Question: $\mathrm{K}_{\mathrm{p}}$ for the following reaction is 3.0 at $1000 \mathrm{~K}$.
$$ \mathrm{CO}_2(\mathrm{~g})+\mathrm{C}(\mathrm{s}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) $$
What will be the value of $\mathrm{K}_{\mathrm{c}}$ for the reaction at the same temperature?
(Given : $\mathrm{R}=0.083 \mathrm{~L}$ bar K $^{-1} \mathrm{~mol}^{-1}$ )
A) 3.6
B) 0.36
C) $3.6 \times 10^{-2}$
D) $3.6 \times 10^{-3}$
Answer: $3.6 \times 10^{-2}$
Solution:
$\begin{aligned} & \mathrm{CO}2(\mathrm{~g})+\mathrm{C}(\mathrm{s}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) \ & \Delta \mathrm{n}{\mathrm{g}}=2-1=1 \ & \mathrm{~K}{\mathrm{p}}=\mathrm{K}{\mathrm{c}}(\mathrm{RT})^{\Delta \mathrm{n}{\mathrm{g}}} \ & \mathrm{K}{\mathrm{p}}=\mathrm{K}{\mathrm{c}}(\mathrm{RT}) \ & {\left[\because \mathrm{K}{\mathrm{p}}=3\right]} \ & \mathrm{K}{\mathrm{c}}=\frac{\mathrm{K}{\mathrm{p}}}{\mathrm{RT}}=\frac{3}{0.083 \times 1000} \ & =0.036 \ & =3.6 \times 10^{-2}\end{aligned}$
