Chapter 8 - MIDTERM 2(Thursday 12:30 − 2:00 PM Date(Thu...

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Unformatted text preview: MIDTERM 2, April 23, 2009 (Thursday), 12:30 − 2:00 PM Date (Thu) April 23, 09 April 23, 09 April 23, 09 April 23, 09 April 23, 09 April 23, 09 April 23, 09 Time 12:30 – 2:00 PM 12:30 – 2:00 PM 12:30 – 2:00 PM 12:30 – 2:00 PM 12:30 – 2:00 PM 12:30 – 2:00 PM 12:30 – 2:00 PM Room 1310 N 2310 N 148 NE 1141 N 3127 N 2127 N 133NE Course / Section 1. Chem 1 / MW11A 1. Chem 1 / MW11B 1. Chem 1 / MW11C 1. Chem 1 / MW11E 1. Chem 1 / MW11D 1. Chem 1 / MW11G 1. Chem 1 / MW11J Instructor 1. Ortiz 1. Li, H. 1. Ramsook 1. Kahanda 1. Khajo 1. Shakya 1. Abdelhadi Coverage: Ch. 4 (sections 4.5, 4.6), Ch. 5, Ch. 10, Ch. 6, Ch. 7 (sections 7.1-7.6), Ch. 8 (sections 8.1-8.3) Chapter 8 Basic Concepts of Chemical Bonding General Chemistry I, © 2009 Professor Maggie Ciszkowska Chemical Bonds, Lewis Symbols, and the Octet Rule • Chemical bond: attractive force holding two or more atoms together. Types of Chemical Bonds • Ionic bond results from the transfer of electrons from a metal to a nonmetal. • Covalent bond results from sharing electrons between the atoms. Usually found between nonmetals. • Metallic bond: attractive force holding pure metal atoms together. • Coordination bond (Chem 2). General Chemistry I, © 2009 Professor Maggie Ciszkowska Chemical Bonds, Lewis Symbols, and the Octet Rule • Lewis Symbols Pictorial understanding of where the electrons are in an atom. Electrons are represented as dots around the symbol for the element. The number of electrons available for bonding are indicated by unpaired dots. These symbols are called Lewis symbols. We generally place the electrons on four sides of a square around the element symbol. • • • General Chemistry I, © 2009 Professor Maggie Ciszkowska Lewis Symbols Be C F General Chemistry I, © 2009 Professor Maggie Ciszkowska Chemical Bonds, Lewis Symbols The Octet Rule The Octet Rule • All noble gases except He have an s2p6 configuration. • Octet rule: atoms tend to gain, lose, or share electrons until they are surrounded by 8 valence electrons (4 electron pairs). • Caution: there are many exceptions to the octet rule. General Chemistry I, © 2009 Professor Maggie Ciszkowska Ionic Bonding Consider the reaction between sodium and chlorine: Na(s) + ½Cl2(g) → NaCl(s) ΔHºf = −410.9 kJ General Chemistry I, © 2009 Professor Maggie Ciszkowska Ionic Bonding • The reaction is very exothermic. • NaCl is more stable than its constituent elements. Why? • Na has lost an electron to become Na+ and chlorine has gained the electron to become Cl−. Note: Na+ has an Ne electron configuration and Cl− has an Ar electron configuration. • That is, both Na+ and Cl− have an octet of electrons surrounding the central ion. Na + Cl Na+ + [ Cl ]− General Chemistry I, © 2009 Professor Maggie Ciszkowska Ionic Bonding • NaCl forms a very regular structure in which each Na+ ion is surrounded by 6 Cl− ions. • Similarly, each Cl− ion is surrounded by six Na+ ions. • There is a regular arrangement of Na+ and Cl− in 3D (three dimensional) structure. • Note, the ions are packed as closely as possible. • Na+ and Cl− ions form the ionic lattice. General Chemistry I, © 2009 Professor Maggie Ciszkowska Ionic Bonding General Chemistry I, © 2009 Professor Maggie Ciszkowska Ionic Bonding Energetics of Ionic Bond Formation • The formation of Na+(g) and Cl−(g) from NaCl(s) is endothermic: NaCl(s) → Na+(g) + Cl−(g) ΔH = +788 kJ/mol The energy needed to completely separate a mole of a solid ionic compound into its gaseous ions – Lattice Energy. • The reverse process - the formation of a crystal lattice from the ions in the gas phase is exothermic: Na+(g) + Cl−(g) → NaCl(s) ΔH = −788 kJ/mol General Chemistry I, © 2009 Professor Maggie Ciszkowska Ionic Bonding Energetics of Ionic Bond Formation • Lattice energy: the energy required to completely separate an ionic solid into its gaseous ions. • Lattice energy depends on the charges on the ions and the sizes of the ions: Q1Q2 El = κ d κ is a constant (8.99×10 9 J·m/C2), Q1 and Q2 are the charges on the ions, and d is the distance between ions. General Chemistry I, © 2009 Professor Maggie Ciszkowska Ionic Bonding Energetics of Ionic Bond Formation • Lattice energy increases as: • the charges on the ions increase • the distance between the ions decreases Q1Q2 El = κ d General Chemistry I, © 2009 Professor Maggie Ciszkowska Ionic Bonding • Compare crystal lattice energy of MgS with NaCl. General Chemistry I, © 2009 Professor Maggie Ciszkowska Ionic Bonding • Compare crystal lattice energy of MgS with NaCl. • NaCl: Na+, Cl− Q1Q2 El = κ 2+, S2− • MgS: Mg d • Sizes of cations and anions are similar • Charges in MgS are larger: +2 and −2 • Therefore: Elattice(MgS) ≈ 4×Elattice(NaCl) General Chemistry I, © 2009 Professor Maggie Ciszkowska Ionic Bonding • Compare crystal lattice energy of LiF with NaCl. General Chemistry I, © 2009 Professor Maggie Ciszkowska Ionic Bonding • Compare crystal lattice energy of LiF with NaCl. • NaCl: Na+, Cl− Q1Q2 El = κ +, F − • LiF: Li d • Charges of cations and anions are identical • Sizes of cations and anions in LiF are smaller than in NaCl. • Therefore: Elattice(LiF) > Elattice(NaCl) General Chemistry I, © 2009 Professor Maggie Ciszkowska Ionic Bonding General Chemistry I, © 2009 Professor Maggie Ciszkowska Electron Configurations of Ions Electron Configurations of Ions of the Representative Elements • These are derived from the electron configuration of elements with the required number of electrons added or removed from the most accessible orbital. • Electron configurations can predict stable ion formation: • Mg: [Ne]3s2 • Mg+: [Ne]3s1 not stable • Mg2+: [Ne] stable • Cl: [Ne]3s23p5 • Cl−: [Ne]3s23p6 = [Ar] stable General Chemistry I, © 2009 Professor Maggie Ciszkowska Ionic Bonding Transition Metal Ions • Lattice energies compensate for the loss of up to three electrons. • In general, electrons are removed from orbitals in order of decreasing n (i.e. electrons are removed from 4s before the 3d). Polyatomic Ions • Polyatomic ions are formed when there is an overall charge on a compound containing covalent bonds, e.g. SO42−, NO3−. General Chemistry I, © 2009 Professor Maggie Ciszkowska Electron Configurations of Ions Practice electron configurations for ions. General Chemistry I, © 2009 Professor Maggie Ciszkowska MIDTERM 2, April 23, 2009 (Thursday), 12:30 − 2:00 PM Date (Thu) April 23, 09 April 23, 09 April 23, 09 April 23, 09 April 23, 09 April 23, 09 April 23, 09 Time 12:30 – 2:00 PM 12:30 – 2:00 PM 12:30 – 2:00 PM 12:30 – 2:00 PM 12:30 – 2:00 PM 12:30 – 2:00 PM 12:30 – 2:00 PM Room 1310 N 2310 N 148 NE 1141 N 3127 N 2127 N 133NE Course / Section 1. Chem 1 / MW11A 1. Chem 1 / MW11B 1. Chem 1 / MW11C 1. Chem 1 / MW11E 1. Chem 1 / MW11D 1. Chem 1 / MW11G 1. Chem 1 / MW11J Instructor 1. Ortiz 1. Li, H. 1. Ramsook 1. Kahanda 1. Khajo 1. Shakya 1. Abdelhadi Coverage: Ch. 4 (sections 4.5, 4.6), Ch. 5, Ch. 10, Ch. 6, Ch. 7 (sections 7.1-7.6), Ch. 8 (sections 8.1-8.3) Covalent Bonding • When two similar atoms bond, none of them wants to lose or gain an electron to form an octet. • When similar atoms bond, they share pairs of electrons to each obtain an octet. • Each pair of shared electrons constitutes one chemical bond. • Examples: H + H → H2 has two electrons on a line connecting the two H nuclei. Cl + Cl → Cl2 Cl + Cl Cl Cl General Chemistry I, © 2009 Professor Maggie Ciszkowska Covalent Bonding General Chemistry I, © 2009 Professor Maggie Ciszkowska Covalent Bonding Lewis Structures • Covalent bonds can be represented by the Lewis symbols of the elements (only valence e− are shown): Cl + Cl Cl Cl • In Lewis structures, each pair of electrons in a bond is represented by a single line: Cl Cl H F H O H H N H H H H C H H General Chemistry I, © 2009 Professor Maggie Ciszkowska Covalent Bonding Multiple Bonds • It is possible for more than one pair of electrons to be shared between two atoms (multiple bonds): • One shared pair of electrons = single bond (e.g. H2); • Two shared pairs of electrons = double bond (e.g. O2); • Three shared pairs of electrons = triple bond (e.g. N2). H H O O N N • Generally, bond distances decrease as we move from single through double to triple bonds. General Chemistry I, © 2009 Professor Maggie Ciszkowska Bond Polarity and Electronegativity • In a covalent bond, electrons are shared. • Sharing of electrons to form a covalent bond does not imply equal sharing of those electrons. • There are some covalent bonds in which the electrons are located closer to one atom than the other. • Unequal sharing of electrons results in polar bonds. H⎯H Non-polar bond H⎯F Polar bond General Chemistry I, © 2009 Professor Maggie Ciszkowska Bond Polarity and Electronegativity Electronegativity • Electronegativity: The ability of one atoms in a molecule to attract electrons to itself. • Pauling set electronegativities on a scale from 0.7 (Cs) to 4.0 (F). • Electronegativity increases: • across a period (from left to right) • up a group • F – the most electronegative element • Fr (or Cs) – the least electronegative element General Chemistry I, © 2009 Professor Maggie Ciszkowska Bond Polarity and Electronegativity Electronegativity General Chemistry I, © 2009 Professor Maggie Ciszkowska Bond Polarity and Electronegativity Electronegativity and Bond Polarity • Difference in electronegativity is a gauge of bond polarity: • electronegativity differences around 0 result in non-polar covalent bonds (equal or almost equal sharing of electrons); • electronegativity differences around 2 result in polar covalent bonds (unequal sharing of electrons); • electronegativity differences around 3 result in ionic bonds (transfer of electrons). General Chemistry I, © 2009 Professor Maggie Ciszkowska Bond Polarity and Electronegativity Electronegativity and Bond Polarity • There is no sharp distinction between bonding types. • The positive end (or pole) in a polar bond is represented δ+ and the negative pole δ-. General Chemistry I, © 2009 Professor Maggie Ciszkowska Bond Polarity and Electronegativity Dipole Moments • Consider HF: • The difference in electronegativity leads to a polar bond. • There is more electron density on F than on H. • Since there are two different “ends” of the molecule, we call HF a dipole. • Dipole moment, μ, is the magnitude of the dipole: μ = Qr where Q is the magnitude of the charges, and r is the distance between these charges. • Unites: dipole moments are measured in debyes, D. General Chemistry I, © 2009 Professor Maggie Ciszkowska Bond Polarity and Electronegativity Dipole Moments • Compare the dipole moment of these three covalent bonds: H2, HF, HCl General Chemistry I, © 2009 Professor Maggie Ciszkowska Bond Polarity and Electronegativity Dipole Moments • Compare the dipole moment of these three covalent bonds: H2, HF, HCl H2: μ = 0 (non-polar bond) HF: μ ≠ 0 (polar bond) HCl: μ ≠ 0 (polar bond) General Chemistry I, © 2009 Professor Maggie Ciszkowska Bond Polarity and Electronegativity Dipole Moments • Compare the dipole moment of these three covalent bonds: H2, HF, HCl H2: μ = 0 (non-polar bond) HF: μ ≠ 0 (polar bond) HCl: μ ≠ 0 (polar bond) μ(H2) < μ(HCl) < μ(HF) General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures 1. Ionic or covalent? 2. Pick the central atom (usually there is one central atom; CH4, NH3, SO2, H2O) 3. In your Textbook: Add the valence electrons. Do NOT do it now! 4. Draw a trial structure with all atoms bonded to the central atom and with all SINGLE bonds. 5. Complete the octet for the central atom the complete the octets of the other atoms. The octet rule must be satisfied here. (Octet rule: each atom in a molecule is surrounded by 8 electrons.) General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures 6. Add ALL the valence electrons in the molecule coming from all individual atoms, and compare it with the number of electrons in your trial structure. 7. If the trial structure has too many electrons, change single bond(s) to double or triple bond(s) - one bond single → double: 2 fewer electrons - one bond single → triple: 4 fewer electrons 8. If the trial structure does not have enough electrons, add electrons to the central atom. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Determine the Lewis structure for CO2. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Determine the Lewis structure for CO2. 1. Covalent 2. C – central atom 3. Trial structure O⎯C⎯O 4. All atoms have an octet of e 5. Check to number of valence e: from single atoms: 16 e from the trial structure: 20 e 6. We have to remove 4 e; 2 double bond instead of single bonds. O⎯C⎯O General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Determine the Lewis structure for HCN. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Determine the Lewis structure for HCN. H⎯C⎯N General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Determine the Lewis structure for PH3. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Determine the Lewis structure for ClF3. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Determine the Lewis structure for SF4. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Determine the Lewis structure for PCl5. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Determine the Lewis structure for NO3−. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Determine the Lewis structure for O3. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures – Carbon Compounds Determine the Lewis structure for C2H6. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures – Carbon Compounds Determine the Lewis structure for C2H4. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures – Carbon Compounds Determine the Lewis structure for C2H2. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures – Carbon Compounds Determine the Lewis structure for CH4O. (3 H are connected to C, O forms one bond with C and one bond with H; CH3OH) General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Resonance Structures • Some molecules are not well described by Lewis Structures. • Typically, structures with multiple bonds can have similar structures with the multiple bonds between different pairs of atoms General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures – Carbon Compounds Determine the Lewis structure for O3. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Resonance Structures • Example: experimentally, ozone has two identical bonds whereas the Lewis Structure requires one single (longer) and one double bond (shorter). O O O General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Resonance Structures General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Resonance Structures • Resonance structures are attempts to represent a real structure that is a mix between several extreme possibilities. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Resonance Structures • Example: in ozone the extreme possibilities have one double and one single bond. The resonance structure has two identical bonds of intermediate character. O O O O O O • Common examples: O3, NO3−, SO42−, NO2, and benzene. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures – Carbon Compounds Determine the Lewis structure for NO3−. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Resonance in Benzene • Benzene consists of 6 carbon atoms in a hexagon. Each C atom is attached to two other C atoms and one hydrogen atom. • There are alternating double and single bonds between the C atoms. • Experimentally, the C-C bonds in benzene are all the same length. • Experimentally, benzene is planar. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Resonance in Benzene • We write resonance structures for benzene in which there are single bonds between each pair of C atoms and the 6 additional electrons are delocalized over the entire ring: or • Benzene belongs to a category of organic molecules called aromatic compounds (due to their odor). General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Determine the Lewis structure for BF3. We need to use an idea of a formal charge. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Formal Charge • It is possible to draw more than one Lewis structure with the octet rule obeyed for all the atoms. • To determine which structure is most reasonable, we use formal charge. • Formal charge is the charge on an atom that it would have if all the atoms had the same electronegativity (it is NOT a real charge). General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Formal Charge • To calculate formal charge: • • All nonbonding electrons are assigned to the atom on which they are found. Half the bonding electrons are assigned to each atom in a bond. • Formal charge = valence electrons − number of bonds − electrons in lone pairs General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Formal Charge • Consider: • For C: • • There are 4 valence electrons (from periodic table). In the Lewis structure there are 2 nonbonding electrons and 3 from the triple bond. There are 5 electrons from the Lewis structure. Formal charge: 4 − 3−2 = −1. C N • General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Formal Charge • Consider: • For N: • • There are 5 valence electrons. In the Lewis structure there are 2 nonbonding electrons and 3 from the triple bond. There are 5 electrons from the Lewis structure. Formal charge = 5 − 3−2 = 0. C N • • We write: C N General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Formal Charge • The most stable structure has: • the lowest formal charge on each atom, • the most negative formal charge on the most electronegative atoms. General Chemistry I, © 2009 Professor Maggie Ciszkowska Exceptions to the Octet Rule • There are three classes of exceptions to the octet rule: • • • Molecules with an odd number of electrons; Molecules in which one atom has less than an octet; Molecules in which one atom has more than an octet. Odd Number of Electrons • Few examples. Generally molecules such as ClO2, NO, and NO2 have an odd number of electrons. N O N O General Chemistry I, © 2009 Professor Maggie Ciszkowska Exceptions to the Octet Rule Less than an Octet • Relatively rare. • Molecules with less than an octet are typical for compounds of Groups 1A, 2A, and 3A. • Most typical example is BF3. General Chemistry I, © 2009 Professor Maggie Ciszkowska Drawing Lewis Structures Determine the Lewis structure for BF3. General Chemistry I, © 2009 Professor Maggie Ciszkowska Exceptions to the Octet Rule BF3 • 4 possible resonance structures F F B F F F B F F B F F F B F F Use the formal charge concept to determine the most important structure General Chemistry I, © 2009 Professor Maggie Ciszkowska Exceptions to the Octet Rule BF3 • 4 possible resonance structures F F B F F B F F F B F F F F B F Most important – based on formal charges General Chemistry I, © 2009 Professor Maggie Ciszkowska Exceptions to the Octet Rule Less than an Octet • Relatively rare. • Molecules with less than an octet are typical for compounds of Groups 1A, 2A, and 3A. • Most typical example is BF3. • Formal charges indicate that the Lewis structure with an incomplete octet is more important than the ones with double bonds. General Chemistry I, © 2009 Professor Maggie Ciszkowska Exceptions to the Octet Rule Determine the Lewis structure for NCS−. General Chemistry I, © 2009 Professor Maggie Ciszkowska Exceptions to the Octet Rule More than an Octet • This is the largest class of exceptions. • Atoms from the 3rd period onwards can accommodate more than an octet. • Beyond the third period, the d-orbitals are low enough in energy to participate in bonding and accept the extra electron density. General Chemistry I, © 2009 Professor Maggie Ciszkowska Exceptions to the Octet Rule • Determine the Lewis structure for PCl5. General Chemistry I, © 2009 Professor Maggie Ciszkowska Strengths of Covalent Bonds • The energy required to break a covalent bond is called the bond dissociation enthalpy, D. That is, for the Cl2 molecule, D(Cl−Cl) is given by ΔH for the reaction: Cl2(g) → 2Cl(g) Cl⎯Cl ΔH°298 = 242 kJ/mol → Cl + Cl bond energy of Cl−Cl bond General Chemistry I, © 2009 Professor Maggie Ciszkowska Strengths of Covalent Bonds • The energy required to break a covalent bond is called the bond dissociation enthalpy, D. • When more than one bond is broken: CH4(g) → C(g) + 4 H(g) ΔH = 1660 kJ • the bond enthalpy is a fraction of ΔH for the atomization reaction: D(C−H) = ¼ΔH = ¼(1660 kJ) = 415 kJ • Bond enthalpies can either be positive or negative. General Chemistry I, © 2009 Professor Maggie Ciszkowska Strengths of Covalent Bonds General Chemistry I, © 2009 Professor Maggie Ciszkowska Strengths of Covalent Bonds Bond Enthalpies and the Enthalpies of Reactions • We can use bond enthalpies to calculate the enthalpy for a chemical reaction. • We recognize that in any chemical reaction bonds need to be broken and then new bonds get formed. • The enthalpy of the reaction is given by the sum of bond enthalpies for bonds broken less the sum of bond enthalpies for bonds formed. General Chemistry I, © 2009 Professor Maggie Ciszkowska Strengths of Covalent Bonds Bond Enthalpies and the Enthalpies of Reactions • Mathematically, if ΔHrxn is the enthalpy for a reaction, then ΔH rxn = ∑ D(bonds broken ) − ∑ D(bonds formed ) • We illustrate the concept with the reaction between methane, CH4, and chlorine: CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) ΔHrxn = ? General Chemistry I, © 2009 Professor Maggie Ciszkowska Strengths of Covalent Bonds General Chemistry I, © 2009 Professor Maggie Ciszkowska Strengths of Covalent Bonds CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) ΔHrxn = ? In this reaction one C−H bond and one Cl−Cl bond gets broken while one C−Cl bond and one H−Cl bond gets formed. ΔH rxn = {[D(C − H ) + D(Cl − Cl )] − [D(C − Cl ) + D(H − Cl )]} ΔH rxn = −104 kJ • The overall reaction is exothermic which means than the bonds formed are stronger than the bonds broken. • The above result is consistent with Hess’s law. General Chemistry I, © 2009 Professor Maggie Ciszkowska Strengths of Covalent Bonds Bond Enthalpy and Bond Length • We know that multiple bonds are shorter than single bonds. • We can show that multiple bonds are stronger than single bonds. D(C⎯C) > D(C⎯C) > D(C⎯C) • As the number of bonds between atoms increases, the atoms are held closer and more tightly together. General Chemistry I, © 2009 Professor Maggie Ciszkowska Strengths of Covalent Bonds General Chemistry I, © 2009 Professor Maggie Ciszkowska ΔH of Chemical Reaction Based on Bond Energy • Use the Table of bond energies to estimate ΔH° for the following reaction: C2H6(g) + 7/2 O2(g) → 2 CO2(g) + 3 H2O(g) • We have to know what kinds of bond are there in these molecules…… General Chemistry I, © 2009 Professor Maggie Ciszkowska ΔH of Chemical Reaction Based on Bond Energy We have to know what kinds of bond are there in these molecules……C2H6(g) + 7/2 O2(g) → 2 CO2(g) + 3 H2O(g) • C2H6: • O2: • CO2: • H2O: one single C−C bond six single C−H bonds one double O−O bond two double C−O bonds two single O−H bonds General Chemistry I, © 2009 Professor Maggie Ciszkowska ΔH of Chemical Reaction Based on Bond Energy ΔH rxn = ∑ D(bonds broken ) − ∑ D(bonds formed ) C2H6(g) + 7/2 O2(g) → 2 CO2(g) + 3 H2O(g) 7 D(bonds broken ) = D(C − C) + 6 ⋅ D(C − H) + ⋅ D(O = O) 2 D(bonds formed ) = 4 ⋅ D(C = O) + 6 ⋅ D (O − H ) D(bonds formed ) = 2 ⋅ 2 ⋅ D(C = O) + 3 ⋅ 2 ⋅ D(O − H ) General Chemistry I, © 2009 Professor Maggie Ciszkowska ΔH of Chemical Reaction Based on Bond Energy C2H6(g) + 7/2 O2(g) → 2 CO2(g) + 3 H2O(g) ΔH rxn 7 = {D(C − C) + 6 ⋅ D(C − H) + ⋅ D(O = O)} − 2 − {4 ⋅ D(C = O) + 6 ⋅ D(O − H )} General Chemistry I, © 2009 Professor Maggie Ciszkowska ΔH of Chemical Reaction Based on Bond Energy kJ kJ 7 kJ }− + 6 ⋅ 413 + ⋅ 495 ΔH rxn = {348 mol mol 2 mol kJ kJ } − {4 ⋅ 799 + 6 ⋅ 463 mol mol kJ kJ kJ }− ΔH rxn = {348 + 2478 + 1732.5 mol mol mol kJ kJ } − {3196 + 2778 mol mol kJ ΔH rxn = −1415.5 mol General Chemistry I, © 2009 Professor Maggie Ciszkowska Strengths of Covalent Bonds General Chemistry I, © 2009 Professor Maggie Ciszkowska Strengths of Covalent Bonds General Chemistry I, © 2009 Professor Maggie Ciszkowska ...
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This note was uploaded on 05/04/2009 for the course CHEM 0207 taught by Professor Cizkowska during the Spring '09 term at CUNY Brooklyn.

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