Question

At ${87}^{\circ }C$, the following equilibrium is established $H_2\left(g\right)+S\left(s\right)\rightleftharpoons H_{2}S\left(g\right)$ $K_p=7\times {10}^{-2}$.
If 0.50 mole of hydrogen and 1.0 mole of sulfur are heated to 87${}^{\circ }C$ in 1.0 L vessel, what will be the partial pressure of $H_{2}S$ at equilibrium?

• A. 0.966 atm
• B. 1.38 atm
• C. 0.0327 atm
• D. 9.66 atm

Answer 1

Answer: (A)

0.966 atm

The equilibrium reaction is $H_2\left(g\right)+S\left(s\right)\rightleftharpoons H_{2}S\left(g\right)$

Initially, 0.50 moles of hydrogen are present.

At equilibrium, $0.50-x$ moles of hydrogen and x moles of hydrogen sulfide are present.

The expression for the equilibrium constant is $K_P=\frac{P_{H_{2}S}}{P_{H_2}}$.

$P_{H_{2}S}=X_{H_{2}S}\times P=\frac{x}{0.5}P$

$P_{H_2}=X_{H_2}\times P=\frac{0.5-x}{0.5}P$

Substitute values in the above expression.

$7\times {10}^{-2}=\frac{x}{0.5-x}$.

The number of moles of hydrogen sulphide $=x=0.0327$.

Total pressure can be calculated from Ideal gas equation-

$P=\frac{nRT}{V}=\frac{0.5\times 0.0821\times 360}{1}=14.77atm$.
Partial pressure of $H_{2}S$, $P_{H_{2}S}=X_{H_{2}S}\times P=\frac{x}{0.5}P=\frac{0.0327}{0.5}\times 14.77=0.966atm$

Hence, the correct option is A.