A solid XY kept in an evacuated sealed container undergoes decomposition to form a mixture of gases X and Y at temperature T. The equilibrium pressure is 10 bar in this vessel. $$K_p$$ for this reaction is?

First we write the balanced chemical equation for the decomposition:

$$XY(s) \rightarrow X(g) + Y(g)$$

Because $$XY$$ is a solid, its activity is taken as 1 and it never appears in the expression of the equilibrium constant.

The definition of the equilibrium constant in terms of partial pressures is

$$K_p \;=\; (P_X)\,(P_Y)$$

where $$P_X$$ and $$P_Y$$ are the equilibrium partial pressures of the gases $$X$$ and $$Y$$ respectively.

The vessel was completely evacuated at the start, so there were no gases present initially. As the reaction proceeds, one mole of $$X(g)$$ and one mole of $$Y(g)$$ are produced simultaneously from every mole of $$XY(s)$$ that decomposes. Hence the number of moles, and therefore the partial pressures, of the two gases will always remain equal:

$$P_X = P_Y$$

Let each of these equal partial pressures be denoted by $$p$$. The total equilibrium pressure inside the container is the sum of the two partial pressures, so

$$P_{\text{total}} \;=\; P_X + P_Y = p + p = 2p$$

According to the problem statement, the total equilibrium pressure is 10 bar. Substituting this value, we get

$$2p = 10 \;\text{bar}$$

$$\Rightarrow\; p = 5 \;\text{bar}$$

Now we substitute $$P_X = P_Y = 5 \;\text{bar}$$ into the expression for $$K_p$$:

$$K_p = (P_X)(P_Y) = (5 \;\text{bar})(5 \;\text{bar}) = 25$$

Hence, the correct answer is Option A.