JEE Main 2016 — Chemical Equilibrium Question with Solution

To determine the equilibrium constant, $K_p$, for the decomposition reaction of the solid compound XY into its gaseous components X and Y, we need to consider the following reaction:

$$\text{XY (s)}\rightleftharpoons \text{X (g)}+\text{Y (g)}$$

For a reaction where a solid decomposes into gases, the equilibrium constant $K_p$ is defined in terms of the partial pressures of the gaseous products. Since XY is a solid, its activity is considered to be 1 and does not appear in the equilibrium expression. Hence the expression for $K_p$ is:

$$K_p=P_X\cdot P_Y$$

Where $P_X$ and $P_Y$ are the partial pressures of the gases X and Y, respectively.

Given that the total pressure at equilibrium is 10 bar, and assuming that X and Y are present in equal amounts due to the stoichiometry of the decomposition reaction (i.e., 1:1), we can write:

$$P_X=P_Y=\frac{10}{2}=5\text{bar}$$

Substituting these partial pressures into the expression for $K_p$ gives:

$$K_p=P_X\cdot P_Y=5\text{bar}\cdot 5\text{bar}=25{\text{bar}}^2$$

Therefore, the equilibrium constant $K_p$ for this reaction is 25. Thus, the correct answer is:

Option C - 25