Question

At ${87}^{o}C,$ the following equilibrium is established $H_2\left(g\right)+S\left(s\right)\rightleftharpoons H_{2}S\left(g\right);K_P=7\times {10}^{-2}$
If 0.50 mole of hydrogen and 1.0 mole of sulphur are heated to ${87}^{o}C$ 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

The correct option is A 0.966 atm

$\begin{array}{ccccc}& H_2\left(g\right)& +& S\left(s\right)\rightleftharpoons & H_{2}S\left(g\right)\\ \text{Concentration at equilibrium}& 0.5-x& & 1-x& x\end{array}$
$K_c=\frac{\left[H_{2}S\right]}{\left[H_2\right]}\Rightarrow 7\times {10}^{-2}=\frac{x}{0.5-x}$
$x=0.0327$
$P_{H_{2}S}=\left(\frac{n_{H_{2}S}}{V}\right)RT\Rightarrow 0.0327\times 0.0821\times 360\Rightarrow 0.966atm.$