Question

When $PC{l_5}$ was heated in a closed vessel, the pressure increased from $1$ atm to $1.5$ atm. Find the degree of dissociation $\alpha$ .

Answer 1

Degree of dissociation is the fraction of a mole of reactant that underwent dissociation. It is represented by $\alpha$ . Initially, in the container only $PC{l_5}$ will be there. When we start heating it, it will dissociate into $PC{l_3}$ and $C{l_2}$ . Initially the pressure will be due to $PC{l_5}$ but after attaining equilibrium, $PC{l_3}$ and $C{l_2}$ will also contribute to pressure change.

Complete step by step solution
Writing the reaction that will take place on heating $PC{l_5}$ : $PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$
Let the pressure exerted by $PC{l_5}$ initially be $P$ . After attaining equilibrium, suppose $\alpha$ be the degree of dissociation. Pressure due to $PC{l_5}$ will become $P$ $- P\alpha$ and due to $PC{l_3}$ and $C{l_2}$ will be $P\alpha$
$P - P\alpha =$ Partial pressure due to $PC{l_5}$
$P\alpha =$ Partial pressure due to $PC{l_3}$ and $C{l_2}$
$PC{l_5}_{(g)} \rightleftharpoons PC{l_3}_{(g)} + C{l_2}_{(g)}$

| $PC{l_5}_{(g)}$| $PC{l_3}_{(g)}$| $C{l_2}_{(g)}$
---|---|---|---
At time, t $= 0$| $P$| $0$| $0$
At time, t| $P$ $- P\alpha$| $P\alpha$| $P\alpha$

$P = 1$ atm (initially due to $PC{l_5}$ )
Final pressure $= 1.5$ atm (due to $PC{l_5}$ , $PC{l_3}$ and $C{l_2}$ )
Taking the summation of partial pressures of $PC{l_5}$ , $PC{l_3}$ and $C{l_2}$ ;
$\Rightarrow (P - P\alpha ) + P\alpha + P\alpha = 1.5$
$\Rightarrow P + P\alpha = 1.5$
$\Rightarrow P(1 + \alpha ) = 1.5$
$\Rightarrow 1 + \alpha = \dfrac{{1.5}}{1}$ ( $P = 1$ )
$\Rightarrow \alpha = 1.5 - 1 = 0.5$
$\therefore \alpha = 0.5$
So, the degree of dissociation will be $0.5$ .

Note
Degree of dissociation can also be taken as measurement of percentage of reactant converted into product. In this question, we got the value of the degree of dissociation, $\alpha = 0.5$ . It means, $50\%$ of $PC{l_5}$ is converted into $PC{l_3}$ and $C{l_2}$ . For a reversible dissociation in a chemical equilibrium $AB \rightleftharpoons A + B$
The dissociation constant ${K_d}$ is the ratio of undissociated compounds to dissociate compounds.
${K_d} = \dfrac{{[A][B]}}{{[AB]}}$ where the bracket represents the equilibrium concentration of the species.