JEE Chemistry 2025 Question with Solution

Given: Hydration energy of $K^+$ is $-x \, \text{kJ/mol}$, hydration energy of $Cl^-$ is $-y \, \text{kJ/mol}$, and lattice energy of KCl is $-z \, \text{kJ/mol}$.

Find: Heat of solution of KCl.

For dissolution of an ionic solid, first the lattice is broken and then the gaseous ions are hydrated. Using Hess's law,

$$\Delta H_{sol} = \Delta H_1 + \Delta H_2$$

Breaking the lattice requires energy equal to the negative of the given lattice energy:

$$\Delta H_1 = -(-z) = +z$$

The total hydration enthalpy is:

$$\Delta H_2 = -x + (-y) = -(x+y)$$

Therefore,

$$\Delta H_{sol} = z + (-(x+y)) = z - (x+y)$$

So, the heat of solution is $z - (x+y)$. The correct option is C.

Given: $\text{Hydration energy of } K^+ = -x$, $\text{hydration energy of } Cl^- = -y$, and $\text{lattice energy of KCl} = -z$.

Find: The heat of solution of KCl.

The enthalpy of solution is obtained by considering two steps:

1. $KCl(s) \rightarrow K^+(g) + Cl^-(g)$
2. $K^+(g) \rightarrow K^+(aq)$ and $Cl^-(g) \rightarrow Cl^-(aq)$

For step 1, the given lattice energy is negative because it refers to lattice formation. Hence, lattice dissociation enthalpy is:

$$\Delta H_1 = +z$$

For step 2, the hydration enthalpy is:

$$\Delta H_2 = -x + (-y) = -(x+y)$$

Applying Hess's law,

$$\Delta H_{sol} = \Delta H_1 + \Delta H_2$$ $$\Delta H_{sol} = z -(x+y)$$

Thus, the result matches option (3). Therefore, the correct option is C.