Question

$2HX\left(g\right)\leftrightharpoons H_2\left(g\right)+X_2\left(g\right);K_C=1.0\times {10}^{-5}$
What is the concentration of $HX$ if the equilibrium concentration of $H_2$ and $X_2$ is $1.2\times {10}^{-3}\text{M}$ and $1.2\times {10}^{-4}\text{M}$ respectively?

• A. $1.2\times {10}^{-4}\text{M}$
• B. $1.2\times {10}^{-3}\text{M}$
• C. $1.2\times {10}^{-2}\text{M}$
• D. $1.2\times {10}^{-1}\text{M}$

Answer 1

The correct option is D $1.2\times {10}^{-1}\text{M}$

Given,
$2HX\leftrightharpoons H_2\left(g\right)+X_2\left(g\right);K_C=1.0\times {10}^{-5}$
$\Rightarrow K_C=\frac{\left[H_2\right]\left[X_2\right]}{[HX{]}^2}$
$\Rightarrow [HX{]}^2=\frac{\left[H_2\right]\left[X_2\right]}{K_C}$
$\Rightarrow [HX{]}^2=\frac{1.2\times {10}^{-3}\times 1.2\times {10}^{-4}}{1.0\times {10}^{-5}}$
$\therefore \left[HX\right]=1.2\times {10}^{-1}M$