Lattice energy of NaCl(s) is -790 kJ `" mol"^(-1)` and enthalpy of hydration is -785 kJ `" mol"^(-1)`. Calculate enthalpy of solution of NaCl(s).

Answer

To calculate the enthalpy of solution of NaCl(s), we can use the Born-Haber cycle, which relates the lattice energy, enthalpy of hydration, and enthalpy of solution. The relationship can be expressed as:

\[ \text{Lattice Energy} = \text{Enthalpy of Hydration} + \text{Enthalpy of Solution} \]

Given:
- Lattice Energy of NaCl(s) = -790 kJ/mol
- Enthalpy of Hydration = -785 kJ/mol

We need to find the Enthalpy of Solution (ΔH_solution).

### Step 1: Write the equation based on the Born-Haber cycle.

\[ \text{Lattice Energy} = \text{Enthalpy of Hydration} + \text{Enthalpy of Solution} \]

### Step 2: Substitute the known values into the equation.

\[ -790 \, \text{kJ/mol} = -785 \, \text{kJ/mol} + \Delta H_{\text{solution}} \]

### Step 3: Rearrange the equation to solve for ΔH_solution.

\[ \Delta H_{\text{solution}} = -790 \, \text{kJ/mol} + 785 \, \text{kJ/mol} \]

### Step 4: Perform the calculation.

\[ \Delta H_{\text{solution}} = -790 + 785 = -5 \, \text{kJ/mol} \]

### Final Answer:
The enthalpy of solution of NaCl(s) is -5 kJ/mol.

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