One mole $\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ and one mole $\mathrm{CO}(\mathrm{g})$ are taken in $1 \mathrm{~L}$ flask and heated to $725 \mathrm{~K}$. At equilibrium, $40 \%$ (by mass) of water reacted with $\mathrm{CO}(\mathrm{g})$ as follows $\mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g})$ Its $\mathrm{K}_{\mathrm{c}}$ value is

• Correct answer
  A

  0.444
• B

  2.220
• C

  0.222
• D

  4.440

Solution

$\Rightarrow$ Number of moles of $\mathrm{H}_2 \mathrm{O}$ reacting $=40 \%$ of 1 mole $=0.4$ moles $($ degree of dissociation $=\alpha)$ $\Rightarrow 0.4$ moles of $\mathrm{CO}$ and 0.4 moles of $\mathrm{H}_2 \mathrm{O}$ react. Let $\mathrm{C}=$ initial concentrations of $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{CO}$. $\Rightarrow$ At equilibrium; $\begin{aligned} & {\left[\mathrm{H}_2 \mathrm{O}\right]=\frac{0.6 \mathrm{~mol}}{1 \mathrm{~L}}=0.6 \mathrm{M}} \\ & {[\mathrm{CO}]=\frac{0.6 \mathrm{~mol}}{1 \mathrm{~L}}=0.6 \mathrm{M}} \\ & {\left[\mathrm{H}_2\right]=\left[\mathrm{CO}_2\right]=\mathrm{c} \alpha=1 \times 0.4=0.4 \mathrm{M}} \\ & \Rightarrow \mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{CO}_2\right]}{\left[\mathrm{H}_2 \mathrm{O}\right]}=\frac{(0.4)(0.4)}{(0.6)(0.6)} \\ & =0.444 . \end{aligned}$

Hence, the correct answer is option: 'A'.