Question

${(1) \,N_2(g) +3H_2(g)<=> 2NH_3(g), K_1}$
${(2) \,N_2(g) +O_2(g)<=> 2NHO(g), K_2}$
${(3) \,H_2(g) +\frac{1}{2}O_2(g)<=> H_2O(g), K_3}$
The equation for the equilibrium constant of the reaction
${2NH_3(g) +\frac{5}{2}O_2(g)<=> 2NO(g), 3H_2O(g)}$, $(K_4)$ in terms of $K_1$ and $K_3$ is:

• A. $\frac{K_{1}.K_{2}}{K_{3}}$
• B. $\frac{K_{1}.K^{2}_{3}}{K_{2}}$
• C. $K_1K_2K_3$
• D. $\frac{K_{2}.K^{3}_{3}}{K_{1}}$

Answer 1

Answer: (D)

$\frac{K_{2}.K^{3}_{3}}{K_{1}}$

Answer 2

To calculate the value of $K_4$ in the given equation we should apply:
$eqn. (2) + eqn.(3) x 3 - eqn. (1)$
hence $K_{4}=\frac{K_{2}K^{3}_{3}}{K_{1}}$