1. Equilibrium Calculations Comparing K to Q

2. Value of the Equilibrium Constant K tells where the equilibrium lies How likely (to what extent) the reaction is to occur Does not give rate (that’s KINETICS) Three possibilities  K = 1 Equal ratio of products to reactants Equilibrium lies in the “middle”

3. Value of the Equilibrium Constant K tells where the equilibrium lies How likely (to what extent) the reaction is to occur Does not give rate (that’s KINETICS) Three possibilities  K >1 Bigger number on top, so… More products Equilibrium favors the PRODUCTS (to the right)

4. Value of the Equilibrium Constant K tells where the equilibrium lies How likely (to what extent) the reaction is to occur Does not give rate (that’s KINETICS) Three possibilities  K <1 Bigger number on bottom, so… More reactants Equilibrium favors the REACTANTS (to the left)

5. Practice Problem #1 H 2(g) + Cl 2(g)  HCl (g) At equilibrium [H 2 ] = 0.42 M, [Cl 2 ] = 0.075 M, and [HCl] = 0.95 M. Determine the equilibrium constant.

6. Practice Problem #2 C (s) + CO 2(g)  CO (g) K c = 6.5 At equilibrium some CO 2, 5.4×10 -5 moles of CO, and 1.88 x 10 -2 moles of C exist in a 1.000- L container. How many moles of CO 2 exist at equilibrium?

7. Practice Problem #3 CO 2(g) + H 2(g)  CO (g) + H 2 O (g) Calculate the value of the equilibrium constant, K c, for the system shown, if 0.1908 moles of CO 2, 0.0908 moles of H 2, 0.0092 moles of CO, and 0.0092 moles of H 2 O vapor were present in a 2.00 L reaction vessel were present at equilibrium.

8. Practice Problem #4 N 2 O 4 (g)  2 NO 2 (g) A sample of N 2 O 4 (g) is allowed to come to equilibrium with NO 2 (g) in a 750 mL flask. At equilibrium, 38.6 g N 2 O 4 remains and 0.0637 moles of NO 2 have been produced. Calculate the equilibrium constant, Kc, for the reaction.

9. Reaction Quotient (Q) Where you are at

10. Reaction Quotient (Q) Snapshot of reaction at some point Same calculation as K c They will give you the K of the reaction Compare Q and K = [C] c [D] d [A] a [B] b

11. Reaction Quotient (Q) Example: 2 NO 2 (g)  N 2 O 4 (g) K p = 11 @ 100°C AT SOME POINT, 0.200 moles of NO 2 are mixed with 0.200 moles of N 2 O 4 in a 4.00 L flask at 100°C. = [P N 2 O 4 ] [P NO 2 ] 2 = 0.654

12. Reaction Quotient (Q) Where am I?? K p = 11, Q = 0.654 Compare Q to K to see where you are 3 situations  Q = K At EQUILIBRIUM!!

13. Reaction Quotient (Q) Where am I?? K p = 11, Q = 0.654 Compare Q to K to see where you are 3 situations  Q > K Bigger number, so… Favors products Must shift left (towards reactants to get to equilibrium)

14. Reaction Quotient (Q) Where am I?? K p = 11, Q = 0.654 Compare Q to K to see where you are 3 situations  Q < K Smaller number, so… Favors reactants Must shift right (towards products to get to equilibrium)