1. EQUILIBRIUM Chapter 13 and 14

2. Homework Chapter 13 pg 628 1-71 odd Chapter 14 pg 688 11-87 odd 2

3. AP Test Historically, the first question on the free response has ALWAYS been an equilibrium question. The new style of test no longer requires that, however, I would be surprised if they didn’t have one on the test. 3

4. Equilibrium A state when two competing reactions are canceling each other out. H 3 O + + OH - ⇌ H 2 O + H 2 O You reach equilibrium when the rate of forward reaction is equal to the rate of backwards reaction. This does NOT mean the concentration of products and reactants are equal. The above reaction is at equilibrium when [H 3 O + ] = [OH - ] = 1x10 -7 mol/L. In 1 L of water there are 55 moles.

5. Macro Micro Macroscopically, anything that is in equilibrium is stable. Meaning the concentrations of the different substances are not changing. Law of mass action- when the ratio of product to reactant is a constant value the system is at equilibrium Microscopically, atoms and molecules are constantly reacting in both reactions http://chemconnections.org/Java/equilibrium/index.html

6. Changes in Concentration N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g)

7. Equilibrium expression and constant K is calculated form EQUILIBRIUM CONCENTRATIONS! Its values may only be calculated experimentally. aA + bB ⇌ cC + dD equilibrium constant = equilibrium expression K = [C] c [D] d [A] a [B] b [ ] means concentration in M

8. Determine the equilibrium expression For the following: Br 2 (g) ⇌ 2Br(g) N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g) H 2 (g) + Br 2 (g) ⇌ 2 HBr(g) HCN(aq) ⇌ H + (aq) + CN - (aq)

9. K values, K is always positive Intermediate K. 0.1<K<10 Significant concentrations of all substances are present. Very Large K. K >> 1 The product concentration is very large with virtually no reactant concentration (the reaction has gone to completion). Very Small K. K<<1 The reactant concentration is very large with virtually no product concentration.

10. Types of K K c = equilibrium constant in terms of concentration. K p = equilibrium concentration in terms of pressure. K a = acid dissociation constant K b = base dissociation constant K w = ion-product constant for water These are listed on your equation sheet

11. Actual equation sheet 11

12. K problem The following equilibrium concentrations were observed for the Haber process at 127 o C. [NH 3 ] = 3.1 x 10 -2 M, [N 2 ] = 8.5 x 10 -1 M, [H 2 ] = 3.1 x10 -3 M Forward Reaction. N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g) Reverse Reaction. 2NH 3 (g) ⇌ N 2 (g) + 3H 2 (g) Multiply by Factor n. 1/2N 2 (g) + 3/2H 2 (g) ⇌ NH 3 (g) K is unitless

13. Summary For forward reaction jA + kB ⇌ lC + mD, K = [C] l [D] m [A] j [B] k For reverse reaction jA + kB ⇌ lC + mD, K’ = K -1 = [A] j [B] k [C] l [D] m For reaction njA + nkB ⇌ nlC + nmD K’’ = K n = [C] nl [D] nm [A] nj [B] nk For an overall reaction of two or more steps, K overall = K 1 x K 2 x K 3 x...

14. Equilibrium of Gases. PV = nRT or P = (n/V)RT n/V is mol/L, or a concentration. Concentration of a gas is C = P/RT Therefore you can determine the K of a gas from its pressure and temperature

15. Problem 2NO(g) + Cl 2 (g) ⇌ 2NOCl(g) The reaction for the formation of nitrosyl chloride was studied at 25 o C. The pressures at equilibrium were found to be P NOCl = 1.2 atm P NO = 5.0 x10 -2 atm P Cl2 = 3.0 x10 -1 atm Calculate the value of K for the reaction at 25 o C. PV = nRT R =.0821 atm L / mol K

16. KpKp Since we are always dividing by the pressure value by RT, there has to be a way to cancel that out. K p is a value that looks at the pressure of the gases involved instead the concentrations To convert K p to K K may be written K c, or K with respect to conc. K p = K(RT)  n Where  n =  coefficients of products -  coefficients of reactants

17. Same Problem 2NO(g) + Cl 2 (g) ⇌ 2NOCl(g) The reaction for the formation of nitrosyl chloride was studied at 25 o C. The pressures at equilibrium were found to be P NOCl = 1.2 atm P NO = 5.0 x10 -2 atm P Cl2 = 3.0 x10 -1 atm Calculate the value of K p for the reaction at 25 o C, and then convert that K p to K.

18. Example N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g)

19. Pressure Problem Dinitrogen tetroxide in its liquid state was used as one of the fuels on the lunar lander for the NASA Apollo missions. In the gas phase it decomposes to gaseous nitrogen dioxide: N 2 O 4 (g) ⇌ 2NO 2 (g) Consider an experiment in which gaseous N 2 O 4 was placed in a flask and allowed to reach equilibrium at a temperature where K p = 0.133. At equilibrium, the pressure of N 2 O 4 was found to be 2.71 atm. Calculate the equilibrium pressure of NO 2 (g). 19

20. Homogeneous Equilibria Homogeneous equilibria – involve the same phase: N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g) HCN(aq) ⇌ H + (aq) + CN - (aq)

21. Heterogeneous Equilibria Heterogeneous equilibria – involve more than one phase: 2KClO 3 (s) ⇌ 2KCl(s) + 3O 2 (g) 2H 2 O(l) ⇌ 2H 2 (g) + O 2 (g)

22. The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.  The concentrations of pure liquids and solids are constant. 2KClO 3 (s) ⇌ 2KCl(s) + 3O 2 (g)

23. Write expressions for K for the following: The decomposition of solid phosphorus pentachloride to liquid phosphorus trichloride and chlorine gas. Deep blue solid copper(II) sulfate pentahydrate is heated to drive off water vapor to form solid green copper(II) sulfate.