Question

Consider the reaction,
${{\text{N}}_{\text{2}}}{\text{(g)}}\,{\text{ + }}\,{\text{3}}{{\text{H}}_{\text{2}}}{\text{(g)}}\, \rightleftharpoons {\text{2}}\,{\text{N}}{{\text{H}}_{\text{3}}}{\text{(g)}}$
The equilibrium constant of the above reaction is ${{\text{K}}_{\text{P}}}$. If pure ammonia is left to dissociated, the partial pressure of ammonia at equilibrium is given by:
(Assume that ${{\text{P}}_{{\text{N}}{{\text{H}}_{\text{3}}}}} < < {{\text{P}}_{{\text{total}}}}$ at equilibrium)
A. $\dfrac{{{{\text{3}}^{3/2}}{{\text{K}}_{\text{P}}}^{1/2}{{\text{P}}^2}}}{4}$
B. $\dfrac{{{{\text{3}}^{3/2}}{{\text{K}}_{\text{P}}}^{1/2}{{\text{P}}^2}}}{{16}}$
C. $\dfrac{{{{\text{K}}_{\text{P}}}^{1/2}{{\text{P}}^2}}}{{16}}$
D. $\dfrac{{{{\text{K}}_{\text{P}}}^{1/2}{{\text{P}}^2}}}{4}$

Answer 1

We will determine the equilibrium concentrations for the reaction. Then we will write equilibrium constant expressions. By substituting the equilibrium concentration we will determine the partial pressure of left ammonia.

Complete answer:
The equilibrium constant ${{\text{K}}_{\text{P}}}$ is defined as the product of the pressure of products having a stoichiometric coefficient as power divided by the product of the pressure of reactants having a stoichiometric coefficient as power.
Consider a general reaction as follows:
${\text{aA}}\, + \,{\text{bB}}\,\, \rightleftharpoons {\text{cC}}\,{\text{ + }}\,{\text{dD}}$

The equilibrium constant expression for the above reaction is as follows:
${{\text{k}}_{\text{P}}}{\text{ = }}\,\dfrac{{{{\left[ {\text{C}} \right]}^{\text{c}}}\,\,{{\left[ {\text{D}} \right]}^{\text{d}}}}}{{{{\left[ {\text{A}} \right]}^{\text{a}}}{{\left[ {\text{B}} \right]}^{\text{b}}}}}\,$
Where, $\left[ {\;} \right]$ represents the pressure.

The given reaction is as follows:
${{\text{N}}_{\text{2}}}{\text{(g)}}\,{\text{ + }}\,{\text{3}}{{\text{H}}_{\text{2}}}{\text{(g)}}\, \rightleftharpoons {\text{2}}\,{\text{N}}{{\text{H}}_{\text{3}}}{\text{(g)}}$
The equilibrium constant expression for the above reaction is as follows:
${{\text{k}}_{\text{P}}}{\text{ = }}\,\dfrac{{{{\left[ {{\text{N}}{{\text{H}}_{\text{3}}}} \right]}^{\text{2}}}\,\,}}{{\left[ {{{\text{N}}_{\text{2}}}} \right]{{\left[ {{{\text{H}}_{\text{2}}}} \right]}^{\text{3}}}}}\,$
Amount is taken in the form of pressure. So,
${{\text{k}}_{\text{P}}}{\text{ = }}\,\dfrac{{{{\left[ {{{\text{P}}_{{\text{N}}{{\text{H}}_{\text{3}}}}}} \right]}^{\text{2}}}\,\,}}{{\left[ {{{\text{P}}_{{{\text{N}}_2}}}} \right]{{\left[ {{{\text{P}}_{{{\text{H}}_2}}}} \right]}^{\text{3}}}}}\,$…..$(1)$
We will write the initial pressure and equilibrium pressure of each species to determine the total pressure at equilibrium as follows:
As the ammonia is left over so,
$\:\:\;\;\:\:\:\:\:\: \:\:\:\: {\text{2}}\,{\text{N}}{{\text{H}}_{\text{3}}}{\text{(g)}}\, \rightleftharpoons {{\text{N}}_{\text{2}}}{\text{(g)}}\,{\text{ + }}\,{\text{3}}{{\text{H}}_{\text{2}}}{\text{(g)}}$
${\text{t = 0,}}\,\,\,\,\,\,\,\,\,{{\text{P}}_{{\text{N}}{{\text{H}}_{\text{3}}}}}\,\,\,\,\,\,\,\:\:\:\:\:\:\:\:0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\:\:\:\:0$
${\text{t = equi,}}\,\,{{\text{P}}_{{\text{N}}{{\text{H}}_{\text{3}}}}}\,\,\,\:\:\:\:\:\:\:\:\:\dfrac{{\text{P}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\:\:\dfrac{{{\text{3P }}}}{2}$
Initial all pressure of reaction mixture will be due to ammonia. At equilibrium, this will be due to all species.
So, the total pressure at equilibrium is,
${{\text{P}}_{{\text{equi}}}} = \,\,\,{{\text{P}}_{{\text{N}}{{\text{H}}_{\text{3}}}}} + \,\dfrac{{\text{P}}}{2}\,\, + \,\dfrac{{{\text{3P }}}}{2}$
It is given that ${{\text{P}}_{{\text{N}}{{\text{H}}_{\text{3}}}}} < < {{\text{P}}_{{\text{total}}}}$ at equilibrium so, total pressure at equilibrium is,
${{\text{P}}_{{\text{equi}}}} = \,\,\dfrac{{\text{P}}}{2}\,\, + \,\dfrac{{{\text{3P }}}}{2}$
Substitute equilibrium pressure of each in equation $(1)$ .
$\Rightarrow {{\text{k}}_{\text{P}}}{\text{ = }}\,\dfrac{{{{\left[ {{{\text{P}}_{{\text{N}}{{\text{H}}_{\text{3}}}}}} \right]}^{\text{2}}}\,\,}}{{\left[ {\dfrac{{\text{P}}}{2}} \right]{{\left[ {\dfrac{{{\text{3P}}}}{2}} \right]}^{\text{3}}}}}\,$
$\Rightarrow {\left[ {{{\text{P}}_{{\text{N}}{{\text{H}}_{\text{3}}}}}} \right]^{\text{2}}} = \,{{\text{k}}_{\text{P}}} \times \left[ {\dfrac{{\text{P}}}{2}} \right]{\left[ {\dfrac{{{\text{3P}}}}{2}} \right]^{\text{3}}}\,$
$\Rightarrow {\left[ {{{\text{P}}_{{\text{N}}{{\text{H}}_{\text{3}}}}}} \right]^{\text{2}}} = \,{{\text{k}}_{\text{P}}} \times {{\text{3}}^3}\, \times {\left[ {\dfrac{{\text{P}}}{4}} \right]^4}\,$
$\Rightarrow {\left[ {{{\text{P}}_{{\text{N}}{{\text{H}}_{\text{3}}}}}} \right]^{\text{2}}} = \,\dfrac{{{{\text{3}}^3}{{\text{k}}_{\text{P}}}{{\text{P}}^4}}}{{256}}$
$\Rightarrow \left[ {{{\text{P}}_{{\text{N}}{{\text{H}}_{\text{3}}}}}} \right] = \dfrac{{{{\text{3}}^{3/2}}{{\text{K}}_{\text{P}}}^{1/2}{{\text{P}}^2}}}{{16}}$
So, the partial pressure of ammonia at equilibrium is given by $\dfrac{{{{\text{3}}^{3/2}}{{\text{K}}_{\text{P}}}^{1/2}{{\text{P}}^2}}}{{16}}$.

**Therefore, option (B) $\dfrac{{{{\text{3}}^{3/2}}{{\text{K}}_{\text{P}}}^{1/2}{{\text{P}}^2}}}{{16}}$ is correct.

Note:**
At equilibrium total pressure of the mixture is due to all species. As the ammonia is dissociating so, the amount of ammonia is decreasing, so we can neglect the pressure of ammonia at equilibrium. We have to determine the partial pressure of leftover ammonia so, we consider ammonia is dissociating into nitrogen and hydrogen to determine the equilibrium concentration. Amount of product is divided by the stoichiometric of the reactant and multiplied by the stoichiometric of the product.