Question

At temperature $T{\text{ K, PC}}{{\text{l}}_5}$ is $50\%$ dissociated at an equilibrium pressure of ${\text{4 atm}}$. At what pressure it would dissociate to the extent of $80\%$ at the same temperature?
A.${\text{0}}{\text{.05 atm}}$
B.${\text{0}}{\text{.60 atm}}$
C.${\text{0}}{\text{.75 atm}}$
D.${\text{0}}{\text{.25 atm}}$

Answer 1

There is a direct relationship between degree of dissociation for dissociation of type $A \to B + C$and total pressure at equilibrium; we are going to use the relation here. We will get two equations one for $50\%$ dissociation and other for $80\%$ dissociation. Degree of dissociation is in fraction so we need to convert the percentage into fraction. Dividing both the equations we will get the value of pressure.

Formula used: ${{\text{K}}_{\text{p}}} = \dfrac{{{\alpha ^2} \times {{\text{P}}_{\text{T}}}}}{{(1 - {\alpha ^2})}}$ for reaction of type $A \to B + C$
Where, ${\alpha}_1$ is degree of dissociation, ${{\text{K}}_P}$ is equilibrium constant in terms of pressure, ${{\text{P}}_{\text{T}}}$ is total pressure at equilibrium.

Complete step by step solution:
α is the degree of dissociation, it is the ratio of dissociated moles to the total no of moles of the reactant. ${{\text{K}}_P}$ is equilibrium constant which is a function of temperature, if temperature remains constant then it will not change. Hence for both the dissociation ${{\text{K}}_P}$ will remain the same as temperature is constant. To convert α into fraction we need to divide it with 100.
So ${\alpha}_1$ will be $\dfrac{{50}}{{100}} = 0.5$ and pressure is given to us that is ${\text{4 atm}}$.
Putting the above value in formula:
${K_p} = \dfrac{{{{(0.5)}^2} \times 4}}{{(1 - {{(0.5)}^2})}} = \dfrac{1}{{0.75}}$
Similarly ${\alpha}_1$ will be $\dfrac{{80}}{{100}} = 0.8$
${K_p} = \dfrac{{{{(0.8)}^2} \times {{\text{P}}_{\text{t}}}}}{{(1 - {{(0.8)}^2})}} = 1.777{{\text{P}}_{\text{t}}}$
Dividing or equating both the equation we will get:
$\dfrac{1}{{0.75}} = 1.7777{{\text{P}}_{\text{t}}}$
Rearranging we will get the value of pressure:
${{\text{P}}_{\text{t}}}{\text{ = 0}}{\text{.75 atm}}$

Hence, option C is correct.

Note: The dissociation of ${\text{PC}}{{\text{l}}_5}$ follows the following reaction: ${\text{PC}}{{\text{l}}_{\text{5}}}{\text{ }} \rightleftharpoons {\text{ PC}}{{\text{l}}_3}{\text{ + C}}{{\text{l}}_2}{\text{ }}$. As we can see that the number of moles of both the product is same, this implies that pressure of these reactants will be same at equilibrium and pressure of ${\text{PC}}{{\text{l}}_5}$ will be different.
${\text{a}}$ is initial moles of reactant and $x$ is amount of reactant dissociated. In this case $\alpha$ will be $\dfrac{x}{{\text{a}}}$.
${\text{time PC}}{{\text{l}}_{\text{5}}}{\text{ }} \rightleftharpoons {\text{ PC}}{{\text{l}}_3}{\text{ + C}}{{\text{l}}_2}{\text{ }}$