An equilibrium is established ${SO}_{2}, O_{2}$ and ${SO}_{3}$ as ${SO}_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ {SO}_{3} (g)$ starting with 2 moles ${SO}_{2}$ and $1.5$ moles $O_{2}$ in 5 liter flask. The equilibrium mixture required $0.4$ moles of $K M n O_{4}$ in acidic medium. The value of $K_{C}^{\circ}$ is :

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$\sqrt 5$

$0.2$

$25$

$5$

answer is B.

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Detailed Solution

t=0 2 1.5

$t = t_{e q} \frac{1}{5} \frac{1}{5} \frac{1}{5}$

${SO}_{2} + {KMnO}_{4} ⟶ {Mn}^{2 +} + {SO}_{4}^{2 -}$

$n_{f} = 2 n_{f} = 5$

$2 \times n_{{SO}_{2}} = 5 \times 0.4$

$n_{{SO}_{2}} = 1$

$K_{C}^{\circ} = \frac{\frac{1}{5}}{\sqrt{\frac{1}{5}} - \frac{1}{5}} = \sqrt{5}$

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