Question

$2$ of $N_2$ is mixed with $6$ of $H_2$ in a closed vessel of one litre capacity. If 50% of $N_2$ is converted into ${NH}_3$ at equiibrium, the value of $K_c$ for the reaction ?
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$

• A. 4/27
• B. 27/4
• C. 1/27
• D. 24

Answer 1

The correct option is A 4/27

$N_2\left(g\right)+3H_2\left(g\right)\leftrightharpoons 2N{H}_3\left(g\right)$

$Att=0\frac{2}{1}\frac{6}{1}0$

$Atequilibrium2-x6-3x2x$

But 2x=$\frac{50}{100}$$\times 2$

$x=\frac{1}{2}$

$\therefore$$K_c$=$\frac{{\left(2x\right)}^2}{(2-x)(6-3x^3)}$

=$\frac{{\left(1\right)}^2}{\left(1.5\right)(6-1.5{)}^3}$

$\therefore K_c=\frac{4}{27}$