1. Lecture 21/21/05

2. Law of Mass Action

3. Example H 2 (g) + I 2 (g) ↔ 2HI (g)

4. Equilibrium Constant (K) Direction of reaction does not matter Doesn’t matter if you start with only H 2, only I 2, or a mixture of both….K is still the same Dependent on temperature not concentrations K is the same if you start with 1 M, 2 M, 3 M etc of H 2 Leave out solids and water from the K expression Try S (s) + O 2 (g) ↔ SO 2 (g) Try NH 3 (aq) + H 2 O (l) ↔ NH 4 + (aq) + OH - (aq)

5. Conversion of K c to K p aA +bB ↔ cC + dD

6. Conversion of K c to K p 2NO 2 (g) ↔ 2NO (g) + O 2 (g)

7. Conversion of K c to K p N 2 (g) + O 2 (g) ↔ 2NO (g)

8. Uses for Equilibrium constant (1) Does the reaction favor the products or reactants? 1. K>>1 : Reaction is product-favored 1. At equilibrium, [products] > [reactants] 2. K<<1 : Reaction is reactant-favored 1. At equilibrium, [products] < [reactants]

9. (1) Does the reaction favor the products or reactants?

11. Reaction quotient (Q) Same formula as K, but not necessarily at equilibrium Q < K Not at equilibrium Reactants  Products Q > K Not at equilibrium Products  Reactants Q = K System is at equilibrium Uses for Equilibrium constant (2) Predict the direction of the reaction.

12. Example Nitrogen dioxide (NO 2 ) forms N 2 O 4 with K = 171 at 298 K. 2NO 2 (g) ↔ N 2 O 4 (g) If [NO 2 ] = 0.015 M and [N 2 O 2 ] = 0.025 M, is the system in equilibrium? If not, which direction is the reaction going?

13. Uses for Equilibrium constant (3) Calculations: equilibrium concentrations  K For the reaction: 2 SO 2 (g) + O 2 (g) ↔ 2 SO 3 (g) If at 852 K, [SO 2 ] = 3.61 x 10 -3 M, [O 2 ]=6.11 x 10 -4 M, [SO 3 ] = 1.01 x 10 -2 M. Calculate K.

14. What if 1.00 mol of SO 2 and 1.00 mol of O 2 put into 1.00-L flask at 1000 K. At equilibrium, 0.925 mol of SO 3 have been formed. Calculate K