Equilibrium Constant Kc or Keq The equilibrium constant

Equilibrium Constant (Kc or Keq): The equilibrium constant (Kc or Keq) is the ratio of product concentrations to reactant concentrations at equilibrium. Each concentration is raised to the power equal to the number of moles of that substance in the balanced chemical equation. a. A + b. B Kc= c. C + d. D

Examples: 1) A liter of the gas mixture at equilibrium contains 0. 0045 mol of N 2 O 4 and 0. 030 mol of NO 2 at 10 o. C. Calculate the value for the constant (Kc) for the reaction: N 2 O 4(g) 2 NO 2(g) Kc= = = Kc= 0. 20 mol/L = 0. 20 mol. L-1

Examples: 2) One mole of H 2(g) and one mole of I 2(g) are sealed in a 1 -L flask and allowed to react at 45 o. C. At equilibrium, 1. 56 mol of HI(g) is present. Calculate the value of Kc. x mol H 2(g) + I 2(g) 1. 56 mol 2 HI(g) 1. 00 mol 2. 00 mol X = = 0. 78 mol I 2 (reacted) = 0. 78 mol H 2 At equilibrium: 1. 00 -0. 78 = 0. 22 mol → = Kc= = [I 2]= 0. 22 mol/L [HI]= 1. 56 mol/L Kc= 50

3) 12. 0 mol of N 2 O(g) is placed in a 5. 0 L reaction vessel and the following equilibrium is established. 2 N 2 O(g) 2 N 2(g) + O 2(g) At equilibrium, 4. 6 mol of N 2 O(g) is present. What is the value of the equilibrium constant? Answer: Kc = 12. 0 mol - 4. 6 mol = 7. 40 mol N 2 O (Reacted) At equilibrium: N 2 O = 4. 6 mol Concentrations: [N 2 O] = 4. 6 mol / 5. 0 L = 0. 92 mol / L N 2 = 7. 40 mol [N 2] = 7. 40 mol / 5. 0 L = 1. 48 mol / L O 2 = 7. 40 / 2 = 3. 7 mol = = [O 2] = 3. 7 mol / 5. 0 L = 0. 74 mol / L Kc= 1. 91 mol/L = 1. 91 mol. L-1

Examples: 4) The equilibrium constant for the reaction is 11. 1. A sample of pure Br. Cl is placed in 1 -L container and allowed to decompose. At equilibrium the mixture contains 4. 00 mol Cl 2. What are the equilibrium concentrations of Br 2 & Br. Cl”? 2 Br. Cl(g) Br 2(g) + Cl 2(g) [Cl 2]= 4. 00 mol/L = [Br 2] [Br. Cl]= ? Kc= 11. 1 Kc= [Br. Cl]2 = 1. 44 mol 2/L 2 = [Br. Cl] = 1. 20 mol/L