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A vessel of 25 litre was filled with 0.01 mole of $S b_{2} S_{3}$ and 0.01 mole of $H_{2}$ to attain the equilibrium at $440 {}^{\circ}C$ as After equilibrium the $H_{2} S$ formed was analysed by dissolving it in water and treating with excess of $P b^{2 +}$ to give 1.195 g of PbS (Molecular weight = 239) precipitate. What is value of $K_{c}$ of the reaction at $440 {}^{\circ}C ?$

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Detailed Solution

$n H_{2} S = \frac{1.198}{239} = 0.005$

$\begin{array}{l} \Rightarrow [H_{2}] = 0.01 - 0.005 = 0.005 \\ [H_{2} S] = 0.005 \Rightarrow K_{C} = 1 \end{array}$

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