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.$${\displaystyle \frac{1}{4}}$$times
C.2 times
D.$${\displaystyle \frac{1}{5}}$$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 $$PC{l}_5$$= $${\displaystyle \frac{1-\alpha }{1+\alpha }}$$
​Mole fraction of $$PC{l}_3$$= $${\displaystyle \frac{\alpha }{1+\alpha }}$$
​Mole fraction of $$C{l}_2$$= $${\displaystyle \frac{\alpha }{1+\alpha }}$$
According to the Dalton’s law of partial pressure,
Partial pressure of $$PC{l}_5$$=$${\displaystyle \frac{1-\alpha }{1+\alpha }}$$P
​Partial pressure of $$PC{l}_3$$= $${\displaystyle \frac{\alpha }{1+\alpha }}$$P
​Partial pressure of $$C{l}_2$$= $${\displaystyle \frac{\alpha }{1+\alpha }}$$P​
 So, equilibrium constant of partial pressure, $$K_P$$= $${\displaystyle \frac{P_{PC{l}_3}{P}_{C{l}_2}}{P_{PC{l}_5}}}$$= $${\displaystyle \frac{{\displaystyle \frac{\alpha }{1+\alpha }}P\times {\displaystyle \frac{\alpha }{1+\alpha }}P}{{\displaystyle \frac{1-\alpha }{1+\alpha }}P}}$$
$$K_P$$= $${\displaystyle \frac{{\alpha }^{2}P}{1-{\alpha }^2}}$$
Or, $$K_P$$= $${\alpha }^{2}P$$ [$$1-{\alpha }^2\approx 1$$]
Degree of dissociation,$$\alpha =\sqrt{{\displaystyle \frac{K_P}{P}}}$$
$$\alpha \propto {}^1/{}_{\sqrt{P}}$$
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]