−(aq) Acids & Bases 30 Problem: Prof. Stanley makes a new organic acid. He prepares a 0.01 M solution and finds that the pH is 4.0. What is the Ka for this new acid? HA(aq) H+(aq) + A−(aq) Problem: What is the pH of a 0.01 M solution of HCN? Ka ≅ 4 × 10−10 Acids & Bases 31 Effect of Structure on Acid/Base Behavior Hydrohalic Acids A combination of factors affects the acid strength of hydrohalic acids: 1) Polarity of the H-X bond Electrostatic 2) Strength of the H-X bond attraction!! 3) Stability of the conjugate base, X− Thus, HF is a weak acid because the rather small fluoride ion (F−) has a concentrated negative charge that very effectively and strongly attracts the H+ cation, not allowing it to dissociate and become a strong acid (like HCl, HBr, or HI). HF HCl HBr HI Acids & Bases 32 Oxyacids Oxyacids are those acids in which the central atom (most commonly N, Cl, S or P) is bonded to at least one, and usually more, oxygen atoms. The resulting negative charge on this unit is balanced by the proper # of H+ that associate with the oxygen atoms (one per oxygen atom). A list of common oxyacids and their names: HNO2 HNO3 Nitrous Nitric HClO HClO2 HClO3 HClO4 Hypochlorous Chlorous Chloric Percloric H2SO3 H2SO4 Sulfurous Sulfuric H3PO2 H3PO3 H3PO4 Hypophosphorous Phosphorus Phosphoric Acids & Bases 33 The strength of oxyacids increases for a series of compounds as follows: 1) given the same central atom, the more oxygens present the stronger the acid Reasoning: the more electronegative oxygen atoms present, the more the negative charge on the anion is spread out over a larger volume. This means that the negative charge will be less concentrated on any single oxygen. Thus, there will be a lower electrostatic attraction to the H+ cations, allowing them to dissociate more easily. 2) for the same # of oxygens, the more electronegative the central atom, the stronger the acid Reasoning: the more electronegative the central atom, the more the negative charge on the anion will be pulled towards the central atom and away from the outlying oxygens. Thus, there will be a lower electrostatic attraction to H+ cations, allowing easier dissociation. 3) For almost all acids, the higher the negative charge on the anion (conjugate base), the lower the acidity of the acid. Reasoning: the higher the negative charge on the anion (mono- or polyatomic) the stronger the electrostatic attraction to the H+ cations. This makes it harder for the H+ cation to dissociate. Acids & Bases 34 4) For almost all acids, the more electronegative atoms present (like O or F) the higher the acidity of that acid. Reasoning: the more electronegative atoms present, the more the negative charge on the anion will be pulled towards these atoms and away from the atom that the H+ is associated with. Example: HClO HClO2 HClO3 HClO4 Hypochlorous Chlorous Chloric Percloric Ka = 3 x 10−8 Ka = 1 x 10−2 Ka = 5 x 102 Ka ≈ 1 x 1010 Electrostatic charge potential (ECP) surface plots for ClO− through ClO4− anions (no H+). The red color (dark) indicates more negative charge and a stronger electrostatic attraction to the H+ cation (weaker acid). Positive charge is indicated by the blue color (darker color on center atom). Acids & Bases 35 Here are the ECP surface plots for the acids: HClO HClO2 HClO3 HClO4 Note that the blue area around the H atom indicating positive charge is increasing as O atoms are added. Problem: Which is the stronger acid of the pair? a) H2SO4 or H3PO4 b) HNO2 or HNO3 c) HOBr or HOI d) H2SeO4 or H2SeO3 e) HI or HF f) H2SO3 or H2SO4 g) O O H H H C C OH or F F F C C OH Acids & Bases 36 Weak Bases These are bases that only react to a relatively small extent with H+ in solution: :NH3(aq) + H+(aq) NH4+(aq) The convention, however, for writing equilibria for weak bases is to react the base with water, which generates a small quantity of OH-: NH4+(aq) + OH−(aq) :NH3(aq) + H2O This is the equilibrium we use to define our base equilibrium constant, Kb: [NH4 + ][ OH− ] = 1.8 × 10 − 5 K= b [NH3] In this equilibrium NH4+(aq) is called the conjugate acid (it acts like a weak acid). Acids & Bases 37 Some weak bases are shown in the table below: Base Formula Structure Kb methylamine NH2CH3 H N CH 3 4.4 x 10-4 H carbonate ion CO32- O 2- 1.8 x 10-4 C O O ammonia NH3 HNH 1.8 x 10-5 H hydrosulfide ion nicotine HS- 1.8 x 10-7 HS C10H14N2 N CH 3 N hydroxylamine NH2OH H N OH 7 x 10-7 1.4 x 10-11 1.1 x 10-8 H pyridine C5H5N N 1.9 x 10-9 Note that all bases have atoms with lone pairs of electrons that can interact with a H+. Remember that a H+ doesn't have any electrons and has a positive charge. It will be attracted to atoms with negative charges and/or lone pairs of electrons. Acids & Bases 38 Example: What is the pH of a 0.1 M solution of ammonia? Kb ≅ 1 × 10−5 Init: 0.1 M 0M :NH3(aq) + H2O 0M NH4+(aq) + OH−(aq) @ Eq: 0.1 - x x x Substitute our x values into the equilibrium expression and solve for x: [NH4+ ][ OH− ] = 1 × 10 − 5 K= b [NH3] ( x )( x ) = 1 × 10 − 5 (0.1 − x ) But, this will be a quadratic expression and the quadratic formula is needed to exactly solve for x (ugh!). But lets use the very good approximation from weak acid equilibria problem solving that will dramatically simplify the algebra. Acids & Bases 39 Because Kb is pretty small (10−5) we know that ammonia is a weak base and that we are only going to make small quantities of OH−. That means that x will be a pretty small number. This in turn means that 0.1 - x will be ∼ 0.1. Our assumption, therefore, will be that x will be much less than 0.1 M (initial concentration of acid). This now really simplifies our algebra: ( x )( x ) = 1 × 10 − 5 (0.1 − x ) assume that x << 0.1 and drop from deonominator Important Approximation! ( x )( x ) −5 = 1 × 10 (0.1) x 2 = 1 × 10− 6 } take square root of each side x = [OH−] = [NH4+] = 1 × 10−3 M pOH = −log(1 × 10−3) = 3.0 pH = 14 − pOH = 11 DANGER!! VERY Common mistake!! Acids & Bases 40 Problem: What is the pH of a 1 M solution of sodium carbonate? Kb ≅ 1 × 10−4 CO32−(aq) + H2O HCO3−(aq) + OH−(aq) Problem: What is the pH of a 0.01 M solution of NH2OH? Kb ≅ 1 × 10−8 Acids & Bases 41 Lewis Acids & Bases Acid electron acceptor electron donor Base Although the Lewis definition includes the “traditional” Arrhenius and Brønsted-Lowry acids & bases mentioned so far, it also encompasses molecules that don’t. Foremost are metal atoms that form bonds to other molecules using empty orbitals on the metal (Lewis acid) and filled lone pairs on the donor atoms (Lewis bases). Transition metal atoms typically form the strongest bonds, followed by actinide and lanthanides. Some examples are shown below: PR3 Cl Cl Os O H H PhO H3C W Cl H2 N R3P Pd PR3 N PR3 Cl O H2 Problem: Identify the Lewis Base and Acid parts of each compound shown above. Extra: What is the oxidation state of the metal in each compound? Acids & Bases 42 Another class of important Lewis acids are trivalent compounds of boron and aluminum that have an empty orbital present. These are typically very reactive to donor molecules including water. BF3, for example, is considered a “superacid.” F Cl B Al F F Cl Cl BCl3 will form a moderately strong bond to :NH3: Cl B Cl Cl H + H H Cl B N H Cl Cl H N H Problem: What is the hybridization of the B in BCl3? What about the B in Cl3B:NH3? Acids & Bases 43 Relationship Between Ka and Kb Consider our ammonia equilibrium: :NH3(aq) + H2O NH4+(aq) + OH−(aq) [NH4+ ][ OH− ] Kb = [NH3] Kw but remember that : [OH ] = [H+ ] substituting this in for [OH− ] we now have : [NH4+ ] Kw Kb = [NH3] [H+] − This expression now corresponds to the following equilibrium multiplied times Kw: :NH3(aq) + H+(aq) NH4+(aq) The reverse of this reaction, however, is the "acid" equilibrium: NH4+(aq) :NH3(aq) + H+(aq) Writing the reaction this way means that we can now set-up a Ka equilibrium expression: Acids & Bases 44 [NH3][H+ ] Ka = [ NH+ ] 4 But note that the reciprocal of this expression is already in the Kb relationship: [NH3][H+ ] Ka = [ NH+ ] 4 [NH4+ ] Kw Kb = [NH3] [H+] Therefore, Ka and Kb are related via Kw. Here are the very important Ka / Kb relationships you need to know/understand: Kw Kb = Ka -or- Kw Ka = Kb Kw = Ka × Kb Acids & Bases 45 Knowing the mathematical relationship between Ka and Kb is very important. Many references only list Ka values for bases and not the Kb values that you would expect. When a Ka value is given for a base it is really the value for the conjugate acid of that base. For example: Kb = 1.8 × 10−5 Base = NH3 Conjugate Acid = [NH4]+ Ka = 5.6 × 10−10 (Kb)(Ka) = (1.8 × 10−5)(5.6 × 10−10) = 1 × 10−14 Kw If a reference gives a Ka value for the base you are looking up, you'll have to convert it to a Kb value in order to calculate the [OH−], and then the pH. You must, therefore, pay close attention to what data the problem is giving you and what you need to use. READ CAREFULLY!! DANGER!! VERY Common mistake!! Acids & Bases 46 Problem: What is the pH of a 0.1 M solution of the weak base triethyl amine (NEt3)? Ka = 1 × 10−11 Problem: What is the pH of a 0.01 M solution of hydroxyamine (NH2OH)? Ka = 1 × 10−6 Acids & Bases 47 Salts of Weak Acids and Bases When a weak acid (e.g., acetic acid) reacts with a strong base (e.g., an alkali hydroxide like NaOH) water and the salt of the weak acid is formed: Note that the acetate anion formed in the reaction is itself a weak base. Thus it can react with water to produce a small amount of [OH−]. Therefore, dissolving sodium acetate (the salt of acetic acid) in water will make a basic solution. Acids & Bases 48 Similarly, the reaction of a weak base with a strong acid will produce the salt of a weak base, which will act as a weak acid. Shown below we have the reaction of ammonia with HCl to produce ammonium chloride (the salt of a weak base): Dissolving ammonium chloride in water, therefore, produces an acidic solution (NH4+ is a weak acid). Acids & Bases 49 The big question is “How the heck do I tell whether a salt will produce an acidic, basic, or neutral solution??” The KEY is to remember the strong acids and bases: Strong Acids: HCl(aq) HBr(aq) HI(aq) HNO3(aq) H2SO4(aq) HClO4(aq) H+(aq) H+(aq) H+(aq) H+(aq) + + + + Cl−(aq) Br−(aq) I−(aq) NO3−(aq) 2H+(aq) + SO4−(aq) * H+(aq) + ClO4−(aq) These are extremely weak conjugate bases -Neutral Anions! Strong Bases: LiOH(aq) NaOH(aq) KOH(aq) RbOH(aq) CsOH(aq) Li+(aq) Na+(aq) K+(aq) Rb+(aq) Cs+(aq) + + + + + OH−(aq) OH−(aq) OH−(aq) OH−(aq) OH−(aq) These are extremely weak conjugate acids -Neutral Cations! Putting these Neutral (neutral here refers to their acid-base properties, NOT their charges!!) anions and cations together generates Neutral (not acidic or basic) salts (Ca2+, Sr2+, Ba2+ salts not shown): Acids & Bases 50 Neutral Cation Li+ Na+ K+ Rb+ Cs+ Neutral Anion Cl− Br− I− NO3− SO42− ClO4− Neutral Salts LiCl, LiBr, LiI, LiNO3, Li2SO4, NaCl, NaBr, NaI, NaNO3, Na2SO4, KCl, KBr, KI, KNO3, K2SO4, RbCl, RbBr, RbI, RbNO3, Rb2SO4, CsCl, CsBr, CsI, CsNO3, Cs2SO4, Perchlorate salts are explosive !!! Solutions of these Neutral Salts are neither acidic nor basic, but rather have a pH = 7 (Neutral!). A simple set of guidelines, therefore, are: CATIONS other than Li+, Na+, K+, Rb+, Cs+ (& Ca2+, Sr2+, Ba2+) will generate ACIDIC solutions (that is, the cation is a good conjugate acid) ANIONS other than Cl−, Br−, I−, NO3−, SO42− will generate a BASIC solution (that is, the anion is a good conjugate base) Acids & Bases 51 SALTS Cation(+)Anion(-) Cation comes from the base Anion comes from the acid Identify the Cation Is the cation from a strong base (Group 1A or Ca 2+, Sr 2+, Ba 2+) YES NO Identify the anion Is it from a strong acid? YES Salt Solution will be Neutral (Salt of a strong acid & base) Cation is from a weak base (cation is a conjugate acid) NO Anion is from a weak acid (anion is a moderate to good conjugate base) Salt solution will be basic (salt of a weak acid) Identify the anion Is the anion from a strong acid? YES Salt solution will be acidic (salt of a weak base) NO Salt of a weak acid & a weak base! Complicated! Don’t worry Acids & Bases 52 Problem: Will the following salts make an acidic, basic or neutral solution when dissolved in water? a) NaF Ka (HF) = 3.5 × 10-4 b) NaCl c) N H+ Cl- Ka (HCl) = 1 × 108 Ka (pyridineH+) = 5 × 10-6 d) C6H5COONa Ka (C6H5COOH) = 6.5 × 10-5 e) RbBr Ka (HBr) = 1 × 1010 f) CH3COOCs Ka (CH3COOH) = 1.7 × 10-5 g) Cs2S Ka (H2S) = 9.1 × 10-8 h) NH4NO3 Ka (NH4+) = 5.6 × 10-10 i) KCN Ka (HCN) = 4.9 × 10-10 j) KNO3 Acids & Bases 53 Example: What is the pH of 1 M NaF? Ka (HF) = 1 × 10−4. Neutral Basic Anion Cation Initial: 0 F−(aq) + H2O @ Eq: 1M HF(aq) + OH−(aq) 1-x 0 x x This is a basic equilibrium, so we need to convert Ka into Kb: − 14 Kw 1 × 10 Kb = = = 1 × 10− 10 Ka 1 × 10 − 4 DANGER!! VERY Common mistake!! Our equilibrium expression, therefore, is: [HF][OH− ] Kb = = 1 × 10 − 10 [F− ] ( x )( x ) = 1 × 10 − 10 (1 − x ) assume that x << 1 and drop from deonominator ( x )( x ) = 1 × 10 − 10 (1) x 2 = 1 × 10 − 10 } take square root of each side x = [OH−] = [HF] = 1 × 10−5 M Acids & Bases 54 So the [OH−] = 1 × 10−5 M. The pOH is: pOH = -log(1 × 10−5) = 5 The pH, then, is given by: pH = 14 - pOH = 14 - 5 = 9 DANGER!! VERY Common mistake!! Acids & Bases 55 Problem: What is the pH of a 0.1 M solution of NH4Cl. Ka (NH4+) = 1 x 10−9. Acids & Bases 56 Reactions of Acids & Bases Strong Acids and Strong Bases: The reaction of a strong acid and strong base is the simplest type of acid-base reaction. Consider the rxn of NaOH and HCl: Na+(aq) + OH−(aq) + H+(aq) + Cl−(aq) H2O + Na+(aq) + Cl−(aq) The rxn above shows all the aqueous species. Note that NaCl(aq) is one of the products. When acids react with bases, water and the salt of the acid/base are formed (the base provides the cation and acid the anion). The net ionic equation is shown below: H2O OH−(aq) + H+(aq) A key point to remember is that when we react acids and bases we are usually reacting solutions of acids and bases. When one mixes two solutions together the overall volume increases and the concentrations of all species will decrease! Because of the change in volume due to the mixing together of two solutions one has to take this into account. DANGER!! VERY Common mistake!! We do this by dealing directly with MOLES and not Molarity. So you have to: Acids & Bases 57 convert molarity into moles, • do your calculations, then • convert back to molarity using the NEW total (combined) solution volume Example: What is the pH if we mix 100 mL of 0.1 M HCl with 50 mL of 0.1 M NaOH? First, convert concentrations into moles: (100 mL)(0.1 M) = 10 mmol HCl (50 mL)(0.1 M) = 5 mmol NaOH Now for the reaction: 5 mmol NaOH will react with only 5 mmol HCl, leaving behind 5 mmol HCl. But now we have 150 mL of solution, so [H+] is: • DANGER!! VERY Common mistake!! 5 mmol = 0.033 M H+ 150 mL } p H = 1.5 Acids & Bases 58 Problem: What will be the pH when 100 mL of 0.1 M HNO3 is mixed with 300 mL of 0.2 M KOH? Problem: What will be the pH when 100 mL of 0.1 M HNO3 is mixed with 200 mL of 0.05 M KOH? Acids & Bases 59 Strong Acids and Weak Bases: Strong Bases and Weak Acids: Things get more complicated (well, not too bad) when you react a strong acid with a weak base or a strong base with a weak acid. For this case, however, we will simplify things a little by only using equivalent amounts of the acid and base. The key thing to remember when you react a weak acid or base with a strong base or acid is that one ends up with the SALT of that weak acid or base. Remember that the SALT of a weak acid is actually a weak base and gives a basic solution. Similarly, the SALT of a weak base generates an acidic solution. DANGER!! VERY Common mistake!! Acids & Bases 60 Example: What is the pH if 150 mL of 0.5 M acetic acid is mixed with 150 mL of 0.5 M NaOH? Ka (HOAc) ≅ 1 × 10−5. H2O + OAc−(aq) HOAc(aq) + OH−(aq) The OAc− produced is a weak base and will realizing this DANGER!! produce a weakly basic solution: Not ERY Common- mistake!! V H2O + OAc−(aq) HOAc(aq) + OH−(aq) Since this is a base equilibrium, we need to use Kb: Kw 1 × 10 − 14 Kb = = = 1 × 10 − 9 Ka 1 × 10 − 5 Not realizing this - DANGER!! Kb = [HOAc][OH− ] [OAc − ] VERY Common mistake!! = 1 × 10 − 9 Calculate the concentration of OAc− : (150 mL)(0.5 M) = 75 mmol HOAc (150 mL)(0.5 M) = 75 mmol OH− Acids & Bases 61 Thus, we will produce 75 mmol of OAc−. The concentration of OAc−, therefore, will be: Volume Change DANGER!! VERY Common mistake!! 75 mmol = 0.25 M OAc − 300 mL Now we can set-up our initial and @eq conditions: Initial: 0.25 M H2O + OAc−(aq) @ Eq: 0 0 HOAc(aq) + OH−(aq) 0.25 - x x x [HOAc][OH− ] = 1 × 10 − 9 Kb = [OAc − ] ( x )( x ) = 1 × 10 − 9 (0.25 − x ) assume that x << 0.25 and drop from deonominator Approximation ( x )( x ) −9 = 1 × 10 (0.25) x 2 = 2.5 × 10 − 10 } take square root of each side x = 1.6 × 10 − 5 = [OH− ] pOH = 4.6 } so the pH = 14 - 4.6 = 9.4 Forgetting to convert pOH to pH - DANGER!! VERY Common mistake!! Acids & Bases 62 Problem: What is the pH of the reaction of 500 mL of 2 M NH3 and 500 mL of 2 M HCl? Ka (NH4+) ≅ 1 × 10−10. Acids & Bases 63 Titrations A titration is the careful measured addition of a known concentration of one substance that will react with another unknown material in order to determine the concentration of the unknown material. Determining the concentrations of unknown materials is a routine procedure in chemistry. Titrations are commonly used to determine the concentrations of acids and bases in solution, as well as many other chemicals. Titrations are also often the simplest and least expensive way of determining concentrations of unknown materials. If we are titrating an unknown acid with a known amount of base with a known concentration, we can use the (M1)(V1) = (M2)(V2) relationship to find the unknown's concentration at the equivalence point (the point at which we've added just enough base to react with all the acid). In order to determine when we have reached the equivalence point in an acid base reaction, we generally use an indicator. Acids & Bases 64 Indicators An indicator is a weak organic acid or base that has sharply different colors in its associated and dissociated forms: HIn(aq) H+(aq) + In−(aq) red acidic blue basic Indicators usually have very intense colors so one only has to use a very small amount (a few drops) so it will not affect the titration of the solution. Remember that the indicator is an acid or base so if you add a lot it will affect the titration!! OH Phenophtalein O O OH O + H+ HO HO acidic form - colorless anionic basic form - red Methyl Red O O OH NN acidic form - red O N NN N anionic basic form - yellow + H+ Acids & Bases 65 Some Common Acid-Base Indicators Name pH color change region Acid color Base color Methyl violet 0-2 yellow violet methyl yellow 1.2 - 2.3 red yellow methyl orange 2.9 - 4.0 red yellow methyl red 4.2 - 6.3 red yellow bromthymol blue 6.0 - 7.6 yellow blue thymol blue 8.0 - 9.6 yellow blue phenolphthalein 8.3 - 10 colorless pink Alizarin yellow 10.1 - 12 yellow red HIn(aq) H+(aq) + In−(aq) [H+][In− ] Ka = [HIn] At the color change [H+] = [In−] = [HIn], so: Acids & Bases 66 Titrating an unknown strong acid with a known amount of strong base: Acids & Bases 67 Titrating an unknown strong base with a known amount of strong acid: Acids & Bases 68 Titrating an unknown weak base with a known amount of strong acid: Acids & Bases 69 Titrating an unknown weak acid with a known amount of strong base: Acids & Bases 70 Buffer Solutions Consider an acetic acid solution: HOAc(aq) + − H (aq) + OAc (aq) If we add enough NaOAc (the salt of acetic acid) to increase the OAc− concentration roughly equal to HOAc, we now form the following mixture: HOAc(aq) + H (aq) + OAc−(aq) The added OAc−, which is a weak base, will consume some of the free H+ causing the pH to rise (become less acidic). Note that if we add H+ to this solution, it will react with the large pool of weak base OAc− to form HOAc. The H+ concentration, therefore, will stay about the same. Similarly, if we add some OH− it will react with the H+ present. Since we have an equilibrium, however, some HOAc will, in turn, dissociate to replace the missing H+, keeping it about the same. A solution of a weak acid and the salt (conjugate base) of a weak acid – or – a weak base and the salt (conjugate acid) of a weak base is called a BUFFER! Acids & Bases 71 If the concentration of the two components is high enough and you don’t add too much acid or base, the buffer solution will be very resistant to changes in the pH due to added acid or base. Buffer solutions are critically important to biological systems (i.e., keeping us alive!). For a solution of a weak acid or base and their salt, one can write the following equations for calculating the [H+] and [OH−] of the buffer solution generated: [OH− ] ⎛ [Base] ⎞ = Kb ⎜ ⎟ [acidic salt] ⎠ ⎝ ⎛ [Acid] ⎞ = Ka ⎜ [basic salt] ⎟ ⎝ ⎠ These are called the Henderson-Hasselbalch equations. The log form of these equations: [H+] ⎛ [basic salt] ⎞ pH = pK a + log ⎜ ⎟ [acid] ⎠ ⎝ ⎛ [acidic salt] ⎞ pOH = pKb + log ⎜ ⎟ [base] ⎠ ⎝ Acids & Bases 72 If the acid/base concentration is the same as the salt concentration, then we can write: [OH−] = Kb [H+] = Ka –– or –– pOH = pKb pH = pKa By varying the weak acid or base and the salt being used to make the buffer solution, as well as their concentration ratio, one can set the pH of the buffer to almost anything you want. The more concentrated the buffer components, the more effective the buffer solution will be at resisting pH changes. But remember that you can always overload a buffer by adding too much acid or base to it. One of the trickiest things for you to determine is just what salts will work to make a buffer. The salts of strong acids and bases (i.e., NaCl, KBr, Na2SO4, CsNO3) do NOT usually make buffers, nor do mixtures of strong acids and bases with their salts! Acids & Bases 73 Note, however, that reacting ½ equivalent of a strong base with a weak acid, generates a buffer solution! This is the region where we have a buffer solution present Acids & Bases 74 Problem: Which of the following solutions is a traditional buffer prepared from a weak acid/base and a conjugate salt? a) 0.1 M NaOH + 0.1 M NaOAc b) 0.01 M HCl + 0.01 M NaCl c) 0.5 M NH4Cl + 0.5 M NH3 d) 2 M NaOAc + 2 M HOAc e) 0.01 M citric acid + 0.01 M sodium citrate f) 0.3 M H3PO4 + 0.3 M NaH2PO4 g) 2 M HNO3 + 2 M NaNO3 h) 0.05 M H2CO3 + 0.05 M KHCO3 i) 0.5 M HI + 0.5 M CsI j) 0.2 M benzoic acid + 0.2 M cesium benzoate k) 1 M NaOAc + 1 M KBr l) 0.001 M HCl + 0.001 M KOH m) 0.1 M KOAc + 0.1 M NH3 n) 0.2 M HOAc + 0.2 M KHCO3 b) Which will be the most effective buffer solution at maintaining a given pH? c) What are the pH’s of the buffer solutions? REDOX 1 Chapter 21 REDOX & Electrochemistry REDOX 2 There are two types of chemical reactions: 1) acid-base reactions (Lewis) 2) electron-transfer reactions Oxidation/oxidize: when you remove electrons from a material Reduction/reduce: when you add electrons to a material You cannot have oxidation without reduction; you cannot generally reduce a molecule without oxidizing another molecule (and the other way around). Oxidizing agent: a chemical that causes another material to be oxidized (the oxidizing agent is reduced!) Reducing agent: a chemical that causes another material to be reduced (the reducing agent is oxidized!) REDOX REDOX 3 Oxidation States In order to understand redox reactions, we first need to be able to figure out what the oxidation state of an element is. The oxidation state is a method to indicate how many electrons are "assigned" to a particular element. For this we use a +/- system: +n indicates that an atom has lost electrons and now has a positive charge −n indicates that an atom has gained electrons and now has a negative charge indicates that an atom has its elemental number of electrons assigned to it and, therefore, has no charge 0 Oxidation state is a “formalism”, that is, is may or may not reflect the actual charge on an atom. Common reference atoms & their oxidations states: Alkali metals = +1, alkaline earths = +2 O = −2 (exception = peroxides, H2O2, −1) Halides = −1 (exception = oxyhalides) REDOX 4 The key to being able to figure out the oxidation state of an element in a molecule is to note its electronegativity: • The higher the electronegativity the more the element likes to add electrons to its valence shell. • The lower the electronegativity the more likely an element will lose electrons. Let's dissect an example reaction: Reaction: PH 3 + O 2 P2 O5 + H2 O Electronegativities: PH 3 + O 2 H = 2.1 P = 2.1 O = 3.5 P2O5 + H2O -3 +1 0 +5 -2 +1 -2 Atoms that gain electrons usually gain enough to fill their valence shell (octet rule). Atoms that lose electrons only lose enough to get them down to the next lowest filled valence shell (but not always all the way!). REDOX 5 When two elements have the same electronegativity (as with phosphorus and hydrogen) the element that is furthest to the upper right hand side of the periodic table is the one that gets the electrons. 1 1 H Hydrogen 3 2 13 4 11 Sodium 19 K Potassium 37 20 55 5 6 7 8 9 21 22 23 24 25 26 27 Ca Sc Calcium 38 Rb Sr Rubidium 4 Strontium 56 Scandium 39 Y V Vanadium Chromium Manganese 41 42 43 40 Zirconium 57 72 Niobium 73 Hf Ta Cesium Hafnium Lanthanum Iron 44 Cobalt 45 Tantalum Molybdenum 74 Technetium Ruthenium 75 76 W Re Os Tungsten Rhenium Osmium Rhodium 77 Ir Iridium He 9 10 8 Helium C N O F Ne Nitrogen Oxygen Fluorine Neon 28 Nickel 46 11 29 12 Al Aluminum 30 31 Cu Zn Copper 47 Zinc 48 Palladium 78 Silver 79 Cadmium 80 Pt Au Platinum Gold 49 Indium 81 Tl Mercury 14 15 16 17 Thallium 18 Si P S Cl Ar Silicon Phosphorus Sulfur Chlorine Argon 32 33 34 35 Ge As Se Galium Zr Nb Mo Tc Ru Rh Pd Ag Cd In Yttrium Cs Ba La Barium 10 Cr Mn Fe Co Ni Ti Titanium 7 17 Carbon 13 3 15 16 B 12 Magnesium 6 2 Boron Group 8 Berylium Na Mg 14 5 Li Be Lithium 18 Periodic Table of the Elements Germanium 50 Arsenic 51 Selenium 52 Sn Sb Te Tin 82 Antimony 83 Tellurium 84 36 Kr Bromine 53 Krypton 54 I Xe Iodine Xenon 85 86 Pb Bi Po At Rn Lead Bismuth Polonium Astatine Radon It is important to remember that the bond between phosphorus and hydrogen is, in reality, pretty much neutral and covalent in character (although the hydrogen atoms actually do have a small amount of positive charge on them). Only when there is a considerable difference in the electronegativities of two atoms does one see formal charge separations and polar bonds occurring. REDOX 6 Problem: What are the oxidation state assignments for the following compounds: a) CH4 b) CO2 c) KMnO4 d) FeS e) H2SO4 f) HCN g) Na2CrO4 h) N2 i) MnO2 j) H2C2O4 REDOX 7 Balancing Redox Equations The best way of balancing redox reactions is the half cell method. Break up the overall rxn into 2 half-cell rxns: one for the reduction and one for the oxidation. We then want to multiply each half cell reaction to make the overall # of electrons the same so they cancel out. Note that this fairly simple rxn (and many redox rxns) could have been balanced the “normal” way. But if it is a redox rxn and looks “hard”, use the half cell method. REDOX 8 Let's do another balancing act: Note that in the last two examples, one of the products or reactants (O2, P2O5, and Al2O3) had two redox active atoms present (O, P, and Al) and that had to be accounted for when writing out the half cell reaction. Also note that when balancing half cell rxns, we mainly worry about the redox active atoms and not the H and O atoms (assuming they are not changing oxidation states). These will be balanced at a later point (see next example). REDOX 9 Another EXAMPLE: This is the “core” redox balanced reaction. Now, you have to check the oxygen, and then hydrogen atom balance: REDOX 10 To balance the rxn in basic solution you need to balance the rxn first as if it was in acidic solution (as shown above). Then you “get rid” of the H+ by adding as many OH− to each side as there are protons and reacting the H+ and OH− together to make waters: REDOX 11 Problem: Balance the following (in water): a) Cr2O72− + Cl− Cr3+ + Cl2(g) (acidic) b) MnO4- + CN− MnO2(s) + CNO− (basic) REDOX 12 c) S2O32−(aq) + OCl− (s) Cl− (aq) + S4O62− (aq) (acidic) d) Au(s) + CN−(aq) + O2(g) Au(CN)4−(aq) + OH−(aq) (basic) REDOX 13 Electrochemical/Voltaic/Galvanic Cells Consider the reaction of Zn(s) and Cu2+(aq): Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) This reaction involves the transfer of 2 e- between the Zn(s) and Cu2+. The Cu2+ oxidizes the Zn, and is in turn reduced to form metallic Cu, while the Zn metal dissolves into solution as Zn2+. This reaction is spontaneous with a ΔGº = − 212 kJ/mol. Mixing Zn(s) and Cu2+ together in a beaker causes a spontaneous reaction that gives off all the energy as heat, warming the solution. The transferring of the electrons between the Zn and Cu2+, however, can accomplish much more useful work if we can figure out a way to get the eto flow through a wire as electricity, where the work represented by −ΔGº (free energy) can now be used as electrical work. REDOX 14 To harness the intrinsic chemical energy stored in these oxidizing and reducing agents, we must separate them to force the electrons to flow though an external circuit and do some work. flow of electrons salt bridge or semi-permeable membrane Zn Zn 2+ Cl Zn Cl - Zn 2+ Cl 2+ Cl - - Cl - Cl Cu Cl - Cu 2+ Cl - 2+ Cu Cu Cu 2+ Cl - Cl Zn2+ Cl - Cl 2+ 2+ 2+ Zn Cl Cl - Cu Cu Cu Cu Cu Cu Cu Cu ANODE CATHODE oxidation reduction Anion flow Red Cat = Reduction occurs at Cathode REDOX 15 There are different configurations of liquid galvanic cells that one can setup: flow of electrons salt bridge or semi-permeable membrane Zn Cl - Zn 2+ Zn 2+ Cl - Cl - Cu 2+ Cu Cl - Cl - Cu2+ 2+ Cu Cl - Cl - 2+ Cl Cl - Cl Zn 2+ - Cu2+ Cu2+ Zn Cl Cl Cl 2+ Zn Cl ANODE oxidation Anion flow Cu Cu Cu Cu semipermeable membrane Cu Cu Cu CATHODE reduction flow of electrons flow of electrons ANODE oxidation Zn Zn Cu ANODE oxidation Anion flow CATHODE reduction salt bridge CATHODE reduction Daniell Cell REDOX 16 Measuring Cell Potentials (Voltages) M salt bridge or semi-permeable membrane H2 H+ X− H+ Pt H+ X− X− H+ + X− X− −M M+ + X H − X− −X X − X− X M+ X− M+ + H X− X− + M+ X− H − M+ X + + X− H X− M X− M+ M+ On a platinum electrode, H2(g) and H+(aq) are in redox equilibrium with one another (1 atm H2, 1M H+). This is called a hydrogen electrode. H2 2H+ + 2e- By setting this electrochemical potential to 0.0 V, we have a reference electrode to which we can measure the innate ability of a material in the other electrode compartment of the electrochemical cell to either accept (be reduced/cathode) or give up (be oxidized/anode) electrons from/to H2/H+. REDOX 17 Standard Reduction Potentials at 25°C Half Cell Rxn F2(g) + 2e2F-(aq) O3(g) + 2H+(aq) + 2eO2(g) + H2O Co3+(aq) + eCo2+(aq) Cl2(g) + 2e2Cl-(aq) O2(g) + 4H+(aq) + 4e2H2O Ag+(aq) + eAg(s) Cu+(aq) + eCu(s) Cu2+(aq) + 2eCu(s) AgCl(s) + eAg(s) + Cl−(aq) Cu2+(aq) + eCu+(aq) 2H+(aq) + 2eH2(g) Pb2+(aq) + 2ePb(s) V3+(aq) + eV2+(aq) Zn2+(aq) + 2eZn(s) Al3+(aq) + 3eAl(s) H2(g) + 2e2H-(aq) Mg2+(aq) + 2eMg(s) Na+(aq) + eNa(s) Li+(aq) + eLi(s) E°red (V) +2.87 +2.07 +1.81 +1.36 +1.23 +0.80 +0.52 +0.34 +0.22 +0.15 0.00 -0.13 -0.24 -0.76 -1.66 -2.25 -2.36 -2.71 -3.05 It is critically important that you learn how to read, interpret, understand, and properly use this table. Voltage (or potential) represents the driving force for “pushing” the electrons from one location to another (for example, from a reducing agent to an oxidizing agent). The higher the voltage the more REDOX 18 strongly the electrons will be pushed through the wire (or solution). A positive cell potential (voltage) indicates a spontaneous electrochemical reaction. A negative cell potential (voltage) indicates a non-spontaneous reaction (the opposite reaction will, therefore, be spontaneous!). Note that the sign notation here is opposite that we learned for ΔG! Problems: Which of the following substances is the best reducing agent? a) Cu b) Zn c) Pb d) Ag e) H2 Which of the following substances is the best oxidizing agent? c) O2 d) Na+ e) Li+ a) Cu+ b) O3 REDOX 19 To calculate the voltage of an electrochemical cell one can simply add together the two half-cell reactions that make up the overall galvanic (electrochemical) cell. Let’s calculate the spontaneous cell potential (voltage) for our Cu/Zn cell. First look up the two half cell reactions from the table of Standard Reduction Potentials: Cu2+(aq) + 2eZn2+(aq) + 2e- Cu(s) Zn(s) +0.34 -0.76 The more positive potential (relative to the hydrogen electrode) for the Cu2+ reduction compared to the Zn2+ reduction means that Cu2+ is a stronger oxidizing agent (wants to be reduced more) than Zn2+. REDOX 20 To add these half-cell rxns together to give us our overall spontaneous net rxn and cell potential, we must switch one of the two half-cell rxns around so that it is written as an oxidation rxn (a compound giving up electrons on the product side of the rxn). The half-cell rxn with the smaller positive, or more negative potential gets flipped around (reversed). This would be the Zn rxn: Cu2+(aq) + 2eCu(s) Zn2+(aq) + 2eZn(s) +0.34 +0.76 When one flips one of the half-cell rxns around, one also flips (reverses) the sign of the cell potential!! The next step is to multiply each of the half cell rxns by the appropriate # to balance the # of electrons involved so that they cancel out. For the Cu/Zn cell, there are 2e- on each side of the two half-cell rxns, so everything is already balanced: Cu2+(aq) + 2eCu(s) Zn2+(aq) + 2e+ Zn(s) +0.34 V +0.76 V Cu2+(aq) + Zn(s) +1.1 V Zn2+(aq) + Cu(s) REDOX 21 So the net reaction has a voltage of +1.1 V, the positive potential indicates that the rxn is spontaneous. Note that the voltage calculated for the cell is for standard conditions (1 atm, 1 M). Changing the concentrations away from 1 M will change the cell potential! Similarly, changing the temperature will also affect the voltage. The Nernst Equation allows one to calculate the voltage (potential) of a electrochemical reaction when one is not working under standard conditions! Also note that for galvanic cells (batteries) we always want to use a spontaneous reaction that gives off energy that we can use to perform work. But on homeworks or tests I may give you an overall balanced reaction that is non-spontaneous. In this case you need to write down the half-cell rxns in the right order to give the overall balanced rxn and calculate the cell potentials from the way that these half-cell reactions are written. REDOX 22 Example: What is the potential for the following reaction? 2Na+(aq) + 2Cl−(aq) Cl2(g) + 2Na(s) Reduction half cell: 2[Na+(aq) + 1eOxidation half cell: 2Cl−(aq) This half-cell was the one that was flipped around (reversed) Na(s)] Cl2(g) + 2e- −2.71 V −1.36 V sum = −4.07 V Non-spontaneous The reverse reaction is very spontaneous with a cell potential of +4.07 V. REDOX 23 Example: What is the potential for the electrochemical cell composed of the following 2 this, we are looking for half cell rxns? When phrased like a positive cell potential!a spontaneous reaction that gives Cr+3(aq) + 3eMnO2 + 4H+ + 2e- Cr(s) Mn2+(aq) + 2H2O -0.74 +1.28 The Cr+3(aq) + 3e- → Cr(s) rxn has the more negative potential, so it gets flipped around: Cr(s) Cr+3(aq) + 3eMnO2 + 4H+ + 2eMn2+(aq) + 2H2O +0.74 +1.28 Now we have to balance the # of electrons in each rxn so we can add them together. 6 e- is the common factor, so we need to multiply the Cr rxn by 2 and the MnO2 rxn by 3: 2[Cr(s) Cr+3(aq) + 3e- ] 3[MnO2 + 4H+ + 2e- Mn2+(aq) + 2H2O] +0.74 +1.28 2Cr(s) + 3MnO2 + 12H+ 2Cr+3(aq) + 3Mn2+(aq) + 6H2O +2.02V Note that you do NOT multiply the cell potentials by the numerical factors (2 or 3) used to balance the # of electrons in each half-cell rxn!! DANGER!! Common mistake!! REDOX 24 Example: Zn metal reacts with HCl, but Cu metal doesn’t. Calculate the cell potentials for these rxns to see if they fit the experimental data. Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) Splitting this rxn into the two half cell rxns, we have: Zn(s) Zn2+(aq) + 2eH2(g) 2H+(aq) + 2eZn(s) + 2H+(aq) Zn2+(aq) + H2(g) +0.76 0.00 +0.76V Note that the Zn half-cell rxn is flipped around from how it is written in the Standard Reduction Table, since it is acting as a reducing agent here. The calculated positive cell potential indicates a spontaneous rxn. REDOX 25 For the copper rxn, we have: Cu(s) + 2H+(aq) Cu2+(aq) + H2(g) Cu2+(aq) + 2eH2(g) 2H+(aq) + 2e- Cu(s) Cu(s) + 2H+(aq) Cu2+(aq) + H2(g) −0.34 0.00 −0.34V Once again, note that the copper half-cell rxn is flipped around from how it is written in the Standard Reduction Table, since it is acting as a reducing agent here. So here we calculate that the cell potential is negative, indicating a nonspontaneous rxn – or one that should not occur normally, which fits the experimental data. This is because copper metal is a more inert material relative to zinc metal. One could also note the position of copper above the hydrogen half cell rxn, which indicates that the reverse rxn of Cu+2 with H2 gas would be spontaneous: Cu2+(aq) + H2(g) Cu(s) + 2H+(aq) +0.34V Example: What about the reaction of Cu with nitric acid? Will this be a spontaneous rxn? REDOX 26 Nitric acid is not like the other strong acids (HCl, HBr, HI, H2SO4) in that it is a good oxidizing acid due to the presence of the NO3− anion, which is not just a simple inert counter-anion for H+. The combination of NO3− and H+ makes for a rather strong oxidizing mixture. Cu2+(aq) + 2e-] 2[NO3−(aq) + 4H+(aq) + 3eNO(g) + 2H2O] 3[Cu(s) 3Cu(s) + 8H+(aq) + 2NO3−(aq) 3Cu2+(aq) + 2NO(g) + 4H2O −0.34 +0.96 +0.62V If you get nitric acid on your skin, you will not only feel the burning of the acid (H+), but your skin will be oxidized to a yellow-brown color! So concentrated nitric acid is doubly dangerous! REDOX 27 Although one can add half-cell rxns to yield overall redox equations, one can not simply add two halfcell rxns to yield another half-cell rxn. For example, consider the addition of the following two half cell rxns to generate a third half cell rxn: Cu2+(aq) + 1eCu+(aq) + 1eCu2+(aq) + 2e- Cu+(aq) Cu(s) Cu(s) +0.16 V +0.52 V +0.68 V The correct half-cell potential for this rxn is 0.34 V, which is exactly half of what we incorrectly attempted to calculate above. This factor of ½ comes from the fact that all the half-cell potential values are normalized to a single e- value even if multiple e- are used in the half cell rxn. As you might expect, the electrochemical potential for a rxn is directly related to the ΔG for a reaction (only with an opposite sign relationship!): ΔGº = −nFEº n = # of electrons being transferred, F = Faraday’s constant (96.5 kJ/Vmol), Eº = standard potential. Problems: Calculate the cell potentials for the following reactions. Are they spontaneous or not? REDOX 28 a) F2(g) + 2Na(s) b) Cu(s) + Zn2+(aq) c) 2Al3+(aq) + 3Pb(s) 2Na+(aq) + 2F−(aq) Cu2+(aq) + Zn(s) 2Al(s) + 3Pb2+(aq) d) 2Co3+(aq) + 2Cl-(aq) 2Co2+(aq) + Cl2(g) e) 2Mg(s) + O2(aq) + 4H+(aq) 2Mg2+(aq) + 2H2O(aq) f) 2Al(s) + 6H+(aq) 2Al3+(aq) + 3H2(g) g) 2Cu(s) + Cl2(g) 2Cu+(aq) + 2Cl−(aq) REDOX 29 Non-Standard Conditions: Nernst Equation Eº values are for standard conditions (1 M or 1 atm). If the reactant/product concentrations are different the cell potential, E, will differ from Eº. One can calculate non-standard potentials using the Nernst equation: 2.303 RT E= E − log Q nF R = gas constant, 8.314 J/mol●K T = absolute temperature, K n = # of moles of electrons transferred F = faraday, 96,485 C/mol●eQ = reaction quotient 2.303 RT = 0.0592 V imol At room temp, 25 C , F substituting this into the Nernst eq gives : 0.0592 E= E − log Q n REDOX 30 Example: Calculate the potential for the Fe3+/Fe2+ electrode when the concentration of Fe2+ is five times greater than that of Fe3+. Look up potential in a half cell table: Fe3+ + eFe2+ Eº = +0.771 V The Fe2+ concentration is five times greater than that of Fe3+, so the Q expression is: [products] p [Fe2+ ] Q= =5 r= 3+ [reactants] [Fe ] Assume room temp (25ºC = 298K) and solve using the Nernst equation: 0.0592 E= E − log Q n 0.0592 E = 0.771 − log(5) 1 E = 0.771 − (0.0592)(0.7) = 0.730 V Problem: The potential for this is lower than Eº where both concentrations are 1 M. This should make sense based on Le Chatelier’s principle. Why? REDOX 31 Problem: A cell is constructed at 25ºC as follows. One half-cell consists of Cu2+/Cu, [Cu2+] = 0.4 M. The other half-cell involves Zn2+/Zn with [Zn2+] = 0.4 M. Apply the Nernst equation to the overall cell reaction to determine the cell potential. Problem: Consider the following half-cell rxn: Pb(s) + SO42−(aq) PbSO4(s) + 2e- Eº = +0.13 V Can the Nernst equation help us figure out how to change concentrations to increase the cell potential? Why? REDOX 32 Lead-Acid Storage Battery Anode Rxn: Pb(s) + SO42−(aq) PbSO4(s) + 2e- Eº = +0.13 V Cathode Rxn: PbO2(s) + 4H+(aq) + SO42−(aq) + 2ePbSO4(s) + 2H2O Eº = +1.69 V Overall Rxn: Pb(s) + PbO2(s) + 4H2SO4(aq) 2PbSO4(s) + 2H2O Eº = +1.82 V REDOX 33 Dry Cell (regular battery) Graphite Cathode Moist Paste of NH4Cl, MnO2 & carbon Porous separator Zinc anode Note that the production of NH3(g) in this “regular” cell produces an insulating layer between the cathode and the MnO2/NH4+. This causes these batteries to fade relatively quickly after steady use. Letting them sit for a while, allows the NH3(g) to diffuse away & allow fresh reagents to come in contact with the cathode, thus the battery becomes rejuvenated. Anode Rxn: Zn Zn2+ + 2e- Eº = 0.76 V Cathode Rxn: 2MnO2 + 8NH4+ 2Mn3+ + 4H2O + 8NH3 Eº = 1.21 V Overall Rxn: Zn + 2MnO2 + 8NH4+ 2Mn3+ + Zn2+ + 4H2O + 8NH3 Real batteries only produce 1.5 V because the real chemistry is far more complicated and less efficient. Eº = 1.97 V REDOX 34 Alkaline Cells Graphite Cathode Moist Paste of KOH, MnO 2 & graphite Porous separator Zinc anode Anode Rxn: Zn + 2OH− Cathode Rxn: 2MnO2 + 2H2O The simple replacement of NH4Cl electrolyte with a KOH gel eliminates the production of the problematic NH3 gas that occurs in the regular dry cell. This battery has much improved stability for delivering electrical current until all the reagents are used up. Zn(OH)2 + 2e2MnO(OH) + 2OH− Overall Rxn: Zn + 2MnO2 + H2O 2MnO(OH) + Zn(OH)2 Eº = 1.5 V REDOX 35 Corrosion of Steel/Iron Cathode O2 Fe Steel 2+ O2 Anode Cathode: ½O2 + H2O + 2eFe2+ + 2eAnode: Fe Sum: Water Fe 2+ ½O2 + Fe + H2O 2OH− Fe(OH)2 Subsequent Redox rxn: 4Fe(OH)2 + O2 2Fe2O3 • 4H2O RUST The presence of NaCl (salt) in the water makes it more conductive to e- flow, making it easier for the O2 to oxidize the iron without having to come directly in contact with it. Instead it can react anywhere at the surface of the water drop. REDOX 36 Electrolysis – Electrolytic Cells The process by which a chemical reaction is made to occur by the passage of an electric current through a solution is call electrolysis. -+ power source 99.999% pure Cu plates out Cu 2+ Cu Cu Cu 2+ Cu Cu 2+ impure copper dissolves 2+ Cu Cu Cu 2+ 2+ Cu 2+ 2+ Cu 2+ Anode sludge Ag, Au, Pt, Rh REDOX 37 The amount of material plated out in an electrochemical cell is related to the # of electrons that pass through the cell: # moles deposited ∝ # moles of electrons 1 Faraday = 96,485 C/mol e1 Amp = 1 C/sec The overall formula for calculating the number of moles of material deposited in a electrochemical cell given a certain current and time is: (#amps)(time in seconds) # moles = (96,485 C / mole e-)(#e−) Remember that if the time is given in minutes or hours that you have to first convert it into seconds. Also remember to convert amps to C/sec for the units to work out correctly. REDOX 38 Example: How many grams of Cu(s) will be deposited from a solution of Cu2+ by a current of 1.5 amps flowing for 2 hours? Cu2+ + 2eCu(s) It takes 2 moles of electrons to produce 1 mole of Cu(s) from one mole of Cu2+. So n = 2. (1.5C/sec) (7200 sec) # moles Cu = (96,500 C/mol e-)(2 e-) 10,800 C # moles Cu = = 0.056 mol 193,000 C/mol # grams of Cu is equal to the # moles times the molecular (or atomic) weight of Cu: (5.6 × 10−2 mol Cu) (63.5 g/mol Cu) = 3.55 g Cu REDOX 39 Problem: How many moles of Al metal will be electrodeposited if 9.6 amps is passed through a solution of Al+3 for 10,000 seconds? Problem: How long (in seconds) will it take to electrodeposit 0.25 moles of Ti metal by passing a current of 96.48 amps through a solution of Ti+4 ? REDOX 40 Standard Reduction Potentials - 25ºC REDOX 41 REDOX 42 Tables from Chemistry, 8th Ed, Whitten, Davis, Peck & Stanley (Thomson-Brooks/Cole) CHEM 1422 Honors: General Chemistry Section 01 Previous 2 Years of Blank Homeworks & Exams Prof. George G. Stanley Department of Chemistry CHEM 1202 - Homework # 1 Background Due January 22, 2009 (2 PM) Name: ___________________________ Signature: ________________________ Group: __________________________ Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 1. (5 pts) Why do or don't you like chemistry as it has been taught to you so far in high school and college? What did or didn’t you like? Don't try to suck up to me by writing only good things (unless they are true). I'm looking for critical comments and reasoning. Tell me about things you have liked (if any) and disliked. Short mindless answers will get no credit. This question is an individual, not group, effort. Please type up your answer neatly, include your name, and E-mail it to me separately (gstanley@lsu.edu) as a Word or PDF file (or body of the E-mail message). 2. (5 pts) Define electronegativity in your own words (don’t just copy from a book). Give a couple of elemental examples to illustrate your definition (i.e., elements with high or low values and how their electronegativity affects their ability to gain or lose electrons). 3. (5 pts) Look up, list, and explain the trend in melting points in terms of intermolecular attractive forces for the following compounds: CsI, LiF, BaS, MgO, and Al2O3. Make use of Coulomb’s Law (in a qualitative sense) in your explanation. CHEM 1422 - HW#1 Background (2009) 4. (5 pts) Consider the reaction: 2C(s) + 2Br2 2 Br2C=CBr2 If 120 g of carbon reacts with 160 g of Br2 and the reaction goes in 50% yield, how many grams of Br2C=CBr2 (C2Br4) are produced? Clearly show and explain all your work. 5. (10 pts) Sketch out the Lewis dot structures for the following molecules (use lines for bonds and pairs of dots for lone pairs). Indicate the formal charges, if any, on the appropriate atoms. a) H2SO4 b) C2H2 c) AlCl3 d) NO3 e) P(CH3)3 CHEM 1202 - Homework # 2 Thermodynamics Due Sept 19th, 2006 Name _____________________________ Signature: _________________________ Group Name: ______________________ Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 1. (12 pts) From the thermodynamic data given at the end of your lecture notes calculate ΔHºrxn, ΔSºrxn and ΔGºrxn for the following reactions. For ΔGºrxn please use the ΔGºrxn = ΔHºrxn − TΔSºrxn formula with T = 298K. Indicate whether the reactions are spontaneous or non-spontaneous. Show your work. a) 12NH3(g) + 21O2(g) b) 2O3(g) 8HNO3(g) + 4NO(g) + 14H2O(g) 3O2(g) C6H12O6(s) (glucose) + 6O2(g) } this is photosynthesis! c) 6CO2(g) + 6H2O(l) ΔHf° (C6H12O6) = −1274 KJ/mol ΔGf° (C6H12O6) = −910 KJ/mol S° (C6H12O6) = 212 J/Kmol 2. (8 pts) For the following processes, is the entropy of reaction (ΔSrxn) increasing, decreasing or staying about the same? Use the qualitative rules about entropy discussed in lecture to determine the answer. a) Ca(s) + H2SO4(aq) b) 4Fe(s) + 3O2(g) c) N2(g) + 3H2(g) CaSO4(s) + H2(g) 2Fe2O3(s) (rust) 2NH3(g) d) setting off a fire cracker (illegal in East Baton Rouge Parish) e) AgCl(s) + Br−(aq) f) Mg(s) + CO(g) g) Cu+2(aq) + 2OH−(aq) h) Ag+(aq) + Cu(s) AgBr(s) + Cl−(aq) MgO(s) + C(s) Cu(OH)2(s) Ag(s) + Cu+(aq) 3. (5 pts) A typical “instant” hot pack uses 55 g of CaCl2(s) and 180 g of H2O, which react together to make hydrated Ca2+(aq). Use the following formula to calculate the temperature (ºC) of the heat pack after it has been activated (mixed together). Assume that the initial temperature of the water is 25°C (room temperature), Cp (water) = 75 J/mol K (watch your units!). ΔHrxn = −60 kJ/mol ΔT = − nsalt ( ΔHrxn) nwater (Cpwater ) nwater = # of moles of water used ΔT = Tfinal - Tinital nsalt = # of moles of salt dissolved Cp = heat capacity of solvent ΔHrxn = enthalpy of dissolving salt in solution (also called ΔHsol) 4. (5 pts) (a) Using the Gibbs Free Energy formula to plot the value of ΔGºrxn from 500 K to 1200 K for the reaction: 2HgO(s) 2Hg(l) + O2(g) (use table at the end of the chapter -- you can assume that ΔHºf and Sº do not vary much with temperature) The vertical axis is energy in kJ/mol -- you have to fill in the range of units for this axis based on your calculations. (HINT: you only have to calculate 3 to 5 data points to make the plot, but you can also do this in Excel and let it calculate ΔGºrxn at many temps). (b) What is the temperature range for which this reaction is spontaneous? Indicate on plot. (c) Mark and indicate the temperature at which ΔGrxn = 0? CHEM 1422 - Homework # 2 Thermodynamics Due January 29, 2009 Name: ____________________________ Signature: _______________________ Group Name: _____________________ Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 1. (12 pts) From the thermodynamic data given at the end of your lecture notes calculate ΔHºrxn, ΔSºrxn and ΔGºrxn for the following reactions. For ΔGºrxn please use the ΔGºrxn = ΔHºrxn − TΔSºrxn formula with T = 298K. Indicate whether the reactions are spontaneous or non-spontaneous. Show your work. a) Cl2(g) + H2(g) b) Fe(s) + 5CO(g) c) SiO2(s) + 6HF(g) d) CaCO3(s) 2HCl(g) Fe(CO)5(l) H2SiF6(aq) + 2H2O(l) CaO(s) + CO2(g) [glass is mainly SiO2] HW#2 – Thermodynamics (2009) 2 2. (8 pts) Is the entropy increasing, decreasing or staying about the same? Use the qualitative entropy rules discussed in lecture to determine the answer. Write the answer to the right of each process. a) Fe2O3(s) + Al(s) b) raw egg Fe(s) + Al2O3(s) hard boiled egg c) CaCl2(s) + 6H2O(l) d) C6H12(l) + 9O2(g) e) H2CO(aq) + H2O(l) [Ca(H2O)6]Cl2(s) 6CO2(g) + 6H2O(g) 2H2(g) + CO2(g) f) mowing the lawn g) AgCl(s) + I−(aq) h) Sr(s) + 2H2O(l) AgI(s) + Cl−(aq) Sr2+(aq) + 2OH−(aq) + H2(g) 3. (2 pts) Circle the compound that has the highest entropy. Give a brief reason explaining your answer. a) Hg(l) b) H2O(l) c) Pb(s) d) C2H5OH(l) e) CCl4(l) 4. (4 pts) Why does Al2O3(s) have a lower entropy than Fe2O3(s)? There are two primary qualitative reasons for this. You may have to use the chemistry library to get more information (i.e., properties) on these two common compounds to answer the question. 5. (4 pts) a) Small amounts of Fe(CO)5(l) usually form in steel tanks containing pressurized CO(g). You worked out the thermodynamics of this in question 1b. At what temperature (ºC) will the formation of Fe(CO)4 become non-spontaneous? Show your work. b) A similar reaction occurs to make Ni(CO)4(l) with ΔGºrxn = −38 kJ/mol, ΔHºrxn = −230 kJ/mol, and ΔSºrxn = −480 J/Kmol. High pressure reactors use a thin disk of metal as a safety mechanism that will rupture and release gasses if the pressure in the reactor gets too high. If one was using CO gas, which disk (Fe or Ni) would be more likely to prematurely fail due to the metal being dissolved away by CO? Briefly explain why. CHEM 1202 - Homework # 3 Chemical Kinetics Due Tuesday, Sept 26, 2006 by 1 PM Name _____________________________ Signed Name ______________________ Group Name: _____________________ 1. (5 pts) Which of the following energy diagrams best represents a reaction which will be the fastest and most spontaneous. Circle your choice and include your reasoning below (brief statement). a) b) c) d) R P ΔG ΔG ΔG R P R Rxn Coordinate ΔG R P Rxn Coordinate P Rxn Coordinate Rxn Coordinate 2. (5 pts) Consider the energy diagram to the right: Circle the following diagram below that best represents the effect of adding a catalyst to the above reaction. Include your reasoning below (brief statement). a) b) c) ΔG R P d) Rxn Coordinate P ΔG ΔG ΔG R P R Rxn Coordinate ΔG R R P Rxn Coordinate Rxn Coordinate P Rxn Coordinate 3. (5 pts) Consider the following reaction that is quite important for the manufacturing of many chemicals: + 3H2 cyclohexane benzene ΔG = +200 kJ/mol Activation Energy = +400 kJ/mol Circle the energy curve shown below (R = reactants, P = products) that best represents the reaction described above? Include your reasoning below (brief statement). a) ΔG b) R ΔG P Rxn Coordinate c) R ΔG P Rxn Coordinate d) ΔG R P Rxn Coordinate P R Rxn Coordinate HW #3 – Kinetics (2006) 2 4. (5 pts) Consider the following reaction and kinetic data. Circle the correct kinetic rate expression for this reaction. Clearly show and briefly discuss your reasoning. A+B Exp # [A] [B] Initial Rate (Msec−1) 1 1.5 M 1.5 M 0.2 2 1.5 M 3.0 M 0.8 3 3.0 M 3.0 M 0.8 4 6.0 M 3.0 M C 0.8 a) rate = k[A][B] b) rate = k[A] c) rate = k[B] d) rate = k[B]2 e) not enough data 5. (4 pts) Consider the following reaction and kinetic data. What is the rate constant for the reaction? Clearly show all your work and reasoning. H2NCH2CH2CH2CH2NH2 + 2HCl Exp # [(N)2] [HCl] Initial Rate (Msec−1) 1 0.1 0.1 0.2 2 0.2 0.1 0.4 3 0.4 0.2 3.2 4 0.1 0.2 [H3NCH2CH2CH2CH2NH3]2+(Cl−)2 0.8 6. (6 pts) The reaction in question 3 is non-spontaneous at room temperature. + 3H2 cyclohexane benzene ΔG = +200 kJ/mol Activation Energy = +400 kJ/mol Discuss the one thing that you can do that will increase the rate of reaction and make the reaction more spontaneous. Clearly discuss how this change will influence the rate of reaction and why it will affect the thermodynamics (spontaneity) of the reaction. CHEM 1422 - Homework # 3 Chemical Kinetics Due Feb 12, 2009 (2 PM) Name: ___________________________ Signature: _______________________ Group Name: ____________________ Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 1. (3 pts) Which of the following energy diagrams best represents the slowest spontaneous reaction? Circle your choice. Give a brief, but clear, explanation for your answer below the diagrams. a) b) c) d) R P G G R Rxn Coordinate P R Rxn Coordinate G G R P Rxn Coordinate P Rxn Coordinate 2. (5 pts) a) Describe in your own words and terms where the origin of the activation barrier comes from and what it represents in a chemical reaction. b) Given the same thermodynamic factors, consider the reaction of two small molecules or two large molecules with one another. Which pair should have the higher activation energy? Why? HW#3 – Kinetics (2009) 2 3. (3 pts) Consider the following reaction and information: O HO C H2 CH3 H2C G = 100 kJ/mol CH2 + CO + H2O Activation Energy = +400 kJ/mol Circle the energy curve shown below (R = reactants, P = products) that best represents the reaction described above? Give a brief, but clear, explanation for your answer below the diagrams. a) b) c) d) R G R G G P Rxn Coordinate P Rxn Coordinate G R P Rxn Coordinate P R Rxn Coordinate 4. (5 pts) Consider the following reaction and kinetic data. Circle the correct kinetic rate expression for this reaction. Show all your work and/or discuss your reasoning. 2A + B a) rate = k[A][B] b) rate = k[A]2 c) rate = k[B] Exp # [A] [B] Initial Rate (Msec1) 1 0.2 M 0.1 M 0.02 2 0.4 M 0.1 M 0.04 3 0.2 M 0.3 M 0.18 4 0.4 M 0.3 M 0.36 C+D d) rate = k[B]2 e) rate = k[A][B]2 HW#3 – Kinetics (2009) 3 5. (5 pts) Consider the following reaction and kinetic data. Circle the correct rate constant for this reaction. Clearly show all your work including the rate law that you determine. A + 2B a) 2.2 x 106 M1sec1 b) 22 M1sec1 c) 220 M1sec1 Exp # [A] [B] 0.2 M 0.1 M 0.2 M 0.2 M 0.002 3 0.4 M 0.2 M 0.008 4 0.8 M 0.4 M e) not enough data 0.002 2 d) 0.05 M1sec1 Initial Rate (Msec1) 1 C+D 0.032 6. (4 pts) Catalysts can be used on non-spontaneous reactions to lower the activation barrier. If a catalyst lowers the activation barrier too much, however, a serious problem can arise. Consider the diagrams shown below. What is the problem for the catalyzed rxn with the lower activation energy? Why can a “substantial” activation barrier actually help an “uphill” chemical reaction if one wants to make as much product as possible? 7. (5 pts) A reaction has a initial rate of reaction of 0.001 Msec-1 at 70°C. This increases to 0.100 Msec-1 at 90°C. Calculate the activation energy for this reaction? CHEM 1202 - Homework # 4 Chemical Equilibrium # 1 Due Thursday, Oct 19, 2006 Name ___________________________ Signed Name: ____________________ Group Name: ____________________ Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 2NO(g) Keq = 4 1. (3 pts) Given the following information: N2(g) + O2(g) Calculate the equilibrium constants for the equations shown below (you don’t have to show your work): a) 2NO(g) N2(g) + O2(g) NO(g) 4NO(g) b) ½N2(g) + ½O2(g) c) 2N2(g) + 2O2(g) Keq = Keq = Keq = 2. (2 pts) Circle the equilibrium expression listed below that is the correct one for the following reaction: SCl2(g) + H2O(g) a) [SCl2] [H2O] Keq = [HCl] [SO] b) 2HCl(g) + SO (g) [SCl2] [H2O] Keq = [HCl]2 [SO] c) [HCl] [SO] Keq = [SCl ] [H O] 2 2 d) [HCl]2 [SO] Keq = [SCl ] [H O] 2 2 3. (4 pts) The initial concentrations for the following reaction are [CH3I] = [Cl−] = 0 M, and [CH3Cl] = [I−] = 2 M. What is the concentration of the reactant [Cl−] at equilibrium? Show your work! Soln = solution Initial: @Eq: CH3I(soln) + Cl−(soln) CH3Cl(soln) + I−(soln) Keq = 9 4. (6 pts) The initial concentrations of reactants = 0.2 M and products = 1.6 M. What is the concentration of methanol (CH3OH) at equilibrium for the following reaction? Clearly show all your work! Initial: @Eq: a) 0.1 M CH3OH(g) + HI (g) b) 0.3 M CH3I(g) + H2O(g) c) 0.6 M d) 1.2 M Keq = 25 e) 1.5 M f) 3.0 M Homework # 4 (Equilibrium #1) 2006 2 5. (2 pts) Consider the potential energy diagram to the right: Which of the following Keq values and relative reaction rates fits best? Briefly and clearly give your reasoning in the space below. a) Keq = 1.5, fast rxn c) Keq = 1 x 10−56, fast rxn b) Keq = 200, slow rxn ΔG P R Rxn Coordinate e) Keq = 1 x 10−4, slow rxn d) Keq = 1 x 106, very fast rxn 6. (4 pts) Consider the following reaction shown below. We start with [Rh4(CO)12] = 0.2 M, [H2] = 1 M, and [HRh(CO)3] = 0 M. When the reaction reaches equilibrium there is 0.02 M HRh(CO)3. Calculate Keq for this reaction. Clearly show all your work. Initial: Rh4(CO)12 + 2 H2 4 HRh(CO)3 @ Eq: 7. (4 pts) Consider the following reaction: P4(soln) + 5 NaClO2(soln) P4O10(soln) + 5 NaCl(soln) Keq = 1 × 1052 The initial concentrations of P4 and NaClO2 are both 0.5 M (no products present). Circle the concentration of P4O12 at equilibrium? Clearly and briefly explain your answer below. a) 0 M b) 0.05 M c) 0.1 M d) 0.5 M e) 0.75 M f) 2.5 M 8. (5 pts) The initial concentration of [CO2] = 2 M, CaO(s) is present in excess, and there is no CaCO3(s). Calculate the equilibrium concentration of [CO2]. Initial: @ Eq: CaO(s) + CO2(g) CaCO3(s) Keq = 10 CHEM 1202 - Homework # 5 Chemical Equilibrium # 2 Homework due Oct 26, 2006 Name ___________________________ Signature: _______________________ Group Name: ____________________ Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 1. (15 pts) Consider the following reactions: A) 4Na(s) + O2(g) 2Na2O(s) B) H2S(g) + Fe2+(aq) ΔHrxn = −600 KJ/mol FeS(s) + 2H+(aq) C) 3NO2(g) + H2O(g) 2HNO3(g) + NO(g) D) NH4NO3(s) NH4+(aq) + NO3−(aq) E) 2Fe(CO)5(l) Fe2(CO)9(s) + CO(g) F) sugar(s) G) H2O(l) + CO2(g) H) Cl2(g) + H2(g) sugar(soln) ΔHrxn = +30 KJ/mol ΔHrxn = 0 KJ/mol H2CO3(aq) 2HCl(g) exothermic Based on the information above, which of the equilibria will (there can be more than one correct answer!!): • produce more products when heated? _____________ • produce more products when the pressure is raised? _____________ • produce more products when the pressure is lowered? _____________ • be unaffected by adding or subtracting some product (so long as some remains)? ___________ • produce more reactants when heated? _____________ • produce more reactants when the pressure is raised? ___________ • be generally unaffected by temperature? ________________ • be unaffected by pressure? __________________ 2. (5 pts) The initial concentrations for the following reaction are [AgCl(s)] = excess present, [NH3] = 2M, [Cl−] = 0.01 M, and [Ag(NH3)2+(aq)] = 0.01 M. What is the concentration of [NH3] at equilibrium? Initial: AgCl(s) + 2NH3(aq) Ag(NH3)2+(aq) + Cl−(aq) Keq = 0.01 @Equilibrium: 3. (5 pts) Prof. Stanley’s hydroformylation catalyst produces a 30:1 ratio of linear aldehyde product to branched aldehyde product. If we assume that this represents an equilibrium ratio (i.e., Keq = 30 favoring the linear aldehyde product), what is the ΔGº energy difference between the linear and branched products? Which is more stable (lower in energy)? Please give your answer in KJ/mol. 4. (5 pts) What is the equilibrium concentration for [SO42−] in the following reaction. Initial: Al2(SO4)3 (s) @Equilib: 2Al+3(aq) + 3SO42−(aq) Ksp = 108 x 10−15 CHEM 1422 - Homework # 4 Equilibrium Due Tuesday, March 3, 2009 (2 PM) Name: ___________________________ Signature: _______________________ Group Name: ____________________ Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 1. (3 pts) A reaction has an equilibrium constant of 1 × 10−6 and reaches equilibrium very slowly. Circle the following potential energy diagram that best fits this data. Briefly and clearly discuss your reasoning. a) b) c) d) R P ΔG ΔG ΔG P R P R Rxn Coordinate ΔG R Rxn Coordinate P Rxn Coordinate . Rxn Coordinate 2. (3 pts) Consider the following equilibrium: Keq = 1 × 1018 If one starts with 6 M acetylene (C2H2) and lets the reaction reach equilibrium, what will be the equilibrium concentration of benzene (C6H6)? Circle the answer and clearly discuss your reasoning. 3C2H2(g) a) 0 M C6H6(g) b) 0.6 M c) 1 M d) 2 M e) 6 M 3. (3 pts) The initial concentrations of reactants and products are all 2 M. What is the concentration of methanol (CH3OH) at equilibrium for the following reaction? Keq = 25 Circle the answer and clearly show your work!! CH3OH(g) + HI (g) a) 0 M b) 0.33 M c) 0.66 M CH3I(g) + H2O(g) d) 1.00 M e) 1.33 M f) 2.66 M CHEM 1422 – HW # 4 – Equilbrium 2 4. (4 pts) Calculate the concentrations for all species at equilibrium for the following reaction. The initial concentrations are [H2] = [I2] = 0 M, [HI] = 4 M. Clearly show your work. Initial: H2(g) + I2(g) 2HI(g) Keq = 36 (@ 1200 K) @ eq: 5. (6 pts) Consider the following reactions: A) Br2(aq) + 2Cl− (aq) Cl2(g) + 2Br−(aq) ΔHrxn = +68 kJ/mol B) AgCl(s) + 2NH3(aq) Ag(NH3)2+(aq) ΔHrxn = −13 kJ/mol C) 2N2O(g) ΔHrxn = −30 kJ/mol 2N2(g) + O2(g) 2H2O(l) + SiF62− (aq) + 2H+(aq) D) SiO2(s) + 6HF(aq) E) Rh(H)(CO)(PPh3)2(aq) + CO(g) Rh(H)(CO)2(PPh3)(aq) + PPh3(aq) F) hemoglobin(aq) + 4O2(g) G) 2H2O2(aq) H) S2−(aq) + Fe2+(aq) I ) H2(g) + D2(g) hemoglobin(O2)4(aq) 2H2O(l) + O2(g) exothermic FeS(s) 2HD(g) ΔHrxn = 0 kJ/mol Based on the information above, which of the equilibria will: • produce more products when heated? _____________ • produce more products when the pressure is raised? _____________ • be unaffected by adding or substracting some product (so long as some remains)? ___________ • produce more reactants when heated? _____________ • produce more reactants when the pressure is raised? ___________ • be unaffected by temperature? _______ • be unaffected by pressure? __________________ CHEM 1422 – HW # 4 – Equilbrium 3 6. (4 pts) The initial concentrations for the following reaction are [CH3I] = [F−] = 1 M, and [CH3F] = [I−] = 9 M. What will be the concentrations of each species at equilibrium? Clearly show all your work. Initial: CH3I(aq) + F−(aq) CH3F(aq) + I−(aq) Keq = 16 @Eq: 7. (4 pts) Which of the following salts is the least soluble (i.e., will give the lowest Pb+2(aq) concentration)? Circle your answer. Calculate the concentration of [Pb+2] for the answer and put it and the calculation details below. a) PbCO3 (Ksp = 1 × 10−13) b) Pb3(AsO4)2 (Ksp = 1.1 × 10−36) c) Pb(CrO4) (Ksp = 1 × 10−14) d) Pb(OH)2 (Ksp = 4 × 10−16) f) PbS (Ksp = 1 × 10−24) e) Pb3(PO4)2 (Ksp = 1.1 × 10−44) 8. (3 pts) What is the equilibrium concentration of Ag+(aq) in the presence of 1 M CrO42− (aq)? Clearly show all your work. Initial: excess Ag2CrO4(s) @Equilibrium: 2Ag+(aq) + CrO42−(aq) Ksp = 4 x 10−12 CHEM 1202 - Homework # 6 Acids & Bases # 1 Due Nov 14th, 2006 by 2PM Name ___________________________ Signature: ____________________________ Group Name: _________________________ Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 1. (5 pts) What are the pH’s for the following solutions? a) 1 M HI = b) 0.1 M HClO4 = d) 0.001 M NaOH = c) 10 M LiOH = e) 1 × 10−20 M Ba(OH)2 = 2. (5 pts) Why are strong acids strong acids. Specifically discuss the series HF, HCl, HBr, and HI and why HF is a weak acid and HI is one of the strongest acids known. 3. (5 pts) Calculate the Ka value of stanoic acid (a monoprotic acid) if a 0.01 M solution has a pH = 4. Clearly show all your work. Homework # 6 – Acids & Bases 1 (2006) 2 4. (5 pts) What is the pH of a 1 M solution of the base amyl amine. pKb = 8 Clearly show all your work. Circle the correct answer from those given below. No credit will be given if work is not shown. a) -2 b) 0 c) 4 d) 7 e) 10 5. (5 pts) Calculate the pH of a 0.1 M solution of an acid that has a pKa of 5.0. Clearly show all your work. Circle the correct answer from those given below. No credit will be given if work is not shown. a) − 1 b) 1 c) 3 d) 7 e) 11 6. (5 pts) An unknown acid solution has a pH of 5. What important, but simple, piece of information do you need to tell if the solution is from a strong or weak acid (aside from the Ka value)? CHEM 1202 - Homework # 7 & 8 Acids & Bases 2 Due Tuesday, Nov 28th, 2006 by Noon Printed Name: _____________________ Signature: ________________________ Group Name: _____________________ Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: Table 1. Dissociation Constants for some Acids. Acid pKa Value Acid pKa Value Acid pKa Value NH4+ 10 HBF4 −9 formic 4 HClO 8 H2CO3 7 benzoic 5 1. (5 pts) Which of the acids listed in Table 1, given a 0.01 M solution in water, will have a pH closest to 2? b) benzoic c) H2CO3 d) formic e) HBF4 a) NH4+ 2. (5 pts) Which of the acids listed in Table 1, when reacted with an equivalent amount of NaOH, will form a solution with the highest pH? a) NH4+ b) benzoic c) H2CO3 d) formic e) HBF4 3. (5 pts) a) b) c) d) e) Order the acids in Table 1 from strongest to weakest. Circle the correct choice. benzoic > formic > H2CO3 > HBF4 > NH4+ > HClO HBF4 > formic > benzoic > H2CO3 > HClO > NH4+ NH4+ > benzoic > HBF4 > formic > H2CO3 > HClO benzoic > HBF4 > NH4+ > formic > H2CO3 > HClO NH4+ > HClO > H2CO3 > benzoic > formic > HBF4 4. (5 pts) What is the pH of a 0.01 M solution of the weak base benzylamine (C6H5CH2NH2)? pKa = 8. Circle the answer below and clearly show all your work. a) 4 b) 5 c) 9 d) 10 e) 13 5. (10 pts) Will FeCl3 generate an acidic, neutral, or basic solution when dissolved in water. Clearly discuss your reasoning. Homework # 7&8 – Acids 2 (2006) 6. (5 pts) What is the pH of a 1 M solution of KClO? See Table 1 for pKa values. Clearly show all your work. 7. (10 pts) Consider the following list of salts: A) NH4Cl E) MoCl4 I) KClO4 B) KI C) CsF D) potassium benzoate F) BaI2 J) NaClO G) AlBr3 H) LiNO3 Which salts will generate an acidic solution? ________________ Which salts will generate a basic solution? _________________ Which salts will generate a neutral solution? _________________ 8. (5 pts) Calculate the pKb of the weak base phenylamine if a 1 M solution has a pH = 10. 9. (10 pts) What is the pH if 800 mL of 0.125 M KOH is added to 200 mL of 0.5 M sucoloic acid (a monoprotic acid)? pKa = 11 (clearly show all your work) CHEM 1202 - Homework # 5 Acids & Bases # 1 Due Thursday, March 19 (by 2:00 PM) Name: ___________________________ Signature: _______________________ Group Name: ____________________ Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 1. (3 pts) Consider the following weak acids and bases and their pKa values: O A) H C O H (pKa = 3.8) D) N(CH3)3 (pKa = 12) B) HCN (pKa = 9.3) C) H2CO3 (pKa = 6.4) E) H2SO3 (pKa = 1.8) F) NH3 (pKa = 9) Which compound is the strongest acid (use letter) ? ___________ Which compound is the strongest base (use letter) ? ___________ Which compound has the strongest conjugate base (use letter) ? ____________ 2. (5 pts) What are the pH’s for the following solutions? = b) 10 M H2SO4 = d) 0.1 M NaOH = c) 1 × 10−10 M HNO3 = e) 10 M CsOH = a) 0.1 M HBr 3. (4 pts) What is the pKa value of a 0.1 M solution of palmetic acid (HA) that has a pH of 5? Clearly show all your work and put a box around your answer. 4. (3 pts) Calculate the pH of a 0.01 M solution of acid that has a pKa of 6.0. Clearly show all your work and put a box around your answer. CHEM 1422 – HW # 5 – Acids & Bases #1 2 5. (5 pts) What is the pH of a 0.1 M solution of the base ethyl amine (CH3CH2NH2). Ka = 1 × 10−11 Clearly show all your work and put a box around your answer. 6. (5 pts) What is the Kb of a 0.01 M solution of a base that has a pH of 10? Clearly show all your work and put a box around your answer. 7. (5 pts) Which of the following acids is the strongest based on its structure and atoms present? Clearly discuss your reasoning. CH F H+ F F P F F F 3 O - or - P O O O CH3 H+ CHEM 1422 - Homework # 6 Acids & Bases # 2 Due Tuesday, April 14, 2009 Name: ___________________________ Signature: _______________________ Group Name: ____________________ 1. (5 pts) 250 mL of 0.2 M acetic acid (HOAc) reacts with 250 mL of 0.2 M NaOH. What is the pH of the resulting solution? pKa (HOAc) = 5 2. (5 pts) Identify whether the following 1:1 solutions will be acidic, basic, or neutral: a) NaOAc/HOAc (pKa = 4.7) b) NH3/NH4NO3 (pKa = 9.3) c) NaCl/KNO3 d) H3PO4/NaH2PO4 (pKa = 2.1) e) CH3NH2/CH3NH3Cl (pKa = 10.6) Table 1. Indicators Name p Ka Acid color Base Color Name p Ka Acid color Base Color Methyl violet 1 yellow violet bromthymol blue 7 yellow blue methyl yellow 1.7 red yellow thymol blue 8.8 yellow blue methyl orange 3.5 red yellow phenolphthalein 9 colorless pink 5 red yellow Alizarin yellow 11 yellow red methyl red 3. (3 pts) What is the approximate pH of a colorless solution that turns yellow if a small amount of methyl yellow, or alizarin yellow is added to it, but turns pink if phenolphthalein is added. See Table 1 for information about indicators. Circle your answer. a) 0 to 1 b) 4 to 5 c) 7 to 8 d) 9 to 10 e) 13 to 14 4. (2 pts) Which indicator (see Table 1) would work best to indicate the equivalence point for the titration 12 curve shown to the right (circle answer): 10 a) methyl violet b) methyl red c) thymol blue pH d) phenolphtalein e) alizarin yellow 8 6 4 2 0 0 10 20 30 40 m of HCl added L 50 CHEM 1422 – HW # 6 – Acids/Bases 2 2 5. (3 pts) What is the pH of a 0.01 M solution of NaHCO3? pKb = 4. Clearly show all your work. 6. (4 pts) Consider the following list of salts: A) CsNO3 B) TiCl4 C) Na(SH) D) KF E) Ca(OH)2 F) Ti(ClO4)4 G) [HN(CH3)3]Cl H) LiHCO3 I) NaPF6 J) RbBr Which salts will generate acidic solutions? ________________ Which salts will generate basic solutions? _________________ Which salts will generate neutral solutions? _________________ 7. (4 pts) What is the pH if 500 mL of 0.2 M HCl is added to 500 mL of 0.2 M ammonia (NH3, pKb = 5) (clearly show all your work) 8. (4 pts) What is the pH of the following 0.5 M aqueous solutions containing equal amounts of the two components shown (find pKa/b or Ka/b values in your textbook appendix or lecture notes): a) NH4Cl + NH3 b) HNO2 + KNO2 c) H2S + CsHS d) C5H4N (pyridine) + [C5H4NH]Br CHEM 1202 - Homework # 9 & 10 Redox & Electrochemistry Due Friday, Dec 8th, 2006 by Noon Printed Name: ____________________ Signature: ________________________ Group Name: _____________________ Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 1. (5 pts) Which of the following substances is the best reducing agent? a) F− b) Mg c) Li+ d) Ag+ e) Zn 2. (5 pts) Which of the following substances is the best oxidizing agent? a) F− b) Mg2+ c) O3 d) Ag+ e) Cu 3. (10 pts) Balance the following reaction in acidic solution. S2O32−(aq) + OCl−(aq) S4O62−(aq) + Cl−(aq) 4. (10 pts) Balance the following reaction in basic solution: BiO3−(aq) + Cr3+(aq) Bi3+(aq) + CrO42−(aq) 5. (5 pts) Write the oxidation state for the underlined element in the box following each compound. a) LiAlH4 b) Ba3(AsO4)2 d) CaSO3 e) H2O2 c) Na2NiCl4 Homework 9&10 – Redox/Electrochemistry (2006) 6. (15 pts) Calculate the redox potentials for the following reactions. Show the two half cell reactions, written in the proper direction and their potentials used to calculate your answer. a) H2(g) + 2Li(s) 2H−(soln) + 2Li+(soln) b) 4H+(aq) + O2(g) + 2Cu(s) c) F2(g) + 2Cl−(aq) 2H2O + 2Cu2+(aq) 2F−(aq) + Cl2(g) Eº = Eº = Eº = d) Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s) Eº = e) 3Pb2+(aq) + 2Al(s) 3Pb(s) + 2Al+3(aq) Eº = 7. (10 pts) Library/web research topic: Describe in your own words the chemistry (with formulas) involved in a lithium-ion battery. Is lithium metal used? What is the voltage of this electrochemical reaction? List two main advantages and two main disadvantages of lithium-ion batteries with BRIEF explanations. DO NOT COPY DIRECTLY FROM ANY REFERENCE (except for chemical formulas). List your primary reference used at the end. CHEM 1422 - Homework # 7 Redox & Electrochemistry Due Tuesday, April 21 (4 PM) Name: ___________________________ Signature: _______________________ Group: ___________________________ Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 1. (1 pt) Which of the following substances is the best reducing agent? Briefly explain your answer. a) Na+ b) Zn c) Li+ d) Ag e) Al 2. (1 pt) Which of the following substances is the best oxidizing agent? Briefly explain your answer. a) O2 b) Li+ c) Cl2 d) Ag+ e) F 3. (1 pt) Which of the following substances is the best reducing agent? Briefly explain your answer. a) F2 b) Mg c) Li+ d) Na e) Zn2+ 4. (1 pt) Which of the following substances is the best oxidizing agent? Briefly explain your answer. a) H+ b) Al3+ c) Ag+ d) Li e) O3 5. (3 pts) Balance the following reaction in acidic solution (add water or H+ as needed). Clearly show your work. I2(aq) + MnO2(s) I (aq) + MnO4(aq) CHEM 1422 – HW # 7 – Redox & Electrochemistry 2 6. (3 pts) Balance the following reaction in basic solution (add water or OH as needed). Clearly show your work. Cr2O72(aq) + ClO(aq) Cr3+(aq) + ClO4(aq) 7. (5 pts) Write the oxidation state for the underlined element in the box following each compound. a) NaH b) KNO3 d) Ca3(PO3)2 c) Na2PtCl6 e) Na(NCS) 8. (5 pts) Calculate the redox potentials for the following reactions. Show the two half cell reactions used to calculate the overall potential. a) 3H2(g) + 2Al3+(aq) 6H(aq) + 2Al(s) b) 2AgCl(s) + Mg(s) 2Ag(s) + Mg2+(aq) + 2Cl(aq) c) 2F2(g) + 2H2O(l) d) Mg2+(aq) + Cu(s) e) Li(s) + Ag+(aq) 4F(aq) + 4H(aq) + O2(g) Mg(s) + Cu2+(s) Li+(aq) + Ag(s) CHEM 1422 – HW # 7 – Redox & Electrochemistry 3 9. (4 pts) A MgCl2 solution containing a Mg electrode is connected by means of a salt bridge to a CuCl2 solution containing a copper electrode. Sketch out a Galvanic Cell showing this and clearly indicate the movement of anions, cations, electrons, which electrode is dissolving, and which is forming a metallic deposit. Label the anode and cathode and show the cell potential. 10. (3 pts) How long (in hours) will it take to electrodeposit 1 mole of Al metal by passing a current of 9.65 amps through a solution of Al+3 ? Please clearly show all your work and put a box around your final answer. 11. (3 pts) What is the concentration of [Ag+] in a half-cell if the reduction potential of the Ag+/Ag couple is observed to be 0.40 V? Clearly show all your work and put a box around your final answer. CHEM 1422 - Homework # 8 Organic & ChemDraw Due Friday, May 1 (by 3:00 PM) Please download ChemDraw from Tigerware (located under Scientific Software, Chemistry software) and install (along with the included Chem3D program) on your Windows or Macintosh computer. You need to register with CambridgeSoft using an LSU e-mail address in order to get a license key (serial #) for installing ChemDraw. I sent instructions on this via e-mail. Study groups can work together, as usual, but each student needs to E-mail their own typed report with ChemDraw structures to Prof. Stanley as a Word or PDF file. Prof. Stanley is available to answer questions about ChemDraw. 1. (20 pts) Sketch out nice ChemDraw structures for the following molecules. Each should be shown in standard organic line notation and with all atoms identified. Example: H O O H H2C tetrahydrofuran: C CH2 H2C CH2 H O H C C H Or ganic line notation H C H H Tw o possible w ay s of drawing "complete" str uct ur es a) 2-hexanone b) heptanal c) Z-2-octene d) E-2-octene e) dimethylformamide f) 2-pentyne g) benzoic acid h) 6-ethyl-3-methylnonane i) 1,3-diethylbenzene j) pyridine 2. (10 pts) Name and redraw the following molecules: a) b) e) d) NH c) CHEM 1422 - Homework # 9-10 Gaussian & GaussView Due Thursday, April 30 (by 4:00 PM) Please download Gaussian 03 and GaussView 4 from Tigerware (located under Scientific Software, Chemistry software) and install on your Windows computer. Up to 3 people can work together on this assignment. Macintosh users should pair up with those with PC/Windows computers. Submit one copy of your report and don’t forget to put all the names of those working together on the report. All reports should be typed, formatted nicely, and include color images (when possible). Prof. Stanley is available to answer questions about Gaussian & GaussView. Separate instructions for using GaussView and Gaussian have been posted by Prof. Stanley. The last part of this assignment is the same as that given to the 1431 Honors Chemistry Lab students. This double assignment will count for 60 pts. 1. (40 pts) Do DFT molecular orbital calculations on perchloric, sulfuric, and phosphoric acids (optimize, DFT, B3LYP, 6-311G(d) basis set). Make sure you select a single d function on your basis set. Questions: a) Capture and display the optimized molecules with their Mulliken charges. b) Show the electrostatic surface potential plots using a density value of 0.04. Adjust the color scale for perchloric acid to best represent the atomic charges, then use the same numerical ranges for displaying the other two molecules. c) Do the charges tell you anything about the acidity of the molecule? Note that the calculation is done in “vacuum” with no solvent molecules around – a proton will not dissociate without water molecules to interact with. Discuss any correlation of the charges calculated with the acidity of the molecules. d) List and compare the bond distances for the 3 molecules in table format and discuss how they compare to the “localized” structures drawn above. Do the bond distances fit any periodic trends? e) Show the highest occupied molecular orbital (HOMO) for each molecule along with its energy in eV. Discuss and explain the trend (if any) in the energies of the HOMO’s (lower is more stable) for these three molecules with periodic properties and what we discussed in class for oxyacids. Discuss what kind of bonding, non-bonding (lone pair), or antibonding interactions are occurring within the HOMO for each molecule. Although one might expect hybrid orbitals (e.g., sp3) for some of these, MO calculations often tend to display orbitals as more or less pure s, p, etc. For example, an oxygen lone pair is more likely to show up as a p orbital. 2. (20 pts) Shirakura & Suginome published a paper in the Journal of the American Chemical Society (2008, 130, 5410-5411) on nickel catalyzed coupling of silylacetylenes with dienes: The first step in the catalysis is the oxidative addition (breaking) of the alkyne C-H bond to the nickel atom to make a nickel-hydride-alkynyl complex: H Me3Si C C H + Ni Ni C C SiMe3 In order for a C-H bond to do an oxidative addition to a Ni metal center the orbital associated with the C-H bond must be reactive enough to want to participate in this reaction. The higher the energy of the orbital the more reactive it will be (usually). This reaction apparently does not work for regular alkynes that do not have silyl groups attached. Use Gaussian (optimization, DFT, B3LYP method, 6-311G basis set) to perform calculations on the following alkyne molecules: Questions: a) Identify the highest energy (C≡)C-H bonding orbital (# from calculation), list the energy of each in eV, and show a picture of this molecular orbital for each alkyne. b) How does the energy of the orbital from question # 2 affect the C-H bond strength? How does the group attached to the other side of the alkyne affect this energy? c) List and compare the energies (in eV) of the alkyne π-bonding orbitals for each system and show a picture of one of these molecular orbitals for each alkyne. Although these aren’t directly involved in the C-H bond breaking step discussed above, what factor seems to be affecting the energy of these orbitals? CHEM 1202 – Exam # 1A (Sept 28, 2006) Name (printed):_________________________ Thermo & Kinetics Name (signed):_________________________ Put a big X this box if you want your graded exam put out in the public racks outside Prof. Stanley’s office after grading. If you don’t check it, Prof. Stanley will keep your exam and you will have to stop by to pick it up from him personally. Answer Sheet -- Please Hand In! 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ 7. _____ 8. _____ 9. _____ 10. _____ 11. _____ 12. _____ 13. _____ 14. _____ 15. _____ 16. 17. 18. 19. Bonus: CHEM 1202 – Exam # 1A Sept 28, 2006 Please Answer All Questions on the Answer Sheet 1. (4 pts) Qualitatively, for which of the following reactions will Srxn increase the most? a) Mg(s) + 2H+(aq) Mg2+(aq) + H2(g) c) 2Al3+(aq) + 3SO42(aq) Al2(SO4)3 (s) b) 2FBr(g) F2(g) + Br2(g) d) H2C2O4(aq) H2(g) + 2CO2(g) 2. (4 pts) Which of the following substances has the highest Entropy? a) CH4(g) b) C3H8(g) c) CH3OH(l) d) PbCl2(s) e) LiOH(s) 3. (4 pts) Which of the following statements about G is FALSE? a) b) c) d) e) G = H TS G° is at standard conditions (1 atm, 1 M) spontaneous reactions have + Sºrxn G represents the “free” or available energy available in a reaction to perform work spontaneous reactions with negative Sºrxn must be exothermic 4. (4 pts) Calculate G°rxn for the following reaction: N2(g) + O2(g) Error! Objects cannot be created from editing field codes. 2NO(g) G°f (NO(g)) = +90 kJ/mol a) 180 kJ/mol b) 90 kJ/mol c) +90 kJ/mol d) +180 kJ/mol e) +270 kJ/mol 5. (4 pts) Consider the reaction: 2H2S(g) + 3O2(g) Error! Objects cannot be created from editing field codes.2SO2(g) + 2H2O(l) Hºrxn = 1126 kJ/mol and Gºrxn = 1006 kJ/mol. Which of the following best describes the reason for the difference between Hºrxn and Gºrxn? a) b) c) d) e) because the surroundings are absorbing the energy difference because the energy difference is going into the Sºrxn because there is an element in its natural state as one of the reactants because water has a high heat capacity because some of the energy is given off as light, not heat 6. (4 pts) Calculate Srxn for the following reaction: 2CH4(g) + O2(g) Error! Objects cannot be created from editing field codes.2CH3OH(g) Sº (CH4) = 186 J/molK, Sº(O2) = 205 J/molK, Sº(CH3OH) = 238 J/molK c) +101 J/molK d) +153 J/molK a) 101 J/molK b) 153 J/molK e) +358 J/molK 7. (4 pts) How much energy is needed to raise the temperature of 1000 gm of water (1 liter) from 25ºC to 100ºC? Specfic heat capacity of water 4 Jgm1ºC1. a) 3 kJ b) 30 kJ c) 300 kJ d) 30,000 kJ The following potential energy diagrams are to be used for problems 8-11. e) 300,000 kJ 8. (4 pts) Which potential energy diagram best represents the fastest spontaneous reaction? 9. (4 pts) Which potential energy diagram best represents the slowest spontaneous reaction? 10. (4 pts) Which potential energy diagram best represents a non-spontaneous reaction that is proceeding through two intermediate species? 11. (4 pts) For which spontaneous potential energy diagram will the rate have the greatest temperature dependence? 12. (4 pts) What is the kinetic rate law for the following data: Exp # [A] [B] Rate (Msec1) a) rate = k[A]2 b) rate = k[A]2[B] 1 1.5 1.5 0.2 c) rate = k[A][B]2 d) rate = k[A] 2 1.5 3.0 0.4 e) rate = k[B]2 f) rate = k[A][B] 3 3.0 3.0 1.6 4 3.0 6.0 3.2 13. (4 pts) The rxn: CO + 2H2Error! Objects cannot be created from editing field codes.CH3OH has the following rate law: rate = k[CO]1[H2]. Calculate the rate constant k if rate = 4 Msec1 when [CO] = 2 M and [H2] = 4 M. a) 0.5 Msec1 b) 1 Msec1 c) 2 Msec1 d) 4 Msec1 e) 8 Msec1 14. (4 pts) Which of the following rate laws is consistent with a bimolecular reaction between A & B? a) rate = k[A][B] b) rate = k[A]2[B] c) rate = k[A][B]2 d) rate = k[A]0.5[B]1.5 e) rate = k[B]2 15. (4 pts) Which of the following statements about the activation energy (or barrier), Ea, is FALSE? a) The activation energy is related to the probability of the reactant molecules reacting. b) Catalysts lower the activation energy by providing a better mechanism for the reactants. c) The activation energy is related to the size/shape of the reacting molecule and strength of the chemical bonds. d) Raising the temperature lowers the activation energy. e) A spontaneous reaction with no Ea will be extremely fast and have little temperature dependence. 16. (10 pts) Consider the following reaction: 2NO(g) + N2(g) Error! Objects cannot be created from editing field codes. 2N2O(g) At 298 K Gºrxn = +35 kJ/mol. Hºf (NO) = +90 kJ/mol, Hºf (N2O) = +82 kJ/mol, Sº(NO) = 210 J/molK, Sº(N2) = 190 J/molK, and Sº(N2O) = 220 J/molK. (a) At what temperature will Gº = 0 and below which the reaction will be spontaneous? Clearly show all your work. (b) Why does one have to go to a lower temperature to get a spontaneous reaction? 17. (10 pts) Sº for sucrose (sugar), C12H22O11(s), is 392 J/molK. Sº for butane, C4H10(g), is 310 J/molK. (a) Why does butane, which is a gas, have a lower entropy than sucrose, which is a solid? (b) How should you modify the statement “that gases have higher entropies than solids” to make it consistent with the above example? 18. (10 pts) Calculate the rate constant k for the reaction 2NO(g) + N2(g) Error! Objects cannot be created from editing field codes. 2N2O(g) using the data below. Clearly show all your work and include the correct units on k. Exp # [NO] [ N2 ] Initial Rate (Msec1) 1 1M 2M 0.1 2 4M 2M 1.6 3 4M 4M 1.6 19. (10 pts) (a) Sketch out a potential energy (G) diagram with the following information: Grxn = 40 kJ/mol, one intermediate with an energy 10 kJ/mol lower than the reactant energy, Ea1 = 10 kJ/mol, Ea2 = 20 kJ/mol. Label the product with a “P”, and the transition state energies Ea1 and Ea2. (b) Indicate with an arrow and label with “RDS” the rate determining point (step) of this energy diagram. I indicated the energy of the reactant (R) on the plot, you need to draw out the rest of the diagram. BONUS (5 pts): Why are perpetual motion machines impossible? CHEM 1422 – Exam # 1 (Feb 19, 2009) Name (printed):_________________________ Thermo, Kinetics & 1st part of Equilibrium Name (signed):_________________________ Put a big X this box if you want your graded exam put out in the public racks outside Prof. Stanley’s office after grading. If you don’t check it, Prof. Stanley will keep your exam and you will have to stop by to pick it up from him personally. 1. (5 pts) Qualitatively, for which of the following reactions will Srxn increase the most? Circle the correct answer and provide clear justification/reasoning for your choice in the space to the left of the answers. a) Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq) b) AgCl(s) + Br(aq) c) 3C2H2(g) d) 2HgO(s) e) 2SO2(g) + O2(g) AgBr(s) + Cl(aq) C6H6(l) 2Hg(l) + O2(g) 2SO3(g) 2. (5 pts) Which of the following substances has the highest absolute Entropy? Circle the correct answer and provide clear justification/reasoning for your choice in the space to the right of the answers. a) H2Te(g) b) H2S(g) c) NH3(g) d) H2O(l) e) LiBH4(s) 3. (5 pts) What is H°rxn for the following reaction? H°f (PCl3) = 320 kJ/mol; P4 = elemental P. Circle the correct answer and provide clear justification/reasoning for your choice in the space to the right of the answers. P4(s) + 6Cl2(g) 4PCl3(l) a) 1280 kJ/mol b) 320 kJ/mol c) 0 kJ/mol d) 320 kJ/mol e) 1280 kJ/mol 5. (5 pts) Calculate Srxn for: 3O2(g) 2O3(g) Sº(O2) = 205 J/mol.K, Sº(O3) = 240 J/mol.K Circle the correct answer and provide clear justification/reasoning for your choice in the space to the right of the answers. a) 620 J/mol.K b) 310 J/mol.K c) 135 J/mol.K d) 310 J/mol.K e) 620 J/mol.K CHEM 1422 - Exam # 1 (Thermo, Kinetics, Equilibrium) 2 The following potential energy (PE) diagrams are to be used for problems 6 - 8. They can be used more than once. 6. (5 pts) Which potential energy diagram best represents a spontaneous reaction that will have the largest temperature dependence with regards to its rate of reaction? Indicate the letter for the diagram and provide clear justification/reasoning for your choice. 7. (5 pts) Consider the following reaction mechanism for converting reactants A + B + C into product F. Indicate the letter for the potential energy diagram that best matches this spontaneous reaction and provide clear justification/reasoning for your choice. Step # 1: A + B D (slow step) Step # 2: D + C E (fast; species E observable) Step # 3: E F 8. (5 pts) Which potential energy diagram best represents the reaction that will rapidly form an equilibrium mixture with roughly equal amounts of reactants and products? Indicate the letter for the PE diagram and provide clear justification/reasoning for your choice. 9. (5 pts) What is the kinetic rate law for the following data. Circle the correct answer and provide clear justification/reasoning for your choice. Exp # [A] [B] Rate (Msec1) 1 1.5 1.5 0.2 a) rate = k[A][B] b) rate = k[A]2[B] 2 1.5 3.0 0.8 c) rate = k[A][B]2 d) rate = k[A] 3 3.0 4.5 3.6 e) rate = k[B]2 f) rate = k[A]2 4 3.0 3.0 1.6 CHEM 1422 - Exam # 1 (Thermo, Kinetics, Equilibrium) 3 10. (5 pts) Consider the following rxn: 4C3H5N3O9(l) 6N2(g) + 12CO2(g) + 10H2O(g) + O2(g) H°rxn = 5646 kJ/mol and Ea = 10 kJ. Based on this thermodynamic and kinetic data what can you tell me about the reaction: energy given off (or absorbed), likely form of the energy (heat, light, nuclear, electrical, etc), spontaneity, and likely speed of the reaction. Would you want to be around 10 moles of the reactant when it “reacts” ? 11. (10 pts) The second law of thermodynamics says that the entropy of the universe always increases for a spontaneous reaction. But a spontaneous reaction can have a Srxn term that is negative, i.e., the entropy of the reaction decreases. Explain why this does not violate the second law of thermodynamics. MgO(s) + C(s) 12. (10 pts) I did the burning Mg demonstration outside for the class: Mg(s) + CO2(g) H°rxn = 811 kJ/mol. When I used the propane torch to light the Mg(s) sitting on the paper on top of the dry ice, CO2(s) (temp = 79°C; CO2(s) CO2(g)), the Mg barely started burning. But as soon as I dropped more dry ice onto the Mg pile it really took off and reacted vigorously. There are two kinetic factors that work in opposite directions that affect this rxn, but one is clearly more important. What are these two kinetic factors and how do they affect the rxn rate? CHEM 1422 - Exam # 1 (Thermo, Kinetics, Equilibrium) 4 13. (10 pts) Calculate the rate constant for the following reaction. Clearly show all your work. 2C2H4(OH)2(aq) + 5O2(g) 4CO2(g) + 6H2O(g) Exp # [C2H4(OH)2] [O2] Rate (Msec1) 1 0.1 0.01 2.1 × 104 2 0.1 0.02 8.4 × 104 3 0.1 0.03 1.9 × 103 4 0.4 0.03 7.6 × 103 14. (10 pts) Consider the following equilibrium: N2(g) + 3H2(g) 2NH3(g) The initial concentrations are: [N2] = 2 M; [H2] = 1 M, [NH3] = 10 M. At equilibrium [N2] = 3 M? Calculate Keq for this reaction. Clearly show all your work. 15. (10 pts) Consider the reaction: F2(g) + Cl2(g) 2FCl(g) Keq = 64 If one starts with [F2] = [Cl2] = 0, and [FCl] = 10 M, what will be the concentration of FCl @ equilibrium? Clearly show all your work! CHEM 1202 – Exam # 2 – Make-UP Equilibrium & Acids/Bases Name (printed) _____________________ Name (signed) _____________________ Put a big X this box if you want your graded exam put out in the public racks outside Prof. Stanley’s office after grading. If you don’t check it, Prof. Stanley will keep your exam in his office and you will have to stop by to pick it up from him personally. Answer Sheet -- Hand In! Please use CAPITAL letters for your multiple choice answers! 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ 7. _____ 8. _____ 9. _____ 10. _____ 11. _____ 12. _____ 13. _____ 14. _____ 15. _____ 16. 17. Exam # 2A (2006) 18. 19. 2 Exam # 2A (2006) 2 CHEM 1202 – Exam # 2 - Makeup November 2006 Please Answer All Questions on the Answer Sheet 1. (4 pts) Calculate Keq for the following reaction: 2NH3(g) + 2H2O(g) 2NO(g) + 5H2(g) Equilibrium concentrations: [NH3] = 4 M, and [H2O] = 4 M, [NO] = 0.4 M, [H2] = 2 M a ) 2 × 1 0− 5 b) 0.02 c) 1 d) 20 e) 50 2. (4 pts) For which of the following reactions will decreasing the overall pressure make more products? a) N2(g) + 2H2(g) b) Br2(g) + I2(g) ΔHrxn = −2 KJ/mol 2BrI(g) 3H2O(l) + IO3−(aq) c) 3H2O2(aq) + I−(aq) Ag(NH3)2+(aq) + Cl−(aq) d) AgCl(s) + 2NH3(aq) e) CH3CH3(g) ΔHrxn = −95 KJ/mol H2N=NH2(g) H2C=CH2(g) + H2(g) ΔHrxn = +10 KJ/mol ΔHrxn = +120 KJ/mol 3. (4 pts) Heating which of the reactions listed in #2 above by the same amount will cause the largest shift in equilibrium favoring products? 4. (4 pts) Consider the equilibrium: 2Rh3+(aq) + 3Cl−(aq) + 3H2(g) 2Rh(s) + 6HCl(aq). 3. Which of the following statements about this reaction is FALSE? Keq = 1 × 10 a) The reaction favors the reactants. b) This reaction is spontaneous under standard conditions. c) Increasing the pressure will make more product. d) ΔGº is negative for this reaction. e) The addition of Rh(s) will have no effect on the reaction so long as some is present. 5. (4 pts) What is the ΔGº value that corresponds to Keq = 2.44 × 10−3 at 300 K. R = 8.314 J/molK a) −150 kJ/mol b) −75 kJ/mol c) −25 kJ/mol 6. (4 pts) Calculate Keq for this reaction: 3NO2(g) + H2O(g) d) 0 kJ/mol e) 15 kJ/mol 2HNO3(g) + NO(g). Initially we start with [NO2] = 4 M and [H2O] = 1 M (no products). When the reaction reaches equilibrium there is 0.5 M NO product. a) 3.2 × 10–4 b) 6.4 × 10–2 7. (4 pts) Consider the reaction: 3SO3(g) c) 0.4 3SO2(g) + O3(g) d) 40 e) 3.2 × 106 Keq = 1 x 10−25 The initial concentrations of SO2 = O3 = SO3 = 2 M. What is the [SO2] concentration at equilibrium? a) 0 M b) 1 M 8. (4 pts) F2(g) + Cl2(g) c) 2 M 2FCl(g) d) 4 M e) 6 M Keq = 64 If one starts with [F2] = [Cl2] = 0, and [FCl] = 10 M, what will be the concentration of FCl @ equilibrium? a) 1 M b) 4 M c) 8 M d) 10 M e ) 12 M Exam # 2A (2006) 2 9. (4 pts) 2NH3(aq) + 3I2(s) 2NI3(s) + 3H2(g) Keq = 16 Initial concentrations: [NH3] = 2 M, [I2] = excess, [NI3] = excess, [H2] = 4 M. What is the concentration of [H2] at equilibrium? a) 0.5 M c) 6 Μ b) 4 M d) 8 M e ) 10 M 10. (4 pts) Consider the potential energy diagram to the right. Which of the following equilibrium constants best fits this reaction? a) 1 × 1010 b) 0.01 d) 1 × 1069 c) 10 11. (4 pts) Consider the reaction: 2HRe(CO)5(aq) Re2(CO)10(aq) + H2(g) Initially, [Re2(CO)10] = [H2] = 1 M (no reactant). What is HRe(CO)5 at equilibrium? Keq = 4 a) 0 M b) 0.2 M c) 0.4 M d) 0.8 M e) 2 M 12. (4 pts) Which of the following salts will be the least soluble in H2O (give the lowest concentration of [Ag+])? a) AgCl (Ksp = 1 × 10–10 ) b) AgCN (Ksp = 1 × 10–16 ) c) AgI (Ksp = 1 × 10–16 ) d) AgBr (Ksp = 1 × 10–13 ) e) Ag2S (Ksp = 1 × 10–49) f) AgF (Ksp = 1 × 10–6 ) 13. (4 pts) Which of the following is the weakest base? a) KOH b) Sr(OH)2 c) NaOH d) LiOH e) Be(OH)2 14. (4 pts) If water has a pO2 = 7, what is the O2 concentration represented by this? a) 1 x 10−13 M b) 1 x 10−7 M c) 1 x 10−3 M d) 0.3 M e) 3 M c) 1 d) 3 e) 10 15. (4 pts) What is the pH of 10 M HNO3? a) −1 b) 0 2NH3(g) Keq = 36 16. (10 pts) Consider the reaction: H2(g) + H2N-NH2(g) Initially, [H2] = [H2N-NH2] = 0.2 M, and [NH3] = 2 M. What is the concentration of [NH3] at equilibrium? Clearly show all your work and put a box around your final answer. Ksp = 4 × 10−12 a) Assuming you have excess Mg(OH)2(s) present, what will be the equilibrium concentration of [OH−] in the solution? b) What is the pH of the solution? Show all your work and put boxes around your final answers. 17. (10 pts) Consider the reaction: Mg(OH)2(s) Mg2+(aq) + 2OH−(aq) 18. (10 pts) a) What is the concentration of Tl+(aq) from TlCl(s) in pure water? Ksp (TlCl) = 1 × 10−14 b) What is [Tl+(aq)] from TlCl(s) in 1 M KCl(aq)? Clearly show all your work and box your final answers. 19. (10 pts) CO2(g) + H2O(l) H2CO3(aq) Keq ≅ 1 × 10−3 Ea = only slightly larger than ΔGrxn H2CO3 (carbonic acid) is an unstable molecule that can not be isolated, but does exist in water when CO2 is present. Use the Keq and Ea values to sketch out a simple qualitative one-step potential energy diagram (ΔG vs. rxn coordinate, numerical scale not needed, indicate reactants with R and products with P) and clearly explain why carbonic acid can not be isolated based on the Keq and Ea values. CHEM 1422 – Exam # 2 (March 26, 2009) Name (printed):_________________________ Equilibrium & Acids/Bases Name (signed):_________________________ Put a big X this box if you want your graded exam put out in the public racks outside Prof. Stanley’s office after grading. If you don’t check it, Prof. Stanley will keep your exam and you will have to stop by to pick it up from him personally. 1. (5 pts) For which of the following reactions will increasing the overall pressure make more products? Circle your answer and briefly discuss your reasoning in the space below. a) 3NO2(g) + H2O(l) 2HNO3(aq) + NO(g) b) 3H2S(g) + 2Fe(s) c) MgO(s) + C(s) Fe2S3(s) + 3H2(g) Mg(s) + CO2(g) d) 2HCl(g) Cl2(g) + H2(g) e) AgCl(s) + 2NH3(aq) Ag(NH3)2+(aq) + Cl(aq) Re2(CO)10(soln) + H2(g) Initially, [Re2(CO)10] = 2. (5 pts) Consider the reaction: 2HRe(CO)5(soln) [H2] = 1 M. What is the concentration of HRe(CO)5 at equilibrium? Keq = 4 Circle your answer and show your work in the space below. a) 0 M b) 0.2 M c) 0.4 M d) 0.8 M e) 2 M 3. (5 pts) Which of the following salts will be the least soluble in H2O? Circle your answer and show your work or reasoning below. a) AgCl (Ksp = 1 × 10–10) b) AgF (Ksp = 1 × 10–6) c) AgI (Ksp = 1 × 10–16) d) AgBr (Ksp = 1 × 10–13) e) Ag3PO4 (Ksp = 1 × 10–20) f) AgCN (Ksp = 1 × 10–12) CHEM 1422 - Exam # 2 (Equilibrium & Acids/Bases) 2 4. (5 pts) What is the Gº value (in kJ/mol) that corresponds to a Keq = 0.09 at 500 K. R = 8.314 J/molK Circle your answer and show your work or reasoning below. a) 1.0 kJ/mol b) 10 kJ/mol c) 10 kJ/mol d) 100 kJ/mol e) 10,000 kJ/mol 5. (5 pts) Which of the following salts will produce an acidic solution when dissolved in water? Briefly discuss your reasoning in the space below. a) KClO4 b) CsBr c) Li2CO3 d) FeCl3 e) Ca(OH)2 6. (5 pts) I was at an agonomy talk and the speaker was discussing soil nutrients. The free Fe3+ concentration in soil was expressed on his overhead as: p[Fe+3] = 17. Which of the following is the Fe+3 concentration represented by p[Fe+3] = 17. Briefly discuss your answer below. a) 1 × 1017 M b) 1.7 × 1010 M c) 1.7 × 107 M d) 0.17 M e) 1.7 M 7. (5 pts) What is the pH of a 1 M solution of the weak base nicotine? Show your work or discuss your reasoning below. a) 2 b) 4 c) 8 d) 14 e) Can’t calculate from information given 8. (5 pts) What is the pH of a 1 M solution of a compound that has a pKb = 10 ? Briefly explain your answer. a) 0 b) c) 10 d) 14 e) Can’t calculate from information given CHEM 1422 - Exam # 2 (Equilibrium & Acids/Bases) 3 9. (5 pts) What is the pH of a 1 × 1010 M solution of CsOH? Briefly explain your answer. a) 4 b) c) 10 d) 14 e) 18 10. (5 pts) Calculate the pH of a 0.001 M solution of an acid with a pKa = 5. Show your work. 11. (10 pts) Consider the following reaction: Fe2(CO)8(s) + 2CO(g) 2Fe(CO)5(soln) Keq = 25 There is initially excess Fe2(CO)8(s), [CO] = 1 M, and Fe(CO)5 = 11 M. What will be the concentration of Fe(CO)5 at equilibrium? Clearly show all your work. Put a box around your final answer. 12. (10 pts) What is the concentration of Ag(aq) in a 1 M K2SO4(aq) solution to which excess Ag2SO4(s) has been added. Ksp (Ag2SO4) = 4 × 1010 Clearly show all your work. Put a box around your final answer. CHEM 1422 - Exam # 2 (Equilibrium & Acids/Bases) 4 13. (10 pts) What is the pH of a 0.1 M solution of a base whose conjugate acid has a pKa value of 9. Clearly show all your work. Put a box around your final answer. 14. (10 pts) What is the pH of a 0.01 M solution of KOCl. pKa (HOCl) = 8. Clearly show all your work. Put a box around your final answer. 15. (10 pts) Consider the following three bases: (CH3)3P H Which is the strongest and why? Clearly discuss your reasoning. (CF3)2N CHEM 1202 – Exam # 3A (11/30/2006) Acids/Bases Name (printed) _____________________ Name (signed) _____________________ Put a big X this box if you want your graded exam put out in the public racks outside Prof. Stanley’s office after grading. If you don’t check it, Prof. Stanley will keep your exam in his office and you will have to stop by to pick it up from him personally. Answer Sheet -- Please Hand In! Please use CAPITAL letters for your multiple choice answers! 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ 7. _____ 8. _____ 9. _____ 10. _____ 11. _____ 12. _____ 13. _____ 14. _____ 15. _____ 16. 17. Exam 3A – Acids/Bases (2006) 18. 19. BONUS QUESTION: Exam 3A – Acids/Bases (2006) CHEM 1202 – Exam # 3A November 30, 2006 Please Answer All Questions on the Answer Sheet 1. (4 pts) What is the pH of 1 M RbOH? a) 1 b) 5 c) 9 d) 14 e) 20 c) 7 d) 10 e) 14 c) 5 d) 7 e) 11 d) H2SO4 e) H2S 2. (4 pts) What is the pH of 1 × 10−10 M HNO3? a) 1 b) 5 3. (4 pts) What is the pH of 1 × 10−3 M HClO4? a) 1 b) 3 4. (4 pts) Which of the following is NOT a strong acid? a) HCl b) HNO3 c) HI 5. (4 pts) Which of the following statements is FALSE? a) Weak acids have moderately strong conjugate bases. b) Fe3+(aq) is considered a Lewis Acid. c) HF is a weak acid in large part due to the concentrated negative charge on the small F− anion. O C d) H3C OH is a weak base. e) Compounds with nitrogen atoms that have a lone pair of electrons are weak bases. 6. (4 pts) Which of the following is NOT a strong base? a) Fe(OH)3 b) LiOH c) CsOH d) Ba(OH)2 e) RbOH 7. (4 pts) What is the ΔGº value (in KJ/mol) that corresponds to a pKa = −8 (typical for a strong acid). R = 8.314 J/molK, T = 300 K a) 34 KJ/mol b) 16 KJ/mol d) −2 KJ/mol e) −46 KJ/mol d) PF6− c) 0 KJ/mol e) NO3− 8. (4 pts) Which of the following anions is the most basic? a) Cl− b) c) F F C H3C O O C S F O O O 9. (4 pts) What is the pH of 10 M HPF6? pKa = −10 a) −10 b) −1 c) 7 d) 12 e) Can’t calculate from information given Exam 3A – Acids/Bases (2006) 10. (4 pts) What is the pH of 0.01 M formic acid, HCOOH? a) −1 b) 1 c) 6 d) 9 e) Can’t calculate from information given 11. (4 pts) What is the pH of a 1 × 10−4 M solution of the weak base nicotine? Kb = 1 × 10−6 a) 0 b) 5 c) 7 d) 9 e) 10 12. (4 pts) Calculate the pH of a 0.1 M solution of carbonic acid, H2CO3. Ka1 = 1 × 10−7, Ka2 = 1 × 10−11. b) 2 c) 4 d) 5 e) 10 a) −1 13. (4 pts) I react 333 mL of 0.3 M NaOH with 666 mL of 0.15 M acetic acid. Which of the following statements is TRUE? pKa (acetic acid) = 5 a) You will make a very acidic solution (pH ~ 0) c) You will make a very basic solution (pH ~ 14) e) You will make a neutral solution b) You will make a moderately acidic solution (pH ~ 4) d) You will make a moderately basic solution (pH ~ 9) 14. (4 pts) Which of the following salts will produce a basic solution when dissolved in water? a) Ga(NO3)3 b) Li2SO4 c) BaO d) NH4NO3 e) SrBr2 15. (4 pts) Which of the following salts will produce an acidic solution when dissolved in water? a) Ba(ClO4)2 b) CsI c) Rb2CO3 d) AlCl3 e) NaBr 16. (10 pts) What is the pH of a 0.1 M solution of citric acid? pKa = 5. Clearly show all your work. Put a box around your final answer. 17. (10 pts) What is the pH of a 1 M solution of the base methyl amine? pKa (methyl amineH+) = 10. Clearly show all your work. Put a box around your final answer. 18. (10 pts) Why is triflic acid (F3CSO3H) an even stronger acid than sulfuric acid (H2SO4)? Clearly discuss. There are two or three important factors. 19. (10 pts) What is the pH of a 1 M solution of sodium carbonate (Na2CO3). pKb (CO32−) = 4 Clearly show all your work. Put a box around your final answer. Bonus Question (10 points): Consider an acetic acid/sodium acetate buffer system. pKa (acetic acid) = 5. A) Assuming a 1:1 mixture of acetic acid and sodium acetate, what is the pH of this buffer system? Briefly explain/discuss. If you don’t know how to give a precise # for the pH, at least state whether this buffer will be acidic, basic, or neutral (with reasoning). B) Clearly explain how this buffer system keeps the same approximate pH when one adds small amounts of strong acid or base. CHEM 1422 – Exam # 3 (April 23, 2009) Name (printed):_________________________ Acids/Bases & Redox/Electrochemistry Name (signed):_________________________ Put a big X this box if you want your graded exam put out in the public racks outside Prof. Stanley’s office after grading. If you don’t check it, Prof. Stanley will keep your exam and you will have to stop by to pick it up from him personally. 1. (5 pts) Consider the following anions. Using fundamental concepts we have discussed in class, which one would be the weakest conjugate base and give rise to the strongest acid. Circle your choice and provide clear reasoning for your answer. a) AsO43 b) F c) PH2 d) CO32 e) PF6 2. (5 pts) [N(CH3)2] is a stronger base than OH. If 0.1 moles of [N(CH3)2] is added to 1 L of water, what will the resulting pH be? Circle the answer (yes, the correct answer is there) and clearly explain your reasoning below. a) 0 b) 0.1 c) 7 d) 13 e) 25 3. (5 pts) Identify the following 1:1 solution mixtures as acidic, basic, or neutral. a) NH3 + NH4NO3 (pKa = 9.3) b) NaBr + KClO4 c) H3PO4 + NaH2PO4 (pKa = 2.1) d) C5H4N (pyridine) + [C5H4NH]Br (pKa = 5.2) e) H2S + CsHS (pKb = 7.0) 4. (5 pts) Calculate the pH of a 0.1 M solution of carbonic acid, H2CO3. pKa1 = 7, pKa2 = 11. Circle your choice and clearly show your work (or reasoning) below. a) 1 b) 2 c) 4 d) 5 e) 10 CHEM 1422 - Exam # 3 (Acids/Bases & Redox/Electrochemistry) 2 5. (5 pts) Which of the following salts will produce an acidic solution when dissolved in water? Circle your answer and briefly discuss your reasoning in the space below. a) KClO4 b ) C s2 S c) Li2CO3 d) CrCl3 e) Mg(OH)2 6. (5 pts) What is the pH of a 1.0 M solution of sodium carbonate (Na2CO3). pKb (CO32) = 4 Clearly show all your work. Put a box around your final answer. 7. (5 pts) Which of the following substances is the best reducing agent? Circle your answer and briefly explain your reasoning below. a) Na+ b) Pb c) Li+ d) F e) Mg 8. (5 pts) What is the potential of a battery composed of the following: Al/Al3+(aq, 1 M) and Ag/Ag+(aq, 1 M)? Temperature = 25°C. Clearly show all your work and the overall balanced reaction. 9. (5 pts) Given the data presented on the Standard Reduction Potential Table provided, what is the highest voltage single cell battery that you could make? Write the overall balanced rxn and potential. Briefly discuss why this “ultimate” battery isn’t practical. CHEM 1422 - Exam # 3 (Acids/Bases & Redox/Electrochemistry) 3 10. (5 pts) Qualitatively how would the hydrogen electrode potential change if the [H+] concentration changed from 1.0 M? Why? 11. (10 pts) 100 mL of 0.5 M acetic acid (HOAc) reacts with 400 mL of 0.125 M NaOH. What is the pH of the resulting solution? pKa (HOAc) = 5 Clearly show all your work. Put a box around your final answer. 12. (10 pts) Consider the titration curve shown to the right. 12 10 a) (2 pts) What type of material (acid or base) is being titrated? Brief reasoning? pH 8 6 4 2 b) (2 pts) Is the material from part a) strong or weak? Brief reasoning? 0 0 10 20 30 40 mL of HCl added c) (2 pts) Why is the region between 10 & 20 mL not changing pH? d) (4 pts) What is the approximate pKa/b of the material in question? Brief reasoning? 50 CHEM 1422 - Exam # 3 (Acids/Bases & Redox/Electrochemistry) 13. (10 pts) Balance the following reaction in basic solution (add, CN, water or OH as needed). Clearly show all your work. Au(s) + CN(aq) + O2(g) [Au(CN)4] (aq) + OH (aq) 14. (10 pts) How many moles of Al can be deposited per hour from a molten mass of Al2O3/Na3AlF6 if a current of 30,000 amps is used? 15. (10 pts) Consider the following electrochemical cell at 25°C: H2(g)/2H+(aq) and Ag/Ag+(aq, 1.0 M). A potential of 0.741 V is measured. What is the pH of the cell? Clearly show all your work. 4 CHEM 1422 Honors: General Chemistry Section 01 Previous 2 Years of Homeworks & Exams with Answers Prof. George G. Stanley Department of Chemistry CHEM 1202 - Homework # 1 Background Due January 22, 2009 (2 PM) ANSWER KEY Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 1. (5 pts) Why do or don't you like chemistry as it has been taught to you so far in high school and college? What did or didn’t you like? Don't try to suck up to me by writing only good things (unless they are true). I'm looking for critical comments and reasoning. Tell me about things you have liked (if any) and disliked. Short mindless answers will get no credit. This question is an individual, not group, effort. Please type up your answer neatly, include your name, and E-mail it to me separately (gstanley@lsu.edu) as a Word or PDF file (or body of the E-mail message). 2. (5 pts) Define electronegativity in your own words (don’t just copy from a book). Give a couple of elemental examples to illustrate your definition (i.e., elements with high or low values and how their electronegativity affects their ability to gain or lose electrons). Electronegativity is the property of an atom to want to hang on to its valence electrons and to possess a filled octet of electrons. For example, fluorine is the most electronegative element on the Periodic Table and has 7 valence electrons. It is almost impossible for fluorine to lose an electron to form F+. But it has a very strong tendency to add an electron to form the 8 valence electron F−. The high electronegativity means that F− is very stable and unlikely to lose an electron. On the other hand, Cs is one of the least electronegative elements and has a very high tendency to lose its one valence electron to form Cs+. Elements with low electronegativity (high tendency to lose electrons to form cationic atoms) are called electropositive elements. Electronegative elements tend to want to gain electrons and possess negative charges (or partial negative charges in neutral compounds), while electropositive elements tend to want to lose electrons to form cations. 3. (5 pts) Look up, list, and explain the trend in melting points in terms of intermolecular attractive forces for the following compounds: CsI, LiF, BaS, MgO, and Al2O3. Make use of Coulomb’s Law (in a qualitative sense) in your explanation. These can all be considered ionic compounds. Their melting points generally increase as the magnitude of the positive and negative charges on the ions increase. Coulomb’s law is: force ∝ + qq d2 − M.P. CsI 626°C LiF 845°C BaS 1200°C The force of attraction between charged ions is proportional to the magnitude of the charges times one another divided by the distance between the charges MgO 2852°C squared. So the higher the charges and the smaller the ions (the closer they can get to one another) the stronger the force of attaction. This should relate Al2O3 2015°C directly to physical properties such as melting points, boiling points, hardness, density, etc. CsI has the lowest melting point since the ions only have ±1charges (relative attractive force of 1 x 1 = 1) and Cs+ and I− are rather large ions that have a larger separation distance between them. LiF, on the other hand, has the smallest ±1charged ions and the closer approach leads to a stronger attractive force and higher melting point. The same logic holds for BaS and MgO, both of which are composed of ±2 charged ions (2 x 2 = 4 times the relative attractive force). But Ba2+ and S2− are both larger ions relative to Mg2+ and O2−, thus MgO has the higher melting point. Al2O3 is an exception to these guidelines. It is composed of Al3+ and O2− ions that should have a stronger attraction compared to two ±2 charged ions. But there is a covalent component in the Al-O bonding that decreases the simple ionic charges leading to a reduction in the melting point. CHEM 1422 - HW#1 Background (2009) 2 4. (5 pts) Consider the reaction: 2C(s) + 2Br2 Br2C=CBr2 If 120 g of carbon reacts with 160 g of Br2 and the reaction goes in 50% yield, how many grams of Br2C=CBr2 (C2Br4) are produced? Clearly show and explain all your work. This is a limiting reagent problem and one always needs to first calculate the # of moles of the reagents and figure out from the stoichiometry to determine the limiting reagent. # moles C = (120 g C)/(12 g/mole) = 10 moles C # moles Br2 = (160 g Br2)/(160 g/mole) = 1 mole Br2 So Br2 is the limiting reagent (carbon is present in large excess). The stoichiometry of the reaction is that 2 moles of Br2 will react with 2 moles of carbon to make 1 mole of product (Br2C=CBr2). Because we only have 1 mole of Br2 present only 0.5 moles of product will be made (half as much as the limiting reagent Br2). Thus we need to calculate the molecular weight of the product and multiply this times the 0.5 moles of product that can be produced from 1 mole of Br2 reacting with 1 mole of carbon. MW (Br2C=CBr2) = 344 g/mol. Grams product (Br2C=CBr2) = (344 g/mol)(0.5 moles) = 172 g But that assumes a 100% yield. 50% yield means that I need to multiply the 100% yield amount by 0.5 (50%): 50% yield = (172 g product, 100% yield)(0.5) = 86 g product produced. 5. (10 pts) Sketch out the Lewis dot structures for the following molecules (use lines for bonds and pairs of dots for lone pairs). Indicate the formal charges, if any, on the appropriate atoms. H a) H2SO4 b) C2H2 O O S H O C H H O d) NO3− c) AlCl3 Cl C Al Cl Cl Al and B are the two common elements that often only have a 6 e- valence count (one empty orbital) for their compounds. These are Lewis Acids. H e) P(CH3)3 H HC HCH P H HCH H O N O O When an oxygen atom only has one bond and an octet of e- it has a formal negative charge. When N has 4 bonds it is assigned a formal positive charge. CHEM 1202 - Homework # 2 Thermodynamics Due Sept 19th, 2006 ANSWER KEY Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 1. (12 pts) From the thermodynamic data given at the end of your lecture notes calculate ΔHºrxn, ΔSºrxn and ΔGºrxn for the following reactions. For ΔGºrxn please use the ΔGºrxn = ΔHºrxn − TΔSºrxn formula with T = 298K. Indicate whether the reactions are spontaneous or non-spontaneous. Show your work. a) 12NH3(g) + 21O2(g) 8HNO3(g) + 4NO(g) + 14H2O(g) ΔHrxn = ΣΔHprod − ΣΔHreact = [(8)(−135.1 kJ/mol) + (4)(90.2 kJ/mol) + 14(−241.8 kJ/mol)] − [(12)(−46.1 kJ/mol) + (21)(−0 kJ/mol)] ΔHrxn = (−4105.2 kJ/mol) − (−553.2 kJ/mol) = −3552 kJ/mol ΔSrxn = ΣΔSprod − ΣΔSreact = [(8)(266 J/molK) + (4)(211 J/molK) + 14(189 J/molK)] − [(12)(192 J/molK) + (21)(205 J/molK)] ΔSrxn = 5618 J/molK − 6609 J/molK = −991 J/molK ΔGrxn = ΔHrxn − TΔSrxn = −3552 kJ/mol − (298K)(−0.991 kJ/molK) ΔGrxn = −3552 kJ/mol + 295 kJ/mol = −3257 kJ/mol b) 2O3(g) spontaneous 3O2(g) ΔHrxn = ΣΔHprod − ΣΔHreact = [(3)(0kJ/mol)] − [(2)(143kJ/mol)] ΔHrxn = (0kJ/mol) − (286kJ/mol) = −286 kJ/mol ΔSrxn = ΣΔSprod − ΣΔSreact = [(3)(205 J/molK)] − [(2)(239 J/molK)] ΔSrxn = 615 J/molK − 478 J/molK = 137 J/molK ΔGrxn = ΔHrxn − TΔSrxn = −286 kJ/mol − (298K)(0.137 kJ/molK) ΔGrxn = −286 kJ/mol − 41 kJ/mol = −327 kJ/mol spontaneous C6H12O6(s) (glucose) + 6O2(g) } this is photosynthesis! c) 6CO2(g) + 6H2O(l) ΔHf° (C6H12O6) = −1274 KJ/mol ΔGf° (C6H12O6) = −910 KJ/mol S° (C6H12O6) = 212 J/Kmol ΔHrxn = ΣΔHprod − ΣΔHreact = [(−1274 kJ/mol) + (3)(0 kJ/mol)] − [(6)( −285.8 kJ/mol) + (6)( −393.5 kJ/mol)] ΔHrxn = (−1274 kJ/mol) − (−4075.8 kJ/mol) = 2803 kJ/mol ΔSrxn = ΣΔSprod − ΣΔSreact = [(212 J/molK) + (6)(205 J/molK)] − [(6)(69.9 J/molK) + (6)(213.6 J/molK)] ΔSrxn = 1442.1 J/molK − 1701 J/molK = −258.9 J/molK ΔGrxn = ΔHrxn − TΔSrxn = 2802.8 kJ/mol − (298K)(−0.259 kJ/molK) ΔGrxn = 2803 kJ/mol + 77 kJ/mol = 2880 kJ/mol non-spontaneous 2. (8 pts) For the following processes, is the entropy of reaction (ΔSrxn) increasing, decreasing or staying about the same? Use the qualitative rules about entropy discussed in lecture to determine the answer. a) Ca(s) + H2SO4(aq) b) 4Fe(s) + 3O2(g) c) N2(g) + 3H2(g) INCREASING CaSO4(s) + H2(g) DECREASING 2Fe2O3(s) (rust) 2NH3(g) DECREASING d) setting off a fire cracker (illegal in East Baton Rouge Parish) e) AgCl(s) + Br−(aq) f) Mg(s) + CO(g) g) Cu+2(aq) + 2OH−(aq) h) Ag+(aq) + Cu(s) AgBr(s) + Cl−(aq) INCREASING STAYING ABOUT THE SAME MgO(s) + C(s) DECREASING Cu(OH)2(s) DECREASING Ag(s) + Cu+(aq) STAYING ABOUT THE SAME 3. (5 pts) A typical “instant” hot pack uses 55 g of CaCl2(s) and 180 g of H2O, which react together to make hydrated Ca2+(aq). Use the following formula to calculate the temperature (ºC) of the heat pack after it has been activated (mixed together). Assume that the initial temperature of the water is 25°C (room temperature), Cp (water) = 75 J/mol K (watch your units!). ΔHrxn = −60 kJ/mol ΔT = − nsalt ( ΔHrxn) nwater (Cpwater ) ΔT = Tfinal - Tinital nwater = # of moles of water used nsalt = # of moles of salt dissolved Cp = heat capacity of solvent ΔHrxn = enthalpy of dissolving salt in solution (also called ΔHsol) # of moles of H2O = (180 g)/(18 g/mol) = 10 moles H2O # of moles of CaCl2 = (55 g)/(110g/mol) = 0.5 moles CaCl2 ΔHrxn is given – you don’t have to calculate!!! (0.5mol )( −60kJ / mol ) 30kJ ΔT = − = = 40 K = 40 C (1K = 1°C, same magnitude, different scale) (10mol )(0.075kJ / molK ) 0.75kJ / K Final Temp = Initial Temp + ΔT = 25°C + 40°C = 65°C 4. (5 pts) (a) Using the Gibbs Free Energy formula to plot the value of ΔGºrxn from 500 K to 1200 K for the reaction: 2HgO(s) 2Hg(l) + O2(g) (use table at the end of the chapter -- you can assume that ΔHºf and Sº do not vary much with temperature) The vertical axis is energy in kJ/mol -- you have to fill in the range of units for this axis based ΔHrxn = 181.6 kJ/mol on your calculations. (HINT: you ΔSrxn = 216.4 J/mol K only have to calculate 3 to 5 data points to make the plot, but you can also do this in Excel and let it calculate ΔGºrxn at many temps). (b) What is the temperature range for which this reaction is spontaneous? Indicate on plot. (c) Mark and indicate the temperature at which ΔGrxn = 0? CHEM 1422 - Homework # 2 ANSWER KEY Thermodynamics Due January 29, 2009 Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 1. (12 pts) From the thermodynamic data given at the end of your lecture notes calculate Hºrxn, Sºrxn and Gºrxn for the following reactions. For Gºrxn please use the Gºrxn = Hºrxn TSºrxn formula with T = 298K. Indicate whether the reactions are spontaneous or non-spontaneous. Show your work. a) Cl2(g) + H2(g) 2HCl(g) Hrxn =Hprod Hreact = [(2)(92.3 kJ/mol)] [(0 kJ/mol) + (0 kJ/mol)] Hrxn = (184.6 kJ/mol) (0 kJ/mol) = 184.6 kJ/mol Srxn = Sprod Sreact = [(2)(186.8 J/molK)] [223.0 J/molK + 130.6 J/molK] Srxn = 373.6 J/molK353.6 J/molK = 20 J/molK Grxn = Hrxn TSrxn = 184.6 kJ/mol (298K)(0.02 kJ/molK) Grxn = 184.6 kJ/mol 6.0 kJ/mol = 190.6 kJ/mol b) Fe(s) + 5CO(g) spontaneous Fe(CO)5(l) Hrxn =Hprod Hreact = [(774 kJ/mol)] [(0 kJ/mol) + (5)(110.5 kJ/mol)] Hrxn = (774 kJ/mol) (552.5 kJ/mol) = 221.5 kJ/mol Srxn = Sprod Sreact = [(338 J/molK)] [27.3 J/molK + (5)(197.6 J/molK)] Srxn = 338 J/molK1015 J/molK = 677 J/molK Grxn = Hrxn TSrxn = 221.5 kJ/mol (298K)(0.677 kJ/molK) Grxn = 221.5 kJ/mol 202 kJ/mol = 19.5 kJ/mol c) SiO2(s) + 6HF(g) H2SiF6(aq) + 2H2O(l) spontaneous [glass is mainly SiO2] Hrxn =Hprod Hreact = [(2331 kJ/mol) + (2)(286 kJ/mol)] [(911 kJ/mol) + (6)(271 kJ/mol)] Hrxn = (2903 kJ/mol) (2537 kJ/mol) = 366 kJ/mol Srxn = Sprod Sreact = [(190 J/molK) + (2)(70 J/molK)] [42 J/molK + (6)(174 J/molK)] Srxn = 330 J/molK1086 J/molK = 756 J/molK Grxn = Hrxn TSrxn = 366 kJ/mol (298K)(0.756 kJ/molK) Grxn = 366 kJ/mol 225 kJ/mol = 141 kJ/mol d) CaCO3(s) spontaneous CaO(s) + CO2(g) Hrxn =Hprod Hreact = [(635 kJ/mol) + (393 kJ/mol)] [(1207 kJ/mol)] Hrxn = (1028 kJ/mol) (1207 kJ/mol) = 179 kJ/mol Srxn = Sprod Sreact = [(40 J/molK) + (214 J/molK)] [93 J/molK] Srxn = 254 J/molK93 J/molK = 161 J/molK Grxn = Hrxn TSrxn = 179 kJ/mol (298K)(0.161 kJ/molK) Grxn = 179 kJ/mol 48 kJ/mol = 131 kJ/mol non-spontaneous I provided the S° value for H2SiF6 in class (190 J/mol•K) CHEM 1422 – HW#2 – Thermodynamics (2009) 2 2. (8 pts) Is the entropy increasing, decreasing or staying about the same? Use the qualitative entropy rules discussed in lecture to determine the answer. Write the answer to the right of each process. a) Fe2O3(s) + Al(s) Fe(s) + Al2O3(s) b) raw egg hard boiled egg decreasing (liquid egg going to solid egg) [Ca(H2O)6]Cl2(s) decreasing (7 particles combining into 1 particle) c) CaCl2(s) + 6H2O(l) d) C6H12(l) + 9O2(g) 6CO2(g) + 6H2O(g) increasing (9 gas & 1 liquid making 12 gases) e) H2CO(aq) + H2O(l) f) mowing the lawn about the same 2H2(g) + CO2(g) increasing (2 liquids going to 3 gases) increasing (cutting up longer grass blades into many short pieces) g) AgCl(s) + I(aq) h) Sr(s) + 2H2O(l) AgI(s) + Cl(aq) about the same Sr2+(aq) + 2OH(aq) + H2(g) increasing 3. (2 pts) Circle the compound that has the highest entropy. Give a brief reason explaining your answer. a) Hg(l) b) H2O(l) c) Pb(s) d) C2H5OH(l) e) CCl4(l) If you look up the entropies, Sº[CCl4(l)] = 216 J/molK, while Sº[C2H5OH(l)] = 161 J/molK. Although C2H5OH (ethanol) has more atoms and has a somewhat more complex structure, the heavier Cl atoms in CCl4 end up contributing more to the entropy. It was not easy, therefore, to qualitatively order these two choices so I’ll give full credit for either one, but you also have to have some qualitative reasoning and not just that you looked up the entropies and CCl4 had the largest value. 4. (4 pts) Why does Al2O3(s) have a lower entropy than Fe2O3(s)? There are two primary qualitative reasons for this. You may have to use the chemistry library to get more information (i.e., properties) on these two common compounds to answer the question. Al2O3(s), alumina, has a lower entropy (Sº = 51 J/molK) vs. Fe 2O3(s), rust, (Sº = 87 J/molK) because it is has stronger bonding. This is indicated by the higher melting point of alumina (2015°C) vs. rust (1565°C). Alumina is one of the harder common materials and the main component in rubies and sandpaper, while rust is a relatively soft, flaky solid. Hardness and melting point of a material is usually related to the strength of the bonds that connect the solid together. The Al-O-Al bonding network is very strong relative to iron. We discussed this in class with the difference in entropy between graphite and diamond. The second main reason is that Al2O3 has a lower MW, but the same relative complexity as Fe2O3, which usually indicates a lower entropy. Each of these factors counts for about 50% of the entropy difference. 5. (4 pts) a) Small amounts of Fe(CO)5(l) usually form in steel tanks containing pressurized CO(g). You worked out the thermodynamics of this in question 1b. At what temperature (ºC) will the formation of Fe(CO)5 become non-spontaneous? Show your work. b) A similar reaction occurs to make Ni(CO)4(l) with Gºrxn = 38 kJ/mol, Hºrxn = 230 kJ/mol, and Sºrxn = 480 J/Kmol. High pressure reactors use a thin disk of metal as a safety mechanism that will rupture and release gasses if the pressure in the reactor gets too high. If one was using CO gas, which disk (Fe or Ni) would be more likely to prematurely fail due to the metal being dissolved away by CO? Briefly explain why. a) We can find this temperature by using the Gºrxn = Hºrxn - TSºrxn formula, setting Gºrxn = 0 (the point at which the reaction shifts from negative G to positive G), and solving for the temperature: T H rxn 221 kJ /mol 327 K S rxn 0.677 kJ /molK or 54C b) Gºrxn for the nickel reaction is more spontaneous (38 kJ/mol for Ni vs. 19 kJ/mol for Fe) meaning that it will be more likely to occur and release more “free” energy in doing so. That means from a thermodynamic viewpoint that it is more favorable for the Ni disk to be dissolved away by CO gas. Although that does NOT mean that the Ni reaction will be any faster (reaction rates are governed by kinetics and not thermodynamics), this does turn out to be the case and this actually happened repeatedly for one of our high pressure reactors a number of years ago. CHEM 1202 - Homework # 3 Chemical Kinetics Due Tuesday, Sept 26, 2006 by 1 PM ANSWER KEY 1. (5 pts) Which of the following energy diagrams best represents a reaction which will be the fastest and most spontaneous. Circle your choice and include your reasoning below (brief statement). a) b) c) d) R P ΔG ΔG ΔG R P R Rxn Coordinate ΔG R P Rxn Coordinate P Rxn Coordinate Rxn Coordinate d) is a spontaneous reaction (downhill from reactants to products) and has the smallest activation energy. 2. (5 pts) Consider the energy diagram to the right: Circle the following diagram below that best represents the effect of adding a catalyst to the reaction. Include your reasoning below (brief statement). a) b) c) ΔG d) R P P ΔG ΔG ΔG R P R Rxn Coordinate Rxn Coordinate ΔG R R P Rxn Coordinate Rxn Coordinate P Rxn Coordinate c) has the same reactant/product energies (ΔG), while all the others have changed. ΔG is a thermodynamic parameter and catalysts never change the thermodynamics of a reaction. They just lower the activation energy for that reaction. 3. (5 pts) Consider the following reaction that is quite important for the manufacturing of many chemicals: + 3H2 cyclohexane benzene ΔG = +200 kJ/mol Activation Energy = +400 kJ/mol Circle the energy curve shown below (R = reactants, P = products) that best represents the reaction described above? Include your reasoning below (brief statement). a) ΔG b) R ΔG P Rxn Coordinate c) R ΔG P Rxn Coordinate d) ΔG R P Rxn Coordinate Ea P R ΔG Rxn Coordinate d) is the correct answer because it is non-spontaneous (product energy higher than reactant) and the Ea is about twice as big as ΔG. HW #3 – Kinetics (2006) 2 4. (5 pts) Consider the following reaction and kinetic data. Circle the correct kinetic rate expression for this reaction. Clearly show and briefly discuss your reasoning. A+B Exp # [A] [B] Initial Rate (Msec−1) 1 1.5 M 1.5 M 0.2 2 1.5 M 3.0 M 0.8 3 3.0 M 3.0 M 0.8 4 6.0 M 3.0 M 0.8 a) rate = k[A][B] b) rate = k[A] C In Exp # 1 & 2 we double the concentration of B (while holding A at the same concentration), which increases the rate by a factor of 4. Thus we have square power relationship: rate ∝ [B]2. In Exp 2 & 3 (or 3 & 4), doubling the concentration of A has NO EFFECT on the rate. Therefore, A is NOT part of the kinetic rate expression (it has a order of zero!). c) rate = k[B] d) rate = k[B]2 e) not enough data 5. (4 pts) Consider the following reaction and kinetic data. What is the rate constant for the reaction? Clearly show all your work and reasoning. First you have to determine the rate law from the exp data!! H2NCH2CH2CH2CH2NH2 + 2HCl Exp # [(N)2] [HCl] Initial Rate (Msec−1) 1 0.1 0.1 0.2 2 0.2 0.1 0.4 3 0.4 0.2 3.2 4 0.1 0.2 0.8 [H3NCH2CH2CH2CH2NH3]2+(Cl−)2 Order on [(N)2]x is determined from experiments 2 & 1: (0.2/0.1)x = (0.4/0.2) (2)x = 2 x=1 Order on [HCl]y is determined from experiments 4 & 1: (0.2/0.1)y = (0.8/0.2) (2)y = 4 y=2 RATE LAW: rate = k[(N)2] [HCl]2 To calculate the rate constant, plug in one of the experimental sets of data (#1, for example) and solve for k : k = rate / [(N)2] [HCl]2 = 0.2 Msec-1 / (0.1 M)(0.1 M)2 k = 0.2 M sec-1 / (0.001 M3) = k = 200 M-2sec-1 6. (6 pts) The reaction in question 3 is non-spontaneous at room temperature. + 3H2 cyclohexane benzene ΔG = +200 kJ/mol Activation Energy = +400 kJ/mol Discuss the one thing that you can do that will increase the rate of reaction and make the reaction more spontaneous. Clearly discuss how this change will influence the rate of reaction and why it will affect the thermodynamics (spontaneity) of the reaction. Increase the temperature! Raising the temperature increases the rate of all chemical reactions – this is a classic kinetic effect. It will also make this particular reaction more spontaneous, that is, make ΔG less positive by magnifying the effect of the entropy term in the ΔG = ΔH – TΔS formula. At room temperature, this reaction has a positive ΔG, making it non-spontaneous. But the generation of 4 particles from one will definitely increase the entropy of the reaction, giving us a fairly large positive ΔS entropy term. ΔSrxn is multiplied by the temperature magnifying its effect as the temperature is increased. So eventually, as we raise the temperature the entropy term will dominate making ΔG go negative, thus giving us a spontaneous reaction. CHEM 1422 - Homework # 3 Chemical Kinetics Due Feb 12, 2009 (2 PM) ANSWER KEY Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 1. (3 pts) Which of the following energy diagrams best represents the slowest spontaneous reaction? Circle your choice. Give a brief, but clear, explanation for your answer below the diagrams. a) b) c) d) R P ΔG ΔG R Rxn Coordinate P R Rxn Coordinate ΔG ΔG R P Rxn Coordinate P Rxn Coordinate A spontaneous rxn is one that is thermodynamically “downhill” (products are lower in energy than reactants, negative ΔG value). Only c) and d) are spontaneous rxns. The slowest spontaneous rxn will have the largest activation energy, which narrows the choice to c). 2. (5 pts) a) Describe in your own words and terms where the origin of the activation barrier comes from and what it represents in a chemical reaction. b) Given the same thermodynamic factors, consider the reaction of two small molecules or two large molecules with one another. Which pair should have the higher activation energy? Why? The activation energy represents the probability that two molecules (for a bimolecular rxn) will react when they have a collision. The more reactive the two molecules the more likely they will react when they collide (smaller activation energy, higher probability of reaction). Since most rxns involve the breaking and making of chemical bonds between two parts of two different molecules, the smaller the molecule the fewer “nonreactive” bonds present and the higher the probability that when the two molecules collide the reactive portions of the molecules will come into contact and actually react. For two large molecules, there are many more bonds present that will not react, thus reducing the odds that the two reactive portions of the molecule will come together in the right way to react. Thus two larger molecules will have a lower probability of getting the right portions together to react and thus will have a higher activation energy relative to two small molecules. In biological systems that are composed of large complicated molecules, Mother Nature counters this effect to some extent by designing channels that guide small molecules to the reactive portion of the large enzyme. The use of opposite charges on two large proteins can also help guide the reactive portions of these molecules together, thus increasing the odds of proper rxn occurring. HW#3 – Kinetics (2009) 2 3. (3 pts) Consider the following reaction and information: O HO C H2 CH3 H2C ΔG = +100 kJ/mol CH2 + CO + H2O Activation Energy = +400 kJ/mol Circle the energy curve shown below (R = reactants, P = products) that best represents the reaction described above? Give a brief, but clear, explanation for your answer below the diagrams. a) b) c) d) R ΔG R +400 kJ R to Ea ΔG ΔG P Rxn Coordinate P Rxn Coordinate ΔG R P Rxn Coordinate P +100 kJ uphill R Rxn Coordinate ΔG = +100 kJ/mol indicates an endoergic or uphill rxn. Only (d) has the product energies higher than the reactants. The activation energy barrier of 400 kJ should be about 4 times larger than ΔG as measured from the reactant energy to the top of the activation energy barrier, which is the case for (d). 4. (5 pts) Consider the following reaction and kinetic data. Circle the correct kinetic rate expression for this reaction. Show all your work and/or discuss your reasoning. 2A + B a) rate = k[A][B] b) rate = k[A]2 C+D c) rate = k[B] Exp # [A] [B] 0.2 M 0.1 M 0.02 2 0.4 M 0.1 M 0.04 3 0.2 M 0.3 M 0.18 4 0.4 M 0.3 M e) rate = k[A][B]2 Initial Rate (Msec−1) 1 d) rate = k[B]2 0.36 For experiments #1 & #2, the concentration of [A] doubles, while [B] stays the same. Doubling the conc. of [A], then, causes the initial rate to double. Thus there is an exponent of 1 on [A]. For experiments #1 and #3, the conc. of [B] is tripling (factor of 3), while [A] stays the same. When the conc. of [B] triples, the initial rate increases by a factor of 9. Thus, the exponent on [B] is 2 (i.e., [3]x = 9, therefore x = 2). So the kinetic rate expression is: rate = k[A] [B]2 Overall, this is a third order rxn. HW#3 – Kinetics (2009) 3 5. (5 pts) Consider the following reaction and kinetic data. Circle the correct rate constant for this reaction. Clearly show all your work including the rate law that you determine. A + 2B a) 2.2 x 10−6 M−1sec−1 b) 22 M−1sec−1 C+D c) 220 M−1sec−1 Exp # [A] [B] Initial Rate (Msec−1) 1 0.2 M 0.1 M 0.002 2 0.2 M 0.2 M 0.002 3 0.4 M 0.2 M 0.008 4 0.8 M 0.4 M d) 0.05 M−1sec−1 e) not enough data 0.032 For experiments #1 & #2, the concentration of [A] stays the same, while [B] doubles. But doubling the conc. of [B] does not cause any change in the initial rate of the reaction. Thus there is an exponent of 0 on [B], which means that it is not in the kinetic rate expression. For experiments #2 and #3, the conc. of [A] doubles (factor of 2), while [B] stays the same. When the conc. of [A] doubles, the initial rate increases by a factor of 4. Thus, the exponent on [A] is 2 (i.e., [2]x = 4, therefore x = 2). So the kinetic rate expression is: rate = k[A]2 Overall, this is a second order rxn. Now that we have figured out the kinetic rate expression, we can calculate k the rate constant. You can take any of the four experimental runs and substitute in the various values into the kinetic rate expression and solve for the rate constant. I’ll use experiment # 1: rate = k[A]2 or 0.002 Msec−1 = k [0.2 M]2 or k = (0.002 Msec−1)/(0.04 M2) = 0.05 M−1sec−1 6. (4 pts) Catalysts can be used on non-spontaneous reactions to lower the activation barrier. If a catalyst lowers the activation barrier too much, however, a serious problem can arise. Consider the diagrams shown below. What is the problem for the catalyzed rxn with the lower activation energy? Why can a “substantial” activation barrier actually help an “uphill” chemical reaction if one wants to make as much product as possible? If you lower the activation energy too much for a non-spontaneous rxn the products will react backwards quickly to make reactants. The presence of an activation barrier can help to slow the back-reaction to the more stable reactants. This is NOT a problem for a spontaneous rxn where the products are significantly more stable than the reactants and will not want to back-react. 7. (5 pts) A reaction has a initial rate of reaction of 0.001 Msec-1 at 70°C. This increases to 0.100 Msec-1 at 90°C. Calculate the activation energy for this reaction? First you need to convert the temperatures into kelvin: 70° + 273 = 343 K; 90°C + 273 = 363 K The Arrhenius equation uses rate constants, what are given in this problem are initial reaction rate. BUT, reaction rates are directly proportional to rate constants, so the ratio of either will be the same! k2 ⎛ 0.1 ⎞ 8.314 J / molK × ln ⎜ ⎟ k1 38.29 (8.314) ln(100) (8.314)(4.605) ⎝ 0.001 ⎠ Ea = = = = = −4 −4 1⎞ 1⎞ 1.61 × 10 1.61 × 10 1.61 × 10 − 4 ⎛1 ⎛1 ⎜T −T ⎟ ⎜ 343 − 363 ⎟ ⎝1 ⎝ ⎠ 2⎠ R ln Ea = 2.38 × 105 J / mol = 238 kJ / mol CHEM 1202 - Homework # 4 Chemical Equilibrium # 1 Due Thursday, Oct 19, 2006 ANSWER KEY Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 2NO(g) Keq = 4 1. (3 pts) Given the following information: N2(g) + O2(g) Calculate the equilibrium constants for the equations shown below (you don’t have to show your work): a) 2NO(g) N2(g) + O2(g) b) ½N2(g) + ½O2(g) Keq = 0.25 (switching the rxn around, 1/Keq) Keq = 2 (multiplying by ½, raise Keq to ½ power) NO(g) c) 2N2(g) + 2O2(g) 4NO(g) Keq = 16 (multiplying by 2, raise Keq to 2 power) 2. (2 pts) Circle the equilibrium expression listed below that is the correct one for the following reaction: SCl2(g) + H2O(g) a) [SCl2] [H2O] Keq = [HCl] [SO] b) 2HCl(g) + SO (g) [SCl2] [H2O] [HCl] [SO] c) Keq = [HCl]2 [SO] Keq = [SCl ] [H O] 2 2 d) [HCl]2 [SO] Keq = [SCl ] [H O] 2 2 3. (4 pts) The initial concentrations for the following reaction are [CH3I] = [Cl−] = 0 M, and [CH3Cl] = [I−] = 2 M. What is the concentration of the reactant [Cl−] at equilibrium? Show your work! Soln = solution Initial: @Eq: 0M 0M x x CH3I(soln) + Cl−(soln) 2M 2M 2−x 2−x CH3Cl(soln) + I−(soln) Keq = 9 Since the reactant concentrations are 0 M, the rxn has to shift backwards (to the left) to lose product (−x) and make more reactants (+x). So we have (2 − x) on the product side and 0 + x or just x on the reactant side. Substituting these into the equilibrium expression gives: K eq = [CH3Cl][I − ] (2 − x )(2 − x ) (2 − x )2 = = =9 [CH3I][Cl − ] ( x )( x ) ( x )2 (2 − x )2 =9 ( x )2 or, { take the square root of each side to simplify: (2 − x ) = 3, or: 2 − x = 3 x , 4x = 2, so x = 0.5; [Cl-] = x = 0.5 M (x) 4. (6 pts) The initial concentrations of reactants = 0.2 M and products = 1.6 M. What is the concentration of methanol (CH3OH) at equilibrium for the following reaction? Clearly show all your work! Initial: @Eq: 0.2 M 0.2 M 1.6 M CH3OH(g) + HI (g) 0.2 + x 1.6 M CH3I(g) + H2O(g) 0.2 + x 1.6 – x 1.6 – x Keq = 25 Q= (1.6)(1.6) = 64 (0.2)(0.2) Since none of the concentrations are 0, we need to plug the initial concentrations into the equib expression to solve for Q, (rxn quotient) to see which way the rxn will go to reach equib. Q = 64, which is greater than Keq, so the reaction has to lose products (−x) and make more reactants (+x,go backwards) to reach equib! Keq = (1.6 − x ) (1.6 − x )(1.6 − x ) (1.6 − x )2 = 5 , multiply through each side = = 25 , take the square root of each side: 2 (0.2 + x ) (0.2 + x )(0.2 + x ) (0.2 + x ) by (0.2 + x): 1.6 – x = 1 + 5x, or: 6x = 0.6, x = 0.1. [MeOH] = 0.2 + x = 0.3 M a) 0.1 M b) 0.3 M c) 0.6 M d) 1.2 M e) 1.5 M f) 3.0 M Homework # 4 (Equilibrium #1) 2006 2 5. (2 pts) Consider the potential energy diagram to the right: Which of the following Keq values and relative reaction rates fits best? Briefly and clearly give your reasoning in the space below. a) Keq = 1.5, fast rxn c) Keq = 1 x 10−56, fast rxn b) Keq = 200, slow rxn ΔG P R Rxn Coordinate e) Keq = 1 x 10−4, slow rxn d) Keq = 1 x 106, very fast rxn The products are higher in energy than reactants, so that indicates a +ΔGrxn and Keq < 1. Of the two choices with Keq < 1.0 (c & e), the relatively high activation energy and not large energy difference between R & P indicates a moderately small Keq and a slow reaction rate which fits e) best. 6. (4 pts) Consider the following reaction shown below. We start with [Rh4(CO)12] = 0.2 M, [H2] = 1 M, and [HRh(CO)3] = 0 M. When the reaction reaches equilibrium there is 0.02 M HRh(CO)3. Calculate Keq for this reaction. Clearly show all your work. Initial: 0.2 M 1M 0M Rh4(CO)12 + 2 H2 @ Eq: 0.2 – x 4 HRh(CO)3 1 – 2x 0.02 M = 4x But I told you that when the rxn reaches equilibrium there is 0.02 M HRh(CO)3, so 4x = 0.02 M, or x = 0.005 M We can now substitute x = 0.005 into the other @Equilbrium values & get their numerical values: [Rh4(CO)12] = 0.2 – x = 0.2 – 0.005 = 0.195 M [H2] = 1 – 2x = 1 – 0.01 M = 0.99 M K eq = Substituting these equilibrium values into the equilibrium expression allows us to calculate Keq: [HRh(CO)3]4 (0.02)4 1.6 × 10 − 7 = = = 8.38 × 10 −7 [Rh4(CO)12][H2]2 (0.195)(0.99)2 0.191 7. (4 pts) Consider the following reaction: P4(soln) + 5 NaClO2(soln) 0.5 – x P4O10(soln) + 5 NaCl(soln) 0.5 – 5x x Keq = 1 × 1052 5x The initial concentrations of P4 and NaClO2 are both 0.5 M (no products present). Circle the concentration of P4O10 at equilibrium? Clearly and briefly explain your answer below. a) 0 M b) 0.05 M c) 0.1 M d) 0.5 M e) 0.75 M f) 2.5 M Because Keq is so huge, we can assume that essentially all the reactants will react to make products. But due to the equal concentrations of reactants BUT unequal coefficents, we have a limiting reagent problem. The NaClO2 has a coefficient of 5 and will disappear 5 times faster than the P4 reagent. So set that algebraic equation = 0: 0.5 – 5x = 0. Add 5x to each side and divide through by 5 to get x = 0.1. This is the [P4O10] concentration @equilibrium and our answer. 8. (5 pts) The initial concentration of [CO2] = 2 M, CaO(s) is present in excess, and there is no CaCO3(s). Calculate the equilibrium concentration of [CO2]. Initial: @ Eq: solid 2M CaO(s) + CO2(g) solid 2–x 0 CaCO3(s) solid Keq = 10 The key here is to realize that solids do not appear in the equilibrium expression, so: K eq = Substituting in the algebraic terms we get: K eq = 1 [CO2] 1 1 = = 10 ; multiplying each side by 2 – x we get: [CO2] 2 − x 1 = 20 – 10x, solving for x we get: 10x = 19, or x = 1.9. Subtituting this into the @eq concentration for [CO2] we get: [CO2] = 2 – x = 2 – 1.9 = 0.1 M CHEM 1202 - Homework # 5 Chemical Equilibrium # 2 Homework due Oct 26, 2006 ANSWER KEY Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 1. (15 pts) Consider the following reactions: A) 4Na(s) + O2(g) 2Na2O(s) B) H2S(g) + Fe2+(aq) ΔHrxn = −600 KJ/mol FeS(s) + 2H+(aq) C) 3NO2(g) + H2O(g) 2HNO3(g) + NO(g) D) NH4NO3(s) NH4+(aq) + NO3−(aq) E) 2Fe(CO)5(l) Fe2(CO)9(s) + CO(g) F) sugar(s) G) H2O(l) + CO2(g) H) Cl2(g) + H2(g) sugar(soln) ΔHrxn = +30 KJ/mol ΔHrxn = 0 KJ/mol H2CO3(aq) 2HCl(g) exothermic Based on the information above, which of the equilibria will (there can be more than one correct answer!!): • D produce more products when heated? _____________ • A, B, C, G produce more products when the pressure is raised? _____________ • produce more products when the pressure is lowered? _____________ E • be unaffected by adding or subtracting some product (so long as some remains)? ___________ A • A, H produce more reactants when heated? _____________ • produce more reactants when the pressure is raised? ___________ E • F be generally unaffected by temperature? ________________ • D, F, H be unaffected by pressure? __________________ Homework # 5 (Equilibrium #2) 2006 2 2. (5 pts) The initial concentrations for the following reaction are [AgCl(s)] = excess present, [NH3] = 2M, [Cl−] = 0.01 M, and [Ag(NH3)2+(aq)] = 0.01 M. What is the concentration of [NH3] at equilibrium? Initial: solid 2M AgCl(s) + 2NH3(aq) @Equilibrium: less solid 2 – 2x 0.01M 0.01M Ag(NH3)2+(aq) + Cl−(aq) 0.01 + x Keq = 0.01 0.01 + x The first thing to figure out was which way the reaction was going to shift to reach equilibrium. Since none of the initial concentrations was = 0, we have to calculate the Q, the reaction quotient, and compare it to Keq to see which way the reaction will shift to reach equilibrium. Q = (0.01)2/(2)2 = 2.5 x 10−5, which is less than Keq, so the reaction has to make more product to reach equilibrium. That allows us the set the @equilibrium values shown above. Now we can plug these values into the equilibrium expression and solve for x: 0.01 + x (0.01 + x )(0.01 + x ) (0.01 + x )2 = = 0.01 taking the square root of each side gives: = 0 .1 (2 − 2 x )2 (2 − 2 x )2 2 − 2x multiplying each side by 2-2x gives: 0.01 + x = 0.2 – 0.2x, subtracting 0.01 and adding 0.2x to each side gives: 1.2x = 0.19, dividing through by 1.2 gives: x = 0.19/1.2 = 0.16. But x is not our answer, I asked for the concentration of NH3 at equilibrium, which is [NH3] = 2 – 2x = 2 – 0.32 = 1.68 M. So [NH3] = 1.68 M at equilibrium. 3. (5 pts) Prof. Stanley’s hydroformylation catalyst produces a 30:1 ratio of linear aldehyde product to branched aldehyde product. If we assume that this represents an equilibrium ratio (i.e., Keq = 30 favoring the linear aldehyde product), what is the ΔGº energy difference between the linear and branched products? Which is more stable (lower in energy)? Please give your answer in KJ/mol. If Keq = 30 favoring the linear aldehyde product, then the linear product is the more stable product. To calculate the ΔGº value we simply need to plug our Keq value of 30 into the ΔGº formula and assume that the temperature = room temperature or 298 K: ΔGº = −RT(ln Keq) = −(8.31 J/mol K)(298 K)(ln(30)) = − (2476 J/mol)(3.4) = −8418 J/mol ΔGº = −8.42 KJ/mol 4. (5 pts) What is the equilibrium concentration for [SO42−] in the following reaction. Initial: @Equilib: solid Al2(SO4)3 (s) less solid 0M 0M +3(aq) + 3SO 2−(aq) 2Al 4 2x 3x Ksp = 108 x 10−15 Since Al2(SO4)3 (s) is a SOLID, it does not appear in the equilibrium expression. Since I didn’t give you any initial concentrations you can assume that there is excess Al2(SO4)3 (s) present and NO Al+3(aq) and SO42−(aq). So we know which way the equilibrium is going to shift – to fill in the zero concentrations and form more products. Now we can set up our equilibrium expression and substitute in our @equilibrium x values: Ksp = [Al+3]2 [SO42−]3 = 108 x 10−15 substituting in our x values we get: (2x)2(3x)3 = (4x2)(27x3) = 108x5 108x5 = 108 x 10−15 dividing through by 108 and taking the fifth root of each side gives: x = 1 x 10−3. But x is NOT our answer, the [SO42−] = 3x = 3 x 10−3 M CHEM 1422 - Homework # 4 Equilibrium Due Tuesday, March 3, 2009 (2 PM) ANSWER KEY Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 1. (3 pts) A reaction has an equilibrium constant of 1 × 10 6 a nd reaches equilibrium very slowly. Circle the following potential energy diagram that best fits this data. Briefly and clearly discuss your reasoning. a) b) c) d) R P G G G P R P R Rxn Coordinate G R Rxn Coordinate P Rxn Coordinate . Rxn Coordinate The small K eq value means that this reaction is non-spontaneous (G = positive, reactants are lower in energy than products). Thus, you can rule out c) a nd d ) s ince these are both s pontaneous reactions that will have K eq > 1. a ) and b ) are both non-spontaneous reactions. The height of the activation barrier has nothing to do with the thermodynamic s or Keq, but does control the speed or rate of reaction. The fact that the rxn reaches equilibrium slowly implies the larger activation barrier, which fits b). But if your reasoning is OK you could also select a ) and get full credit. 2. (3 pts) Consider the following equilibrium: 3C2H2(g) K eq = 1 × 1018 C6H6(g) I f one starts with 6 M a cetylene (C 2 H2) and lets the reaction reach equilibrium, what will be the equilibrium concentration of benzene (C 6H6)? Circle the answer and clearly discuss your reasoning. a) 0 M b) 0.6 M c) 1 M d) 2 M e) 6 M Always think qualitatively about the problem and what the K eq value is telling you. The huge K eq value means that this reaction goes essentially to completion. So we will have all products and no reactants at equilibrium. The stoichiometry of the reaction, however, is that 3 molecules of acetylene combine to make one molecule of benzene. So if we start with 6 M acetylene, we will end up with 2 M benzene (3:1 reduction ratio). 3. (3 pts) The initial concentrations of reactants and products are all 2 M. What is the concentration of methanol (CH 3OH) at equilibrium for the following reaction? K eq = 25 Circle the answer and clearly show your work!! CH3OH(g ) + HI (g) a) 0 M Initial: b) 0.33 M 2M 2M CH3 OH(g ) + HI (g) @ eq: 2x 2x CH3 I(g) + H2O(g ) c) 0.66 M 2M d) 1.00 M 2M CH3I(g ) + H2O(g) 2+x 2+x e ) 1.33 M f) 2.66 M If all the concentrations are initially equal and Keq = 25, then the reaction has to mak e more products so the ratio of products over reactants is > 1. (Q < K eq) (2 x ) [CH3I ][H 2O] (2 x )2 5 Keq 25 now take the square root of both sides: [CH3OH ][HI ] (2 x )2 (2 x ) rearrange and solve for x: (2 + x) = (5)(2 – x ); 2 + x = 10 – 5x; 6x = 8; x = 8/6 or x = 1.33 (but this isn’t the answer!!) The concentration of MeOH at equilibrium, therefore, is: 2 – x or 0.67 M CHEM 1422 – HW # 4 – Equilbrium 2 4. (4 pts) Calculate the concentrations for all species at equilibrium for the following reaction. The initial concentrations are [H2 ] = [I 2] = 0 M, [HI] = 4 M. Clearly show your work. Initial: 0M 0M 4M H2 (g) + I 2(g ) @ e q: x 2HI(g) K eq = 36 (@ 1200 K) 4 – 2x x plug the algebraic expressions into the equilibrium formula: (4 2x ) [HI ]2 ( 4 2x )2 6 Keq 36 now take the square root of both sides: x [H 2][I 2] ( x )2 rearrange and solve for x: 4 – 2x = 6x; 4 = 8x; x = 0.5. Now plug x into the @eq conditions to solve for the numerical values: so, [H2] = [I 2] = 0.5 M and [HI] = 3 M 5. (6 pts) Consider the following reactions: A) Br 2(aq) + 2Cl (aq) C l2 (g) + 2Br (aq) Hrxn = 68 kJ/mol B) AgCl(s ) + 2NH3 (aq) Ag(NH 3) 2+ (aq) Hrxn = 13 kJ/mol C) 2N 2O(g) Hrxn = 30 kJ/mol 2N2(g) + O 2(g) 2H2O(l) + SiF6 2 (aq) + 2H+ (aq) D) SiO 2 (s) + 6HF(aq ) E) Rh(H)(CO)(PPh3) 2(aq) + CO(g) Rh(H)(CO)2(PPh3)(aq ) + PPh3(aq ) F) hemoglobin(aq) + 4O 2(g) G) 2H 2O 2 (aq) H) S2 (aq) + Fe2+(aq) I ) H 2(g ) + D 2(g ) hemoglobin(O 2 )4 (aq) 2H2O( l) + O2 (g) exothermic FeS(s) Hrxn = 0 kJ/mol 2HD(g) Based on the information above, which of the equilibria will: produce more products when heated? _____________ A E, F produce more products when the pressure is raised? _____________ be unaffected by adding or substracting some product (so long as some remains)? ___________ H B, C, G produce more reactants when heated? _____________ produce more reactants when the pressure is raised? ___________ A, C, G be unaffected by temperature? _______ I be unaffected by pressure? __________________ B, D, H, I I f you don ’t have any Hrxn information (or k now that it is exo- o r endothermic, you can NOT s tate that a reaction is unaffected by temperature. CHEM 1422 – HW # 4 – Equilbrium 3 6. (4 pts) The i nitial concentrations for the following reaction are [CH3 I] = [F ] = 1 M, and [CH3F] = [I ] = 9 M. What will be the concentrations of each species at equilibrium? Clearly show all your work. Initial: 1M 1M 9M CH3I(aq) + F(aq ) @Equilibrium: 1+x 9M CH3F(aq) + I (aq) 9–x 1+x Keq = 16 9 –x (9 x )2 , take the square root of each side to give: (9 x ) 4 Keq 2 16 (1 x ) (1 x ) multiply out to get: 9 – x = 4 + 4x, rearrange to get: 5x = 5, or x = 1. Plug x = 1 back into the equilibrium conditions to get: [CH3I] = [F ] = 2 M , [CH3F] = [I ] = 8 M First you have to calculate Q (rxn quotient) to figure out which way the rxn will go to reach equilib . Q = 81, which is larger than Keq, so the reaction has to go BACKWARDS to reach equilib . 7. (4 pts) Which of the following salts is the l east soluble (i.e., will give the lowest Pb+2(aq ) concentration)? Circle your answer. Calculate the concentration of [Pb +2] for the answer and put it and the calculation details below. a) PbCO 3 (K sp = 1 × 1013) c) Pb(CrO 4) (K sp = 1 × 1014) b) Pb3(AsO 4 )2 (Ksp = 1.1 × 1036) d) Pb(OH) 2 (K sp = 4 × 1016) e) Pb3(PO 4 )2 (K sp = 1.1 × 1044) f) PbS (K sp = 1 × 1024) a) x 2 = 1 × 1013 ; x = [Pb2+] = 3.2 × 107 M b) 108x 5 = 1.1 × 1036 ; 3x = [Pb2+] = 7.5 × 108 M c ) x 2 = 1 × 1014 ; x = [Pb2+] = 1 × 107 M d) 4x 3 = 4 × 1016 ; x = [Pb2+] = 4.6 × 106 M e) 108x 5 = 1.1 × 1044 ; 3x = [Pb2+] = 1.9 × 109 M f) x 2 = 1 × 10 24 ; x = [Pb2+] = 1 × 10 12 8. (3 pts) What is the equilibrium concentration of Ag(aq ) in the presence of 1 M CrO 42 (aq)? Clearly show all your work. Initial: excess Ag2CrO 4 (s) @Equilibrium: less excess 0 1M 2Ag+ (aq) + CrO 4 2 (aq) 2x Ksp = 4 × 1012 1+x Ksp = [Ag+]2 [CrO42] = (2x)2 (1 + x) = 4 × 1012 : make the approximation that x << 1 M to simplify the algebra (2x)2 (1) = 4 × 1012 4x 2 = 4 × 1012 , now divide each side by 4 to give: x 2 = 1 × 1012 ; take the square root of eac h side: x = 1 × 106 ; but the concentration of A g+ is 2x , not just x, so: [Ag+] = 2x = 2 × 10 6 M ANSWER KEY CHEM 1202 - Homework # 6 Acids & Bases # 1 Due Nov 14th, 2006 by 2PM 1. (5 pts) What are the pH’s for the following solutions? a) 1 M HI = 0 b) 0.1 M HClO4 = d) 0.001 M NaOH = 11 1 c) 10 M LiOH = 15 e) 1 × 10−20 M Ba(OH)2 = 7 1 x 10−20 M is much, much lower than the 1 x 10−7 M OH− present in water that it makes no contribution to the pH. It doesn’t really matter whether Ba(OH)2 is a strong or moderate base in this case. 2. (5 pts) Why are strong acids strong acids. Specifically discuss the series HF, HCl, HBr, and HI and why HF is a weak acid and HI is one of the strongest acids known. Strong acids are strong acids because their conjugate bases (the counter-anions) are extremely weak conjugate bases. The Cl−, Br− and I− anions have decreasing affinities for binding H+ in aqueous solution (or in the molecular gas phase species for that matter). The larger size of the anions as we go from Cl− to I− means that the negative charge on the anion is spread out over a larger surface area and there is less electrostatic attraction to the H+ cation. Remember that the extremely small H+ cation has a very concentrated positive charge and will interact the strongest with a small concentrated negative (or multiple negative) charge. The smallest common anion is F−, which does have a fairly strong electrostatic attraction to the H+ cation, even in water. This makes HF a weak acid compared to HCl, HBr, and HI. 3. (5 pts) Calculate the Ka value of stanoic acid (a monoprotic acid) if a 0.01 M solution has a pH = 4. Clearly show all your work. Initial: 0.01M HA(aq) @ Equib: 0.01 − x 0M 0M H+(aq) + A−(aq) x x BUT, before you go any further, I’ve given you the pH of the solution in the problem!! So you don’t have to solve for x, you already know it!! So all we have to do is take the anti-log of pH 4 to get the concentration of the H+ (and A−) in solution: [H+] = antilog(−4) = 1 × 10−4 M Ka = Substituting this into the equlibrium expression gives us: [H + ][OH − ] (1× 10 − 4 )(1× 10 − 4 ) 1× 10 − 8 = ≈ = 1× 10 − 6 −4 ) [HA] (0.01 − 1× 10 0.01 Note that we can approximate 0.01 − 1 x 10−4 M as ~ 0.01. So the answer is: Ka = 1 x 10−6 4. (5 pts) What is the pH of a 1 M solution of the base amyl amine. Kb = 1 x 10−8 Clearly show all your work. Circle the correct answer from those given below. No credit will be given if work is not shown. Initial: 1M pure liq 1−x 0M BaseH+(aq) + OH−(aq) Base(aq) + H2O(l) @ Equib: 0M pure liq x x Since I was nice and gave you Kb for this basic equilibrium, you can go directly to setting up your equilibrium expression: Kb = ( x )( x ) = 1× 10− 8 (1 − x ) ( x )( x ) = 1× 10 − 8 (1) assume that x << 1, or x2 = 1 x 10−8, or x = [OH−] = 1 x 10−4 pOH = −log(1 x 10−4) = 4. BUT THIS IS NOT YOUR ANSWER, since I asked for the pH!! pH = 14 − pOH = 10 a) -2 b) 0 c) 4 d) 7 e) 10 5. (5 pts) Calculate the pH of a 0.1 M solution of an acid that has a pKa of 5.0. Clearly show all your work. Circle the correct answer from those given below. No credit will be given if work is not shown. Initial: HA(aq) @ Equib: 0.1M 0M 0M H+(aq) + A−(aq) 0.1 − x x x Convert the pKa of 5.0 into a Ka value: Ka = antilog(−pKa) = 1 x 10−5 solve for the [H+] and the pH. Ka = ( x )( x ) = 1× 10 − 5 ( 0 .1 − x ) or x = [H+] = 1 x 10−3 a) − 1 b) 1 assume that x << 0.1, x2 = 1× 10 − 5 , or x2 = 1 x 10−6, (0.1) So the pH = −log(1 x 10−3) = 3 c) 3 Now, we can set-up and pH = 3 d) 7 e) 11 6. (5 pts) An unknown acid solution has a pH of 5. What important, but simple, piece of information do you need to tell if the solution is a strong or weak acid (aside from the Ka value)? You need to know the molarity (or concentration) of the solution! The pH by itself only tells you the H+ concentration, NOT whether the acid it came from is a strong or weak acid. For example, a 1 x 10−6 M solution of HCl has a pH of only 6, which is just barely acidic, but HCl is a strong acid. So unless you know both the pH and concentration of the added acid, you can not directly tell if you are dealing with a strong or weak acid. From the pH and concentration of added acid you can calculate the Ka value, which will tell you if you have a strong, medium or weak acid. Of course, if you have an extremely low pH, like 0 to –1, you have to be dealing with a strong acid since even very concentrated weak acids can’t produce that kind of H+ concentration. ANSWER KEY Name ___________________________ CHEM 1202 - Homework # 7 & 8 Acids & Bases 2 Due Tuesday, Nov 28th, 2006 by Noon Table 1. Dissociation Constants for some Acids. Acid pKa Value Acid pKa Value Acid pKa Value NH4+ 10 HBF4 −9 formic 4 HClO 8 H2CO3 7 benzoic 5 1. (5 pts) Which of the acids listed in Table 1, given a 0.01 M solution in water, will have a pH closest to 2? a) NH4+ b) benzoic c) H2CO3 d) formic e) HBF4 2. (5 pts) Which of the acids listed in Table 1, when reacted with an equivalent amount of NaOH, will form a solution with the highest pH? b) benzoic c) H2CO3 d) formic e) HBF4 a) NH4+ 3. (5 pts) a) b) c) d) e) Order the acids in Table 1 from strongest to weakest. Circle the correct choice. benzoic > formic > H2CO3 > HBF4 > NH4+ > HClO HBF4 > formic > benzoic > H2CO3 > HClO > NH4+ NH4+ > benzoic > HBF4 > formic > H2CO3 > HClO benzoic > HBF4 > NH4+ > formic > H2CO3 > HClO NH4+ > HClO > H2CO3 > benzoic > formic > HBF4 4. (5 pts) What is the pH of a 0.01 M solution of the weak base benzylamine (C6H5CH2NH2)? pKa = 8. Circle the answer below and clearly show all your work. Initial: 0.01M pure liq 0M BaseH+(aq) + OH−(aq) Base(aq) + H2O(l) @ Equib: 0.01 − x 0M pure liq x x You first need to convert the pKa into a Ka, then into a Kb for this basic equilibrium: K 1× 10 − 14 ( x )( x ) Kb = w = = 1× 10 − 6 now you can setup your equilib: Kb = = 1 × 10 − 6 −8 (0.01 − x ) Ka 1× 10 assume that x << 0.01, ( x )( x ) = 1× 10 − 6 or x2 = 1 × 10−8, or x = [OH−] = 1 × 10−4 pOH = 4. (0.01) BUT THIS IS NOT YOUR ANSWER, since I asked for the pH!! pH = 14 − 4 = 10 a) 4 b) 5 c) 9 d) 10 e) 13 5. (10 pts) Will FeCl3 generate an acidic, neutral, or basic solution when dissolved in water. Clearly discuss your reasoning. FeCl3 is composed of a Fe3+ cation and three Cl− anions. The Cl− anions are “do nothing” or “neutral” (acidbase property, extremely poor conjugate base) so will not make a basic solution. The Fe3+ cation, on the other hand, is a Lewis acid and will react with water to release H+: Fe3+(aq) + H2O [Fe(OH)]2+ + H+(aq). This is called hydrolysis and most +2 or higher cations can interact with water in a similar fashion to release H+ and make an acidic solution. Note that the [Fe(OH)]2+ is not a typical hydroxide base as the Fe-OH bond is reasonably strong and will not dissociate OH−. Homework # 7&8 - Acids/Bases 2 (2006) 6. (5 pts) What is the pH of a 1 M solution of KClO? See Table 1 for pKa values. Clearly show all your work. The first thing to realize is that this is a basic salt that will generate a basic solution! K+ is a do-nothing cation, while ClO− is an active anion that acts as a weak base in solution. Initial: 1M pure liq 0M Base−(aq) + H2O(l) @ Equib: 1−x 0M BaseH (aq) + OH−(aq) pure liq x x Next you need to convert the pKa value I’ve given you into a Ka, then Kb for the conjugate base in this basic equilibrium: K 1× 10 − 14 ( x )( x ) Kb = w = = 1× 10 − 6 now you can setup your equilib: Kb = = 1× 10 − 6 assume that x << 1, (1 − x ) Ka 1× 10 − 8 2 −6 −3 pOH = 3. pH = 14 − pOH = 11 so the pH = 11 − ( x )( x ) − 6 or x = 1 × 10 , or x = [OH ] = 1 × 10 (1) = 1× 10 7. (10 pts) Consider the following list of salts: A) NH4Cl E) MoCl4 B) KI D) potassium benzoate F) BaI2 J) NaClO I) KClO4 C) CsF G) AlBr3 H) LiNO3 Which salts will generate an acidic solution? ________________ A, E, G Which salts will generate a basic solution? _________________ C, D, J Which salts will generate a neutral solution? _________________ B, F, H, I 8. (5 pts) Calculate the pKb of the weak base phenylamine if a 1 M solution has a pH = 10. Initial: 1M pure liq 0M BaseH+ (aq) + OH−(aq) Base(aq) + H2O(l) @ Equib: 1−x 0M pure liq x x Kb = [baseH+][OH− ] [base] But I have given you the pH, so you know the OH− concentration (and [BaseH+]) at equilibrium. If pH = 10, pOH = 4 and the [OH−] = 1 × 10−4 M. This is also small enough relative to 1 – x, that we can drop the x here and really make a very simple expression to solve for Kb: Kb = [baseH+][OH− ] (1× 10 -4)(1× 10 -4) = = 1× 10 − 8 [base] 1 so pKb = 8 9. (10 pts) What is the pH if 800 mL of 0.125 M KOH is added to 200 mL of 0.5 M sucoloic acid (a monoprotic acid)? pKa = 11 (clearly show all your work) The first thing to realize is that sucoloic acid is a weak acid and that we are titrating it with a strong base. If there are equal amounts of each, we will be making the salt of a weak acid, which is a weak base! So we may be generating a basic solution depending on the amount of acid and base reacting. Convert the # of mL and molarity of each into moles: # moles base = (800 mL OH−)(0.125 M OH−) = 100 mmoles OH− # moles weak acid = (200 mL weak acid)(0.5 M weak acid) = 100 mmoles weak acid So we have an equal amount of each. Don’t forget that we are adding 800 mL to 200 mL to make 1000 mL total solution volume. This will generate a 100 mmoles/1000 mL = 0.1 M solution that will act as a weak base: Initial: @ Equib: 0.1M pure liq −(aq) + H O(l) Base 2 0.1 − x pure liq 0M 0M BaseH (aq) + OH−(aq) x x Next you need to convert the pKa value I’ve given you into a Ka, then Kb for the conjugate base in this basic equilibrium: 1× 10 − 14 K ( x )( x ) Kb = w = = 1× 10 − 3 now you can setup your equilib: Kb = = 1 × 10 − 3 assume that x << 1, (0.1 − x ) Ka 1× 10 − 11 ( x )( x ) pH = 12 − 3 or x2 = 1 × 10−4, or x = [OH−] = 1 × 10−2 pOH = 2. pH = 14 − pOH = 12 so the (0.1) = 1× 10 CHEM 1202 - Homework # 6 Acids & Bases # 1 Due Thursday, March 19, 2009 ANSWER KEY These compounds are the bases – you need to be able to tell acids from bases 1. (3 pts) Consider the following weak acids and bases and their pKa values: O A) H C O H (pKa = 3.8) D) N(CH3)3 (pKa = 12) B) HCN (pKa = 9.3) C) H2CO3 (pKa = 6.4) E) H2SO3 (pKa = 1.8) F) NH3 (pKa = 9) The strongest acid has the lowest pKa value Which compound is the strongest acid (use letter) ? _______ E Which compound is the strongest base (use letter) ? _______ D Which compound has the strongest conjugate base (use letter) The strongest base has the highest pKa value. Note that the pKa actually refers to the strength of the conjugate acid derived from the base The weakest acid has the strongest ? ______ conjugate base. The weakest acid has the B largest pKa value. You need to be able to tell an acid from a base. 2. (5 pts) What are the pH’s for the following solutions? a) 0.1 M HBr = d) 0.1 M NaOH = 1 b) 10 M H2SO4 = 13 e) 10 M CsOH = c) 1 × 10−10 M HNO3 = 7 −1 15 3. (4 pts) What is the pKa value of a 0.1 M solution of palmetic acid (HA) that has a pH of 5? Clearly show all your work and put a box around your answer. Initial: 0M HA(aq) @ Equib: 0.1 M H+(aq) + A−(aq) 0.1 − x 0M x x BUT, before you go any further, I’ve given you the pH of the solution in the problem!! So you don’t have to solve for x, you already know it!! So all we have to do is take the anti-log of pH 5 to get the concentration of the H+ (and A−): [H+] = antilog(−5) = 1 x 10−5 M Ka = Substituting this into the equlibrium expression gives us: (1 × 10 − 5 ) 2 1× 10 − 10 = = 1× 10 − 9 0.1 − 1 × 10 − 5 0 .1 so, Ka = 1 x 10 −9 pKa = −log(Ka) = 9 Note that it is fine to drop the x in the 0.1 − x expression because x is much smaller than 0.1. So the answer is: 4. (3 pts) Calculate the pH of a 0.01 M solution of acid that has a pKa of 6.0. Clearly show all your work and put a box around your answer. Initial: 0M HA(aq) @ Equib: 0.01 M 0M H+(aq) + A−(aq) 0.01 − x x x First, convert the pKa of 6 into a Ka value: Ka = antilog(−pKa) = 1 x 10−6 Now, we can solve for the [H+] and the pH. x2 ( x )( x ) = 1× 10 − 6 , or x2 = 1 x 10−8, = 1× 10 − 6 assume that x << 1, Ka = (0.01) (0.01 − x ) or x = [H+] = 1 x 10−4 So the pH = −log(1 x 10−4) = 4 pH = 4 CHEM 1422 – HW # 5 – Acids & Bases #1 2 5. (5 pts) What is the pH of a 0.1 M solution of the base ethyl amine (CH3CH2NH2). Ka = 1 × 10−11 Clearly show all your work and put a box around your answer. Initial: 0.1 M pure liq 0M BaseH+(aq) + OH−(aq) Base(aq) + H2O(l) @ Equib: 0.1 − x 0M pure liq x x You first need to convert the Ka into a Kb for this basic equilibrium: K 1× 10 − 14 ( x )( x ) Kb = w = = 1× 10 − 3 -- now you can setup your equilib: Kb = = 1× 10 − 3 − 11 ( 0 .1 − x ) Ka 1× 10 ( x )( x ) assume that x << 1, = 1× 10 − 3 or x2 = 1 x 10−4, or x = [OH−] = 1 x 10−2 pOH = 2. BUT THIS IS NOT YOUR (0.1) ANSWER, since I asked for the pH!! pH = 14 − 2 = 12 pH = 12 6. (5 pts) What is the Kb of a 0.01 M solution of a base that has a pH of 10? Clearly show all your work and put a box around your answer. Initial: @ Equib: 0.01 M pure liq −(aq) + H O(l) Base 2 1−x pure liq @ Equib: 10–4 0.01 – 1 x Plug into the Kb expression and solve: K b = 0M 0M BaseH (aq) + OH−(aq) x 1x 10–4 x A pH of 10 means that the pOH is 4. We do NOT use the pH directly in this problem because we are dealing with a basic equilibrium!! Thus, we know that the [OH–] = 1 x 10–4 M. 1 x 10–4 (1× 10 − 4 )2 = 1× 10 − 6 0.01 − 1× 10 − 4 Kb = 1 x 10–6 You can drop this since it is small compared to 0.01 M 7. (5 pts) Which of the following acids is the strongest based on its structure and atoms present? Clearly discuss your reasoning. CH H+ F O 3 F F HPF6 is an extremely strong acid, while HPO2(OCH3)2 - or P P O CH3 is a rather weak acid. The two main reasons is that the F F O negative charge for the [PF6]- anion is spread out evenly O F H+ over the 6 flourine atoms making for a very “dilute” charge. In contrast, the negative charge in the [PO2(OCH3)2]- anion is localized (concentrated) on one (or at most two) oxygen atoms. More localized negative charges will more strongly attract the H+ cation. This is why F- (small concentrated charge) by itself is a moderately good conjugate base for the weak acid HF. But the [PF6]- anion is not a simple F- anion, its charge is spread out over 6 flourine atoms. The other significant factor is that fluorine has a valence of 1, that is, it only wants to make one chemical bond to another atom. In the [PF6]- anion each fluorine is already coordinated to the central phosphorus atom and won’t want to make another covalent bond to a H+. The [PO2(OCH3)2]- anion, on the other hand, has one oxygen atom with only a single bond to the central phosphorus atom (and a negative charge). Oxygen atoms like to have two chemical bonds in general (H2O !!) so it will have a fairly strong tendency to want to make a chemical bond to the H+. Remember that OH- loves to coordinate to H+. The PO3 group attached to the O- is electron-withdrawing, so the P-O- group will not be as electron-rich as OH- and will not want to bond to the H+ nearly as strongly. CHEM 1422 - Homework # 6 Acids & Bases # 2 Due Tuesday, April 14, 2009 ANSWER KEY 1. (5 pts) 250 mL of 0.2 M acetic acid (HOAc) reacts with 250 mL of 0.2 M NaOH. What is the pH of the resulting solution? pKa (HOAc) = 5 The first thing to realize is that one is reacting a weak acid (acetic acid) with a strong base (NaOH). This will generate the salt of a weak acid (sodium acetate, NaOAc), which is a weak base. There are equivalent amounts of acid and base present (same volume and same concentration) and since one is doubling the solution volume (250 to 500 mL) the concentration is being halved to 0.1 M. So what I am asking you to calculate is the pH of a 0.1 M solution of sodium acetate (the salt of a weak acid). This will be a basic equilibrium for which you need to use a Kb value. Initial: 0.1 M -0 0 pKb = pKw – pKa = 14 – 5 = 9 OAc (aq) + H2O(l) HOAc(aq) + OH (aq) Kb = antilog(9) = 1 × 109 @eq: 0.1 – x -x x Plug the @equilibrium values into the equilbrium expression and solve for x: ( x )( x ) ( x )( x ) 2 10 1 10 9 or x = 1 × 10 , Kb 1 10 9 assume that x << 0.1 since Kb is so small, (0.1) (0.1 x ) or x = [OH] = 1 × 105 so the pOH = 5. BUT THIS IS NOT YOUR ANSWER, since I asked for the pH!! pH = 14 pOH = 9 so the pH = 9 2. (5 pts) Identify whether the following 1:1 solutions will be acidic, basic, or neutral: a) b) c) d) e) ACIDIC (buffer solution, pH = 4.7) BASIC (buffer solution, pH = 9.3) Neutral (not a buffer!! neutral salts!!) ACIDIC (buffer solution, pH = 2.1) BASIC (buffer solution, pH = 10.6) NaOAc/HOAc (pKa = 4.7) NH3/NH4NO3 (pKa = 9.3) NaCl/KNO3 H3PO4/NaH2PO4 (pKa = 2.1) CH3NH2/CH3NH3Cl (pKa = 10.6) Table 1. Indicators Name pKa Acid color Methyl violet 1 yellow violet bromthymol blue methyl yellow 1.7 red yellow thymol blue methyl orange 3.5 red yellow 5 red yellow methyl red Base Color Name pKa Acid color Base Color 7 yellow blue 8.8 yellow blue phenolphthalein 9 colorless pink Alizarin yellow 11 yellow red 3. (3 pts) What is the approximate pH of a colorless solution that turns yellow if a small amount of methyl yellow, or alizarin yellow is added to it, but turns pink if phenolphthalein is added. See Table 1 for information about indicators. Circle your answer. a) 0 to 1 b) 4 to 5 c) 7 to 8 d) 9 to 10 e) 13 to 14 12 4. (2 pts) Which indicator (see Table 1) would work best to indicate the equivalence point for the titration curve shown to the right (circle answer): a) methyl violet b) methyl red d) phenolphtalein c) thymol blue e) alizarin yellow 10 pH 8 6 4 Equivalence point around pH 4, need to use an indicator with a color change around 4. 2 0 0 10 20 30 40 m of HCl added L 50 CHEM 1422 – HW # 6 – Acids/Bases 2 2 5. (3 pts) What is the pH of a 0.01 M solution of NaHCO3? pKb = 4. Clearly show all your work. The first thing to realize is that this is a basic salt that will generate a basic solution! Na is a “do-nothing” cation, while HCO3 (bicarbonate) is a basic anion that acts as a weak base in solution, even though it has a proton attached to it. With very strong bases like hydroxide, the HCO3 (bicarbonate) can act as a proton donor, but typically not by itself. Initial: 0.01 M pure liq 0M 0.01 x Since I gave you a pKb, which is appropriate for a base equilibrium, you only have to convert it to a Kb value: Kb = antilog(4) = 1 × 104 H2CO3 (aq) + OH(aq) HCO3(aq) + H2O(l) @ Eq: 0M pure liq Your equilib setup should look like this: Kb x x ( x )( x ) 1 10 4 assume that x << 0.01, (0.01 x ) ( x )( x ) 1 10 4 or x2 = 1 × 106, or x = [OH] = 1 × 103 pOH = 3. BUT THIS IS (0.01) NOT YOUR ANSWER, since I asked for the pH!! pH = 14 pOH = 11 so the pH = 11 Note that the “x” approximation almost doesn’t work here. But it is an order of magnitude smaller than 0.01, so it is just barely reasonable to drop it. 6. (4 pts) Consider the following list of salts: A) CsNO3 B) TiCl4 C) Na(SH) D) KF E) Ca(OH)2 F) Ti(ClO4)4 G) [HN(CH3)3]Cl H) LiHCO3 I) NaPF6 J) RbBr B, F, G Which salts will generate acidic solutions? ________________ C, D, E, H Which salts will generate basic solutions? _________________ A, I, J Which salts will generate neutral solutions? _________________ 7. (4 pts) What is the pH if 500 mL of 0.2 M HCl is added to 500 mL of 0.2 M ammonia (NH3, pKb = 5) (clearly show all your work) The first thing to realize is that one is reacting a weak base (ammonia) with a strong acid (HCl). This will generate the salt of a weak base (ammonium chloride, NH4Cl), which is a weak acid. There are equivalent amounts of acid and base present (same volume and same concentration) and since one is doubling the solution volume (500 to 1000 mL) the concentration is being halved to 0.1 M. So what I am asking you to calculate is the pH of a 0.1 M solution of ammonium chloride (the salt of a weak base). This will be an acidic equilibrium for which you need to use a Ka value. Initial: 0.1 M 0 0 pKa = pKw – pKb = 14 – 5 = 9 +(aq) +(aq) + NH (aq) NH4 H 3 Ka = antilog(9) = 1 × 109 @eq: 0.1 – x x x Plug the @equilibrium values into the equilbrium expression and solve for x: ( x )( x ) ( x )( x ) 2 10 1 10 9 or x = 1 × 10 , Kb 1 10 9 assume that x << 0.1 since Ka is so small, (0.1) (0.1 x ) or x = [H+] = 1 × 105 so the pH = 5. pH = 5 8. (4 pts) What is the pH of the following 0.5 M aqueous solutions containing equal amounts of the two components shown (find pKa/b or Ka/b values in your textbook appendix or lecture notes): a) NH4Cl + NH3 pH = 9.3 b) HNO2 + KNO2 pH = 3.3 c) H2S + CsHS pH = 7.0 d) C5H4N (pyridine) + [C5H4NH]Br pH = 5.2 (pyridine is a weak enough base that its conjugate acid “takes over” and makes the solution acidic) 1:1 mixtures of weak acids or bases and their salts form buffers whose pH = pKa of the weak acid or the acidic salt of a weak base. For weak bases buffers the pH = 14 – pKb. See the Henderson-Hasselbach equations near the end of the Acid-Base chapter!! CHEM 1202 - Homework # 9 & 10 Redox & Electrochemistry Due Friday, Dec 8th, 2006 by Noon ANSWER KEY Check the box to the right if you want your graded homework to be placed out in the public rack outside Prof. Stanley’s office. Otherwise you will have to pick up your homework from Prof. Stanley in person: 1. (5 pts) Which of the following substances is the best reducing agent? a) F− c) Li+ b) Mg d) Ag+ e) Zn 2. (5 pts) Which of the following substances is the best oxidizing agent? a) F− b) Mg2+ d) Ag+ c) O3 e) Cu Two of the sulfur atoms have +2 oxidation state while the other two have +3, only the two that are oxidized should be used in figuring the # of e- transferred. 3. (10 pts) Balance the following reaction in acidic solution. +2, +3 −2 −1 S4O62−(aq) + Cl−(aq) Use the half-cell method. Write out the oxidation and reduction half cells and balance based on # of electrons: 1st determine ox states: +2 − 2 − 2 +1 2 S2O32−(aq) + OCl−(aq) Oxidation half cell: 2S2O32−(aq) Reduction half cell: OCl−(aq) + 2e- S4O62−(aq) + 2eCl−(aq) We need two S2O32- to produce one S4O62-. At least four redox active sulfur atoms on each side are needed to balance the rxn Since we have the same # of electrons on each side of the reaction, we can simply add them together to get our core redox-balanced rxn (otherwise we would have to multiply each rxn by an integer # to get the same # of e- on each side): 2S2O32−(aq) + OCl−(aq) S4O62−(aq) + Cl−(aq) Now we need to balance the oxygen atoms on each side by adding H2O’s to the side missing oxygen atoms and the appropriate # of H+ to the opposite side of the rxn. The product side is missing 1 oxygen atom so we need to add one water and then 2 H+ to the reactant side. This gives us our final balanced equation: 2H+(aq) + 2S2O32−(aq) + OCl−(aq) S4O62−(aq) + Cl−(aq) + H2O 4. (10 pts) Balance the following reaction in basic solution: +5 − 2 +3 +3 +6 − 2 −(aq) + Cr3+(aq) 3+(aq) + CrO 2−(aq) BiO3 Bi 4 Write out the oxidation and reduction half cells and balance based on # of electrons: 1st determine ox states: Oxidation half cell: Cr3+(aq) Reduction half cell: BiO3−(aq) + 2e- CrO42−(aq) + 3eBi3+(aq) Since we do NOT have the same # of electrons on each side of the reaction, we to multiply each rxn by an integer # to get the same # of e- on each side. 6e- is the common factor, so we need to multiply the Cr3+ half cell by 2 and the BiO3− half cell by 3 and add them together to get the core redox-balanced rxn: 3BiO3−(aq) + 2Cr3+(aq) 3Bi3+(aq) + 2CrO42− (aq) We now need to balance the oxygen atoms and H+, then convert to basic solution by adding the same # of OH− anions to each side of the rxn as we have H+, reacting them to make 2 H2O molecules, then canceling out the redundant H2O’s: ACID solution: 3Bi3+(aq) + 2CrO42− (aq) + H2O 2H+ (aq) + 3BiO3−(aq) + 2Cr3+(aq) BASIC Solution: 3Bi3+(aq) + 2CrO42− (aq) + 2OH− (aq) H2O (aq) + 3BiO3−(aq) + 2Cr3+(aq) 5. (5 pts) Write the oxidation state for the underlined element in the box following each compound. a) LiAlH4 d) CaSO3 −1 +4 b) Ba3(AsO4)2 +5 e) H2O2 −1 c) Na2NiCl4 +2 Homework 9&10 – Redox/Electrochemistry (2006) 6. (15 pts) Calculate the redox potentials for the following reactions. Show the two half cell reactions, written in the proper direction and their potentials used to calculate your answer. 2H−(soln) + 2Li+(soln) a) H2(g) + 2Li(s) H2 + 2e2Li 2H− Εº = −2.25 V Εº = +3.05 V 2Li+ + 2e- b) 4H+(aq) + O2(g) + 2Cu(s) 4H+ + O2 + 4e2Cu c) F2(g) + 2Cl−(aq) F2 + 2e2Cl− Eº = + 0.80 V 2H2O + 2Cu2+(aq) Eº = + 0.89 V 2H2O Εº = +1.23 V Εº = −0.34 V 2Cu2+ + 4e- 2F−(aq) + Cl2(g) 2F− Εº = +2.87 V Cl2 + 2eΕº = −1.36 V Eº = + 1.51 V d) Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s) 2Ag Εº = +0.80 V 2Ag+ + 2eCu Cu2+ + 2eΕº = −0.34 V Eº = + 0.46 V e) 3Pb2+(aq) + 2Al(s) 3Pb(s) + 2Al+3(aq) 2+ + 6e3Pb 3Pb Εº = −0.13 V 3+ + 6e2Al 2Al Εº = +1.66 V Eº = + 1.53 V 7. (10 pts) Library/web research topic: Describe in your own words the chemistry (with formulas) involved in a lithium-ion battery. Is lithium metal used? What is the voltage of this electrochemical reaction? List two main advantages and two main disadvantages of lithium-ion batteries with BRIEF explanations. DO NOT COPY DIRECTLY FROM ANY REFERENCE (except for chemical formulas). List your primary reference used at the end. Lithium ion rechargeable batteries have anodes composed of carbon/graphite (represented by C6 in the formula below) while the cathode is composed of Li+[CoO2]− (Co3+ oxidation state) or LiMn2O4 (mixed Mn3+/Mn4+). When the battery is charged some of the carbon anode is reduced to [C6]x− (where x = # of electrons transferred) and some of the the Co3+ in the cathode is oxidized to a +4 oxidation state. Enough Li+ cations migrate to the anode to balance the negative charge build-up. This forms a mixture of [Li+]x[C6]x− at the anode (electron source) and oxidized CoO2 at the cathode (electron acceptor). Li metal is NOT formed unless something goes wrong during the charging process! When the battery discharges (produces electricity) the electrons flow from the reduced anode to the oxidized cathode through the external circuit (wire) reducing the CoO2 to [CoO2]−. The Li+ cations migrate back from the carbon anode to the cathode to reform LiCoO2. There is an electrolyte typically composed of either a polar organic gel and a lithium salt or a nano-porous polymer (Li-polymer-ion). This makes the Li+ cation migration possible between the anode and cathode. Reaction: [Li+]x[C6]x− + xCoO2 C6 + xLiCoO2 The voltage produced is about 3.6 V. Advantages: light weight & high power density (voltage & amps; very good for small electronic devices); no memory effect (doesn’t matter how often you charge or discharge, charging after light use does not reduce the battery capacity); low self-discharge (built-in computer chip for monitoring battery stats does slowly drain battery). Disadvantages: limited lifetime (aging will gradually deactivate regardless of level of use, typically 3 years); heat sensitive (faster deactivation at higher temps); not good for high power drain devices; built-in computer chip and protection circuit to monitor charging and discharging, temp, etc. (adds to cost); internal short-circuit can cause fire due to high power density (recent fires and recalls for notebook computer batteries). Refs: http://electronics.howstuffworks.com/lithium-ion-battery1.htm http://www.buchmann.ca/Article5-Page1.asp http://en.wikipedia.org/wiki/Lithium_ion_battery CHEM 1422 - Homework # 7 Redox & Electrochemistry Due Tuesday, April 21 (4 PM) ANSWER KEY IMPORTANT: for questions 1-4, it is very important to think qualitatively and understand what oxidizing agents (they get reduced) and reducing agents (they get oxidized) do. You should understand from the periodic table trends that Li+, for example, can’t act as a reducing agent. Nor can F act as an oxidizing agent. 1. (1 pt) Which of the following substances is the best reducing agent? Briefly explain your answer. a) Na+ c) Li+ b) Zn d) Ag e) Al It is IMPOSSIBLE for Na+ and Li+ to act as reducting agents – that would produce Na2+ and Li2+. So you can limit your consideration to Zn, Ag, and Al. Of these three, Al has the most positive oxidation potential (reverse of reduction potential) = +1.66 V 2. (1 pt) Which of the following substances is the best oxidizing agent? Briefly explain your answer. b) Li+ a) O2 e) F d) Ag+ c) Cl2 It is IMPOSSIBLE for F to act as an oxidizing agent – that would produce F2. Li+ is very happy where it is, and has a very negative (non-spontaneous) reduction potential, so that can be ruled out as well. That leaves O2, Cl2, and Ag+. Cl2 has the most positive reduction potential (+1.36 V), so that is your best answer. 3. (1 pt) Which of the following substances is the best reducing agent? Briefly explain your answer. a) F2 c) Li+ b) Mg e) Zn2+ d) Na It is IMPOSSIBLE for F2, Li+, and Zn2+ to act as reducting agents – that would produce F+, Li2+, and Zn3+. So you can limit your consideration to Mg and Na. Of these three, Na has the most positive oxidation potential (reverse of reduction potential) = +2.71 V 4. (1 pt) Which of the following substances is the best oxidizing agent? Briefly explain your answer. a) H+ b) Al3+ c) Ag+ d) Li e) O3 It is IMPOSSIBLE for Li to act as an oxidizing agent – that would produce Li. Of the others, O3 has the most positive reduction potential (+2.07 V), so that is your best answer. 5. (3 pts) Balance the following rxn in acidic solution (add water or H+ as needed). Clearly show your work. Oxidation states: 1 I (aq) + +7 2 MnO4(aq) The two half cell rxns are: Oxidation: 2I I2 + 2e + 3eReduction: MnO4 MnO2 Note that you have to have at least 2I- in order to make one I2. It is important to perform this internal balancing within the half cell first. The common factor is 6e–, so we need to multiply the first half reaction by 3 and the second by 2, then add the reactions together and cancel out the 6e– on each side to give the overall core redox balanced reaction: 6I + 2MnO4 3I2 + 2MnO2 0 +4 2 I2(aq) + MnO2(s) Now, check that the # of oxygen atoms on each side matches! There are 8 oxygens on the reactant side, and 4 oxygens on the product side. Add 4 water molecules to the product side that is missing oxygens, then add 8H+ to balance the hydrogens on the reactant side to produce our final balanced equation. Note that the spectator cations are not included. 8H+ + 6I + 2MnO4 3I2 + 2MnO2 + 4H2O CHEM 1422 – HW # 7 – Redox & Electrochemistry 2 6. (3 pts) Balance the following rxn in basic solution (add water or OH as needed). Clearly show your work. Oxidation states: +6 2 +1 2 ClO Cr2O72 + 6e- ClO4 + 6e2Cr3+ Note that we need to add 2Cr3+ because the Cr2O72 has two redox active Cr6+ atoms in it. Each Cr6+ reacts with 3e- for a total of 6e-. Thus, each half cell uses 6e- and we are balanced: Cr2O72 + ClO Cr3+(aq) + ClO4(aq) 8H+ + Cr2O72 + ClO Write out the REDOX half cells and balance: Ox half rxn: Red half rxn: +7 2 +3 Cr2O72(aq) + ClO(aq) 2Cr3+ + ClO4 2Cr3+ + ClO4 + 4H2O Now, convert to basic solution by adding as many OH– to each side as there are H+(you need to add 8OH–). React the added OH– with the H+ on the product side to make H2O's: 8H2O + Cr2O72 + ClO 2Cr3+ + ClO4 + 4H2O + 8OH Finally, cancel out excess waters (4 on each side): Now, check that the # of oxygen atoms on each side matches! There are 8 oxygens on the reactant side, and 4 oxygens on the product side. Add 4 water molecules to the product side that missing 4 oxygens, then add 8H+ to the reactant side to balance the hydrogens: 4H2O + Cr2O72 + ClO 2Cr3+ + ClO4 + 8OH 7. (5 pts) Write the oxidation state for the underlined element in the box following each compound. a) NaH 1 b) KNO3 +5 d) Ca3(PO3)2 +3 e) Na(NCS) 2 c) Na2PtCl6 +4 8. (5 pts) Calculate the redox potentials for the following reactions. Show the two half cell reactions used to calculate the overall potential. 6H(aq) + 2Al(s) a) 3H2(g) + 2Al3+(aq) 2H+ 2eAl H2 Al+3 + 3e– Eº = 0.00 V Eº = 1.66 V Ag + Cl Mg2++ 2e- 2 F O2 + 4H++ 4e- d) Mg2+(aq) + Cu(s) Mg2++ 2eCu 2.58 V Ag Li++ 1e- Eº = 2.87 V Eº = 1.23 V 1.64 V Eº = 2.36 V Eº = 0.34 V 2.70 V Eº = 0.80 V Eº = 3.05 V 3.85 V Mg(s) + Cu2+(s) Mg Cu2++ 2e- e) Li(s) + Ag+(aq) Ag+ + 1eLi Eº = 0.22 V Eº = 2.36 V 4F(aq) + 4H(aq) + O2(g) c) 2F2(g) + 2H2O(l) F2 + 2e2H2O Nonspontaneous 2Ag(s) + Mg2+(aq) + 2Cl(aq) b) 2AgCl(s) + Mg(s) AgCl + 1eMg 1.66 V Li+(aq) + Ag(s) Nonspontaneous CHEM 1422 – HW # 7 – Redox & Electrochemistry 3 9. (4 pts) A MgCl2 solution containing a Mg electrode is connected by means of a salt bridge to a CuCl2 solution containing a copper electrode. Sketch out a Galvanic Cell showing this and clearly indicate the movement of anions, cations, electrons, which electrode is dissolving, and which is forming a metallic deposit. Label the anode and cathode and show the cell potential. The spontaneous rxn is: Mg(s) + Cu2+ Mg2+ + Cu(s) Mg(s) Mg2+(aq) + 2e- E° = 2.36 V Cu2+(aq) + 2eCu(s) E° = 0.34 V Cell potential = 2.70 V 10. (3 pts) How long (in hours) will it take to electrodeposit 1 mole of Al metal by passing a current of 9.65 amps through a solution of Al+3 ? Please clearly show all your work and put a box around your final answer. (# moles)(96, 485 C/mole e-)(# e ) (1)(96, 485 C/mole e-)(3e-) (# amps) (9.65 C/sec) # sec 29, 995 30, 000 30, 000 sec # hrs 8.33 hr 3600 sec/hr # sec 11. (3 pts) What is the concentration of [Ag+] in a half-cell if the reduction potential of the Ag+/Ag couple is observed to be 0.40 V? Clearly show all your work and put a box around your final answer. Ag+(aq) + e- Ag(s) +0.80 V Q 1 [ Ag] Rearrange the Nernst equation and solve for Q, from which we can calculate [Ag+]: E E 0.052 n(E E) (1)(0.4) log(Q) ; or: log(Q) 6.757 n 0.0592 0.0592 Take the antilog (10x) of each side to get the value of Q: Q = 106.767 = 5.71 × 106 [Ag+] 1 1 1.75 10 7 M Q 5.71106 CHEM 1422 - Homework # 8 Organic & ChemDraw Due Friday, May 1 (by 3:00 PM) ANSWER KEY Please download ChemDraw from Tigerware (located under Scientific Software, Chemistry software) and install (along with the included Chem3D program) on your Windows or Macintosh computer. You need to register with CambridgeSoft using an LSU e-mail address in order to get a license key (serial #) for installing ChemDraw. I sent instructions on this via e-mail. Study groups can work together, as usual, but each student needs to E-mail their own typed report with ChemDraw structures to Prof. Stanley as a Word or PDF file. Prof. Stanley is available to answer questions about ChemDraw. 1. (20 pts) Sketch out nice ChemDraw structures for the following molecules. Each should be shown in standard organic line notation and with all atoms identified. Example: H O O H H2C tetrahydrofuran: C CH2 H2C CH2 H O H C C H Or ganic line notation H C H H Tw o possible w ay s of drawing "complete" str uct ur es a) 2-hexanone b) heptanal c) Z-2-octene d) E-2-octene e) dimethylformamide f) 2-pentyne g) benzoic acid h) 6-ethyl-3-methylnonane i) 1,3-diethylbenzene j) pyridine 2. (10 pts) Name and redraw the following molecules: a) vinylcyclohexane b) 5,6-dimethyloctan-3-imine d) (1s,4s)-bicyclo[2.2.1]hepta-2,5diene e) but-2-yn-1-amine c) naphthalene CHEM 1422 - Homework # 9-10 Gaussian & GaussView Due Thursday, April 30 (by 4:00 PM) ANSWER KEY Please download Gaussian 03 and GaussView 4 from Tigerware (located under Scientific Software, Chemistry software) and install on your Windows computer. Up to 3 people can work together on this assignment. Macintosh users should pair up with those with PC/Windows computers. Submit one copy of your report and don’t forget to put all the names of those working together on the report. All reports should be typed, formatted nicely, and include color images (when possible). Prof. Stanley is available to answer questions about Gaussian & GaussView. Separate instructions for using GaussView and Gaussian have been posted by Prof. Stanley. This double assignment will count for 60 pts. 1. (40 pts) Do DFT molecular orbital calculations on perchloric, sulfuric, and phosphoric acids (optimize, DFT, B3LYP, 6-311G(d) basis set). Make sure you select a single d function on your basis set. Questions: a) Capture and display the optimized molecules with their Mulliken charges. Perchloric Acid Sulfuric Acid Phosphoric Acid b) Show the electrostatic surface potential plots using a density value of 0.04. Adjust the color scale for perchloric acid to best represent the atomic charges, then use the same numerical ranges for displaying the other two molecules. Perchloric Acid Sulfuric Acid Phosphoric Acid charge range = -0.02 to +0.4 charge range = -0.02 to +0.4 charge range = -0.02 to +0.4 CHEM 1422 Honors General Chemistry – Gaussian Homework 2 c) Do the charges tell you anything about the acidity of the molecule? Note that the calculation is done in “vacuum” with no solvent molecules around – a proton will not dissociate without water molecules to interact with. Discuss any correlation of the charges calculated with the acidity of the molecules. There appears to be some correlation between the acidity and the atomic charges, but not a very strong one. Consider the charge data in table format showing the comparisons: Charges Atom HClO4 H2SO4 H3PO4 H 0.45 0.47 0.45 O(-H) -0.54 -0.64 (avg) -0.69 (avg) O(=X) -0.35 (avg) -0.44 (avg) -0.56 1.15 1.24 1.28 X No correlation with the H positive charge, which is where one might expect the largest effect. The positive charge on the central X atom does increase as one might expect from electronegativity effects: P is the least electronegative and should lose the most electron-density to the electronegative oxygen atoms. Directly related to this is the fact that the negative charge on the O-H oxygen atoms increases as one goes from HClO4 to H2SO4 to H3PO4. The higher the negative charge on the O-H oxygen atom, the stronger the electrostatic attraction to the H atom, and the harder it will be for the H+ to dissociate. The lower negative charge on the HClO4 oxygen (O=X) atoms means that they should be able to pick up and spread out the negative charge when the H+ dissociates. This will also help make HClO4 the strongest acid. But you can’t carry this analysis too far as H2SO4 looks fairly close to H3PO4, but they are, in fact, quite far apart with respect to their acidities. d) List and compare the bond distances for the 3 molecules in table format and discuss how they compare to the “localized” structures drawn above. Do the bond distances fit any periodic trends? Distances (Å) HClO4 H2SO4 H3PO4 O-H 0.974 0.969 (avg) 0.965 (avg) X-O(-H) 1.720 1.628 (avg) 1.608 (avg) 1.456 (avg) 1.440 (avg) 1.467 O(=X) The X=O and X-OH bond distances correlate extremely well with the double and single bonds drawn at the beginning of this assignment. Double bonds between two atoms are almost always shorter than single bonds. The X=O bond distances are shorter than the single bonds by amounts ranging from 0.141 Å for H3PO4, to 0.264 Å for HClO4, which is typical for double vs. single bonds. There isn’t any recognizable trend in bond distances. The double bond distances decrease as one goes from P=O to S=O, which does fit with the periodic trend of decreasing atom sizes as one goes from P to S. But the Cl=O bond distance increases, even through Cl is even smaller than S. This indicates that the O=Cl double bond is getting weaker, which would increase the bond distance. This would be consistent with periodic trends as Cl does not typically like expanded valence and high oxidation state situations The X-OH bond distances, also, do not fit the pattern of decreasing covalent radii for P-S-Cl as their bond distances increase as one goes from P to Cl. No clear reason for this that you should know of. But, X-OH bond strengths decrease as one goes from P to Cl, so the increasing bond distances do seem to clearly support the trend of weaker σ-bonding between the X and OH atoms. CHEM 1422 Honors General Chemistry – Gaussian Homework 3 e) Show the highest occupied molecular orbital (HOMO) for each molecule along with its energy in eV. Discuss and explain the trend (if any) in the energies of the HOMO’s (lower is more stable) for these three molecules with periodic properties and what we discussed in class for oxyacids. Discuss what kind of bonding, non-bonding (lone pair), or antibonding interactions are occurring within the HOMO for each molecule. Although one might expect hybrid orbitals (e.g., sp3) for some of these, MO calculations often tend to display orbitals as more or less pure s, p, etc. For example, an oxygen lone pair is more likely to show up as a p orbital. Perchloric Acid Sulfuric Acid Phosphoric Acid −9.58 eV −8.98 eV −8.37 eV All three HOMO’s are mainly oxygen pure p lone pair type orbitals. They are non-bonding (no bonding or antibonding with respect to the X-O or O-H bonds). They have weak antibonding “through-space” interactions with adjacent oxygen atoms. This weak repulsion changes the shape from that of a pure isolated p orbital to the that shown where the opposite color orbital lobes repel one another. There may be a very small amount of X=O π-bonding present for H2SO4 and H3PO4, as indicated by the same color orbital connection between the doubly bound oxygen atoms and the central atom. The larger orbital size on the X=O atoms indicates that the p-orbitals (lone pairs) on these atoms are higher in energy relative to the lone pairs on the OH oxygen atoms, at least for this specific orbital. The trend in the orbital energies is exactly what one expects based on the increasing electronegativity of the central atom. HClO4 has the lowest energy because Cl is the most electronegative central atom. S is next, followed by P. There is a fairly uniform shift of approximately 0.6 eV to higher energy as one goes from Cl to S to P in the HOMO energy, which follows the periodic electronegativity trend. CHEM 1422 Honors General Chemistry – Gaussian Homework 4 2. (20 pts) Shirakura & Suginome published a paper in the Journal of the American Chemical Society (2008, 130, 5410-5411) on nickel catalyzed coupling of silylacetylenes with dienes: The first step in the catalysis is the oxidative addition (breaking) of the alkyne C-H bond to the nickel atom to make a nickel-hydride-alkynyl complex: In order for a C-H bond to do an oxidative addition to a Ni metal center the orbital associated with the C-H bond must be reactive enough to want to participate in this reaction. The higher the energy of the orbital the more reactive it will be (usually). This reaction apparently does not work for regular alkynes that do not have silyl groups attached. Use Gaussian (optimization, DFT, B3LYP method, 6-311G basis set) to perform calculations on the following alkyne molecules: Although I didn’t ask for it here are the various angles and bond distances for these three optimized molecules from Gaussian: C≡C H3X-C X-H (avg) C≡C-H X-C≡C H-X-C (avg) 1.203 Å --- --- 180.00° 180.00° --- 1.064 Å 1.213 Å 1.461 Å 1.097 Å 180.00° 180.00° 111.36° 1.066 Å Experimental (≡)C-H 1.061 Å Molecule 1.219 Å 1.854 Å 1.500 Å 180.00° 180.00° 109.85° 0.950** 1.194 Å 1.83 Å 179.9° 176.7° 106.9° to 109.4° ** X-H bond distances (X = almost any element) from X-Ray structures are almost always too short due to the small amount of electron density around a H atom. X-Ray diffraction relies on diffraction of the X-Rays off the electrons around an atom. Most X-H bonds have most of the electron density around the X atom, which is also generally more electronegative and pulls the single electron around the H atom towards the X atom. This results in X-H bond distances that are too short. Neutron diffraction gives accurate X-H bond distances since the neutrons diffract off the atomic nuclei, not the electrons around each atom. CHEM 1422 Honors General Chemistry – Gaussian Homework 5 Questions: a) Identify the highest energy (C≡)C-H bonding orbital (# from calculation), list the energy of each in eV, and show a picture of this molecular orbital for each alkyne. Molecule MO # E (eV) #5 −13.79 #7 −12.34 # 11 Highest Energy (C≡)C-H MO Orbital −11.01 b) How does the energy of the orbital from question # 2 affect the C-H bond strength? How does the group attached to the other side of the alkyne affect this energy? The higher the energy of an orbital (less negative) the less stable that orbital is and the less it contributes to the bonding between atoms. Although the three orbitals shown are NOT the only C-H bonding orbitals, the trend in this three is a steady decrease in the MO energy. This is definitely related to the group change from H to CH3 to SiH3. Exactly how, is a more complicated answer, which I don’t expect you to know and are discussed below. There are several factors that can play a role here. Typically as you add more bonded atoms to a molecule, there is less of any one bonding interaction present in any one MO. The bonding present in the CH3 and SiH3 groups spreads out the orbital density away from the (≡)C-H bond and shifts it towards the CH3 and SiH3 groups. There is also something called the “trans effect” where a strong bond on one side of an atom can weaken another bond on the opposite side. The C≡C triple bond is a good transmitter of this type of effect, so the stronger donating CH3 and SiH3 groups can potentially weaken the C-H bond a small amount. The other orbital that plays an important role in the C-H bond breaking is the lowest energy empty C-H antibonding orbital. Generally speaking, the higher the energy of the bonding orbital, the lower the energy of the antibonding analog. The lower the energy of the C-H antibonding orbital, the easier it is for the Ni atom to donate some of its electrons into that antibonding orbital and help break the C-H bond. The last effect concerns the energy of the C≡C π-bonding orbitals. The donating CH3 and SiH3 groups make the triple bond more electron-rich & higher in energy (see next question) and that, in turn, makes it a better donor to the Ni metal center. The alkyne has to first coordinate to the Ni atom via the C≡C πbonding MO before the C-H bond can be broken. CHEM 1422 Honors General Chemistry – Gaussian Homework 6 c) List and compare the energies (in eV) of the alkyne π-bonding orbitals for each system and show a picture of one of these molecular orbitals for each alkyne. Although these aren’t directly involved in the C-H bond breaking step discussed above, what factor seems to be affecting the energy of these orbitals? Molecule MO # E (eV) #7 −8.13 # 11 −7.13 # 15 Highest Energy (C≡)C-H MO Orbital −7.69 The two of the C-H orbitals on the CH3 and SiH3 groups can weakly interact with the C≡C π-bonding MO in an antibonding fashion to push it up in energy relative to the H-C≡C-H “baseline” case. The spherically symmetric H s orbital cannot interact with the C≡C π-bonding MO when it is collinear with the C≡C bond axis. BUT, when two of the H-atoms on the CH3 or SiH3 groups are oriented up and down from the C≡C bond axis their s-orbitals can now weakly interact with the C≡C π-bonding orbitals in both a bonding and antibonding fashion. In the orbitals above, we are seeing the weak antibonding interaction that affects and pushes up the energy of the C≡C π-bonding MO. If you look through the lower energy orbitals you will find one that looks similar, but has a π-type bonding interaction between the X-H bonds (X = C or Si) and the C≡C π-system (shown below for H3C-C≡C-H & H3Si-C≡C-H). −11.7 eV −10.0 eV The CH3 C-H orbitals are close enough to the C≡C π-bonding MO to have a stronger bonding/antibonding interaction compared to the SiH3 group, which is further away and has longer X-H bonds: C-C(≡C) = 1.46 Å, Si-C(≡C) = 1.85 Å, H2C-H = 1.1 Å, H2Si-H = 1.5 Å. The further away orbitals are from one another the weaker their interactions (either bonding or antibonding). This is why the HOMO in H3Si-C≡C-H is lower in energy (more stable) relative to the same orbital for H3C-C≡C-H. ...
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This note was uploaded on 11/12/2011 for the course CHEM 1422 taught by Professor Staff during the Fall '08 term at LSU.

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