Question

One mole of $N_2\left(g\right)$ is mixed with $2$ mole of $H_2\left(g\right)$ in a $4L$ vessel. If $50\%$ of $N_2\left(g\right)$ is converted to $N{H}_3\left(g\right)$ by the following reaction:
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3$
What will be the value of $K_c$ for the following equilibrium?
$N{H}_3\left(g\right)\rightleftharpoons \frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)$

• A. $256$
• B. $16$
• C. $\frac{1}{16}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{16}$

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
$$\begin{array}{cccc}\text{Initial moles}& 1& 2& 0\\ \text{at eqm}& 1-x& 2-3x& 2x\end{array}$$
where $x=0.5$
$K_c=\frac{{\left(\frac{1}{4}\right)}^2}{\left(\frac{0.5}{4}\right){\left(\frac{0.5}{4}\right)}^3}=256$
$N{H}_3\left(g\right)\rightleftharpoons \frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)K_c^{\prime }=\frac{1}{\sqrt{K_c}}=\frac{1}{16}$