2. Lecture 9 :  G, Q, and K The Meaning of  G Reading: Zumdahl 10.10, 10.11, 10.12 Outline –Relating  G to Q –Relating  G to K –A descriptive example of  G

3. The Second Law The Second Law: there is always an increase in the entropy of the universe. From our definitions of system and surroundings:  S universe =  S system +  S surroundings

4. Defining  G Recall, the second law of thermodynamics:  S univ =  S total =  S system +  S surr Also recall:  S surr = -  H sys /T Then  S total =  S system -  H sys /T  S total = -T  S system +  H sys

5. Defining  G (cont.) We then define:  G = -T  S total Then:  Or  G =  H - T  S  G = -T  S sys +  H sys  G = The Gibbs Free Energy w/ P const

6. Relating  G to Q Recall from Lecture 6:  S = R ln (  final /  initial ) For the expansion of a gas  final  Volume

7. Relating  G to Q (cont.) Given this relationship  S = R ln (V final /V initial )

8. Relating  G to Q (cont.) This equation tells us what the change in entropy will be for a change in concentration away from standard state. Entropy change for process occurring under standard conditions Additional term for change in concentration. (1 atm, 298 K)(P ≠ 1 atm)

9. Relating  G to Q (cont.) How does this relate to  G?

10. Relating  G to Q (cont.) Generalizing to a multicomponent reaction: Where

11. Relating  G to Q (cont.)

12. An Example Determine  G rxn at 298 K for: C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l) where P C2H4 = 0.5 atm (others at standard state)  G° rxn = -6 kJ/mol (from Lecture 9)

13. An Example (cont.) C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l)  G rxn = -6 kJ/mol + (8.314 J/mol.K)(298K)ln(2) = -4.3 kJ/mol

14.  G and K The Reaction Quotient (Q) corresponds to a situation where the concentrations of reactants and products are not those at equilibrium. At equilibrium, we have K. What is the relationship between  G and K?

15.  G and K (cont.) At equilibrium,  G rxn = 0 0K 0 =  G° rxn +RTln(K)  G° rxn = -RTln(K)

16.  G and K (cont.) Let’s look at the interaction between  G° and K  G° rxn = -RTln(K) If  G° < 0 then  > 1 Products are favored over reactants

17.  G and K (cont.) Let’s look at the interaction between  G° and K  G° rxn = -RTln(K) If  G° = 0 then  = 1 Products and reactants are equally favored

18.  G and K (cont.) Let’s look at the interaction between  G° and K  G° rxn = -RTln(K) If  G° > 0 then  < 1 Reactants are favored over products

19. Temperature Dependence of K We now have two definitions for  G°  G° rxn = -RTln(K)=  H° - T  S° Rearranging: y = m x + b Plot of ln(K) vs 1/T is a straight line

20. T Dependence of K (cont.) If we measure K as a function of T, we can determine  H° by determining the slope of the line slope intercept

21. T Dependence of K (cont.) Once we know the T dependence of K, we can predict K at another temperature: - the van’t Hoff equation.

22. An Example For the following reaction : CO(g) + 2H 2 (g) CH 3 OH(l)  G° = -29 kJ/mol What is K at 340 K? First, what is K eq when T = 298 K?  G° rxn = -RTln(K) = -29 kJ/mol ln(K 298 ) = (-29 kJ/mol) -(8.314 J/mol.K)(298K) = 11.7 K 298 = 1.2 x 10 5

23. An Example (cont.) Next, to use the van’t Hoff Eq., we need  H° CO(g) + 2H 2 (g) CH 3 OH(l)  H f °(CO(g)) = -110.5 kJ/mol  H f °(H 2 (g)) = 0  H f °(CH 3 OH(l)) = -239 kJ/mol  H° rxn =  H° f (products) -  H° f (reactants) =  H° f (CH 3 OH(l)) -  H° f (CO(g)) = -239 kJ - (-110.5 kJ) = -128.5 kJ

24. An Example (cont.) With  H°, we’re ready for the van’t Hoff Eq. K 340 = 2.0 x 10 3 Why is K reduced? Reaction is Exothermic. Increase T, Shift Eq. To React. K eq will then decrease

25. An Example For the following reaction at 298 K: HBrO(aq) + H 2 O(l) BrO - (aq) + H 3 O + (aq) w/ K a = 2.3 x 10 -9 What is  G° rxn ?  G° rxn = -RTln(K) = -RTln(2.3 x 10 -9 ) = 49.3 kJ/mol

26. An Example (cont.) What is  G rxn when pH = 5, [BrO - ] = 0.1 M, and [HBrO] = 0.2 M ? HBrO(aq) + H 2 O(l) BrO - (aq) + H 3 O + (aq)

27. An Example (cont.) Then: = 49.3 kJ/mol + (8.314 J/mol.K)(298 K)ln(5 x 10 -6 ) = 19.1 kJ/mol  G rxn <  G° rxn “shifting” reaction towards products

28. A Descriptive Example Consider the following gas-phase equilibrium: OClO (g) ClOO (g) Now,  G rxn =  G° rxn +RTln(Q) = -RTln(K) + RTln(Q) = RTln(Q/K)

29. A Descriptive Example (cont.) OClO (g)ClOO (g) From inspection,  G rxn = RTln(Q/K) If Q/K < 1  G rxn < 0 If Q/K > 1  G rxn > 0 OClO (g)ClOO (g) OClO (g)ClOO (g) If Q/K = 1  G rxn = 0 equilibrium

30. A Descriptive Example (cont.) K = 0.005  G rxn = RTln(Q/K)  G° rxn = -RTln(K) = 13.1 kJ/mol Let’s start the rxn off with 5 atm of OClO and 0.005 atm of ClOO. = RTln(.001/.005) = RTln(.2) = -1.6RT < 0  G rxn < 0, reaction is proceeding towards products Q = P ClOO /P OClO =.001

31. A Descriptive Example (cont.)  G rxn = RTln(Q/K) Let’s start the rxn off with 5 atm. of ClOO and 0.5 atm of OClO. = RTln(10/0.005) = RTln(2000) >> 0  G rxn > 0, reaction is proceeding towards reactants Q = P ClOO /P OClO = 10

32. A Descriptive Example (cont.) Where is equilibrium? Back to Chem 142 OClO (g)ClOO (g) P init (atm)5 0.005 P equil. (atm)5 - x 0.005 + x x = 0.02 P OClO = 5 -.02 = 4.98 atm P ClOO = 0.025 atm

33. A Descriptive Example (cont.)  G rxn = RTln(Q/K) K = RTln(1) = 0 equilibrium. Pictorally: “Amount” of OClO Present