Question

The degree of dissociation of $PC{l_5}$ (g) obeying the equilibrium, $PC{l_5}{\text{ }} \rightleftharpoons {\text{ }}PC{l_3} + {\text{ }}C{l_2}$, is approximately related to the pressure at equilibrium by:
A $\alpha {\text{ }}\infty {\text{ }}P$
B $\alpha {\text{ }}\infty \dfrac{1}{{\sqrt P }}$
C $\alpha {\text{ }}\infty \dfrac{1}{{{P^2}}}$
D $\alpha {\text{ }}\infty \dfrac{1}{{{P^4}}}$

Answer 1

At equilibrium the forward and backward reaction rates become the same. As a result, equilibrium constant can be written as the ratio of product side concentration to reactant side concentration.

Complete step by step answer:
For a reversible reaction at a situation when the amount of product formed is equal to the amount of reactant is formed then it is called equilibrium. At equilibrium the amount of product and reactant concentration become constant.
Now for the reaction $PC{l_5}{\text{ }} \rightleftharpoons {\text{ }}PC{l_3} + {\text{ }}C{l_2}$, let the degree of dissociation of $PC{l_5}$ is ${\text{\alpha }}$. And let the initial mole of $PC{l_5}$ is 1 mole. Therefore, at equilibrium the number of moles of $PC{l_5}$ and $PC{l_3}$ and $C{l_2}$are $1 - \alpha$,$\alpha$and $\alpha$ respectively . P is the total pressure in equilibrium.
At equilibrium the total number of moles is

$$1 - \alpha + \alpha + \alpha \\\ = 1 + \alpha \\\$$

Now at equilibrium the mole fractions of $PC{l_5}$ and $PC{l_3}$, $C{l_2}$are$\dfrac{{1 - \alpha }}{{1 + \alpha }}$,$\dfrac{\alpha }{{1 + \alpha }}$and $\dfrac{\alpha }{{1 + \alpha }}$ respectively.
Now according to the Dalton’s law of partial pressure is ${P_i} = {x_i}P$
At equilibrium The partial pressures of $PC{l_5}$ and $PC{l_3}$ ,$C{l_2}$are${P_{PC{l_5}}} = \dfrac{{1 - \alpha }}{{1 + \alpha }}P$,${P_{PC{l_3}}} = \dfrac{\alpha }{{1 + \alpha }}P$and ${P_{C{l_2}}} = \dfrac{\alpha }{{1 + \alpha }}P$ respectively. Where ${\alpha ^2} < 1$.
There fore the equilibrium constant is,

$${K_P} = \dfrac{{\left[ {PC{l_3}} \right]\left[ {C{l_2}} \right]}}{{\left[ {PC{l_5}} \right]}} \\\ {K_P} = \dfrac{{\left[ {\dfrac{\alpha }{{1 + \alpha }}P} \right]\left[ {\dfrac{\alpha }{{1 + \alpha }}P} \right]}}{{\left[ {\dfrac{{1 - \alpha }}{{1 + \alpha }}P} \right]}} \\\ {K_P} = \dfrac{{{{\left[ {\dfrac{\alpha }{{1 + \alpha }}} \right]}^2}{{\left[ P \right]}^2}}}{{\left[ {\dfrac{{1 - \alpha }}{{1 + \alpha }}} \right]P}} \\\$$ $${K_P} = \dfrac{{{\alpha ^2}\left[ P \right]}}{{(1 - \alpha )(1 + \alpha )}} \\\ {K_P} = \dfrac{{{\alpha ^2}\left[ P \right]}}{{(1 - {\alpha ^2})}} \\\ {K_P} = \dfrac{{{\alpha ^2}\left[ P \right]}}{{(1)}} \\\ {\alpha ^2} = \dfrac{{{K_P}}}{P} \\\ \alpha = \sqrt {\dfrac{{{K_P}}}{P}} \\\ \alpha \infty \dfrac{1}{{\sqrt P }} \\\$$

So, the correct option is B.

Note:
For a reaction, $A + 2B \rightleftharpoons 2C$ let, the rate constant of forward reaction is ${K_f}$ and the rate constant for backward reaction is $\;{k_b}$. therefore, the rates of forward and backward reactions are,
${R_f} = {k_f}\left[ A \right]{\left[ B \right]^2}$and ${\text{ }}{R_b} = {k_b}{\left[ C \right]^2}$respectively. Now, at equilibrium the forward and backward reaction rates become the same. Therefore, the equilibrium constant is ,

$${R_f} = {\text{ }}{R_b} \\\ or,{\text{ }}{k_f}\left[ A \right]{\left[ B \right]^2} = {k_b}{\left[ C \right]^2} \\\ or,\dfrac{{{k_f}}}{{{k_b}}} = \dfrac{{{{\left[ C \right]}^2}}}{{\left[ A \right]{{\left[ B \right]}^2}}} \\\ {k_{eq}} = \dfrac{{{{\left[ C \right]}^2}}}{{\left[ A \right]{{\left[ B \right]}^2}}} \\\$$