Question

What is the total pressure over the solid mixture of ammonium sulphate and ammonium selenide at the equilibrium? Ammonium sulphate and ammonium selenide on heating dissociates as
${{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{S(s)}\,\,\rightleftharpoons \,\text{2N}{{\text{H}}_{\text{3}}}\text{(g)}\,\text{+}\,{{\text{H}}_{\text{2}}}\text{S(g);}\,{{\text{K}}_{\text{p}}}\text{=}\,\text{9 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{at}{{\text{m}}^{\text{3}}}$
${{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{Se(s)}\,\,\rightleftharpoons \,\text{2N}{{\text{H}}_{\text{3}}}\text{(g)}\,\text{+}\,{{\text{H}}_{\text{2}}}\text{Se(g);}\,{{\text{K}}_{\text{p}}}\text{=}\,4.5\,\text{ }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{at}{{\text{m}}^{\text{3}}}$

Answer 1

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 $\text{p}=\dfrac{\text{moles}\,\,\text{of}\,\,\text{species}\,\,\text{at}\,\text{equilibrium}}{\text{total}\,\text{moles}}\text{ }\\!\\!\times\\!\\!\text{ }\,\text{total}\,\text{pressure}$
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.

Complete step-by-step answer: 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 ${{\text{p}}_{2}}\,\text{atm}$
$\begin{aligned} & {{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{S(s)}\,\,\rightleftharpoons \,\text{2N}{{\text{H}}_{\text{3}}}\text{(g)}\,\text{+}\,{{\text{H}}_{\text{2}}}\text{S(g);}\,{{\text{K}}_{\text{p}}}\text{=}\,\text{9 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{at}{{\text{m}}^{\text{3}}} \\\ & \,\,\,\,\,\,\,\,\,\,\,\,{{\text{p}}_{\text{1}}}\,\,\,\,\,\,\,\,\,\,\,\,\,{{\text{p}}_{\text{1}}} \end{aligned}$
$\begin{aligned} & {{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{Se(s)}\,\,\rightleftharpoons \,\text{2N}{{\text{H}}_{\text{3}}}\text{(g)}\,\text{+}\,{{\text{H}}_{\text{2}}}\text{Se(g);}\,{{\text{K}}_{\text{p}}}\text{=}\,4.5\,\text{ }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{at}{{\text{m}}^{\text{3}}} \\\ & \,\,\,\,\,\,\,\,{{\text{p}}_{\text{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\text{p}}_{\text{2}}} \end{aligned}$
For equation first we will calculate the pressure equilibrium constant-
$\begin{aligned} & {{\text{K}}_{{{\text{p}}_{1}}}}\,\text{=}\,\text{p}_{_{\text{N}{{\text{H}}_{\text{3}}}}}^{\text{2}}\text{.}\,\,{{\text{p}}_{{{\text{H}}_{\text{2}}}\text{S}}} \\\ & =\,{{\text{(}{{\text{p}}_{\text{1}}}\text{+}\,{{\text{p}}_{\text{2}}}\text{)}}^{\text{2}}}\text{.}{{\text{p}}_{\text{1}}}\,\,\,\,\,\,\text{ }\\!\\!\\{\\!\\!\text{ }\,{{\text{p}}_{\text{N}{{\text{H}}_{\text{3}}}}}\text{= }{{\text{p}}_{\text{1}}}\text{+}\,{{\text{p}}_{\text{2}}}\,\,\,\,\,\,\text{due}\,\text{to}\,\,\text{combined}\,\text{moles}\,\text{in}\,\text{the}\,\text{container}\\} \\\ & 9\times {{10}^{-3}}=\,{{\text{(}{{\text{p}}_{\text{1}}}\text{+}\,{{\text{p}}_{\text{2}}}\text{)}}^{\text{2}}}\text{.}{{\text{p}}_{\text{1}}}...........(i) \end{aligned}$
In the same manner we will calculate the pressure equilibrium constant –
$\begin{aligned} & {{\text{K}}_{{{\text{p}}_{2}}}}\,\text{=}\,\text{p}_{_{\text{N}{{\text{H}}_{\text{3}}}}}^{\text{2}}\text{.}\,\,{{\text{p}}_{{{\text{H}}_{\text{2}}}\text{Se}}} \\\ & =\,{{\text{(}{{\text{p}}_{\text{1}}}\text{+}\,{{\text{p}}_{\text{2}}}\text{)}}^{\text{2}}}\text{.}{{\text{p}}_{2}}\,\,\,\, \\\ & 4.5\times {{10}^{-3}}=\,{{\text{(}{{\text{p}}_{\text{1}}}\text{+}\,{{\text{p}}_{\text{2}}}\text{)}}^{\text{2}}}\text{.}{{\text{p}}_{2}}...........(ii) \end{aligned}$
After dividing equation (I) by equation (II) we get

& \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{aligned}$$ After putting the value of equation (III) into equation (I) we get $\begin{aligned} & 9\times {{10}^{-3}}=\,{{\text{(}{{\text{p}}_{\text{1}}}\text{+}\,{{\text{p}}_{\text{2}}}\text{)}}^{\text{2}}}\text{.}{{\text{p}}_{\text{1}}} \\\ & 9\times {{10}^{-3}}=\,{{\text{(2}{{\text{p}}_{2}}\text{+}\,{{\text{p}}_{\text{2}}}\text{)}}^{\text{2}}}\text{.2}{{\text{p}}_{2}} \\\ & 18\text{p}_{2}^{3}=\,9\times {{10}^{-3}} \\\ & {{\text{p}}_{2}}=\,0.8\times {{10}^{-1}}=\,0.0\text{8}\,\text{atm} \end{aligned}$ So, pressure over the solid mixture is - $$\begin{aligned} & {{\text{p}}_{\text{T}}}=\,2({{\text{p}}_{1}}+{{\text{p}}_{2}}) \\\ & {{\text{p}}_{\text{T}}}=\,2(3{{\text{p}}_{2}})\,\,\,\,\,\,\,\\{\because \,{{\text{p}}_{1}}=\,2{{\text{p}}_{2}}\\} \\\ & {{\text{p}}_{\text{T}}}=\,6\times 0.08 \\\ & {{\text{p}}_{\text{T}}}=\,0.4\text{5}\,\text{atm} \end{aligned}$$ **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.