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A mixture of 3 mol Cl2 and 3 mol of co is enclosed in a 5l flask at 600 degrees c. When the reaction reaches equilibrium, only 3.30% of the cl2 has been consumed. Calculate the kc for the reaction with detailed steps?
Accepted Answer
Step 1: Write the balanced chemical equation
Cl2(g) + CO(g) 2COCl2(g)
Step 2: Calculate the initial concentrations of the reactants and products
[Cl2]0 = 3 mol / 5 L = 0.60 M
[CO]0 = 3 mol / 5 L = 0.60 M
[COCl2]0 = 0 M
Step 3: Calculate the change in concentration of Cl2
At equilibrium, 3.30% of the Cl2 has been consumed, so the change in concentration is:
Δ[Cl2] = -0.0330 * 0.60 M = -0.0198 M
Step 4: Calculate the equilibrium concentrations of the reactants and products
[Cl2]eq = 0.60 M - 0.0198 M = 0.5802 M
[CO]eq = 0.60 M
[COCl2]eq = 2 * 0.0198 M = 0.0396 M
Step 5: Calculate the equilibrium constant
Kc = [COCl2]eq / [Cl2]eq[CO]eq
Kc = 0.0396 M / (0.5802 M)^2 * (0.60 M)
Kc = 0.0245 M^-1
Cl2(g) + CO(g) 2COCl2(g)
Step 2: Calculate the initial concentrations of the reactants and products
[Cl2]0 = 3 mol / 5 L = 0.60 M
[CO]0 = 3 mol / 5 L = 0.60 M
[COCl2]0 = 0 M
Step 3: Calculate the change in concentration of Cl2
At equilibrium, 3.30% of the Cl2 has been consumed, so the change in concentration is:
Δ[Cl2] = -0.0330 * 0.60 M = -0.0198 M
Step 4: Calculate the equilibrium concentrations of the reactants and products
[Cl2]eq = 0.60 M - 0.0198 M = 0.5802 M
[CO]eq = 0.60 M
[COCl2]eq = 2 * 0.0198 M = 0.0396 M
Step 5: Calculate the equilibrium constant
Kc = [COCl2]eq / [Cl2]eq[CO]eq
Kc = 0.0396 M / (0.5802 M)^2 * (0.60 M)
Kc = 0.0245 M^-1
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