Question

The equilibrium constant, Kp, for the following reaction is 0.636 at 600K.

Answer 1

Answer: The equilibrium partial pressure of $Cl_{2}$ is 0.964 atm.

Explanation:

Given: $K_{p} = 0.636$

$P_{COCl_{2}} = 0.836 atm$

$P_{CO} = 0.551 atm$

The given reaction equation is as follows.

$COCl_{2}(g) \rightleftharpoons CO(g) + Cl_{2}(g)$

Formula used to calculate the partial pressure of $Cl_{2}$ is as follows.

$K_{p} = \frac{P_{CO} \times P_{Cl_{2}}}{P_{COCl_{2}}}$

Substitute the values into above formula as follows.

$K_{p} = \frac{P_{CO} \times P_{Cl_{2}}}{P_{COCl_{2}}}\\0.636 = \frac{0.551 atm \times P_{Cl_{2}}}{0.836 atm}\\P_{Cl_{2}} = 0.964 atm$

Thus, we can conclude that the equilibrium partial pressure of $Cl_{2}$ is 0.964 atm.