Question

What will be the amount of dissociation is if the volume is increased 16 times of the initial volume in the reaction\[PC{{l}_{5}}\rightleftharpoons PC{{l}_{3}}+C{{l}_{2}}\]?
A.4 times
B.\[\dfrac{1}{4}\]times
C.2 times
D.\[\dfrac{1}{5}\]times

Answer 1

Hint: The amount of dissociation (\[\alpha \]) is inversely proportional to the square root of the pressure in the reaction.
The universal gas equation PV = nRT tells that the pressure of the gas is inversely proportional to the volume of the gas.
Here, P = pressure of the gas, V = volume of the gas, n = number of moles of the gas, R = Universal gas constant, T = temperature of the gas.

Complete step by step answer:
\[PC{{l}_{5}}\rightleftharpoons PC{{l}_{3}}+C{{l}_{2}}\]
\[\alpha \]= degree of dissociation of \[PC{{l}_{5}}\]
P = total pressure at the equilibrium condition
Let, ​the number of moles of \[PC{{l}_{5}}\]present is 1mole.
The number of moles of \[PC{{l}_{5}}\]dissociated to form \[PC{{l}_{3}}\]and \[C{{l}_{2}}\]be \[\alpha \].
So, the number of moles of \[PC{{l}_{5}}\]remained at equilibrium = (\[1-\alpha \])
Total number of moles = \[(1-\alpha )+\alpha +\alpha \]= (\[1+\alpha \])
Mole fraction of \[PC{{l}_{5}}\]= \[\dfrac{1-\alpha }{1+\alpha }\]
​Mole fraction of \[PC{{l}_{3}}\]= \[\dfrac{\alpha }{1+\alpha }\]
​Mole fraction of \[C{{l}_{2}}\]= \[\dfrac{\alpha }{1+\alpha }\]
According to the Dalton’s law of partial pressure,
Partial pressure of \[PC{{l}_{5}}\]=\[\dfrac{1-\alpha }{1+\alpha }\]P
​Partial pressure of \[PC{{l}_{3}}\]= \[\dfrac{\alpha }{1+\alpha }\]P
​Partial pressure of \[C{{l}_{2}}\]= \[\dfrac{\alpha }{1+\alpha }\]P​
 So, equilibrium constant of partial pressure, \[{{K}_{P}}\]= \[\dfrac{{{P}_{PC{{l}_{3}}}}{{P}_{C{{l}_{2}}}}}{{{P}_{PC{{l}_{5}}}}}\]= \[\dfrac{\dfrac{\alpha }{1+\alpha }P\times \dfrac{\alpha }{1+\alpha }P}{\dfrac{1-\alpha }{1+\alpha }P}\]
\[{{K}_{P}}\]= \[\dfrac{{{\alpha }^{2}}P}{1-{{\alpha }^{2}}}\]
Or, \[{{K}_{P}}\]= \[{{\alpha }^{2}}P\] [\[1-{{\alpha }^{2}}\approx 1\]]
Degree of dissociation,\[\alpha =\sqrt{\dfrac{{{K}_{P}}}{P}}\]
\[\alpha \propto {}^{1}/{}_{\sqrt{P}}\]
Now, PV = nRT
\[P\propto {}^{1}/{}_{V}\]
So, \[\alpha \propto \sqrt{V}\]
Given, the volume is increased 16 times of the initial volume in the reaction
\[\alpha \propto \sqrt{16}\]
\[\Rightarrow \] 4 times
The amount of dissociation is if the volume is increased 16 times of the initial volume in the reaction\[PC{{l}_{5}}\rightleftharpoons PC{{l}_{3}}+C{{l}_{2}}\] will be 4 times.
So, the correct option is A.

Note: According to the Dalton’s Law of Partial Pressure in terms of mole fraction, mathematically we can say,
\[{{p}_{i}}={{\chi }_{i}}{{p}_{total}}\] [\[{{p}_{i}}\]= partial pressure of a gas, \[{{\chi }_{i}}\]= mole fraction of the gas, \[{{p}_{total}}\]= total pressure in the system]