Question

The equilibrium constant of the following are:

$N_2+3H_2\rightleftharpoons 2N{H}_3;K_3$

$N_2+O_2\rightleftharpoons 2NO;K_2$

$H_2+\frac{1}{2}{O}_2\rightleftharpoons H_{2}O;K_1$

The equilibrium constant (K) for the reaction,

$2N{H}_3+5/2O_2\stackrel{K}{\rightleftharpoons }2NO+3H_{2}O$ will be:

Answer 1

$K_3$ is the equilibrium constant of first reaction given. But to achieve the final reaction this reaction has to be reversed. Thus forming the reaction, $2N{H}_3\rightleftharpoons N_2+3H_2$

Therefore, the equilibrium constant of the above reaction is $\frac{1}{K_3}$

_{For second reaction, $N_2+O_2\rightleftharpoons 2NO$ the equilibrium constant is $K_2$.}

_{For third reaction, the equilibrium constant is $K_1$. On multiplying this reaction by 3 we get}

_{$3H_2+\frac{3}{2}{O}_2\to 3H_{2}O$}

_{The equilibrium constant for this reaction is $K_1^3$.}

_{On adding these three reactions we get the final reaction $2N{H}_3+\frac{5}{2}{O}_2\rightleftharpoons 2NO+3H_{2}O$}

_{Equilibrium constant for this reaction is $K=\frac{K_1^3{K}_2}{K_3}$.}