Question

The degree of dissociation of ${\text{PC}}{{\text{l}}_{\text{5}}}{\text{ (\alpha )}}$ obeying the equilibrium ${\text{PC}}{{\text{l}}_{\text{5}}}{\text{ }} \rightleftharpoons {\text{ PC}}{{\text{l}}_3}{\text{ + C}}{{\text{l}}_2}$ is related to the equilibrium pressure by:
A.$\alpha \propto \dfrac{1}{{{P^4}}}$
B.$\alpha \propto \dfrac{1}{{\sqrt P }}$
C.$\alpha \propto \dfrac{1}{{{P^2}}}$
D.$\alpha \propto P$

Answer 1

Calculate the number of moles at equilibrium in terms of ${{\alpha }}$ and find the pressure at equilibrium using Dalton law.
Formula used: Dalton law of partial pressure: ${P_A} = {\chi _A} \times {P_T}$ where, ${P_A}$ is the partial pressure of component A, ${\chi _A}$ is mole fraction of component A and ${P_T}$ is the total pressure of the system.

Complete step by step answer:
${{\alpha }}$ is the degree of dissociation. It is the ratio of dissociated moles to the total no. of moles of reactant. Since it is not directly related to the pressure, we need to derive the relation. For this, let us assume that we have $a$ moles of ${\text{PC}}{{\text{l}}_{\text{5}}}$ initially when time ${\text{t = 0}}$. We only have reactants at this stage. But as reaction proceeds, equilibrium will be attained at time ${\text{t = }}{{\text{t}}_{eq}}$. Some amount of ${\text{PC}}{{\text{l}}_{\text{5}}}$ will get converted into ${\text{PC}}{{\text{l}}_3}$ and ${\text{C}}{{\text{l}}_2}$. Let $x$ moles of ${\text{PC}}{{\text{l}}_{\text{5}}}$ be dissociated. So, $x$ moles of ${\text{PC}}{{\text{l}}_3}$ and ${\text{C}}{{\text{l}}_2}$ will be formed. The reaction is given below as:

{\text{t = 0 }}a{\text{ }} \\\ {\text{t = }}{{\text{t}}_{eq}}{\text{ }}a - x{\text{ }}x{\text{ }}x \\\\$$ Total no. of moles can be calculated by adding the number of moles of each reactant and product in the given chemical equation. Total number of moles before equilibrium is $$a{\text{ }} + {\text{ }}0{\text{ }} + {\text{ }}0{\text{ }} = {\text{ }}a$$ moles Total number of moles after equilibrium is $$a{\text{ }} - {\text{ }}x{\text{ }} + {\text{ }}x{\text{ }} + {\text{ }}x{\text{ }} = {\text{ }}a{\text{ }} + {\text{ }}x$$ moles Now, we need to calculate the mole fraction of each of the components at equilibrium to know the partial pressure of each of the components at equilibrium. $$ {\text{mole fraction of PC}}{{\text{l}}_5}{\text{ = }}\dfrac{{{\text{no}}{\text{. of moles of PC}}{{\text{l}}_{\text{5}}}}}{{{\text{total no}}{\text{. of moles}}}} \\\ {\text{ = }}\dfrac{{a{\text{ }} - {\text{ }}x}}{{a{\text{ }} + {\text{ }}x}} \\\\$$ $${\text{mole fraction of PC}}{{\text{l}}_3}{\text{ = }}\dfrac{{{\text{no}}{\text{. of moles of PC}}{{\text{l}}_3}}}{{{\text{total no}}{\text{. of moles}}}} \\\ {\text{ = }}\dfrac{x}{{a{\text{ }} + {\text{ }}x}} \\\\$$ $${\text{mole fraction of C}}{{\text{l}}_2}{\text{ = }}\dfrac{{{\text{no}}{\text{. of moles of C}}{{\text{l}}_2}}}{{{\text{total no}}{\text{. of moles}}}} \\\ {\text{ = }}\dfrac{x}{{a{\text{ }} + {\text{ }}x}} \\\\$$ Now, we will calculate the partial pressure using the formula given: $${\text{Partial pressure of PC}}{{\text{l}}_5}{\text{ = }}{\chi _{PC{l_5}}} \times {\text{ }}{{\text{P}}_T} \\\ {\text{ = }}\dfrac{{a{\text{ }} - {\text{ }}x}}{{a{\text{ }} + {\text{ }}x}} \times {{\text{P}}_T} \\\\$$ $${\text{Partial pressure of PC}}{{\text{l}}_3}{\text{ = }}{\chi _{PC{l_3}}} \times {\text{ }}{{\text{P}}_T} \\\ {\text{ = }}\dfrac{x}{{a{\text{ }} + {\text{ }}x}} \times {{\text{P}}_T} \\\\$$ $${\text{Partial pressure of C}}{{\text{l}}_2}{\text{ = }}{\chi _{C{l_2}}} \times {\text{ }}{{\text{P}}_T} \\\ {\text{ = }}\dfrac{x}{{a{\text{ }} + {\text{ }}x}} \times {{\text{P}}_T} \\\\$$ Now, since this is an equilibrium reaction, we have to write either value of equilibrium constant in terms of pressure. $${K_p} = \dfrac{{{P_{PC{l_3}}} \times {P_{C{l_2}}}}}{{{P_{PC{l_5}}}}} \\\ = \dfrac{{\left( {\dfrac{x}{{{\text{a}} + x}}} \right) \times {P_T}\left( {\dfrac{x}{{{\text{a}} + x}}} \right) \times {P_T}}}{{\left( {\dfrac{{{\text{a}} - x}}{{{\text{a}} + x}}} \right) \times {P_T}}} \\\\$$ Solving the above equation we will get: $${K_p} = \dfrac{{{x^2}{P_T}}}{{{{\text{a}}^2} - {x^2}}}$$ The relation with $$\alpha $$ from the above definition we will get: $$\alpha = \dfrac{x}{{\text{a}}}$$ . This makes $$x = \alpha \times {\text{a}}$$. Now, substituting the value of $$x$$, we will get $${K_p} = \dfrac{{{\alpha ^2} \times {{\text{a}}^2} \times {P_T}}}{{{{\text{a}}^2} - {\alpha ^2}{{\text{a}}^2}}} \\\ = \dfrac{{{\alpha ^2} \times {{\text{a}}^2} \times {P_T}}}{{{{\text{a}}^2}(1 - {\alpha ^2})}} \\\ = \dfrac{{{\alpha ^2} \times {P_T}}}{{(1 - {\alpha ^2})}} \\\\$$ Now, taking $$\alpha $$ on one side and rest of the terms on other side: $$\sqrt {\dfrac{{{K_p} \times (1 - {\alpha ^2})}}{{{P_T}}}} = \alpha $$ It is clear from the above formula that $$\alpha \propto \dfrac{1}{{\sqrt {{P_T}} }}$$. **Hence, option B is correct.** **Note:** For using $$\alpha $$ we need to keep certain things in mind. If we are using a degree of dissociation then we can assume the initial number of moles as one. $$\alpha $$ is basically used for decomposition reaction, where one reactant decomposes into one or more than one compound. If it is very important to use $$\alpha $$ we need to convert the equation into a decomposition reaction by reversing it. We can even neglect $$\alpha $$ if it is less than 5 percent because in that case the dissociation of substance will be very less.