3. Solids are not written in expressions. The concentration of a solid is unchanged and doesn’t affect the rate. Water is not written in expressions. The concentration is much higher for water, so it essentially stays the same.

4. Write equilibrium constant expressions for the following: N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g) H 2 CO 3 (aq) + H 2 O( l ) ⇌ HCO 3 - (aq) + H 3 O + (aq)

5. There are various equilibrium constants, but the two we will look at now are about concentration and pressure. K c should be used if given the concentrations of substances. If you get pressures, you will use K p instead. They are not the same.

6. What values would you expect for K if the reaction is product favored? What if the reaction is reactant favored?

7. Page 729 Exercise 16.1 Page 752 2

9. Nitrogen dioxide can exist in equilibrium with N2O4. 2 NO 2 (g) ⇌ N 2 O 4 (g) K=170 If the concentration of NO 2 is 0.015 M and the concentration of N 2 O 4 is 0.025 M. Is Q larger than, smaller than, or equal to K? If not at equilibrium, what direction does the reaction need to proceed?

10. We will often get initial concentration when we need equilibrium concentrations. The E values go into the equilibrium expression. EquationH 2 (g) + I 2 (g) ⇌ 2 HI(g) I = initial0.0175 0 C = change E = equilibrium0.0037

11. An aqueous solution of ethanol and acetic acid, each at 0.810 M initially, is heated to 100 °C. At equilibrium, the acetic acid concentration is 0.748 M. Calculate K. C 2 H 5 OH(aq) + CH 3 CO 2 H(aq) ⇌ CH 3 CO 2 C 2 H 5 (aq) + H 2 O( l )

12. Page 734 Exercise 16.3 and 16.4 Page 736 Exercise 16.5 Page 752 6, 10, 12

13. K c = 55.64 for the equation H 2 (g) + I 2 (g) ⇌ 2 HI (g). If 1.00 mol of each reactant are placed in a 0.500 L flask. What are the concentrations for everything at equilibrium?

14. At a different temperature, K c = 33 for the reaction H 2 (g) + I 2 (g) ⇌ 2 HI (g). If the initial concentrations are 6.00 X 10 -3 M, find the equilibrium concentrations for all substances.

15. Some equations will give an expression that is quadratic. The quadratic formula can be used to solve. If the K is small, more than 100 times smaller than the initial concentration, then the x on the bottom can be cancelled out. If K is small, then the equilibrium is strongly reactant favored, so x will be infinitesimally small.

16. N 2 (g) + O 2 (g) ⇌ 2 NO(g). At 1500 K, K = 1 X 10 -5. If a sample is 0.80 M for nitrogen and 0.20 M for oxygen, what are the equilibrium concentrations of reactants and products?

17. Page 741 Exercise 16.7 Page 753 14, 15, 16

18. C(s) + ½ O 2 (g) ⇌ CO(g) For this equation, K = 4.6 X 10 23 at 25 °C. What would K be if we instead balanced the equation as 2C(s) + O 2 (g) ⇌ 2CO(g)?

19. HCO 2 H(aq) + H 2 O( l ) ⇌ HCO 2 - (aq) + H 3 O + (aq) If the K for the forward reaction is 1.8 X 10 -4, what is the K for the reverse reaction?

20. N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g) K = 3.5 X 10 8 ½ N 2 (g) + 3/2 H 2 (g) ⇌ NH 3 (g) K = 2 NH 3 (g) ⇌ N 2 (g) + 3 H 2 (g) K =

21. When two equations are added, their Ks are multiplied to get the K for the combined equation. AgCl(s) ⇌ Ag + (aq) + Cl - (aq) K 1 = 1.8 X 10 -10 Ag + (aq) + 2 NH 3 (aq) ⇌ [Ag(NH 3 ) 2 ] + (aq) K 2 = 1.6 X 10 7 AgCl(s) + 2 NH 3 (aq) ⇌ [Ag(NH 3 ) 2 ] + (aq) + Cl - (aq) K = K 1 X K 2 = 1.8 X 10 -10 X 1.6 X 10 7 = 2.9 X 10 -3

22. Page 743 Exercises 16.8 and 16.9 Page 754 20, 22, 24

23. When a factor like temperature, concentration, or volume is changed, the equilibrium conditions will shift to counteract the change. The concentrations of all reactants and products will change to make the reaction quotient equal to K again.

24. If you increase the concentration of a reactant, the equilibrium will shift and form more of the products. If you add more of a product, it will shift and form more reactants. When the shift occurs, the concentration of the substance you added more of will decrease to restore the equilibrium.

25. At equilibrium, the concentration of butane is 0.500 M and the concentration of isobutene is 1.25 M. If you add 1.50 mol butane, what are the new equilibrium concentrations?

26. Changing the volume may or may not affect the equilibrium. We can determine if it will or not based on the equation. 2 NO 2 (g) ⇌ N 2 O 4 (g) The above equation will shift to the right if we decrease the volume. It will favor the side with less gas molecules. It will shift to the left if we increase the volume. It will favor the side with more gas molecules. If the number of molecules are the same on each side, the equilibrium won’t change.

27. To determine which way temperature shifts an equilibrium, we need to know if the reaction is endothermic or exothermic. If the reaction is endothermic, it will shift toward the products when temperature increases. It will shift towards the reactants if it is exothermic.

28. Changing the temperature changes the K. When temperature increases, K increases for an endothermic reaction since more products are formed. When temperature increases, K decreases for an exothermic reaction since more reactants are formed.

29. Page 746 Exercise 16.10 Page 747 Exercise 16.11 Page 750 Exercise 16.12 Page 754 26, 28, 30, 32, 36, 64, 68