Question

COCl2(g) decomposes according to the equation above. When pure COCl2(g) is injected into a rigid, previously evacuated flask at 690 K, the pressure in the flask is initially 1.0 atm. After the reaction reaches equilibrium at 690 K, the total pressure in the flask is 1.2 atm. What is the value of Kp for the reaction at 690 K?

Answer 1

Answer: The value of $K_p$ for the reaction at 690 K is 0.05

Explanation:

We are given:

Initial pressure of $COCl_2$ = 1.0 atm

Total pressure at equilibrium = 1.2 atm

The chemical equation for the decomposition of phosgene follows:

                  $COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)$

Initial:            1                    -         -

At eqllm:       1-x                 x        x

We are given:

Total pressure at equilibrium = [(1 - x) + x+ x]

So, the equation becomes:

$[(1 - x) + x+ x]=1.2\\\\x=0.2atm$

The expression for $K_p$ for above equation follows:

$K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}$

$p_{CO}=0.2atm\\p_{Cl_2}=0.2atm\\p_{COCl_2}=(1-0.2)=0.8atm$

Putting values in above equation, we get:

$K_p=\frac{0.2\times 0.2}{0.8}\\\\K_p=0.05$

Hence, the value of $K_p$ for the reaction at 690 K is 0.05