JEE Main 2023 — Chemical Equilibrium Question with Solution

$$\begin{gathered} \mathrm{CO}\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{CO}}_2( \mathrm{g})+{\mathrm{H}}_2( \mathrm{g}) \\ \mathrm{initial} \mathrm{moles} 1 1 \\ \mathrm{moles} \mathrm{at} \mathrm{equilibrium} 1-0.4 1-0.4 0.4 0.4 \\ \mathrm{Molarity} 0.6/10 0.6/10 0.4/10 0.4/10 \end{gathered}$$

Now, the equilibrium constant ${\mathrm{K}}_{\mathrm{C}}=\frac{\left[{\mathrm{CO}}_2\right]\left[{\mathrm{H}}_2\right]}{\left[\mathrm{CO}\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}$

$\Rightarrow {\mathrm{K}}_{\mathrm{C}}=\frac{0.04\times 0.04}{0.06\times 0.06}=0.44$

$\therefore {\mathrm{K}}_{\mathrm{c}}\times {10}^2=44$