Question

a’ moles of ${\text{PC}}{{\text{l}}_{\text{5}}}$ are heated in a closed container to equilibrate ${\text{PC}}{{\text{l}}_{\text{5}}}\left( {\text{g}} \right) \rightleftharpoons {\text{PC}}{{\text{l}}_{\text{3}}}\left( {\text{g}} \right) + {\text{C}}{{\text{l}}_{\text{2}}}\left( {\text{g}} \right)$ at a pressure of $P{\text{ atm}}$. If $x{\text{ moles}}$ of ${\text{PC}}{{\text{l}}_{\text{5}}}$ dissociate at equilibrium, then:
A. $\dfrac{x}{a} = {\left( {\dfrac{{{K_P}}}{P}} \right)^{\dfrac{1}{2}}}$
B. $\dfrac{x}{a} = \dfrac{{{K_P}}}{{{K_P} + P}}$
C. $\dfrac{x}{a} = {\left( {\dfrac{{{K_P}}}{{{K_P} + P}}} \right)^{\dfrac{1}{2}}}$
D. $\dfrac{x}{a} = {\left( {\dfrac{{{K_P} + P}}{{{K_P}}}} \right)^{\dfrac{1}{2}}}$

Answer 1

Hint: $\dfrac{x}{a}$ is the degree of dissociation. We know that the reaction is at equilibrium. Thus, we can write the equation for the equilibrium constant in terms of pressure.

Complete step by step solution:
We have the reaction,
${\text{PC}}{{\text{l}}_{\text{5}}}\left( {\text{g}} \right) \rightleftharpoons {\text{PC}}{{\text{l}}_{\text{3}}}\left( {\text{g}} \right) + {\text{C}}{{\text{l}}_{\text{2}}}\left( {\text{g}} \right)$
In the reaction, ${\text{PC}}{{\text{l}}_{\text{5}}}\left( {\text{g}} \right)$ dissociates into ${\text{PC}}{{\text{l}}_{\text{3}}}\left( {\text{g}} \right)$ and ${\text{C}}{{\text{l}}_{\text{2}}}\left( {\text{g}} \right)$.
Thus, the equilibrium table is as follows:
${\text{PC}}{{\text{l}}_{\text{5}}}\left( {\text{g}} \right) \rightleftharpoons {\text{PC}}{{\text{l}}_{\text{3}}}\left( {\text{g}} \right) + {\text{C}}{{\text{l}}_{\text{2}}}\left( {\text{g}} \right)$
$\;\;\;\;\; 1 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 0 \;\;\;\;\;\;\;\;\;\;\;\;\; 0 $ …… Initial moles
$1 - \alpha \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \alpha \;\;\;\;\;\;\;\;\;\;\;\; \alpha $ …… Moles at equilibrium
Thus, the total moles of the reactant and product at equilibrium are as follows:
Total moles $ = 1 - \alpha + \alpha + \alpha $
Total moles $ = 1 + \alpha $
Where, $\alpha $ is the degree of dissociation. And $\alpha = \dfrac{x}{a}$
The partial pressure exerted by ${\text{PC}}{{\text{l}}_{\text{5}}}\left( {\text{g}} \right)$ at equilibrium is,
${P_{{\text{PC}}{{\text{l}}_{\text{5}}}}} = \left( {\dfrac{{1 - \alpha }}{{1 + \alpha }}} \right)P$
The partial pressure exerted by ${\text{PC}}{{\text{l}}_{\text{3}}}\left( {\text{g}} \right)$ at equilibrium is,
${P_{{\text{PC}}{{\text{l}}_{\text{3}}}}} = \left( {\dfrac{\alpha }{{1 + \alpha }}} \right)P$
The partial pressure exerted by ${\text{C}}{{\text{l}}_{\text{2}}}\left( {\text{g}} \right)$ at equilibrium is,
${P_{{\text{C}}{{\text{l}}_{\text{2}}}}} = \left( {\dfrac{\alpha }{{1 + \alpha }}} \right)P$
Thus, the equilibrium constant for the reaction in terms of partial pressure can be written as follows:
$\Rightarrow {K_P} = \dfrac{{{P_{{\text{PC}}{{\text{l}}_3}}}{P_{{\text{C}}{{\text{l}}_2}}}}}{{{P_{{\text{PC}}{{\text{l}}_5}}}}}$
Where, ${K_P}$ is the equilibrium constant in terms of partial pressure.
$\Rightarrow {K_P} = \dfrac{{\left( {\dfrac{\alpha }{{1 + \alpha }}} \right)P \times \left( {\dfrac{\alpha }{{1 + \alpha }}} \right)P}}{{\left( {\dfrac{{1 - \alpha }}{{1 + \alpha }}} \right)P}}$
$\Rightarrow {K_P} = \dfrac{{{\alpha ^2}}}{{1 - {\alpha ^2}}}P$
$\Rightarrow {K_P}\left( {1 - {\alpha ^2}} \right) = {\alpha ^2}P$
$\Rightarrow \alpha = {\left( {\dfrac{{{K_P}}}{{{K_P} + P}}} \right)^{\dfrac{1}{2}}}$
But $\alpha = \dfrac{x}{a}$. Thus,
$\Rightarrow \dfrac{x}{a} = {\left( {\dfrac{{{K_P}}}{{{K_P} + P}}} \right)^{\dfrac{1}{2}}}$
Thus, if $x{\text{ moles}}$ of ${\text{PC}}{{\text{l}}_{\text{5}}}$ dissociate at equilibrium, then $\dfrac{x}{a} = {\left( {\dfrac{{{K_P}}}{{{K_P} + P}}} \right)^{\dfrac{1}{2}}}$.

Thus, the correct option is (C) $\dfrac{x}{a} = {\left( {\dfrac{{{K_P}}}{{{K_P} + P}}} \right)^{\dfrac{1}{2}}}$.

Note:
The rate of forward and the backward reactions are equal at equilibrium. The equilibrium constant expresses the relationship between pressure or concentration of products and reactants present when the reaction is at equilibrium. Larger the value of the equilibrium constant higher is the formation of the product. The equilibrium constant is specific for a reaction and is constant at constant temperature. For the same reaction, the equilibrium constant has different values at different temperatures. If concentration, pressure and temperature is changed the equilibrium is affected and the reaction can be a forward or backward reaction.