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Standard XII

Chemistry

Derivation of Kp and Kc

A vessel of ...

Question

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 $440^{o} C$ 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 analysed by dissolved it in water and treating with excess of ${P b}^{2 +}$ to give $1.19$ g of $P b S$ as precipitate. What is the value of $K_{c}$ at $440^{o} C$ .

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Solution

The correct option is A $1$
The molar mass of PbS is 239.3 g/mol.

1.19 g of PbS correspond to $\frac{1.19}{239.3} = 0.00497$ moles of Pbs.

This is also equal to the number of moles of $H_{2} S$

${S b}_{2} S_{3} (s) + 3 H_{2} (g) ⇌ 2 S b (s) + 3 H_{2} S (g)$
0.01 - 3x 3x

The total number of moles of $H_{2} S$ at equilibrium is
$3 x = 0.00497$
$x = 0.001657$

The equilibrium number of moles of hydrogen are
$0.01 - 3 x = 0.01 - 3 (0.001657) = 0.005027$

The equilibrium constant expression is $K_{c} = \frac{[H_{2} S]^{3}}{[H_{2}]^{3}}$
$K_{c} = \frac{(\frac{0.004973}{250})^{3}}{(\frac{0.0052027}{250})^{3}} \approx 1$

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