The reaction H2O(g)+CO(g)⇌H2(g)+CO2(g) is initiated with 1 mol of each of H2O and CO in a flask of volume 10 L. If at equilibrium 40 mass per cent of water reacts, the equilibrium constant of the reaction will be

Mass of water reacted $= (\frac{40 g}{100 g} \times 18 g) = 7.2 g$ Amount of water reacted $= \frac{7.2 g}{18 g {mol}^{- 1}} = 0.4 mol$

We will have $\begin{array}{rr} H_{2} O (g) + CO (g) ⇌ H_{2} (g) + {CO}_{2} (g) \\ (1 - 0.4) mol (1 - 0.4) mol 0.4 mol 0.4 mol \end{array}$

Since $Δ v_{g} = 0$ We will have $K_{c} = K_{n}$ Hence

$K_{c} = \frac{(0.4 mol) (0.4 mol)}{(0.6 mol) (0.6 mol)} = 0.44$

The reaction $H_{2} O (g) + CO (g) ⇌ H_{2} (g) + {CO}_{2} (g)$ is initiated with 1 mol of each of $H_{2} O$ and $CO$ in a flask of volume 10 L. If at equilibrium 40 mass per cent of water reacts, the equilibrium constant of the reaction will be