Question

$7.07g$ of $N{H}_{4}HS(s)$ is placed in $1d{m}^3$ flask at $300K$. Following equilibrium is set up. $N{H}_{4}HS(s)\rightleftharpoons N{H}_3(g)+H_{2}S(g);K_p=0.25at{m}^2$
How much $N{H}_{4}HS(s)$ (in grams) is left in the flask after equilibrium is set up?

Answer 1

Answer: 5

$N{H}_{4}HS(s)\rightleftharpoons N{H}_3(g)+H_{2}S(g)$

Pressure is due to $N{H}_3(g)$ and $H_{2}S(g)$

Also, $p_{N{H}_3}=p_{H_{2}S}=p$

$K_p=p_{N{H}_3}\cdot p_{H_{2}S}=p^2$

$p^2=0.25$

$p=0.50atm$

Thus, total pressure $=p_{N{H}_3}+p_{H_{2}S}$

$=1.0atm=pV=nRT$

$n(N{H}_{4}HS$, decomposed $)=\frac{pV}{RT}$

$n=\frac{1atm\times 1d{m}^3}{0.0821dm{}^{3}atmmo{l}^{-1}{K}^{-1}\times 300K}$

$=0.0406mol$

$N{H}_{4}HS$ decomposed $=0.0406\times 51=2.07g$

$N{H}_{4}HS$ left unreacted $=7.07-2.07=5.0g$