Question

At ${400}^{\circ }C$, the following equilibrium is established : $H_2\left(g\right)+S\left(s\right)\rightleftharpoons H_{2}S\left(g\right)$$K_p=6.8\times {10}^{-2}$. If 0.2 mol of hydrogen and 1.0 mol of sulphur are heated to ${90}^{\circ }C$ in a 1.0 litre vessel, what will be the partial pressure of $H_{2}S$ at equilibrium?

• A. $0.013$ atm
• B. $0.345$ atm
• C. $0.456$ atm
• D. $0.564$ atm

Answer 1

The correct option is A $0.013$ atm

$H_2\left(g\right)+S\left(s\right)\rightleftharpoons H_{2}S\left(g\right)$, as for this reaction $\delta n=0$ so $K_p=K_c$. And
$H_2\left(g\right)+S\left(s\right)\rightleftharpoons H_{2}S\left(g\right)$
$0.2-x$ $1-x$ $x$
$K_c=x/(0.2-x)$

$=6.8\times {10}^{-2}$

$=P_{H_{2}S}/P_{H_2}$, so partial pressure of $H_{2}S$ is $0.013$ atm.