Question

At $298\ K$
$N_2( g)+3H_2( g)⇌2NH_3( g),K_1=4×10^5$
$N_2( g)+O_2( g)⇌2NO(g),K_2=1.6×10^{12}$
$H_2( g)+1 2O_2( g)⇌H_2O(g),K_3=1.0×10^{−13}$
Based on above equilibria, the equilibrium constant of the reaction, 
$2NH_3( g)$+$\frac5 {2}$ $O_2( g)⇌2NO(g)+3H_2O(g)$ is _______ $×10^{−33 }$. (Nearest integer)

Answer 1

Correct Answer: 4

The equilibrium constant for the given reaction is calculated by combining the given equilibria: $$K_{\text{eq}} = \frac{K_2 \times K_3^3}{K_1}$$ Substituting the values: $$K_{\text{eq}} = \frac{(1.6 \times 10^{12}) \times (1.0 \times 10^{13})^3}{4 \times 10^5}$$ $$K_{\text{eq}} = \frac{1.6 \times 10^{12} \times 10^{39}}{4 \times 10^5} = 4 \times 10^{33}$$