A student ran the following reaction in the laboratory at 600 K: COCl2 (g) CO (g) + Cl2 (g) When he introduced COCl2 (g) at a pressure of 0.822 atm into a 1.00 L evacuated container, he found the equilibrium partial pressure of COCl2 (g) to be 0.351 atm. Calculate the equilibrium constant, Kp, he obtained for this reaction.

the Kp=0.171 atm

Explanation:

For the reaction

COCl₂ (g) → CO (g) + Cl₂ (g)

since the temperature remains constant, the Kp will remain constant.

Kp = pCO*pCl₂ / pCOCl₂

since for every mole that is converted, a mole of CO (g) and a mole of Cl₂ (g) are generated, then

denoting 1 as COCl₂, 2 as CO, 3 as Cl₂, 0 as initial state and X as conversion, then the moles of each component will be given by

n₁ = n₀ * (1-X), n₂ = n₀*X, n₃ = n₀*X

the final number of total moles will be

nf = n₁ + n₂ + n₃ = n₀ * (1-X) + n₀*X + n₀*X = n₀ + n₀*X = n₀ (1+X)

then assuming ideal gas behaviour and that the container is rigid (V=constant)

P₀*V = n₀*R*T and Pf*V = nf*R*T

dividing both equations

Pf / P₀ = n₀ * (1+X) / n₀ = (1+X)

X = 1 - Pf / P₀

replacing values

X = 1 - Pf / P₀ = 1 - 0.351 atm/0.822 atm = 0.573

thus using Dalton's law

p₁ = Pf*x₁ = Pf*n₀ * (1-X) / [n₀ (1+X) } = Pf * (1-X) / (1+X)

p₂ = Pf*x₂ = Pf*n₀*X/[n₀ (1+X) } = Pf*X / (1+X)

p₃ = Pf*x₃ = Pf*n₀*X/[n₀ (1+X) } = Pf*X / (1+X)

then the equilibrium constant will be

Kp = p₂*p₃/p₁ = Pf*X²/[ (1-X) * (1+X) ] = Pf*X² / (1 - X²)

replacing values

Kp = Pf*X² / (1 - X²) = 0.351 atm * 0.573² / (1-0.573²) = 0.171 atm