Question

Ammonium sulphate and ammonium selenide on heating dissociates as
${{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{S(s)}\,\,\rightleftharpoons \,\text{2N}{{\text{H}}_{\text{3}}}\text{(g)}\,\text{+}\,{{\text{H}}_{\text{2}}}\text{S(g);}\,{{\text{K}}_{\text{p}}}\text{=}\,\text{9 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{at}{{\text{m}}^{\text{3}}}$
${{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{Se(s)}\,\,\rightleftharpoons \,\text{2N}{{\text{H}}_{\text{3}}}\text{(g)}\,\text{+}\,{{\text{H}}_{\text{2}}}\text{Se(g);}\,{{\text{K}}_{\text{p}}}\text{=}\,4.5\,\text{ }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{at}{{\text{m}}^{\text{3}}}$
The total pressure over the solid mixture at the equilibrium is
(A) $\text{0}\text{.15}\,\text{atm}$
(B) $\text{0}\text{.3}\,\text{atm}$
(C) $\text{0}\text{.45}\,\text{atm}$
(D) $\text{0}\text{.6}\,\text{atm}$

Answer 1

Hint: The active mass of solid and pure liquids is constant quantity (unity) because it is an intensive property and does change its concentration with time.
- According to the law of mass action, the rate of a chemical reaction is directly proportional to the active masses of the reacting substances raised to the power equal to the stoichiometry coefficient in the balance chemical reaction.
${{\text{K}}_{\text{p}}}$ In this equation represents pressure equilibrium constant. The magnitude of ${{\text{K}}_{\text{p}}}$ is a measure of the extent to which the chemical reaction occurs. So, ${{\text{K}}_{\text{p}}}$ for a reaction is calculated by Applying law of mass action in the following manner –
${{\text{K}}_{\text{p}}}=\dfrac{{{\text{(}{{\text{p}}_{\text{c}}}\text{)}}^{{{\text{n}}_{\text{1}}}}}{{\text{(}{{\text{p}}_{\text{c}}}\text{)}}^{{{\text{n}}_{\text{2}}}}}}{{{\text{(}{{\text{p}}_{\text{c}}}\text{)}}^{{{\text{m}}_{\text{1}}}}}{{\text{(}{{\text{p}}_{\text{c}}}\text{)}}^{{{\text{m}}_{\text{2}}}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\{\text{where,}\,\,\text{p = partial}\,\text{pressure}\}$
Partial pressure of any species of the reactant or product at equilibrium is $\text{p}=\dfrac{\text{moles}\,\,\text{of}\,\,\text{species}\,\,\text{at}\,\text{equilibrium}}{\text{total}\,\text{moles}}\text{ }\!\!\times\!\!\text{ }\,\text{total}\,\text{pressure}$

Complete Solution :
Supposed the partial pressure of $\text{N}{{\text{H}}_{\text{3}}}$ and ${{\text{H}}_{\text{2}}}\text{S}$ due to the dissociation of ${{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{S(s)}$ are ${{\text{p}}_{\text{1}}}\,\text{atm}$ each and partial pressure of $\text{N}{{\text{H}}_{\text{3}}}$ and ${{\text{H}}_{\text{2}}}\text{Se}$ due to the dissociation of compound ${{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{Se(s)}\,$are ${{\text{p}}_{2}}\,\text{atm}$
$\begin{align}
& {{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{S(s)}\,\,\rightleftharpoons \,\text{2N}{{\text{H}}_{\text{3}}}\text{(g)}\,\text{+}\,{{\text{H}}_{\text{2}}}\text{S(g);}\,{{\text{K}}_{\text{p}}}\text{=}\,\text{9 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{at}{{\text{m}}^{\text{3}}} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,{{\text{p}}_{\text{1}}}\,\,\,\,\,\,\,\,\,\,\,\,\,{{\text{p}}_{\text{1}}}
\end{align}$
$\begin{align}
& {{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{Se(s)}\,\,\rightleftharpoons \,\text{2N}{{\text{H}}_{\text{3}}}\text{(g)}\,\text{+}\,{{\text{H}}_{\text{2}}}\text{Se(g);}\,{{\text{K}}_{\text{p}}}\text{=}\,4.5\,\text{ }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{at}{{\text{m}}^{\text{3}}} \\
 & \,\,\,\,\,\,\,\,{{\text{p}}_{\text{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\text{p}}_{\text{2}}}
\end{align}$

For equation first we will calculate the pressure equilibrium constant-
$\begin{align}
& {{\text{K}}_{{{\text{p}}_{1}}}}\,\text{=}\,\text{p}_{_{\text{N}{{\text{H}}_{\text{3}}}}}^{\text{2}}\text{.}\,\,{{\text{p}}_{{{\text{H}}_{\text{2}}}\text{S}}} \\
 & =\,{{\text{(}{{\text{p}}_{\text{1}}}\text{+}\,{{\text{p}}_{\text{2}}}\text{)}}^{\text{2}}}\text{.}{{\text{p}}_{\text{1}}}\,\,\,\,\,\,\text{ }\!\!\{\!\!\text{ }\,{{\text{p}}_{\text{N}{{\text{H}}_{\text{3}}}}}\text{= }{{\text{p}}_{\text{1}}}\text{+}\,{{\text{p}}_{\text{2}}}\,\,\,\,\,\,\text{due}\,\text{to}\,\,\text{combined}\,\text{moles}\,\text{in}\,\text{the}\,\text{container}\} \\
& 9\times {{10}^{-3}}=\,{{\text{(}{{\text{p}}_{\text{1}}}\text{+}\,{{\text{p}}_{\text{2}}}\text{)}}^{\text{2}}}\text{.}{{\text{p}}_{\text{1}}}...........(i)
\end{align}$

- In the same manner we will calculate the pressure equilibrium constant –
$\begin{align}
& {{\text{K}}_{{{\text{p}}_{2}}}}\,\text{=}\,\text{p}_{_{\text{N}{{\text{H}}_{\text{3}}}}}^{\text{2}}\text{.}\,\,{{\text{p}}_{{{\text{H}}_{\text{2}}}\text{Se}}} \\
& =\,{{\text{(}{{\text{p}}_{\text{1}}}\text{+}\,{{\text{p}}_{\text{2}}}\text{)}}^{\text{2}}}\text{.}{{\text{p}}_{2}}\,\,\,\, \\
& 4.5\times {{10}^{-3}}=\,{{\text{(}{{\text{p}}_{\text{1}}}\text{+}\,{{\text{p}}_{\text{2}}}\text{)}}^{\text{2}}}\text{.}{{\text{p}}_{2}}...........(ii)
\end{align}$

After dividing equation (I) by equation (II) we get
\[\begin{align}
& \dfrac{\text{9 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}}{\text{4}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}}\text{=}\dfrac{{{\text{(}{{\text{p}}_{\text{1}}}\text{+}{{\text{p}}_{\text{2}}}\text{)}}^{\text{2}}}\text{.}\,{{\text{p}}_{\text{1}}}}{{{\text{(}{{\text{p}}_{\text{1}}}\text{+}{{\text{p}}_{\text{2}}}\text{)}}^{\text{2}}}\text{.}\,{{\text{p}}_{\text{2}}}} \\
 & \dfrac{{{\text{p}}_{\text{1}}}}{{{\text{p}}_{\text{2}}}}\text{=}\dfrac{\text{9}}{\text{4}\text{.5}}\text{=}\dfrac{\text{2}}{\text{1}} \\
 & \therefore \,\,\,\,\,{{\text{p}}_{\text{1}}}\text{=}\,\,\text{2}{{\text{p}}_{\text{2}}}\,\,\,................(iii)
\end{align}\]
After putting the value of equation (III) into equation (I) we get
$\begin{align}
& 9\times {{10}^{-3}}=\,{{\text{(}{{\text{p}}_{\text{1}}}\text{+}\,{{\text{p}}_{\text{2}}}\text{)}}^{\text{2}}}\text{.}{{\text{p}}_{\text{1}}} \\
& 9\times {{10}^{-3}}=\,{{\text{(2}{{\text{p}}_{2}}\text{+}\,{{\text{p}}_{\text{2}}}\text{)}}^{\text{2}}}\text{.2}{{\text{p}}_{2}} \\
 & 18\text{p}_{2}^{3}=\,9\times {{10}^{-3}} \\
 & {{\text{p}}_{2}}=\,0.8\times {{10}^{-1}}=\,0.0\text{8}\,\text{atm}
\end{align}$
So, pressure over the solid mixture is -
\[\begin{align}
& {{\text{p}}_{\text{T}}}=\,2({{\text{p}}_{1}}+{{\text{p}}_{2}}) \\
& =\,2(3{{\text{p}}_{2}})\,\,\,\,\,\,\,\{\because \,{{\text{p}}_{1}}=\,2{{\text{p}}_{2}}\} \\
& =\,6\times 0.08 \\
& {{\text{p}}_{\text{T}}}=\,0.4\text{5}\,\text{atm}
\end{align}\]
So, the correct answer is “Option C”.

Note: In this given question two solids are taken together in a closed container, and both the solids decompose to gives gases $\text{N}{{\text{H}}_{\text{3}}}$, ${{\text{H}}_{\text{2}}}\text{S}$ and ${{\text{H}}_{\text{2}}}\text{Se}$. As the gas $\text{N}{{\text{H}}_{\text{3}}}$ is the common gas in the dissociation of solids, so the dissociation of both the solids will be suppressed.