Question

At $298K$
$N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g),K_1=4\times 10^5$
$N_2(g)+O_2(g)\rightleftharpoons 2NO(g),K_2=1.6\times 10^{12}$
$H_2(g)+\frac{1}{2}{O}_2(g)\rightleftharpoons H_{2}O(g),K_3=1.0\times 10^{-13}$
Based on above equilibria, the equilibrium constant of the reaction,
$2N{H}_3(g)+\frac{5}{2}{O}_2(g)\rightleftharpoons 2NO(g)+3H_{2}O(g)$
is ___$\times 10^{-33}$
(Nearest integer)

Answer 1

Answer: 4

$N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g),K_1=4\times 10^5\dots (i)$
$N_2(g)+O_2(g)\rightleftharpoons 2NO(g),K_2=1.6\times 10^{12}\dots (ii)$
$H_2(g)+\frac{1}{2}{O}_2(g)\rightleftharpoons H_{2}O(g),K_3=1.0\times 10^{-13}\dots (iii)$
$(ii)+3\times (iii)-(i)$
$2N{H}_3(g)+\frac{5}{2}{O}_2(g)\rightleftharpoons 2NO(g)+3H_{2}O(g)$
$k_{eq}=\frac{k_2\times k_3^3}{k_1}=\frac{1.6\times 10^{12}\times {\left(10^{-13}\right)}^3}{4\times 10^5}$
$=\frac{1.6}{4}\times 10^{-32}=4\times 10^{-33}$