QJEEM16Chd32c: JEE Main 2016 Chemistry - Class 11 JEEM16Chd32c Solution: A solid XY kept in an evacuated sealed container undergoes…

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