Question

The degree of dissociation of $PC{l_5}\left( \alpha \right)$ obeying the equilibrium $PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$ is related to the equilibrium pressure by
(A). $\alpha \propto \dfrac{1}{{{P^4}}}$
(B). $\alpha \propto \dfrac{1}{{\sqrt P }}$
(C). $\alpha \propto \dfrac{1}{{{P^2}}}$
(D). $\alpha \propto P$

Answer 1

Hint: The degree of dissociation is the phenomena of generating current carrying free ions, which are dissociated from the fraction of solute at a given concentration. It is usually indicated by the Greek symbol $\alpha $ .

Complete step by step answer:
We know that,
$PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$
$1 \;\;\;\;\;\;\;\ \;\;\;\; 0 \;\;\;\;\;\;\;\;\ 0 \;\;\;\;\;\;\;\;\; At \;t = 0$
$1 - \alpha \;\;\;\; \;\;\; \alpha \;\;\; \;\;\;\; \alpha\;\;\ \;\;\; At \;\;time =t(Equilibrium)$

We supposed that initially $PC{l_5}$ has $1$ mole.
$\alpha = $ Degree of dissociation of$PC{l_5}$.
$P = $ Total equilibrium pressure.
We have considered (let) that $\alpha $ moles of $PC{l_5}$ have dissociated to form $\alpha $ moles of $PC{l_3}$ and $\alpha $ moles of$C{l_2}$.
So $\left( {1 - \alpha } \right)$ moles of $PC{l_5}$ will remain at equilibrium.
Total numbers of moles at equilibrium=
$\left( {1 - \alpha } \right) + \alpha + \alpha = \left( {1 + \alpha } \right)moles.$
So let us write moles fractions.
Mole fraction of $PC{l_5} = \dfrac{{\left( {1 - \alpha } \right)}}{{\left( {1 + \alpha } \right)}}$
Mole fraction of $PC{l_3} = \dfrac{\alpha }{{\left( {1 + \alpha } \right)}}$
Mole fraction of $C{l_2} = \dfrac{\alpha }{{\left( {1 + \alpha } \right)}}$
Therefore we can consider the partial pressure of the given reactants and products as:
Partial pressure of $PC{l_5} = \dfrac{{\left( {1 - \alpha } \right)}}{{\left( {1 + \alpha } \right)}}P$
Partial pressure of $PC{l_3} = \dfrac{\alpha }{{\left( {1 + \alpha } \right)}}P$
Partial pressure of $C{l_2} = \dfrac{\alpha }{{\left( {1 + \alpha } \right)}}P$
We can write, ${K_p} = \dfrac{{{P_{{P_{C{l_3}}}}}.{P_{C{l_2}}}}}{{{P_{{P_{C{l_5}}}}}}}$
Equilibrium constant in terms of partial pressures.
$\therefore {K_p} = \dfrac{{\dfrac{\alpha }{{\left( {1 + \alpha } \right)}}P \times \dfrac{\alpha }{{\left( {1 + \alpha } \right)}}P}}{{\dfrac{{\left( {1 - \alpha } \right)}}{{\left( {1 + \alpha } \right)}}P}}$
${K_p} = \dfrac{{{\alpha ^2}P}}{{1 - {\alpha ^2}}}$
We know that $\alpha < 1{\text{ }}\operatorname{so} {\text{ }}{\alpha ^2} < < 1$
So taking $1 - {\alpha ^2} \approx 1$
$\therefore {K_p} = {\alpha ^2}P$
Therefore, $\alpha = \sqrt {\dfrac{{{K_p}}}{P}} $
$\alpha \propto \dfrac{1}{{\sqrt P }}$

So, the correct answer is Option B.

Note: When a solid compound dissociates to give one or more gaseous products, the dissociation pressure is the pressure of gas in equilibrium with the solid at given temperature.
Phosphorus pentachloride is a pale-greenish yellow solid. It is known to have salt-like structure in the crystalline state and to be partly dissociated in solution, especially in polar solvents such as nitrobenzene. It can be prepared by the action of dry chlorine on phosphorus trichloride.

If dissolution is exothermic, the dissociation constant will rise with increasing temperature. If dissolution is endothermic, the dissociation constant will fall with increasing temperature.