Lecture 10:  G, Q, and K Reading: Zumdahl 10.10, Outline –Relating  G to Q –Relating  G to K –The temperature dependence of K

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

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

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)

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

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

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)

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

 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?

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

 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

 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

 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

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

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)

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

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

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

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.

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

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)) = 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 - ( kJ) = kJ

An Example (cont.) With  H°, we’re ready for the van’t Hoff Eq. K 340 = 2.0 x 10 2 Why is K reduced? Reaction is Exothermic. Increase T, Shift Eq. To React. K eq will then decrease