Q. For the reactions, $H_{2}(g) + CO_{2}(g) \rightleftharpoons CO(g) + H_{2}O(g)$ if the initial concentration of [$H_{2}\rbrack = \lbrack CO_{2}\rbrack$ and $x$ mol/L of hydrogen is consumed at equilibrium, the correct expression of $K_{p}$ is
A) $\frac{x^{2}}{(1 - x)^{2}}$
B) $\frac{x^{2}}{(2 + x)^{2}}$
C) $\frac{x^{2}}{{1 - x}^{3}}$
D) $\frac{{(1 + x)}^{2}}{(1 - x)^{2}}$
Explanation
(a)
$H_{2}(g) + CO_{2}(g) \rightleftharpoons CO(g) + H_{2}O(l)$
Initial conc. $1\ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ \ \ \ 0$
At equilibrium $(1 - x)(1 - x)\ \ \ \ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ \ \ \ x$
$K_{p} = \frac{p_{CO} \bullet p_{H_{2}O}}{p_{H_{2}} \bullet p_{CO_{2}}} = \frac{x \bullet x}{(1 - x)(1 - x)} = \frac{x^{2}}{(1 - x)^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.