To find the heat of solution for \( \text{KCl} \), we need to understand the relationship between hydration energy, lattice energy, and the enthalpy of solution. This relationship is governed by the Born-Haber cycle, which is expressed as:
\[\text{Heat of Solution} = \text{Lattice Energy} - (\text{Hydration Energy of } K^+ + \text{Hydration Energy of } Cl^-)\]In the given problem, we have:
Substitute these values into the formula:
\[\text{Heat of Solution} = -z - (-x + -y) = -z + (x + y)\]Simplifying the expression, we get:
\[\text{Heat of Solution} = z - (x + y)\]Thus, the correct option is:
Therefore, the heat of solution of KCl is calculated as the lattice energy minus the sum of the hydration energies of \( K^+ \) and \( Cl^- \).
Step 1: Understand the process of dissolution and the associated energy changes.
The dissolution of an ionic compound like KCl in water involves two main steps:
1. Breaking the lattice: The ionic lattice of KCl must be broken down into individual gaseous ions \( K^+(g) \) and \( Cl^-(g) \). The energy required for this process is the lattice energy, \( \Delta H_{lattice} \). Since lattice energy is defined as the energy released when gaseous ions combine to form one mole of a solid ionic compound, the energy required to break the lattice is the negative of the lattice energy given. \[ KCl(s) \rightarrow K^+(g) + Cl^-(g) \quad \Delta H_1 = -(-z) = +z \, kJ/mol \]
2. Hydration of ions: The gaseous ions then get hydrated by water molecules, forming aqueous ions \( K^+(aq) \) and \( Cl^-(aq) \). The energy released in this process is the hydration energy, \( \Delta H_{hydration} \). The hydration energy of \( K^+ \) is \( -x \) kJ/mol, and the hydration energy of \( Cl^- \) is \( -y \) kJ/mol. The total hydration energy is the sum of the hydration energies of the individual ions. \[ K^+(g) + H_2O \rightarrow K^+(aq) \quad \Delta H_{hyd}(K^+) = -x \, kJ/mol \] \[ Cl^-(g) + H_2O \rightarrow Cl^-(aq) \quad \Delta H_{hyd}(Cl^-) = -y \, kJ/mol \] The overall hydration energy is: \[ K^+(g) + Cl^-(g) \rightarrow K^+(aq) + Cl^-(aq) \quad \Delta H_2 = -x + (-y) = -(x + y) \, kJ/mol \]
Step 2: Apply Hess's Law to find the heat of solution.
The heat of solution \( \Delta H_{sol} \) is the enthalpy change when one mole of a substance dissolves in a specified amount of solvent. According to Hess's Law, the overall enthalpy change for a reaction is independent of the path taken. Therefore, the heat of solution of KCl can be found by summing the enthalpy changes of the two steps mentioned above: \[ \Delta H_{sol} = \Delta H_1 + \Delta H_2 \] \[ \Delta H_{sol} = (+z) + (-(x + y)) \] \[ \Delta H_{sol} = z - (x + y) \, kJ/mol \]
Step 3: Match the result with the given options.
The heat of solution of KCl is \( z - (x + y) \) kJ/mol, which matches option (3).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
Match List - I with List - II.

Choose the correct answer from the options given below :
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,