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

# The reaction ${\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{g}\right)+\mathrm{CO}\left(\mathrm{g}\right)⇌{\mathrm{H}}_{2}\left(\mathrm{g}\right)+{\mathrm{CO}}_{2}\left(\mathrm{g}\right)$ is initiated with 1 mol of each of ${\mathrm{H}}_{2}\mathrm{O}$ and $\mathrm{CO}$ 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

1. A

0.66

2. B

0.55

3. C

0.44

4. D

0.22

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### Solution:

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

We will have

Since $\mathrm{\Delta }{v}_{\mathrm{g}}=0$ We will have ${K}_{c}={K}_{n}$ Hence

${K}_{c}=\frac{\left(0.4\mathrm{mol}\right)\left(0.4\mathrm{mol}\right)}{\left(0.6\mathrm{mol}\right)\left(0.6\mathrm{mol}\right)}=0.44$  