Question

$2 HCl ( g ) \rightleftharpoons H _{2}(g)+ Cl _{2}(g)$
For the above reaction, the equilibrium constant is $1.0 \times 10^{-5}$. If the equilibrium concentration of $H _{2}$ and $Cl _{2}$ 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 $K=\frac{\left[ H _{2}\right]\left[ Cl _{2}\right]}{[ HCl ]^{2}}$
Given, $K=1.0 \times 10^{-5},$
$\left[ H _{2}\right]=1.2 \times 10^{-3} M$
$\left[ Cl _{2}\right]=1.2 \times 10^{-4} M ,$
$[ HCl ]=?$
$K=\frac{\left[ H _{2}\right]\left[ Cl _{2}\right]}{[ HCl ]^{2}}$
or $[ HCl ]^{2} =\frac{\left[ H _{2}\right]\left[ Cl _{2}\right]}{K}$
$[ HCl ] =\sqrt{\frac{\left[ H _{2}\right]\left[ Cl _{2}\right]}{K}}$
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}} $
$=\sqrt{\frac{1.2 \times 1.2}{100}}$
$=0.12 \,M =12 \times 10^{-2} \,M$