Question

Equilibrium constants for the following reactions at $1200K$ are given :
$2H_2{O}_{(g)}\rightleftharpoons 2H_2(g)+O_2$${}_{(g)};K_1=6.4\times 10^{-8}$
$2C{O}_2$${}_{(g)}\rightleftharpoons 2C{O}_{(g)}+O_2$${}_{(g)};K_2=1.6\times 10^{-6}$
The equilibrium constant for the reaction
$H_2$${}_{(g)}$ $+C{O}_2$ ${}_{(g)}\rightleftharpoons C{O}_{(g)}+H_2{O}_{(g)}$ at $1200K$ will be

• A. 0.05
• B. 20
• C. 0.2
• D. 5.0

Answer 1

Answer: (D)

For the given reactions,
$2H_{2}O\rightleftharpoons 2H_2+O_2,K_1=6.4\times 10^{-8}$
or, $H_{2}O\rightleftharpoons H_2+\frac{1}{2}{O}_2,K_1^{\prime }=\sqrt{K_1}$
or $H_2+\frac{1}{2}{O}_2\rightleftharpoons H_{2}O,K_1^{\prime \prime }=1/\sqrt{K_1}\dots$ (i)
$2C{O}_2\rightleftharpoons 2CO+O_2,K_2=1.6\times 10^{-6}$
or $C{O}_2\rightleftharpoons CO+\frac{1}{2}{O}_2,K_2^{\prime }=\sqrt{K_2}\dots$ (ii)
Form (i) and (ii) [on adding]
$K=\frac{\sqrt{K_2}}{\sqrt{K_1}}=\frac{\sqrt{1.6\times 10^{-6}}}{\sqrt{6.4\times 10^{-8}}}$
$K=\sqrt{\frac{10^2}{4}}=\sqrt{25}=5$