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Standard XII

Chemistry

Derivation of Kp and Kc

A vessel of ...

Question

A vessel of $2.50 l i t r e$ was filled with $0.01$ mole of $S b_{2} S_{3}$ adn $0.01$ mole of $H_{2}$ to attain the equilibrium at $440$ as
$S b_{2} S_{3} (s) + 3 H_{2} (s) ⇌ 2 S b (s) + 3 H_{2} S (g)$ .
After equilibrium the $H_{2} S$ formed was analysed by dissolving it in water and treating with excess of $P b^{2 +}$ to give $1.029 g$ of $P b S$ as precipitate. The value of $K_{c}$ of the reaction at $440$ is $X$ (At mass of $P b = 206$ ). $100 X$ is___________.

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Solution

$P b^{2 +} + H_{2} S ⇌ P b S + 2 H^{+}$
The molar mass of PbS is $206 + 32 = 238$
1.029 g of PbS corresponds to $\frac{1.029}{238} = 0.00433$ moles. This corresponds to the number of moles of $H_{2} S$ present at equilibrium.
The number of moles of hydrogen present at equilibrium $= 0.01 - \frac{0.00433}{3} = 0.0057$
The equilibrium constant is $(\frac{0.0043}{0.0057})^{3} = 0.43 = X$
$100 X = 43$

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