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