Question

One mole of $N_2$ is mixed with three mole of $H_2$ in a 4 litre vessel. If $0.25\%$ $N_2$ is converted into $N{H}_3$ by the reaction :

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$. $K_c$ for $\left(1/2\right)N_2+\left(3/2\right)H_2\rightleftharpoons N{H}_3$ is $x\times {10}^{-3}$ $L{mol}^{-1}$.

What is the value of $x$ to the nearest integer?

Answer 1

The degree of dissociation is $0.25$% or $0.0025$.

	$N_2$	$H_2$	$N{H}_3$
Initial number of moles	1	3	0
Equilibrium number of moles	$1-0.0025=0.9975$	$3-0.0075=2.9925$	$0.0050$

The equilibrium constant expression for the reaction $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$ is $K_c=\frac{[N{H}_3{]}^2}{\left[N_2\right][H_2{]}^3}=\frac{(\frac{0.0050}{4}{)}^2}{\left(\frac{0.9975}{4}\right)(\frac{2.9925}{4}{)}^3}=\frac{(0.00125{)}^2}{\left(0.249375\right)(0.748125{)}^3}=1.496\times {10}^{-5}$ .
The equilibrium constant for the reaction $\left(1/2\right)N_2+\left(3/2\right)H_2\rightleftharpoons N{H}_3$ is $K_C^{\prime }=\sqrt{1.496\times {10}^{-5}}=3.868\times {10}^{-3}L{mol}^{-1}=x\times {10}^{-3}L{mol}^{-1}$.
Hence, $x=3.868$