Question:
Value of $K_{P}$ for the equilibrium reaction
$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2(\mathrm{~g})}$ at $288 \mathrm{~K}$ is $47.9 .$ The $\mathrm{K}_{\mathrm{C}}$ for this reaction at same temperature is _________.
(Nearest integer)
$\left(\mathrm{R}=0.083 \mathrm{~L}\right.$ bar $\left.\mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)$
Solution:
$\mathrm{K}_{\mathrm{C}}=\frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{RT}}=\frac{47.9}{0.083 \times 288}=2$