Question

At a certain temperature (T), the equilibrium constant $\left(K_c\right)$ is 1 for the reaction $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
If 2 moles of $N_2,$ 4 moles of $H_2,$ 6 moles of $N{H}_3$ and 3 moles of inert gas are introduced into a two litre rigid vessel at constant temperature T. It has been found that equilibrium concentration of $H_2$ and $N{H}_3$ are equal then what is the equilibrium concentration of $N_2$ (in M)?

Answer 1

$Q_c=\frac{{\left(\frac{6}{2}\right)}^2}{\left(\frac{2}{2}\right){\left(\frac{4}{2}\right)}^3}=1.125$
$\because Q_c>K_c$, the reaction will proceed in backward direction.
$\begin{array}{cccccc}& N_2\left(g\right)& +& 3H_2\left(g\right)& \rightleftharpoons & 2N{H}_3\\ Initialmoles:& 2& & 4& & 6\\ Atequilibrium:& 2+x& & 4+3x& & 6-2x\end{array}$

$\because$ At equilibrium $\left[H_2\right]=\left[N{H}_3\right]$
$\frac{4+3x}{2}=\frac{6-2x}{2}\Rightarrow x=0.4$
$\therefore \left[N_2\right]$ at equilibrium $=\frac{2+x}{2}=1.2M$