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

Hint: 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.