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 $440°C$ as $S{b}_2{S}_3\left(s\right)+3H_2\left(g\right)\rightleftharpoons 2Sb\left(s\right)+3H_{2}S\left(g\right)$.
After equilibrium, the $H_{2}S$ formed was analysed by dissolving it in water and treating with excess of $P{b}^{2+}$ to give $1.195g$ of $PbS$ as precipitate. What is the value of $K_C$ at $440°C$? (Molar mass of $Pb=207gmo{l}^{-1}$)

Answer 1

Given reaction is
$S{b}_2{S}_3\left(s\right)+3H_2\left(g\right)\rightleftharpoons 2Sb\left(s\right)+3H_{2}S\left(g\right)$
$0.01-x0.01-3x2x3x$
Again,
$H_{2}S+P{b}^{2+}\to PbS+2H^{+}$
Number of moles of $PbS$ formed = $\frac{1.195}{238}=0.005mol$
So, at equlibrium $H_{2}S$ formed = $3x=0.005mol$
$\therefore K_C=\frac{(\frac{0.005}{250}{)}^3}{(\frac{(0.01-0.005)}{250}{)}^3}=1$