Question

At 600 K, $PC{l_5}$ ​ dissociates to $40\%$. The total pressure exerted by the mixture at equilibrium, when ${{\text{K}}_{\text{p}}}$= $2.4$is:
A)$12.6{\text{atm}}$
B)$1.26{\text{atm}}$
C)$0.126{\text{atm}}$
D)$126{\text{atm}}$

Answer 1

To answer this question, you should recall the concept of dissociation. Write the reaction of dissociation and calculate the remaining percentage of $PC{l_5}$. Now use these values to find the answer to this question.

Complete Step by step solution:
We know that equilibrium constant for any reaction is calculated by the ratio of the concentration of products to that of reactant. The dissociation reaction can be written as:
$PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$.
Assuming that the total pressure is ${\text{p}}$ at ${\text{t = 0}}$
$PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$
Initially, at t=0 ${\text{p}}$ 0 0
Finally, ${\text{p}} - {\text{x}}$ 0 ${\text{x}}$
We are given the dissociation as $40\%$. Hence ${\text{x}}$ can be written as $40\%$.
Thus ${\text{x}} = \dfrac{{40}}{{100}}{\text{p = 0}}{\text{.4p}}$.
The remaining pressure of the components can be written as:
$PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$
${\text{p - 0}}{\text{.4p}}$ ${\text{0}}{\text{.4p}}$ ${\text{0}}{\text{.4p}}$
Using the definition of equilibrium constant, we can write its value as $\dfrac{{{{\text{p}}_{{\text{PC}}{{\text{l}}_{\text{3}}}}} \times {{\text{p}}_{{\text{C}}{{\text{l}}_{\text{2}}}}}}}{{{{\text{p}}_{{\text{PC}}{{\text{l}}_{\text{5}}}}}}}$ where
${{\text{p}}_{{\text{PC}}{{\text{l}}_{\text{3}}}}}$= Partial pressure of $PC{l_5} = {\text{0}}{\text{.4p}}$, ${{\text{p}}_{{\text{C}}{{\text{l}}_{\text{2}}}}}$= Partial pressure of $PC{l_3}{\text{ = 0}}{\text{.4p}}$ and ${{\text{p}}_{{\text{PC}}{{\text{l}}_{\text{5}}}}}$= Partial pressure of $C{l_2} = 0.6{\text{p}}$. Substituting the given value of ${{\text{K}}_{\text{p}}} = 2.4$ in the relation of the equilibrium constant, we get
$\Rightarrow 2.4 = \dfrac{{0.4{\text{p}} \times 0.4{\text{p}}}}{{0.6{\text{p}}}}$
Rearranging and solving, we get:
$\Rightarrow p = 9{\text{atm}}$.
Now the final pressure at equilibrium can be calculated
$\Rightarrow 0.6p + 0.4p + 0.4p = 1.4p = 12.6{\text{atm}}$
Therefore, we can conclude that the correct answer to this question is option A.

Note: You should know the change in concentration, pressure, catalyst, inert gas addition, etc. not affect equilibrium constant. According to Le Chatelier's principle the temperature, concentration, pressure, catalyst, inert gas addition can lead to a shift in equilibrium position only. We know that activation energy is the minimum energy required to start a chemical reaction. Collisions of particles lead to reactions.
Only particles that collide sufficiently, can react. Now comes the important point. From the kinetics of a reaction, we know that the rate of a reaction increases with increase in temperature due to more energy and more collisions. But the extent of increase in this rate depends on the “energy of activation” of the reaction which is different for both - the forward and the backward reaction. So, a given increase in temperature leads to an increase in the rate of forward and backward reactions to different extents. So, the value of the equilibrium constant changes with temperature.
Further, it has been found that the value of the equilibrium constant of an endothermic reaction increases and that of an exothermic reaction decreases with increase in temperature.