Question

$7.07\, g$ of $NH _{4} HS ( s )$ is placed in $1\, dm ^{3}$ flask at $300\, K$. Following equilibrium is set up. $NH _{4} HS ( s ) \rightleftharpoons NH _{3}( g )+ H _{2} S ( g ) ; K_{p}=0.25\, atm ^{2}$
How much $NH _{4} HS ( s )$ (in grams) is left in the flask after equilibrium is set up?

Answer 1

$NH _{4} HS ( s ) \rightleftharpoons NH _{3}( g )+ H _{2} S ( g )$

Pressure is due to $NH _{3}( g )$ and $H _{2} S ( g )$

Also, $p_{ NH _{3}}=p_{ H _{2} S }=p$

$K_{p}=p_{ NH _{3}} \cdot p_{ H _{2} S }=p^{2}$

$p^{2}=0.25$

$p=0.50\, atm$

Thus, total pressure $=p_{ NH _{3}}+p_{ H _{2} S }$

$=1.0\, atm =p V=n R T$

$n\left( NH _{4} HS\right.$, decomposed $)=\frac{p V}{R T}$

$n=\frac{1\, atm \times 1\, dm ^{3}}{0.0821\, dm\,{}^{3}\, atm\, mol ^{-1}\, K ^{-1} \times 300\, K }$

$=0.0406\, mol$

$NH _{4} HS$ decomposed $=0.0406 \times 51=2.07\, g$

$NH _{4} HS$ left unreacted $=7.07-2.07=5.0\, g$