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A vessel of 250 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 $440^{0} C$ as
$S b_{2} S_{3} (s) + 3 H_{2} (g) ⇌ 2 S b (s) + 3 H_{2} S (g)$
After equilibrium the $H_{2} S$ Formed was analyzed by dissolved it in water and treating with excess of $P b^{2 +}$ to give 1.19g of PbS(Mol.mass of PbS = 238 gm/mole) as precipitate. What is the value of $K_{c}$ at $440^{0} C$ ?

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Detailed Solution

$\begin{array}{l} S b_{2} S_{3} (s) + 3 H_{2} (g) \to 2 S b (s) + 3 H_{2} S (g) \\ 0.01 - x 0.01 - x x x \end{array}$

Where x=0.005

$H_{2} S + P b^{2 +} \to P b S + 2 H^{+}$

No. of moles of PbS formed

$= \frac{1.19}{238} = 0.005 m o l e$

At eqm $[H_{2}] = (\frac{0.005}{250});$

$K_{c} = {[\frac{0.005}{0.005}]}^{3} = 1$

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