The enthalpy of solution of sodium chloride is `4 kJ mol^(-1)` and its enthalpy of hydration of ion is `-784 kJ mol^(-1)`. Then the lattice enthalpy of `NaCl` (in `kJ mol^(-1)`) is

Options

A. a. `+"780 kJ mol"^(-1)`
B. b. `+"394 kJ mol"^(-1)`
C. c. `+"788 kJ mol"^(-1)`
D. d. `+"398 kJ mol"^(-1)`

Explanation

To find the lattice enthalpy of sodium chloride (NaCl), we can use the relationship between the enthalpy of solution, the enthalpy of hydration, and the lattice enthalpy. Here’s how to solve the problem step by step:

### Step 1: Understand the Given Values
- Enthalpy of solution of NaCl (ΔH_solution) = +4 kJ/mol
- Enthalpy of hydration of ions (ΔH_hydration) = -784 kJ/mol

### Step 2: Write the Relationship
The relationship between these enthalpies can be expressed as:
\[ \Delta H_{\text{solution}} = \Delta H_{\text{lattice}} + \Delta H_{\text{hydration}} \]

### Step 3: Rearrange the Equation to Find Lattice Enthalpy
To find the lattice enthalpy (ΔH_lattice), we can rearrange the equation:
\[ \Delta H_{\text{lattice}} = \Delta H_{\text{solution}} - \Delta H_{\text{hydration}} \]

### Step 4: Substitute the Given Values
Now, substitute the given values into the equation:
\[ \Delta H_{\text{lattice}} = 4 \, \text{kJ/mol} - (-784 \, \text{kJ/mol}) \]

### Step 5: Simplify the Equation
When you subtract a negative number, it is equivalent to adding the absolute value:
\[ \Delta H_{\text{lattice}} = 4 \, \text{kJ/mol} + 784 \, \text{kJ/mol} \]
\[ \Delta H_{\text{lattice}} = 788 \, \text{kJ/mol} \]

### Step 6: Final Answer
Thus, the lattice enthalpy of NaCl is:
\[ \Delta H_{\text{lattice}} = 788 \, \text{kJ/mol} \]