# Chp9 - Chapter 9 Exercises H S G Problems 9,11,13,15,17,19...

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Chapter 9 Exercises H, S, G Problems 9,11,13,15,17,19 Equilibrium Problems 23,27,29,33,35,39

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Chapter 9 – Reaction Energetics 1 st and 2 nd Laws of Thermodynamics Enthalpy ( H) Entropy ( S) Free Energy ( G) Rate Laws Reaction Equilibrium (K eq )
Reaction Energetics Thermodynamics Study of energy and its transformations Interested only in E initial and E final Can predict whether reaction is favorable or not Kinetics Process by which reactants form products Rates and mechanism

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1 st Law of Thermodynamics States that energy is neither created nor destroyed in ANY process Energy is transferred from system to surroundings or vice versa Universe = system + surroundings what you’re studying everything else E universe = 0 E universe = E system + E surrondings = 0 E sys = – E sur
Heat (q) and Work (w) Heat (q) and work (w) are the most common ways that energy is exchanged between sys and sur E sys = q + w + q Heat absorbed by system (endothermic) - q Heat released by system (exothermic) + w Work done ON system - w Work done BY system

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Enthalpy ( H) Heat of reaction under constant P conditions (assume w = 0; E = q), units are J Measure H (much like we do E, not E) 2 H 2 + O 2 2 H 2 O; H = -237 kJ Energy is released, reaction is exothermic 6 CO 2 + 6 H 2 O C 6 H 12 O 6 + 6 O 2 ; H = +2816 kJ Energy is consumed, reaction is endothermic
R P Energy Energy absorbed by sur (“warms”); H < 0 EXOTHERMIC Weak bonds Strong bonds R P Energy absorbed by sys (“cools”); H > 0 ENDOTHERMIC Weak bonds Strong bonds Energy Diagrams

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H° is the standard enthalpy Most stable form of a substance at 1 atm (gas → P of 1 atm; liquid → 1 M at 1 atm) If T is not defined, assume 25 o C Properties of 129 n 1) Reverse reaction changes sign of Standard State ( H o )
2) H° of the reverse reaction is – Combustion of glucose C 6 H 12 O 6 (l) + 6O 2 (g) 6H 2 O(l) + 6CO 2 (g); H°= –1367 kJ Photosynthetic formation of glucose 6H 2 O(l) + 6CO 2 (g) C 6 H 12 O 6 (l) + 6O 2 (g); H°= +1367 kJ Properties of H o 1) number of moles of each substance (n) 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s); H° = -1648 kJ 8Fe(s) + 6O 2 (g) 4Fe 2 O 3 (s); H° = 2(1648kJ) = -3296 kJ n →

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## This note was uploaded on 10/30/2009 for the course CH Ch101 taught by Professor Bigham during the Fall '08 term at N.C. State.

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Chp9 - Chapter 9 Exercises H S G Problems 9,11,13,15,17,19...

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