Question

The equilibrium constant ($K_c$) for the reaction
$2HCl\left(g\right)\rightleftharpoons H_2\left(g\right)+{Cl}_2\left(g\right)$ is $4\times {10}^{-34}$ at ${25}^{o}C$. What is the equilibrium constant for the reaction?
$\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}{Cl}_2\left(g\right)\rightleftharpoons HCl\left(g\right)$

• A. $2\times {10}^{-17}$
• B. $2.5\times {10}^{33}$
• C. $5\times {10}^6$
• D. None of these

Answer 1

The correct option is A $2\times {10}^{-17}$

Option $A$ is correct.

Let initial number of moles of $HCl$ be $N$ in the container of volume $Vq$ litres.

Let there be $X$ moles $q$ of $H_2$ and $X$ moles of $C{l}_2$ at the equilibrium.

$2HCl\left(g\right)⟺H_2\left(g\right)+C{l}_2\left(g\right)$

$N-2X$ $X$ $X$

Equilibrium constant $Kc=\left[H_2\right]\left[C{l}_2\right]/\left[\right[HCl]{}^2$

$Kc=\left(X/V\right)²÷(N-2X)/V²=4\times 10{}^{-34}$

$=X²/(N-2X)²=4\times 10{}^{{}^{-34}}$

This means that $X/(N-2X)=2\times 10{}^{-17}$

$N-2X=5\times 10{}^{-16}X$

$X\approx 2\times 10{}^{-17}N$

Hence $2\times {10}^{-17}$is the answer.