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 ${440}^{\circ }\text{C}$ as given below.
$S{b}_2{S}_3\left(s\right)+3H_2\left(g\right)\rightleftharpoons 2Sb\left(s\right)+3H_{2}S\left(g\right)$.
In a separate experiment, this $H_{2}S$ is sufficient to precipitate 1.19 g $PbS$ from $P{b}^{2+}$ solution. Calculate $K_c$ of the above reaction.

Answer 1

For the equilibrium, Volume is constant so no of moles are proportional to concentration
$S{b}_2{S}_3\left(s\right)+3H_2\left(g\right)\rightleftharpoons 2Sb\left(s\right)+3H_{2}S\left(g\right)$
Lets say x moles of $S{b}_2{S}_3$ is consumed then at equilibrium the no of moles left are
$S{b}_2{S}_3=(0.01-x),H_2\left(g\right)=(0.01-3x)and{H}_{2}S\left(g\right)=3x$
Now we also Know that this $H_{2}S$ is sufficient to precipitate 1.19 g $PbS$ from $P{b}^{2+}$ solution.so no of moles of $PbS=\frac{1.19}{238}=molesof{H}_{2}S=3x=0.005moles$ $H_2\left(g\right)=(0.01-3x)=0.005molesand{H}_{2}S\left(g\right)=3x=0.005moles$
now $K_c=\frac{[H_{2}S{]}^3}{[H_2{]}^3}=\frac{(0.005/250{)}^3}{(0.005/250{)}^3}$ Hence $K_c=1$