University:
College
Course:
College Chemistry
Academic year:

2022
Views:

470

Pages:

8
Author:

5CelestialChronicle

WORKSHEET: CHEMICAL EQUILIBRIUM Name: Key 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: 1) a) What is the equilibrium constant expression for the reaction: 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g) Ans: [H₂]⁴ / [H₂O]⁴ b) The equilibrium constant, Kc, for the reaction: 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) K = √(1 / 2.4 x 10⁻⁷) K = 1.6 x 10² What is the equilibrium constant, Ke, for the reaction: 1/3 Cl₂(g) + 2/3 NO(g) ⇌ 2/3 NOCl(g) Ans: 1.6 x 10² c) Given the following equilibrium equations and their corresponding equilibrium constants: i) CO₂(g) + H₂(g) ⇌ CO(g) + H₂O(g) Kc = 6.1 x 10⁸ ii) CH₄(g) + 2O₂(g) ⇌ CO₂(g) + 2H₂O(g) Kc = 1.2 x 10¹⁴ Find Kc for the reaction: CH₄(g) + CO₂(g) ⇌ 2CO(g) + 2H₂O(g) Setup: Add equations (i) and (ii). K = (6.1 x 10⁸)(1.2 x 10¹⁴) K = 7.3 x 10²² Ans: 7.3 x 10²² 2) 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) + 2B(g) ⇌ 3C(g) Initial conc. 9.22 10.11 27.83 Change in conc. +4.10 +8.20 -12.31 Equilibrium conc. 13.32 18.32 15.51 8.21 moles B reacted (3 moles C / 2 moles B) = 12.3 moles C "reacted" Kc = [C]³ / [A][B]² = (15.51)³ / (13.32)(18.32)² = 0.832 Ans: 0.832 3) a) 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 with 0.0124 moles/liter of IBr(g)? Setup: | 2 IBr(g) | I₂(g) | Br₂(g) | |---|---|---|---| Initial conc. | 0.0124 | 0 | 0 | Change in conc. | -2x | +x | +x | Equilibrium conc. | 0.0124-2x | x | x | Kc = [I₂][Br₂] / [IBr]^2 4.13 x 10^-2 = (x)(x) / (0.0124 - 2x)^2 x = 2.52 x 10^-3 Ans: 2.52 x 10^-3 M b) What was the initial concentration of IBr before equilibrium was established? Setup: M_Initial - 2x = 0.0124 M_Initial = 0.0124 + 2x = 0.0124 + 2(2.52 x 10^-3) Ans: 0.0174 4) 0.924 mole of A(g) is placed in a 1.00 liter container at 700°C, where it is 38.8% dissociated when equilibrium was established. 3 A(g) ⇌ 5 B(g) + 2 C(g) What is the value of the equilibrium constant, Kc, at the same temperature? Setup: | 3 A(g) | 5 B(g) | 2 C(g) | |---|---|---|---| Initial conc. | 0.924 | 0 | 0 | Change in conc. | -0.359 | +0.598 | +0.239 | Equilibrium conc. | 0.566 | 0.598 | 0.239 | 0.924 mol A (3 mol B / 3 mol A) = 0.924 mol B formed 0.924 mol B (2 mol C / 5 mol B) = 0.359 mol C formed Kc = [B]^5[C]^2 / [A]^3 = (0.598)^5(0.239)^2 / (0.566)^3 = 0.0241 Ans: 0.0241 Джерела й пов’язаний контент