Question

One mole of $H_{2}O$ and one mole of CO are taken in 10L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation, $H_{2}O\left(g\right)+CO\left(g\right)\rightleftharpoons H_2\left(g\right)+C{O}_2\left(g\right)$
What is the value of equilibrium constant $\left(K_c\right)$ for the reaction.

• A. 44
• B. 4.4
• C. 0.44
• D. 2.22

Answer 1

The correct option is C 0.44

The equilibrium reaction is $H_{2}O\left(g\right)+CO\left(g\right)\rightleftharpoons H_2\left(g\right)+C{O}_2\left(g\right)$.

	$H_{2}O$	CO	$H_2$	$C{O}_2$
Initial moles	1	1	0	0
change	-0.4	-0.4	0.4	0.4
Moles at equilibrium	0.6	0.6	0.4	0.4

The expression for the equilibrium constant is $K_C=\frac{0.4\times 0.4}{0.6\times 0.6}=\frac{4}{9}=0.44$.
Hence, option C is the right answer.