Question

$2HCl(g)\rightleftharpoons H_2(g)+C{l}_2(g)$
For the above reaction, the equilibrium constant is $1.0\times 10^{-5}$. If the equilibrium concentration of $H_2$ and $C{l}_2$ is $1.2\times 10^{-3}M$ and $1.2\times 10^{-4}M$, respectively, then the concentration of $HCl$ is :

• A. $12\times 10^{-4}M$
• B. $12\times 10^{-3}M$
• C. $12\times 10^{-2}M$
• D. $12\times 10^{-1}M$

Answer 1

Answer: (C)

We will use expression $K=\frac{\left[H_2\right]\left[C{l}_2\right]}{[HCl{]}^2}$
Given, $K=1.0\times 10^{-5},$
$\left[H_2\right]=1.2\times 10^{-3}M$
$\left[C{l}_2\right]=1.2\times 10^{-4}M,$
$[HCl]=?$
$K=\frac{\left[H_2\right]\left[C{l}_2\right]}{[HCl{]}^2}$
or $[HCl{]}^2=\frac{\left[H_2\right]\left[C{l}_2\right]}{K}$
$[HCl]=\sqrt{\frac{\left[H_2\right]\left[C{l}_2\right]}{K}}$
Substituting the values
$=\sqrt{\frac{1.2\times 10^{-3}\times 1.2\times 10^{-4}}{1.0\times 10^{-5}}}$
$=\sqrt{1.2\times 1.2\times 10^{-7}\times 10^5}$
$=\sqrt{\frac{1.2\times 1.2}{100}}$
$=0.12M=12\times 10^{-2}M$