Question

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

Answer 1

The number of moles of water that have reacted are $\frac{1\times 40}{100}=0.4mol$.

Moles of water unreacted are $1.0-0.4=0.6$ mol.

0.4 moles of water reacts with 0.4 mole of CO to form 0.4 moles of hydrogen and 0.4 moles of carbon dioxide.

The equilibrium molar concentrations are:

$\left[H_2\right]=\frac{0.4}{10}=0.04M$

$\left[C{O}_2\right]=\frac{0.4}{10}=0.04M$

$\left[H_{2}O\right]=\frac{0.6}{10}=0.06M$

$\left[CO\right]=\frac{0.6}{10}=0.06M$

The equilibrium constant is $K=\frac{\left[H_2\right]\left[C{O}_2\right]}{\left[H_{2}O\right]\left[CO\right]}=\frac{0.04\times 0.04}{0.06\times 0.06}=0.444$.