Question

A vessel contains 1 mole ${\text{PC}}{{\text{l}}_5}({\text{g}})$ at $4{\text{ atm}}$ and $0.5$ mole ${\text{PC}}{{\text{l}}_{\text{3}}}$ formed at equilibrium. Now, equilibrium pressure of mixture is (assume ideal behaviour)
A.${\text{16 atm}}$
B.$6{\text{ atm}}$
C.$2{\text{ atm}}$
D.$4.5{\text{ atm}}$

Answer 1

:Hint:To answer this question, you must recall the formula for equilibrium constant of a reaction, in terms of the partial pressures of its constituents $\left( {{{\text{K}}_{\text{P}}}} \right)$. The equilibrium constant of a chemical reaction is the reaction quotient of the reaction at its chemical equilibrium.
Formula used: For a reaction, $aA + bB \rightleftharpoons cC + dD$
${{\text{K}}_{\text{P}}} = \dfrac{{{\text{P}}_{\text{c}}^{\text{c}}{\text{P}}_{\text{d}}^{\text{d}}}}{{{\text{P}}_{\text{a}}^{\text{a}}{\text{P}}_{\text{b}}^{\text{b}}}}$
Where ${\text{P}}_x^{}$ is the partial pressure of the constituent x in the reaction mixture.

Complete step by step answer:
The dissociation reaction of ${\text{PC}}{{\text{l}}_{\text{5}}}$ into ${\text{PC}}{{\text{l}}_{\text{3}}}$ occurs as follows:

Initial	1	0	0
Final	$0.5$	$0.5$	$0.5$

${\text{PC}}{{\text{l}}_{\text{5}}}{\text{(g)}} \rightleftharpoons {\text{PC}}{{\text{l}}_{\text{3}}}{\text{(g)}} + {\text{C}}{{\text{l}}_{\text{2}}}{\text{(g)}}$.
The initial pressure in the container when 1 mole of phosphorus pentachloride is present is given to be $4{\text{ atm}}$.
The total number of moles of gases present at equilibrium$ = 0.5 + 0.5 + 0.5 = 1.5$
Let the total pressure of the system$ = P$. So we can write,
$ \Rightarrow \dfrac{{0.5}}{{1.5}}P = 4$
$ \Rightarrow P = 12$
Now, writing the equilibrium constant for the reaction as, ${K_P} = \dfrac{{{P_{PC{l_3}}}{P_{C{l_2}}}}}{{{P_{PC{l_5}}}}}$
Substituting the values of the partial pressures of each gas, we get,
${K_P} = \dfrac{{\left( {\dfrac{{0.5}}{{1.5}}P} \right)\left( {\dfrac{{0.5}}{{1.5}}P} \right)}}{{\left( {\dfrac{{0.5}}{{1.5}}P} \right)}} = \dfrac{P}{3}$
${K_P} = 4$
Thus, the equilibrium pressure of the reaction is given as ${K_P} = 4 \times 4 = 16{\text{ atm}}$

Thus, the correct answer is A.
Note:
The equilibrium constant is in general numerically calculated by allowing the reaction under consideration to proceed to equilibrium and then measuring the concentrations of each constituent present in the reaction mixture. Since the concentrations of the constituents of the reaction mixture are measured at equilibrium, the equilibrium constant always has the same value for a given reaction irrespective of the initial amount taken of the reactants. Using this knowledge a standard expression was derived that serves as a basis for all reactions. This basic model form of the equilibrium constant is given as ${K_{eq}} = \dfrac{{{{\left( {{a_C}} \right)}^c}}}{{{{\left( {{a_A}} \right)}^a}{{\left( {{a_B}} \right)}^b}}}$ for a reaction: $aA + bB \rightleftharpoons cC$