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Question: What will be the amount of dissociation is if the volume is increased 16 times of the initial volume...

What will be the amount of dissociation is if the volume is increased 16 times of the initial volume in the reactionPCl5⇌PCl3+Cl2PC{{l}_{5}}\rightleftharpoons PC{{l}_{3}}+C{{l}_{2}}?
A.4 times
B.14\dfrac{1}{4}times
C.2 times
D.15\dfrac{1}{5}times

Explanation

Solution

Hint: The amount of dissociation (α\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:
PCl5⇌PCl3+Cl2PC{{l}_{5}}\rightleftharpoons PC{{l}_{3}}+C{{l}_{2}}
α\alpha = degree of dissociation of PCl5PC{{l}_{5}}
P = total pressure at the equilibrium condition
Let, ​the number of moles of PCl5PC{{l}_{5}}present is 1mole.
The number of moles of PCl5PC{{l}_{5}}dissociated to form PCl3PC{{l}_{3}}and Cl2C{{l}_{2}}be α\alpha .
So, the number of moles of PCl5PC{{l}_{5}}remained at equilibrium = (1−α1-\alpha )
Total number of moles = (1−α)+α+α(1-\alpha )+\alpha +\alpha = (1+α1+\alpha )
Mole fraction of PCl5PC{{l}_{5}}= 1−α1+α\dfrac{1-\alpha }{1+\alpha }
​Mole fraction of PCl3PC{{l}_{3}}= α1+α\dfrac{\alpha }{1+\alpha }
​Mole fraction of Cl2C{{l}_{2}}= α1+α\dfrac{\alpha }{1+\alpha }
According to the Dalton’s law of partial pressure,
Partial pressure of PCl5PC{{l}_{5}}=1−α1+α\dfrac{1-\alpha }{1+\alpha }P
​Partial pressure of PCl3PC{{l}_{3}}= α1+α\dfrac{\alpha }{1+\alpha }P
​Partial pressure of Cl2C{{l}_{2}}= α1+α\dfrac{\alpha }{1+\alpha }P​
So, equilibrium constant of partial pressure, KP{{K}_{P}}= PPCl3PCl2PPCl5\dfrac{{{P}_{PC{{l}_{3}}}}{{P}_{C{{l}_{2}}}}}{{{P}_{PC{{l}_{5}}}}}= α1+αP×α1+αP1−α1+αP\dfrac{\dfrac{\alpha }{1+\alpha }P\times \dfrac{\alpha }{1+\alpha }P}{\dfrac{1-\alpha }{1+\alpha }P}
KP{{K}_{P}}= α2P1−α2\dfrac{{{\alpha }^{2}}P}{1-{{\alpha }^{2}}}
Or, KP{{K}_{P}}= α2P{{\alpha }^{2}}P [1−α2≈11-{{\alpha }^{2}}\approx 1]
Degree of dissociation,α=KPP\alpha =\sqrt{\dfrac{{{K}_{P}}}{P}}
α∝1/P\alpha \propto {}^{1}/{}_{\sqrt{P}}
Now, PV = nRT
P∝1/VP\propto {}^{1}/{}_{V}
So, α∝V\alpha \propto \sqrt{V}
Given, the volume is increased 16 times of the initial volume in the reaction
α∝16\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 reactionPCl5⇌PCl3+Cl2PC{{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,
pi=χiptotal{{p}_{i}}={{\chi }_{i}}{{p}_{total}} [pi{{p}_{i}}= partial pressure of a gas, χi{{\chi }_{i}}= mole fraction of the gas, ptotal{{p}_{total}}= total pressure in the system]