Question

Equilibrium constant for following reactions respectively $K_1,K_{2}and{K}_3$

$N_2+{3H}_2\rightleftharpoons 2{NH}_3{K}_1$

$N_2+O_2\rightleftharpoons 2NO{K}_2$

$H_2+\frac{1}{2}{O}_2\rightleftharpoons H_{2}O{K}_3$

$2{NH}_3+\frac{5}{2}{O}_2\rightleftharpoons 2NO+3H_{2}O{K}_4$

Which of the following relation is correct?

• A. $K_1=\frac{K_2\times {\left(K_3\right)}^3}{K_4}$
• B. $K_4=K_1\times K_2/{\left(K_3\right)}^3$
• C. $K_2=\frac{K_4+K_1}{{\left(K_3\right)}^3}$
• D. None of these

Answer 1

Answer: (A)

$K_1=\frac{K_2\times {\left(K_3\right)}^3}{K_4}$

$N_2+3H_2\rightleftharpoons 2N{H}_3{K}_1$.........(1)

$N_2+O_2\rightleftharpoons 2NO{K}_2$........(2)

$H_2+\frac{1}{2}{O}_2\rightleftharpoons H_{2}O{K}_3$........(3)

$2N{H}_3+\frac{5}{2}{O}_2\rightleftharpoons 2NO+3H_{2}O......K_4$

To get $4th\Rightarrow \left(1\right)+\left(2\right)+\left(3\right)$

$\Rightarrow K_4=K_1^{-1}\times K_2^1\times K_3^3$

$\Rightarrow K_1=\frac{K_2{K}_3^3}{K_4}$