Week 3 Lecture Notes

Week 3 Lecture Notes - January 23: Lecture 8 Announcements...

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January 23: Lecture 8 Announcements - Finish reading Ch.7 (if you have not done so yet) - Continue working the Ch. 7 assigned problems: 3, 5, 11, 13, 15, 23, 25, 31, 33, 39, 43, 45, 51, 53a, 55, 65 - Attend your normal Section this week. Chem 6B Helproom is open 4-10 PM through Thursday evening - Review Session 1 for Exam 1 is Saturday 1-4 PM in 108 Peterson Hall. - Work Practice Exam 1 and 2 under Exam-like conditions (2 hours using Eqn Sheet and Calculator; no interruptions). Practice Exam 2 and the Eqn Sheet will be up as soon as possible. - Get a Remote Transmitter (IR Type) if you have not done so Bond Enthalpies or Bond Energies : We often use “average” bond enthalpies for estimating the energy released or absorbed during a reaction. Bond enthalpies of molecules such as diatomics provide direct values for bonds that are often of interest, such as for H-H, F-F, Cl-Cl, Br-Br, I-I, H-F, H-Cl, H-Br or H-I single bonds, O=O double bonds, or N≡N and C≡O triple bonds.
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*Average bond enthalpies determined from averaging the bond of interest from many different compounds. Average bond enthalpies can be used to estimate reaction enthalpies and to predict the stability of molecules. Reaction Enthalpy from Bond Enthalpy Determine which bond is broken and which bond is formed in the reactant and product side of the reaction. ΔHr = ΣΔH products + ΣΔH reactants Note that when bonds are formed, energy is released, and ΔH = - ΔHB o If the reaction involves liquids, include ΔH vaporization of reactants and ΔH condensation of products. Lattice Energies : Formation of gaseous ions from the ionic compound (and hence a positive number): ionic solid (s) → ions (g)
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Lattice Energies are approximated by the absolute value of Coulomb’s Law: E = absolute value (k Q 1 Q 2 / r 12 )
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Variation of Reaction Enthalpy with Temperature: Kirchoff’s Law : if the heat capacity of the reactants and products are different, the enthalpy will vary with temperature. Δ H rxn ° (T 2 ) = Δ H rxn ° (T 1 ) + (T 2 – T 1 ) Δ C P Δ C P = Σ n p Δ C P,m (products) – Σ n r Δ C P,m (reactants) EX : The following heats of reaction were measured at 25 ° C: 2 ClF (g) + O 2 (g) Cl 2 O (g) + F 2 O (g) 167.4 Δ H (kJ/mol) 2 ClF 3 (g) + 2 O 2 (g) Cl 2 O (g) + 3 F 2 O (g) 341.4 Δ H (kJ/mol) 2 F 2 (g) + O 2 (g) 2 F 2 O (g) 43.4 Δ H (kJ/mol) At the same temperature, calculate Δ H in kJ for the following reaction: ClF (g) + F 2 (g) ClF 3 (g) A) -65 kJ B) -217 kJ C) 233 kJ D) -109 kJ E) +217 kJ Answer: D because you divide the first heat of reaction formula by 2 to get ClF, divide the
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Week 3 Lecture Notes - January 23: Lecture 8 Announcements...

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