1. Thermodynamics Chapter 19 Liquid benzene Production of quicklime Solid benzene ⇅ CaCO 3 (s) ⇌ CaO + CO 2

2. Gibbs Energy For a constant-pressure & constant temperature process:  G =  H sys - T  S sys Gibbs energy (G)  G < 0 The reaction is spontaneous in the forward direction  G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction  G = 0 The reaction is at equilibrium

3. Fig 19.17 Analogy between Potential Energy and Free Energy

4. Fig 19.18 Free Energy and Equilibrium

5. aA + bB cC + dD  G° rxn n  G° (products) f =  m  G° (reactants) f  - Standard free-energy of reaction (  G o rxn ) ≡ free-energy change for a reaction when it occurs under standard-state conditions. Standard free energy of formation (  G°) Free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f

6. Fig 19.19 Energy Conversion What’s “Free” About Gibbs Energy? ΔG ≡ the theoretical maximum amount of work that can be done by the system on the surroundings at constant P and T ΔG = − w max

7. What’s “free” about the Gibbs energy? “Free” does not imply that the energy has no cost “Free” does not imply that the energy has no cost For a constant-temperature process, “free energy” For a constant-temperature process, “free energy” is the amount available to do work e.g., Human metabolism converts glucose to CO 2 and H 2 O with a ΔG° = -2880 kJ/mol This energy represents approx. 688 Cal or about two Snickers bars worth...

9. Sample Exercise 19.9 Determining the Effect of Temperature on Spontaneity The Haber process for the production of ammonia involves the equilibrium Assume that ΔH° and ΔS° for this reaction do not change with temperature. (a)Predict the direction in which ΔG° for this reaction changes with increasing temperature. (b) Calculate the values ΔG° of for the reaction at 25 °C and 500 °C.

10. (a)The temperature dependence of ΔG° comes from the entropy term. We expect ΔS° for this reaction to be negative because the number of molecules of gas is smaller in the products. Because ΔS° is negative, the term –T ΔS° is positive and grows larger with increasing temperature. As a result, ΔG° becomes less negative (or more positive) with increasing temperature. Thus, the driving force for the production of NH 3 becomes smaller with increasing temperature.  G =  H sys - T  S sys

11. Sample Exercise 19.9 Determining the Effect of Temperature on Spontaneity The Haber process for the production of ammonia involves the equilibrium Assume that ΔH° and ΔS° for this reaction do not change with temperature. (a)Predict the direction in which ΔG° for this reaction changes with increasing temperature. (b) Calculate the values ΔG° of for the reaction at 25 °C and 500 °C.

12.  G o =  H sys - T  S sys The reaction is nonspontaneous at 500 o C The reaction is spontaneous at 25 o C

13. Gibbs Free Energy and Chemical Equilibrium We need to distinguish between ΔG and ΔG° During the course of a chemical reaction, not all products and reactants will be in their standard states In this case, we use ΔG When the system reaches equilibrium, the sign of ΔG° tells us whether products or reactants are favored What is the relationship between ΔG and ΔG°?

14. Gibbs Free Energy and Chemical Equilibrium ΔG = ΔG° + RT lnQ R ≡ gas constant (8.314 J/Kmol) T ≡ absolute temperature (K) Q ≡ reaction quotient = [products] o / [reactants] o At Equilibrium: ΔG = 0 Q = K 0 = ΔG° + RT lnK ΔG° = − RT lnK When not all products and reactants are in their standard states:

15.  G° = - RT lnK Table 19.5 or

16. Calculate ΔG° for the following process at 25 °C: BaF 2 (s) ⇌ Ba 2+ (aq) + 2 F − (aq); K sp = 1.7 x 10 -6 Example ΔG = 0 for any equilibrium, so: ΔG° = − RT ln K sp Equilibrium lies to the left ΔG° = − (8.314 J/mol∙K) (298 K) ln (1.7 x 10 -6 ) ΔG° = + 32.9 kJ/mol ΔG° ≈ + 33 kJ/mol

17. Thermodynamics in living systems Many biochemical reactions have a positive ΔG o In living systems, these reactions are coupled to a process with a negative ΔG o (coupled reactions) The favorable rxn drives the unfavorable rxn

18. C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) Metabolism of glucose in humans ΔG° = -2880 kJ/mol Does not occur in a single step as it would in simple combustion Enzymes break glucose down step-wise Free energy released used to synthesize ATP from ADP:

19. Fig 19.20 Free Energy and Cell Metabolism ADP + H 3 PO 4 → ATP + H 2 O ΔG° = +31 kJ/mol (Free energy stored)