Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry S H 2 = S H 2 R ln P H2 P NH

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Alfred Chandler
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1 N (g) + 3 H (g) NH 3 (g) S N = S H = S NH 3 = S N R ln P N S H R ln P H S NH 3 R ln P NH3 ΔS rxn = (S Rln P NH 3 NH3 ) (S N Rln P N ) 3 (S H Rln P H ) ΔS rxn = S S NH 3 N 3S H + Rln P P 3 N H ΔS rxn = ΔS rxn + Rln P P 3 N H P NH 3 P NH Winter Page 76 ΔG = ΔH - TΔS rxn TR ln ΔG = ΔG + RT ln ΔG = ΔG + RT ln Q ΔG Under Non-standard Conditions ΔG = ΔH - TΔS ΔS rxn = ΔS rxn + Rln P N P 3 H P NH 3 P P 3 N H P P 3 N H P NH 3 Q is the Reaction Quotient P NH 3 Winter 009 Page 77 1

Winter 009 The Reaction Quotient, Q: Direction of Net Change. Equilibrium can be approached various ways.

2 Equilibrium Constant (K): a A + b B. g G + h H. Equilibrium constant = K c = [G] g [H] h. [A] m [B] n. Thermodynamic Equilibrium constant = K eq = (a G )g (a H ) h. (a A ) a (a B ) b. Winter 009 The Reaction Quotient, Q: Direction of Net Change. Equilibrium can be approached various ways. Qualitative determination of change of initial conditions as equilibrium is approached is needed. [G] Q c = tg [H] h t [A]tm [B] n t At equilibrium Q c = K c Winter 009 Page 79

ΔG and the Equilibrium Constant K eq ΔG = ΔG + RT ln Q If the reaction is at equilibrium then: ΔG = ΔG + RT ln K eq = 0 ΔG = -RT ln K eq Winter 009 Page 80 Criteria for Spontaneous

3 ΔG and the Equilibrium Constant K eq ΔG = ΔG + RT ln Q If the reaction is at equilibrium then: ΔG = ΔG + RT ln K eq = 0 ΔG = -RT ln K eq Winter 009 Page 80 Criteria for Spontaneous Change Every chemical reaction consists of both a forward and a reverse reaction. The direction of spontaneous change is the direction in which the free energy decreases. Winter 009 Page 81 3

4 Significance of the Magnitude of ΔG Winter 009 Page 8 The Thermodynamic Equilibrium Constant: Activities. For ideal gases at 1.0 bar: S = S - R ln P P = S - R ln P 1 PV=nRT or P=(n/V)RT, pressure is an effective concentration Therefore, in solution: S = S - R ln c c = S - R ln a The effective concentration in the standard state for an ideal solution is c = 1 M. Winter 009 Page 83 4

Activities For pure solids and liquids: a = 1 For ideal gases: a = P (in bars, 1 bar = 0.

5 Activities For pure solids and liquids: a = 1 For ideal gases: a = P (in bars, 1 bar = atm) For ideal solutes in aqueous solution: a = c (in mol L -1 ) Winter 009 Page 84 The Thermodynamic Equilibrium Constant, K eq A dimensionless equilibrium constant expressed in terms of activities. Often K eq = K c or K P, but not always. Must be used to determine ΔG. ΔG = ΔG + RT ln a g a h a a a b Winter 009 Page 85 5

6 ΔG and K eq as Functions of Temperature ΔG = ΔH -TΔS ΔG = -RT ln K eq ln K eq = -ΔG RT = -ΔH RT + TΔS RT ln K eq = -ΔH RT + ΔS R Winter 009 Page 86 Van t Hoff Equation If we evaluate this equation for a change in temperature: Winter 009 Page 87 6

ln K eq = -ΔH RT slope = -ΔH R + ΔS R -ΔH = R slope = -8.

7 Winter 009 Page 88 Temperature Dependence of K eq Assume ΔH and ΔS do not vary significantly with temperature. ln K eq = -ΔH RT slope = -ΔH R + ΔS R -ΔH = R slope = J mol -1 K K = kj mol -1 Winter 009 Page 89 7

Coupled Reactions In order to drive a non-spontaneous reactions we changed the

Another method is to couple two reactions.

Winter 009 Page 90 Smelting Copper Ore Δ Cu O(s) Cu(s) + ½ O (g) ΔG 673K = +15 kj

8 Coupled Reactions In order to drive a non-spontaneous reactions we changed the conditions (i.e. temperature or electrolysis). Another method is to couple two reactions. One with a positive ΔG and one with a negative ΔG. Overall spontaneous process. Winter 009 Page 90 Smelting Copper Ore Δ Cu O(s) Cu(s) + ½ O (g) ΔG 673K = +15 kj Non-spontaneous reaction: Spontaneous reaction: Cu O(s) Cu(s) + ½ O (g) C(s) + ½ O (g) CO(g) +15 kj -175 kj Spontaneous reaction! Cu O(s) + C(s) Cu(s) + CO(g) -50 kj Winter 009 Page 91 8

kj mol -1 ΔG = ΔG + RT ln a ATP a HO a ADP a P i a H3O + ΔG = -9. kj mol -1 + 41.

9 Coupled Reactions in Biological Systems Winter 009 Page 9 The Biological Standard State ADP 3- + HPO H + ATP 4- + H O ΔG = -9. kj mol -1 ΔG = ΔG + RT ln a ATP a HO a ADP a P i a H3O + ΔG = -9. kj mol kj mol -1 = +3.4 kj mol -1 = ΔG ' But [H 3 O + ] = 10-7 M not 1.0 M. ADP to ATP is not spontaneous Winter 009 Page 93 9

10 Glucose lactate + H kj ADP 3- + HPO H + ATP 4- + H O +64 kj ADP 3- + HPO glucose ATP 4- + H O + lactate -153 kj Winter 009 Page 94 10

Ch. 19 Entropy and Free Energy: Spontaneous Change

Ch. 19 Entropy and Free Energy: Spontaneous Change 19-1 Spontaneity: The Meaning of Spontaneous Change 19-2 The Concept of Entropy 19-3 Evaluating Entropy and Entropy Changes 19-4 Criteria for Spontaneous