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 ${440}^{\circ }$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 1.195 g of PbS (Molecular weight=239)precipitate.
What is value of$K_c$ of the reaction at ${440}^{\circ }$C?

• A. 1
• B. 2
• C. 4
• D. None of these

Answer 1

The correct option is A 1

1.195 g of PbS (Molecular weight=239) precipitate is formed. Hence, $\left[H_{2}S\right]=\frac{1.195}{239}=0.005moles$. The equilibrium concentration of hydrogen sulfide is $\frac{0.005}{250}M$.
The equilibrium number of moles of hydrogen is $0.01-0.005=0.005moles$.
The equilibrium concentration of hydrogen is$\left[H_2\right]=\frac{0.005}{250}M$.
The expression for the equilibrium constant is $K_c=(\frac{\left[H_{2}S\right]}{H_2}{)}^3=(\frac{\frac{0.005}{250}}{\frac{0.005}{250}}{)}^3=1$.