Question

The equilibrium constant for the following reaction at 1395 K.
$2H_{2}O\rightleftharpoons 2H_2+O_2{K}_1=2.1\times {10}^{13}$
$2C{O}_2\rightleftharpoons 2CO+O_2{K}_2=1.4\times {10}^{12}$
Calculate the value of K for the reaction : $H_2+C{O}_2\rightleftharpoons CO+H_{2}O$

• A. $k=1.29$
• B. $k=0.258$
• C. $k=5.1$
• D. $k=10.7$

Answer 1

The correct option is B $k=0.258$

The given reactions are
$2H_{2}O\rightleftharpoons 2H_2+O_2$ ......(i)
$K_1=2.1\times {10}^{13}$
$2C{O}_2\rightleftharpoons 2CO+O_2$ ......(ii)
$K_2=1.4\times {10}^{12}$
The first reaction is reversed.
$2H_2+O_2\rightleftharpoons 2H_{2}O$......(iii)
$K_3=\frac{1}{K_1}=\frac{1}{2.1\times {10}^{13}}=4.76\times {10}^{-14}$
The reaction (iii) is then added to reaction (ii) to obtain
$2H_2+2C{O}_2\rightleftharpoons 2CO+2H_{2}O$.....(iv)
$K_4=K_3\times K_2=4.76\times {10}^{-14}\times 1.4\times 1012=0.0667$
The reaction (iv) is divided by 2 to obtain
$H_2+C{O}_2\rightleftharpoons CO+H_{2}O$
Hence, the equilibrium constant $K_5=\sqrt{K_4}=\sqrt{0.0667}=0.258$