Question

Ammonium hydrogen sulphide dissociates as follows
$N{H}_{4}HS\left(s\right)\rightleftharpoons N{H}_3\left(g\right)+H_{2}S\left(g\right)$
If solid $N{H}_{4}HS$ is placed in an evacuated flask at certain temperature it will dissociate until the total is $600torr$.
(a) Calculate the value of equilibrium constant for the dissociation reaction.
(b) Additional $N{H}_3$ is until introduced into the equilibrium mixture without changing the temperature until partial pressure of $N{H}_3$ is $750torr$, what is the partial pressure of $H_{2}S$ under these conditions? What is the total pressure in the flask?

Answer 1

$N{H}_{4}HS\rightleftharpoons N{H}_3+H_{2}S$

Initial: $a$ $0$ $0$

At equilibrium $(a-p_1)$ $p_1$ $p_1$

since, total pressure $=600torr$. Therefore $p_1+p_1=600⟹p_1=300$

$K_p=p\left[N{H}_3\right]\times P\left[H_{2}S\right]=p_1\times p_1=90000(torr{)}^2$

If partial pressure of $N{H}_3=750torr$, then partial pressure of $H_{2}S=p_2$,

then $K_p=90000=750\times p_2$. Therefore, $p_2=\frac{k_p}{750}=120torr$

Hence, total pressure$=(p_2+750)=120+750=870torr$.