Chemical Equilibrium Worksheet Answer Key

WORKSHEET: CHEMICAL EQUILIBRIUM Name: Last, First FOR ALL EQUILIBRIUM PROBLEMS, YOU MUST: 1) Write all equilibrium equations 2) Write all equilibrium concentrations 3) Write all equilibrium expressions SET A:   a) What is the equilibrium constant expression for the reaction:   4 Fe(s) + 3 O₂(g) ⇌ 2 Fe₂O₃(s) Ans: [Fe₂O₃]² / [O₂]³ b) The equilibrium constant, Kc, for the reaction: 2 NO(g) ⇌ N₂(g) + O₂(g) is 2.4 x 10⁻³. What is the equilibrium constant, Kc', for the reaction: N₂(g) + O₂(g) ⇌ 2 NO(g) Ans: 1 / (2.4 x 10⁻³) = 4.17 x 10² c) Given the following equilibrium equations and their corresponding equilibrium constants: 2 CO(g) + O₂(g) ⇌ 2 CO₂(g) Kc = 6.1 x 10¹⁴ H₂(g) + CO₂(g) ⇌ H₂O(g) + CO(g) Kc = 1.2 x 10¹⁰ Find Kc for the reaction: 2 H₂(g) + O₂(g) ⇌ 2 H₂O(g) Setup: Kc = (Kc₁)⁻¹(Kc₂)⁻² Kc = (6.1 x 10¹⁴)⁻¹(1.2 x 10¹⁰)⁻² Kc = 7.3 x 10²² d) A mixture of 9.22 moles of A, 10.11 moles of B, and 27.83 moles of C is placed in a one-liter container at a certain temperature. The reaction is allowed to reach equilibrium. At equilibrium, the number of moles   of B is 18.32. Calculate the equilibrium constant for the reaction: A(g)   + B(g) ⇌ 2 C(g) Setup: A + B ⇌ 2C 9.22 10.11 27.83 -x -x +2x E(9.22-x) (10.11-x) (27.83+2x) Kc calc: Kc = [C]² / [A][B] = (27.83 + 2x)² / (9.22 - x)(10.11 - x) = (27.83 + 3(4.10))² / (9.22 + 4.10)(18.32)² = 0.832 3. At a certain temperature, Kc = 4.13 x 10^-2 for the equilibrium: 2 IBr(g) ⇌ I₂(g) + Br₂(g) Assume that equilibrium is established at the above temperature by adding only IBr(g) to the reaction flask. What are the concentrations   of I₂(g) and Br₂(g) in equilibrium when 0.0124 moles of IBr(g) remain? Setup: 2IBr(g) ⇌ I₂(g) + Br₂(g) I 0.0124 C -2x +x +x E (0.0124-2x) x x Kc calc: Kc = [I₂][Br₂] / [IBr]^2 4.13 x 10^-2 = x^2 / (0.0124 - 2x)^2 4.13 x 10^-2 = x^2 / (0.0124 - 2x)^2 x = 2.52 x 10^-3 M Ans: 2.52 x 10^-3 M b. What was the initial concentration of IBr before equilibrium was established? Setup: Initial = 2x = 2(2.52 x 10^-3) = 0.0174 M Ans: 0.0174 M 4. 0.024 mole of A(g) is placed in a 1.00 liter container at 700°C, where it is 38.8% dissociated when equilibrium is established.   A(g) ⇌ 2 B(g) What is the value of the equilibrium constant, Kc, at the same temperature? Setup: A(g) ⇌ 2 B(g) I 0.024 0 C -x +2x E 0.024-x 2x Kc calc: Kc = [B]^2 / [A] = (2x)^2 / (0.024 - x) = (2(0.0120))^2 / (0.024 - 0.0120) = 0.0241 Ans: 0.0241 5. The equilibrium constant for the reaction 2 NO(g) ⇌ N₂(g) + O₂(g) is 2.60 x 10⁻³ at 1100°C. If 0.820 mole of NO(g) and 0.223 mole each of N₂(g) and O₂(g) are mixed in a 1.00 liter container at 1100°C, what are the concentrations of NO(g), N₂(g), and O₂(g) at equilibrium? Setup: 2 NO(g) ⇌ N₂(g) + O₂(g) I 0.820 0.223 0.223 ← only initial! C -2x +x +x ← I need Qc! E (0.820-2x) (0.223+x) (0.223+x) Qc calc: Qc = [N₂]ᵢₙᵢₜ[O₂]ᵢₙᵢₜ / [NO]ᵢₙᵢₜ² = (0.223)(0.223) / (0.820)² = 7.40 x 10⁻³ Qc > Kc 7.40 x 10⁻³ > 2.60 x 10⁻³ ∴ net rxn to the left Kc calc: Kc = [N₂][O₂] / [NO]^2 2.60 x 10⁻³ = (0.223 + x)(0.223 + x) / (0.820 - 2x)^2 x = 0.164 M Ans: [NO] = 0.820 - 2(0.164) = 0.5 M [N₂] = [O₂] = 0.223 + 0.164 = 0.387 M SET B: 1. A mixture of 1.16 mole of A, 1.38 mole of B, and 0.641 mole of C is placed in a one-liter container at a certain temperature. The reaction was allowed to reach equilibrium. At equilibrium,   the number of moles of B is 1.95. Calculate the equilibrium constant, Kc, for the reaction: A + B ⇌ 2C Setup: A + B ⇌ 2C I 1.16 1.38 0.641 C -x -x +2x E (1.16-x) (1.38-x) (0.641+2x) Kc calc: Kc = [C]^2 / [A][B] = (0.641+2x)^2 / (1.16-x)(1.38-x) = (0.641+2(0.40))^2 / (1.16-0.40)(1.38-0.40) = 0.020 Ans: 0.020 2. 0.822 mole of SO3(g) is placed in a 1.00 liter container at 600 K. 36.7% of the SO3(g) are decomposed when equilibrium is established. 2 SO3(g)   ⇌ 2 SO2(g) + O2(g) What is the value of the equilibrium constant, Kc, at the same temperature? Setup: 2 SO3(g) ⇌ 2 SO2(g) + O2(g) I 0.822 0 0 C -2x +2x +x E 0.540 0.302 0.151 Kc calc: Kc = [SO2]^2[O2] / [SO3]^2 = (0.302)^2(0.151) / (0.540)^2 = 5.05 x 10^-2 Ans: 5.05 x 10^-2