Question

The value of $K_C$ for the equilibrium reaction
$\ce{CO_{2(g)} + C_{(s)} <=> 2CO_{(g)}} $
at $T( K )$ is $0.036 .$ If the equilibrium concentration of $CO _{2}(g)$ is $0.004\, M$, the concentration of $CO (g)$ in $mol \,L ^{-1}$ is

• A. $3.6 \times 10^{-2}$
• B. $2.0 \times 10^{-2}$
• C. $1.2 \times 10^{-2}$
• D. $1.2 \times 10^{-3}$

Answer 1

Answer: (C)

For the reaction,

$CO _{2}(g)+ C (s) \rightleftharpoons 2 CO (g)$

Given,

$K_{C}=0.036$

Equilibrium conc. of $\left[ CO _{2}(g)\right]=0.004\, M$

As we know that,

$K_{C}=\frac{[ CO (g)]^{2}}{\left[ CO _{2}(g)\right]} [ C (s)=1] \ldots$ (i)

On substituting the given values in eq. (i), we get,

$0.036 =\frac{[ CO (g)]^{2}}{0.004} $

$[ CO (g)] =\sqrt{0.036 \times 0.004} $

$=\sqrt{1.44 \times 10^{-4}} $

$=1.2 \times 10^{-2} mol / L$