Question

For the reaction: $A{{B}_{\left( g \right)}}\rightleftharpoons {{A}_{\left( g \right)}}+{{B}_{\left( g \right)}},$AB is $33$ dissociated at a total pressure of P, therefore P is related to ${{K}_{p}}$ by one of the following options
A.$P={{K}_{P}}$
B.$P=3{{K}_{P}}$
C.$P=4{{K}_{P}}$
D.$P=8{{K}_{P}}$

Answer 1

We know that equilibrium constant for any reaction is calculated by the ratio of the concentration of products to that of reactant. When a gaseous dissociates into two gases, we can assume that the pressure of both the gases is the same. So, do that and then try to find the total pressure using equilibrium constant using the pressure of gases.

Complete answer:
According to Le Chatelier's principle the temperature, concentration, pressure, catalyst, and 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.
33% dissociation means out of one mole of AB, $\dfrac{1}{3}$ mole is dissociated. So $1-\dfrac{1}{3}=\dfrac{2}{3}$ mole is present. $A{{B}_{\left( g \right)}}\rightleftharpoons {{A}_{\left( g \right)}}+{{B}_{\left( g \right)}},$

| $A{{B}_{\left( g \right)}}$| ${{A}_{\left( g \right)}}$| ${{B}_{\left( g \right)}}$
---|---|---|---
Number of moles at ${{t}_{0}}$| $1$| $0$| $0$
Number of moles at ${{t}_{eq}}$| $1-\dfrac{1}{3}=\dfrac{2}{3}$| $\dfrac{1}{3}$| $\dfrac{1}{3}$

After dissociation, total number of moles $\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{4}{3}.$
${{P}_{AB}}=\dfrac{\left( \dfrac{2}{3} \right)}{\left( \dfrac{4}{3} \right)}\times P=\dfrac{P}{4}.$
thus, $Kp=\dfrac{{{P}_{A}}{{P}_{B}}}{{{P}_{AB}}}=\dfrac{\left( \dfrac{P}{4}+\dfrac{P}{4} \right)}{\left( \dfrac{P}{2} \right)}=\dfrac{P}{8}.$
$\Rightarrow P=8{{K}_{P}}$
Therefore, the correct answer is option D.

Note:
Remember that in the expression of equilibrium constant of the reaction, we can either use molar concentration or partial pressures. We cannot use both at the same time. The equilibrium constant is independent of the inert gases present in the mixture.