1. Heterogeneous Equilibria: A homogenous reaction is one in which all the substances are in the same state. A heterogeneous reaction is one in which all the substances are not in the same state. CaCO 3 (s) CO 2 (g) + CaO (s) Calcium carbonatecarbon dioxidelime When writing equilibrium constant expressions for Heterogeneous equilibria, you don’t include pure solids or pure liquids. Their concentrations don’t change K eq = [CO 2 ][ CaO ] K eq = [CO 2 ] [ CaCO 3 ] Try writing an K eq expression for this reaction: 2H 2 O (l) 2H 2 (g) +O 2(g)

2. K eq expression for this reaction: 2H 2 O (l) 2H 2 (g) +O 2(g) K = [H 2 ] 2 [O 2 ] K eq expression for this reaction: CuSO 4. 5H 2 O (s)  CuSO 4 (s) + 5H 2 O (g) K = [H 2 O] 5

3. This states that when a change is imposed on a system at equilibrium, the position of the equilibrium shifts in a direction that tends to reduce the effect of that change. CONCENTRATION: When a reactant or product is added to a system at equilibrium, the system shifts away from the added component. (moves in a direction that uses up the excess component) But if a reactant or product is removed, the system shifts toward the removed component.

4. N 2 (g) +3H 2 (g)  2NH 3(g) Equilibrium concentrations [N 2 ] = 0.399 M[H 2 ] = 1.197 M [NH 3 ] = 0.202 M What will happen if 1.000 mol L -1 N 2 is added to equilibrium? New Equilibrium concentrations [N 2 ] = 1.348 M[H 2 ] = 1.044 M [NH 3 ] = 0.304 M Position I: Position II:

5. VOLUME: When the volume of a gaseous system at equilibrium is decreased, the system shifts in the direction that gives the smaller number of gas molecules CaCO 3 (s)  CaO (s) + CO 2(g) decreasing the volume of the following reaction will: shifts the equilibrium to the left

6. N 2 (g) +3H 2 (g)  2NH 3(g) PCl 3 (g) + 3NH 3(g)  P(NH 2 ) 3 (g) + 3HCl (g) shifts the equilibrium to the right has no effect on equilibrium position Take the equilibrium between 2NO 2 (g)  N 2 O 4 (g) brown gasclear gas Which direction is favored by increasing the volume? The right!

7. TEMPERATURE: When heat is added to a system at equilibrium, the system shifts in a direction that uses up the excess heat, the exothermic reaction. A reaction that absorbs heat is endothermic, a reaction that produces heat is exothermic. N 2 (g) +3H 2 (g)  2NH 3(g) + 92 kJ CaCO 3 (s) + 556 kJ  CaO (s) + CO 2(g)

8. N 2 O 4 (g) + energy  2NO 2(g) Change: Addition of N 2 O 4 (g) Addition of NO 2(g) Removal of N 2 O 4 (g) Removal of NO 2(g) Decrease in container volume Increase in container volume Increase in temperature Decrease in temperature REVIEW EXERCISE: Shift: Right Left Right Left Right Left

9. Using the Equilibrium Constant: What can the size of K eq tell you? If the value of K eq is one then the equilibrium concentrations of A and B are the same. Take the equilibrium A (g)  B (g) If the value of K eq is less than one, then the reaction at equilibrium consists mainly of reactants - the equilibrium position is far to the left. If the value of K eq is much larger than one, the reaction system mainly consists of products - the equilibrium position is far to the right. A (g)  B (g) A (g) B (g)

10. You can also use the equilibrium constant to find the concentrations of reactants and products. For example if you know the value of K and the concentration of all the reactants and products except one, we can calculate the missing concentration. Gaseous phosphorus pentachloride decomposes to chlorine gas and gaseous phosphorus trichloride. In a certain experiment, at a temperature where K = 8.96  10 -2, the equilibrium concentrations of PCl 5 is 6.70  10 -3 and PCl 3 is 0.300 M. Calculate the concentration of Cl 2 present at equilibrium. [Cl 2 ] = 2.00  10 -3 mol L -1

11. Often there is an equilibrium between a dissolving solid and its aqueous solution. CaF 2 (s)  Ca 2+ (aq) + 2F - (aq) K sp = [Ca 2+ ] [F - ] 2 Where K sp is the solubility product constant, or simply the solubility product.

12. Calculating the solubility product: CuBr has a measured solubility of 2.0  10 -4 at 25  C. Calculate the solids K sp value CuBr (s)  Cu + (aq) + Br - (aq) K sp = [Cu + ] [Br - ] We know that 2.0  10 -4 mol of solid CuBr dissolves per 1.0L of solution to come to equilibrium. CuBr (s)  Cu + (aq) + Br - (aq) K sp = [Cu + ] [Br - ] = (2.0  10 -4 )(2.0  10 -4 ) = 4.0  10 -8