Question

Starting with pure ${COCl}_2$, if the initial concentration of $4{COCl}_2$ is $0.03M$ and the equilibrium concentration of ${Cl}_2$ is $0.02M$,the equilibrium constant $K_c$ for the reaction-
${COCl}_2\left(g\right)\rightleftharpoons CO\left(g\right)+{Cl}_2\left(g\right)$ would be:

• A. ${4\times 10}^{-4}$
• B. ${4\times 10}^{-3}$
• C. $4.0$
• D. ${4\times 10}^{-2}$

Answer 1

The correct option is D ${4\times 10}^{-2}$

$\underset{\text{At equillibrium}:}{\underset{\text{Initially}:}{}}\underset{(0.03-x)M}{\underset{0.03M}{{CO{Cl}_2}_{\left(g\right)}}}\rightleftharpoons \underset{x}{\underset{0}{{CO}_{\left(g\right)}}}+\underset{x}{\underset{0}{{{Cl}_2}_{\left(g\right)}}}$

As given that the equillibrium concentration of ${Cl}_2$ is $0.02M$

$\therefore x=\left[C_2\right]=0.02M$

Therefore, at equillibrium,

$\left[CO\right]=x=0.02M$

$\left[CO{Cl}_2\right]=(0.03-x)M=(0.03-0.02)M=0.01M$

From the given reaction-

$K_c=\frac{\left[CO\right]\left[{Cl}_2\right]}{\left[CO{Cl}_2\right]}$

$K_c=\frac{0.02\times 0.02}{0.01}$

$\Rightarrow K_c=\frac{4\times {10}^{-4}}{{10}^{-2}}=4\times {10}^{-2}$

Hence, $K_c$ for the reaction ${CO{Cl}_2}_{\left(g\right)}⟶{CO}_{\left(g\right)}+{{Cl}_2}_{\left(g\right)}$ is $4\times {10}^{-2}$.