Question

The degree of dissociation of $PC{{l}_ {5}}$​($\alpha$) obeying the equilibrium $PC{{l}_ {5}} \rightleftharpoons PC{{l}_ {3}} +C{{l}_ {2}}$ is approximately related to the pressure at equilibrium by: (given $\alpha <<1$)
(a) $\alpha \propto \frac {1} {{{P}^ {4}}}$
(b) $\alpha \propto \frac {1} {\sqrt{P}}$
(c) $\alpha \propto \frac {1} {{{P}^ {2}}}$
(d) $\alpha \propto P$

Answer 1

The degree of dissociation is the phenomenon of generating current carrying free ions, which are dissociated from the fraction of solute at a given concentration. Also, we will be using the formula: ${{K}_{p}} = {{\alpha} ^ {2}} P$ for finding the relation between pressure and degree of dissociation at equilibrium.

Complete step by step answer:
We have been provided with $PC{{l}_ {5}} \rightleftharpoons PC{{l}_ {3}} +C{{l}_ {2}}$ at equilibrium,
α= the degree of dissociation of $PC{{l}_ {5}}$ P= total equilibrium pressure.
Suppose initially, 1 mole ​$PC{{l}_{5}}$ is present. α moles of$PC{{l}_{5}}$ will dissociate to form α moles of $PC{{l}_{3}}$​ and α moles of $C{{l}_{2}}$. $(1-\alpha )$ moles $PC{{l}_{5}}$of ​ will remain at equilibrium.
So, Total number of moles = $(1-\alpha) +\alpha +\alpha = (1+\alpha)$,
Now, we will be calculating the mole fraction using the formula: mole fraction of solute=moles of solute ÷ moles of solution,
Mole fraction of $PC{{l}_{5}}$= $\frac{\left( 1-\alpha \right)}{\left( 1+\alpha \right)}$ Mole fraction of $PC{{l}_{3}}$= $\frac{\alpha }{\left( 1+\alpha \right)}$ Mole fraction of$C{{l}_{2}}$ ​= $\frac{\alpha }{\left( 1+\alpha \right)}$
Now, we will be calculating the partial pressure using the formula: partial pressure = mole fraction × mole fraction Partial pressure of ​$PC{{l}_{5}}$= $\frac{\left( 1-\alpha \right)}{\left( 1+\alpha \right)}P$ Partial pressure of ​$PC{{l}_{3}}$= $\frac{\alpha }{\left( 1+\alpha \right)}P$ Partial pressure of $C{{l}_{2}}$​= $\frac{\alpha }{\left( 1+\alpha \right)}P$
Now, we will be finding the equilibrium constant:
${{K}_{p}} =\frac{{{P}_{PC{{l}_ {3}}}} \times {{P}_{C{{l}_ {2}}}}} {{{P}_{PC{{l}_ {5}}}}}$,
Now, simplifying the equation we would get:
${{K}_{P}} =\frac {\frac {\alpha} {1+\alpha} P\times \frac {\alpha} {1+\alpha} P} {\frac {1-\alpha} {1+\alpha} P}$,
Solving the above equation, n we will get: ${{K}_{p}} =\frac {{{\alpha} ^ {2}} P} {1- {{\alpha} ^ {2}}}$
Assume, $1- {{\alpha} ^ {2}} =1$ as the degree of dissociation is small.
So, the value comes out to be: ${{K}_{p}} = {{\alpha} ^ {2}} P$,
Simplifying, the above equation we would get: $\alpha =\sqrt{\frac{{{K}_{P}}} {P}}$,
So, the relation comes out to be: $\alpha \propto \frac {1} {\sqrt{P}}$,
Therefore, we can say that option (b) is correct.

Note: Factors affecting Degree of Ionisation:
(I) At normal dilution, the value of is nearly 1 for strong electrolytes, while it is very less than 1 for weak electrolytes.
(ii) Higher the dielectric constant of a solvent more is its ionising power. ...
(iii) Dilution of solution Amount of solvent.