Question

$N{H}_{4}HS\left(s\right)$ dissociates as follows: $N{H}_{4}HS\left(s\right)\rightleftharpoons N{H}_3\left(g\right)+H_{2}S\left(g\right)$. At ${25}^{\circ }C$, the dissociation pressure of pure solid is $200\sqrt{1.75}$ mm of $Hg$. If the flask contains $300$ mm of Hg of $N{H}_3\left(g\right)$ along with solid $N{H}_{4}HS$, calculate the total pressure at equilibrium: (in mm of Hg).

• A. $600$
• B. $200\sqrt{1.75}$
• C. $400$
• D. $432.3$

Answer 1

Answer: (C)

$400$

(i) First equilibrium:
Let P be the pressure of $N{H}_3$ and $H_{2}S$.

Total pressure is $2P=200\sqrt{1.75}\Rightarrow P=100\sqrt{1.75}$

$K_P=P^2=(100\sqrt{1.75}{)}^2=17500$

(ii) Second equilibrium:

$K_p=(300+P)P=17500$

$P^2+300P-17500=0$

$P=50$

Total pressure $=300+2P=300+2\left(50\right)$ $=400mmHg$