Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\begin{array}{c}{\left({\text{NH}}_4\right)}_2\text{S}\left(\text{s}\right)\rightleftharpoons {\text{2NH}}_3\left(\text{g}\right)+{\text{H}}_2\text{S}\left(\text{g}\right)\\ \text{2p}\text{p}\end{array}$

$\begin{array}{c}{\left({\text{NH}}_4\right)}_2\text{Se}\left(\text{s}\right)\rightleftharpoons {\text{2NH}}_3\left(\text{g}\right)+{\text{H}}_2\text{Se}\left(\text{g}\right)\\ \text{2q}\text{q}\end{array}$

${\text{K}}_{{\text{p}}_1}={\left({\text{p}}_{{\text{NH}}_3}\right)}^2·{\text{p}}_{{\text{H}}_2\text{S}}={\left(2\text{p}+2\text{q}\right)}^2·\text{p}=\text{9.0}\times {\text{10}}^{-3}$

${\text{K}}_{{\text{p}}_{\text{2}}}={\left({\text{p}}_{{\text{NH}}_3}\right)}^2·{\text{p}}_{{\text{H}}_2\text{S}\text{e}}={\left(2\text{p}+2\text{q}\right)}^2·\text{q}=\text{4.5}\times {\text{10}}^{-3}$

$\therefore \frac{\text{p}}{\text{q}}=2\Rightarrow \text{p}=2\text{q}$

$\therefore {\text{K}}_{{\text{p}}_1}={\left(\text{2p}+\text{p}\right)}^2·\text{p}=\text{9.0}\times {\text{10}}^{-3}$

$\therefore \text{p}=\text{0.1}\text{atm}\text{and}\text{q}=\frac{\text{0.1}}{2}=\text{0.05}$

Total pressure = 2p + 2q + p + q = 3p + 3q

= 3 × 0.1 + 3 × 0.05 = 0.45 atm