Question

$2N{H}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)$ If 'a' moles of $N{H}_3$ are given initially. Let x moles of $N{H}_3$ are decomposed at equilbrium and $Vlitre$ be the capacity of vessel in which reaction is being studied. Find out the value of $K_c$.

• A. $\frac{27x^4}{16V^2(a-x{)}^2}$
• B. $\frac{27x^2}{16V^2(a-x{)}^2}$
• C. $\frac{27x^2}{16V^2(a-x)}$
• D. $\frac{27x^4}{16V^4(a-x{)}^2}$

Answer 1

The correct option is A $\frac{27x^4}{16V^2(a-x{)}^2}$

$\begin{array}{cccccc}& 2N{H}_3& \rightleftharpoons & N_2& +& 3H_2\\ \text{Initial moles}& a& & 0& & 0\\ \text{Moles at equilbrium}& a-x& & \frac{x}{2}& & \frac{3x}{2}\\ \text{Concentration at equilbrium}& \frac{a-x}{V}& & \frac{x}{2V}& & \frac{3x}{2V}\end{array}$
$K_c=\frac{\left[N_2\right][H_2{]}^3}{[N{H}_3{]}^2}=\frac{\left(\frac{x}{2V}\right){\left(\frac{3}{2}\frac{x}{V}\right)}^3}{{\left(\frac{a-x}{V}\right)}^2}=\frac{27x^4}{16V^2(a-x{)}^2}$