Question

The following equilibrium constants are given:
$N_2+3H_2\rightleftharpoons 2N{H}_3;K_1$
$N_2+O_2\rightleftharpoons 2NO;K_2$
$H_2+1/2O_2\rightleftharpoons H_{2}O;K_3$
The equilibrium constant for the oxidation of the $N{H}_3$ by oxygen to give $NO$ is:

• A. $\frac{K_1{K}_2}{K_3}$
• B. $\frac{K_2{K}_3^3}{K_1}$
• C. $\frac{K_2{K}_3^2}{K_1}$
• D. $\frac{K_2^2{K}_3}{K_1}$

Answer 1

Answer: (B)

$\frac{K_2{K}_3^3}{K_1}$

The given reactions are :-

$\left(I\right)$ $N_2+3H_2\rightleftharpoons 2{NH}_3$

$\therefore K_1=\frac{{\left[{NH}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}\to \left(I\right)$

$\left(II\right)$ $N_2+O_2\rightleftharpoons 2NO$

$\therefore K_2=\frac{{\left[NO\right]}^2}{\left[N_2\right]\left[O_2\right]}\to \left(II\right)$

$\left(III\right)$ $H_2+1/2O_2\rightleftharpoons H_{2}O$

$\therefore K_3=\frac{\left[H_{2}O\right]}{\left[H_2\right]{\left[O_2\right]}^{1/2}}\to \left(III\right)$

Now, for the reaction,

$2{NH}_3+\frac{5}{2}{O}_2\rightleftharpoons 2NO+3H_{2}O$

Now, $K=\frac{{\left[NO\right]}^2{\left[H_{2}O\right]}^3}{{\left[{NH}_3\right]}^2{\left[O_2\right]}^{5/2}}$

$=\frac{\frac{{\left[NO\right]}^2}{\left[N_2\right]\left[O_2\right]}.{\left(\frac{\left[H_{2}O\right]}{\left[H_2\right]{\left[O_2\right]}^{1/2}}\right)}^3}{\frac{{\left({NH}_3\right)}^2}{\left[N_2\right]{\left[H_2\right]}^3}}$

$\Rightarrow K=\frac{K_2{K_3}^3}{K_1}$