Question

$56g$ of $N_2$ and $6g$ of $H_2$ were kept at ${400}^{o}C$ in $1$ litre vessel. The equilibrium mixture contained $27.54g$ of ${NH}_3$. The approximate value fro $K_c$ for the above reaction in ${mol}^2{lit}^2$ is:

• A. $12$
• B. $20$
• C. $30$
• D. $40$

Answer 1

The correct option is A $12$

$Given:$ The reaction is $N_2+3H_2\rightleftarrows 2N{H}_{3}56g6g27.5kg$

Initial no. of moles of $H_2=\frac{6}{2}=3mol{N}_2=\frac{56}{2.8}=2mol$

Initial number of moles $2N_2+$ $33H_2\rightleftarrows$$02N{H}_3$

At $equilibrium,$ $(2-x)(3-3x)$ $2x$

$\therefore$ Number of moles of $N{H}_3=\frac{27.54}{17}=1.62$

i.e. $2x=1.62x=0.81$

$\therefore$ At equilibrium,

No. of moles of $N_2=2-x=2-0.81=1.19$

No. of moles in $H_2=3(1-x)=3(1-0.81)=0.57$

$\therefore$ Equilibrium concentration of ,

$\left[N_2\right]=\frac{1.19}{1}=1.19M$

$\left[H_2\right]=\frac{0.57}{1}=0.57M$

$\left[N{H}_3\right]=\frac{1.62}{1}=1.62M$

$\therefore$ Equilibrium constant,

$K_c=\frac{{\left[N{H}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}=\frac{{1.62}^2}{1.19\times {0.57}^3}=11.9\approx 12mo{l}^{2}li{t}^2$