1. 17.2 The equilibrium law 17.2.1 Solve homogeneous equilibrium problems using the expression for Kc. –The use of quadratic equations will NOT be assessed.

2. In order to calculate the value of K, we need to know the concentration in mol/L or mol dm -3 of all reactants and products at equilibrium. Sometimes we do not immediately know all of the concentrations at equilibrium, but we are given enough information to determine all of the concentrations at equilibrium and then K.

3. For Example 0.100 mol/L Br 2 and 0.100 mol/L F 2 were initially mixed together and allowed to come to equilibrium according to the following reaction: Br 2 (g) + F 2 (g) 2BrF(g) At equilibrium the concentration of Br 2 was found to be 0.00135 mol/L. Calculate the value of K.

4. Fill this in… Br 2 (g)F 2 (g)2BrF(g) Initial Conc.0.100M 0 Change in conc. Final Conc.0.00135M

5. Should Look like: Br 2 (g)F 2 (g)2BrF(g) Initial Conc.0.100M 0 Change in conc. -0.09865M +[2 x 0.09865M] =0.1973M Final Conc.0.00135M 0.1973M

6. Good ol’ Stoichiometry We used stoichiometry (the relationship with the balanced chemical equation to help us solve for the final concentrations) Since Br 2 and F 2 were a 1:1 mole ratio and had the same initial concentration, they would have the same final. But BrF was a 1:2 mole ratio with Br 2, so we had to double the change in concentration and therefore the final answer.

7. We’re not finished! Now we can solve for K because we found the final concentrations (at equilibrium) K=[BrF] 2 / [Br 2 ][F 2 ] =[0.1973M] 2 / [0.00135M][0.00135M] =2.15 x 10 4

8. Sample Problem 2 : A closed container initially had a CO(g) concentration of 0.750 M and a H 2 O(g) concentration of 0.275 M. It was allowed to reach equilibrium. At equilibrium, analysis showed a CO 2 (g) concentration of 0.250 M. The equation for the reaction is: CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) Calculate K Answer: K=5.00

9. SAMPLE PROBLEM 3 A closed chemical system initially contained 6.0 M SO 2 ; 2.5 M NO 2 ; and 1.0 M SO 3. Equilibrium was eventually reached for the reaction SO 2 (g) + NO 2 (g) SO 3 (g) + NO(g) At equilibrium, the container was found to have 3.0 M SO 3 present. Calculate K. Answer: K=3.00

10. A chemist puts 0.085 mol of N 2 and 0.038 mol of O 2 in a 1.0 L container, once equilibrium is reached its equilibrium constant is found to be 4.2 x10 -8. What is the concentration of NO in the mixture at equilibrium? N 2 (g) + O 2 (g) 2NO(g)

11. answer N 2 (g)O 2 (g) 2NO(g) 0.085 M0.038 M0 -x +2x 0.085-x0.038-x2x

12. Assumptions Normally, you would solve this using the quadratic equation, but IB has noted that their students will NOT need to use the quadratic equation. We’re going to do a short cut that we’ll check to make sure is okay later on. Instead of using 0.085-x, we’re going to ignore the x, same for 0.038-x (ignore the x) and solve for unknown

13. N 2 (g)O 2 (g) 2NO(g) 0.085 M0.038 M0 -x (ignore) +2x 0.0850.0382x

14. k = [2x] 2 =4.2 x10 -8 [0.085][0.038] 4x 2 = (4.2x10 -8 ) x [0.085] x [0.038] 4x 2 = 1.357 x 10 -10 x 2 = 3.392 x 10 -11 x = 5.82 x 10 -6 [NO] = 2x = 2 x 5.82 x10 -6 = 1.2 x10 -5 M

15. To show that the short cut was okay, check out your answer if the x wasn’t omitted. Ex: 0.085- 5.82 x 10 -6 = 0.085 (when you only keep 2 sig figs) So it was okay to omit the x. In university you won’t be so lucky!!!

16. Additional practice:

17. Using quadratic: