Question

$1mole$ of nitrogen and $3moles$ of hydrogen are mixed in a $4L$ container. If $0.25$ percent of nitrogen is converted to ammonia by the following reaction:

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$.

Calculate the equilibrium constant ($K_c$) in concentration units. What will be the value of $K$ for the following equilibrium?

$\frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)\rightleftharpoons {NH}_3\left(g\right)$

Answer 1

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$
At equilibrium, $(1-x)$ $(3-3x)$ $2x$

Given, $x=0.0025$

Active conc. $\frac{(1-0.0025)}{4}$ $\frac{(3-0.0075)}{4}$ $\frac{\left(0.0050\right)}{4}$

Applying law of mass action,
$K_c=\frac{{\left[{NH}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}=\frac{{\left(\frac{\left(0.0050\right)}{4}\right)}^2}{\left(\frac{0.9975}{4}\right){\left(\frac{2.9925}{4}\right)}^3}=1.49\times {10}^{-5}{litre}^2{mol}^{-2}$

$K$ for the reaction $\frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)\rightleftharpoons {NH}_3\left(g\right)$ is equal to $\sqrt{K_c}$.

$K=\sqrt{K_c}=\sqrt{1.49\times {10}^{-5}}=3.86\times {10}^{-3}litre{mol}^{-1}$