Question

At ${87}^{o}C,$ the following equilibrium is established:
$H_2\left(g\right)+S\left(s\right)\rightleftharpoons H_{2}S\left(g\right),K_c=8\times {10}^{-2}$
If 0.3 mole of hydrogen and 2 mole of sulphur are heated to ${87}^{o}C$ in a 2 L vessel, what will be the partial pressure of $H_{2}S$ at equilibrium.

• A. 0.32 atm
• B. 1.43 atm
• C. 0.62 atm
• D. 4.0 atm

Answer 1

The correct option is A 0.32 atm

For reaction,
$H_2\left(g\right)+S\left(s\right)\rightleftharpoons H_{2}S\left(g\right)$
$0.3-xx$
$K_C=\frac{\left[H_{2}S\right(g\left)\right]}{\left[H_2\right(g\left)\right]}\Rightarrow 8\times {10}^{-2}=\frac{x}{0.3-x}$
$0.024-0.08x=x$
$0.024=1.08x$
$x=0.022$
$P=\frac{nRT}{V}$
$P_{H_{2}S}=\frac{0.022\times 0.08\times 360}{2}\Rightarrow \approx 0.32atm$