JEE Main 2024 — Chemical Equilibrium Question with Solution

To find the equilibrium constant for the overall reaction $$\mathrm{X}\rightleftharpoons \mathrm{W}$$, we need to combine the equilibrium constants for the individual reactions given. Let's analyze this step-by-step:

The given reactions and their equilibrium constants are:

$$\begin{array}{rl}& \mathrm{X}\rightleftharpoons \mathrm{Y}{K}_1=1.0\\ & \mathrm{Y}\rightleftharpoons \mathrm{Z}{K}_2=2.0\\ & \mathrm{Z}\rightleftharpoons \mathrm{W}{K}_3=4.0\end{array}$$

The equilibrium constant for the overall reaction $$\mathrm{X}\rightleftharpoons \mathrm{W}$$ is the product of the equilibrium constants for the individual steps. This is because the equilibrium constant for a composite reaction is the product of the equilibrium constants for the sequential reactions that lead to the composite reaction.

So, we have:

$$K_{\text{overall}}=K_1\cdot K_2\cdot K_3$$

Now, substituting the given values:

$$K_{\text{overall}}=1.0\cdot 2.0\cdot 4.0=8.0$$

Therefore, the equilibrium constant for the reaction $$\mathrm{X}\rightleftharpoons \mathrm{W}$$ is 8.0, which corresponds to Option B.