The degree of dissociation of PCl5(\alpha) obeying the equilibrium, PCl5(g) \leftrightarrowPCl3(g) + Cl2(g) is approximately related to the pressure at equilibrium by  

  • Option 1)

    \alpha ∝ P

  • Option 2)

    \alpha ∝ \frac{1}{\sqrt{p}}

  • Option 3)

    \alpha ∝\frac{1}{p^{2}}

  • Option 4)

    \alpha ∝ \frac{1}{p^{4}}

 

Answers (1)
P Prateek Shrivastava

As we learnt in

Equilibrium constant in gaseous system -

For equilibrium involving gases, it is usually more convenient to express the equilibrium constant in terms of partial pressure. 

- wherein

The ideal gas equation is written as 

PV=nRT

 

 Let's consider the given reaction

PCl_{5}(g)\leftrightharpoons PCl_{5}(g)+Cl_{2}(g)

1-\alpha               \alpha                    \alpha

Total pressure =P  

P_{PCl _{3}}= \frac{P\alpha }{1+\alpha}, P_{Cl _{2}}= \frac{P\alpha }{1+\alpha}

P_{PCl _{3}}= \frac{P(1-\alpha) }{1+\alpha}

K_{P}= \frac{P_{Cl_{2}}P_{PCl_3}}{P_{PCl_3}}= \frac{P\alpha ^{2}}{1-\alpha ^{2}}

1-\alpha ^{2}\approx 1, \alpha \angle \angle 1

\therefore K_{p}=P\alpha ^{2}

\alpha =\sqrt{\frac{K_{p}}{P}}

\Rightarrow \alpha\ \propto \frac{1}{\sqrt{p}}

 


Option 1)

\alpha ∝ P

This option is incorrect

Option 2)

\alpha ∝ \frac{1}{\sqrt{p}}

This option is correct

Option 3)

\alpha ∝\frac{1}{p^{2}}

This option is incorrect

Option 4)

\alpha ∝ \frac{1}{p^{4}}

This option is incorrect

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