QJEEM20Ch8891: JEE Main 2020 Chemistry - Class 11 JEEM20Ch8891 Solution: If the equilibrium constant for A ⇌ B + C is and that of B + C…

We have two reactions:

$A \;\rightleftharpoons\; B + C \quad\text{with } K_{eq}^{(1)}$

$B + C \;\rightleftharpoons\; P \quad\text{with } K_{eq}^{(2)}$

We want the equilibrium constant for the overall reaction

$A \;\rightleftharpoons\; P.$

How to combine the equilibrium constants

When two reactions are added to yield a net (overall) reaction, the equilibrium constant of the net reaction is the product of the equilibrium constants of the individual reactions. Formally:

$\text{(Reaction 1)} + \text{(Reaction 2)} \;\Rightarrow\; \text{(Net Reaction)},$

and

$K_{\text{net}} \;=\; K_{\text{(Reaction 1)}} \times K_{\text{(Reaction 2)}}.$

Applying it here

Reaction 1: $A \to B + C$ has $K_{eq}^{(1)}.$

Reaction 2: $B + C \to P$ has $K_{eq}^{(2)}.$

When we add them,

$\underbrace{A}_{\text{Reactant 1}} \;\longrightarrow\; \underbrace{B + C}_{\text{Products of Reaction 1}=\text{Reactants of Reaction 2}} \;\longrightarrow\; \underbrace{P}_{\text{Product of Reaction 2}}.$

So the net reaction is $A \to P$ , and its equilibrium constant is

$K_{eq}^{\text{(net)}} \;=\; K_{eq}^{(1)} \times K_{eq}^{(2)}.$

Thus, the correct answer is:

$\boxed{K_{eq}^{(1)} \, K_{eq}^{(2)}}.$

Hence, the answer (Option D) is correct.