Question

Equilibrium constants for the following reaction at 1200 K are given.

$2H_{2}O\left(g\right)\leftrightharpoons 2H_2\left(g\right)+O_2\left(g\right);K_1=6.4\times {10}^{-8}$

$2C{O}_2\left(g\right)\leftrightharpoons 2CO\left(g\right)+O_2\left(g\right);K_2=1.6\times {10}^{-6}$

The equilibrium constant for the reaction

$H_2\left(g\right)+C{O}_2\left(g\right)\leftrightharpoons CO\left(g\right)+H_{2}O\left(g\right)$ at 1200 K will be:

• A. 0.05
• B. 20
• C. 0.2
• D. 5

Answer 1

The correct option is D 5

$\frac{1}{2}\times [2H_{2}O\leftrightharpoons 2H_2+O_2;K_1]$

$H_{2}O\leftrightharpoons H_2+\frac{1}{2}{O}_2;K^{\prime }=\sqrt{K_1}$

$H_2+\frac{1}{2}{O}_2\leftrightharpoons H_{2}O;K"=\frac{1}{\sqrt{K_1}}$ ....... (1)

$\frac{1}{2}\times [2C{O}_2\leftrightharpoons 2CO+O_2;K_2]$

$C{O}_2\leftrightharpoons CO+\frac{1}{2}{O}_2;\sqrt{K_2}=K"$ ....... (2)

From equations (1) and (2).

$H_2+\frac{1}{2}{O}_2\leftrightharpoons H_{2}O;K"=\frac{1}{\sqrt{K_1}}$

$C{O}_2\leftrightharpoons CO+\frac{1}{2}{O}_2;K"=\sqrt{K_2}$

--------------------------------------------------------------------------------

$H_2+C{O}_2\leftrightharpoons H_{2}O+CO$

$K=K"\times K"$

$=\frac{1}{\sqrt{K_1}}\times \sqrt{K_2}$

$=\sqrt{\frac{K_2}{K_1}}=\sqrt{\frac{1.6\times {10}^{-6}}{6.4\times {10}^{-8}}}=\sqrt{\frac{1}{4}\times {10}^2}=\sqrt{25}=5$

Hence, the correct option is $\text{D}$