Chemical and Ionic Equilibrium - Result Question 44

Solution:

44. (i) Mole of solid $NH _4 HS$ taken initially $=\dfrac{3.06}{51}=0.06$

$ \begin{aligned} & \text { At equilibrium } NH _4 HS(s) \rightleftharpoons \underset{0.018}{NH _3(g)}+ \underset{0.018}{H _2 S(g)} \\ & K _c=\left(\dfrac{0.018}{2}\right)^{2}=8.1 \times 10^{-5} \\ & p\left(NH _3\right)=\dfrac{0.018 \times 0.082 \times 300}{2}=0.22 atm \\ & K _p=(0.22)^{2}=4.84 \times 10^{-2} \end{aligned} $

(ii) Addition of solid $NH _4 HS$ will have no effect on equilibrium.