Question

When $3.06g$ of ${NH}_{4}HS\left(s\right)$ is introduced into a $twolitre$ evacuated flask at ${27}^{o}C$, $30$% of the solid decomposes into gaseous ammonia and hydrogen sulphide.
(i) Calculate $K_c$ and $K_p$ for the reaction at ${27}^{o}C$.
(ii) What would happen to the equilibrium when more ${NH}_{4}HS\left(s\right)$ is introduced into the flask?

Answer 1

${NH}_{4}HS\left(s\right)\rightleftharpoons {NH}_3\left(g\right)+H_{2}S\left(g\right)$
Moles of ${NH}_{4}HS=\frac{3.06}{51}=0.06$

Degree of dissociation $=0.3$

(i) At equilibrium
$\left[{NH}_3\right(g\left)\right]=\frac{0.3\times 0.06}{2};\left[H_{2}S\right(g\left)\right]=\frac{0.3\times 0.06}{2}$

$K_c=\left[{NH}_3\right(g\left)\right]\left[H_{2}S\right(g\left)\right]=8.1\times {10}^{-5}$

Now applying,
$K_p=K_c{\left(RT\right)}^{\Delta n}=8.1\times {10}^{-5}\times {(0.082\times 300)}^2=0.049$

(ii) If more amount of $N{H}_{4}HS$ is introduced in the flask at equilibrium then, the reaction will favour forward reaction in order to decrease the concentration of $N{H}_{4}HS$. This is in accordance to Le-chatelier's principle.