Question

If $0.2mol$ of $H_2\left(g\right)$ and $2mol$ of $S\left(s\right)$ are mixed in a $1d{m}^3$ vessel at ${90}^{\circ }C$, the partial pressure of $H_{2}S\left(g\right)$ formed according to the reaction, $H_2\left(g\right)+S\left(s\right)\rightleftharpoons H_{2}S,K_P=6.8\times {10}^{-2}$ would be:

• A. $0.19atm$
• B. $0.38atm$
• C. $0.6atm$
• D. $0.072atm$

Answer 1

The correct option is A $0.19atm$

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

number mole of $H_{2}n=0.2mol$

temp $t={90}^{O}C+273=363K$

volume = $1d{m}^3=1litre$

$P=\frac{nRT}{V}=\frac{0.2\times 0.082\times 363}{1}$

$5.96atM$

$H_2\left(g\right)+S\left(s\right)\rightleftharpoons H_2\left(s\right)$

$5.96-P_0$ $P_0$

$K_p=\frac{P_0}{5.96-P_0}$

$6.8\times {10}^{-2}=\frac{P_0}{5.96-P_0}$

$40.53\times {10}^{-2}-0.068P_0=P_0$

$1.068P_0=40.53\times {10}^{-2}$

$P_0=\frac{40.53\times {10}^{-2}}{1.068}=0.3795$

$\approx 0.38atm$