Question

At ${90}^{o}C$, the following equilibrium is established
$H_2\left(g\right)+S\left(s\right)\rightleftharpoons H_{2}S\left(g\right)$ $K_p=6.8\times {10}^{-2}$
If $0.2$ mol of hydrogen and $1.0$ mol of sulphur are heated to ${90}^{o}C$ in a $1.0$ litre vessel, what will be the partial pressure of $H_{2}S$ at equilibrium?

Answer 1

$K_C=\frac{\left[H_{2}S\right]}{\left[H_2\right]\left[S\right]}$

$\because K_P=K_C(RT{)}^{\Delta ng}$ where $\Delta ng$= Difference in gaseous moles of products and reactants

Here $\Delta ng=\left(1\right)-\left(1\right)=0$

$\therefore K_P=K_C=6.8\times {10}^{-2}=\frac{\left[H_{2}S\right]}{0.2\times 1}$

$\therefore \left[H_{2}S\right]=1.36\times {10}^{-2}$= Number of moles of $H_{2}S$

Partial pressure of $H_{2}S=$ (mole fraction of $H_{2}S$) $\times$ Total pressure

$=X_{H_{2}S}\times P_t.....\left(1\right)$

$P_t=1atm$

$P_{H_{2}S}=\frac{1.36\times {10}^{-2}}{1.36\times {10}^{-2}+0.2+1}=0.0112atm$