Consider the following reaction:

Br2 (g) + Cl2 (g) ⇌ 2BrCl (g), Kp=1.112 at 150 K.

A reaction mixture initially contains a Br2 partial pressure of 751 torr and a Cl2 partial pressure of 737 torr at 150 K.

Calculate the equilibrium partial pressure of BrCl.

the equilibrium partial pressure of BrCl is pBC = 784.52 torr

Explanation:

Since

Br₂ (g) + Cl₂ (g) ⇌ 2BrCl (g), Kp=1.112 at 150 K

denoting BC as BrCl, B as Br₂, C as Cl₂, p as partial pressure, then

Kp = pBC²/[pB*pC]

solving for pBC

pBC = √ (Kp*pB*pC)

replacing values

pBC = √ (Kp*pB*pC) = √ (1.112*751 torr*737 torr) = 784.52 torr

pBC = 784.52 torr

then the equilibrium partial pressure of BrCl is pBC = 784.52 torr