The lattice energy of solid `NACl` is `180K. Cal mol^(-1)`. The dissolution of the solid in water in the form of ions is endothermic to the extent of `1 K.cal mol^(-1)`. If the hydration energies of `NA^(+)` and `Cl^(-)` are in ratio `6 : 5`, what is the enthalpy of hydration of `NA^(+)` ion

Options

A. `-"8.5 kcal mol"^(-1)`
B. `-"97.64 kcal mol"^(-1)`
C. `+"82.6 kcal mol"^(-1)`
D. `+"100 kcal mol"^(-1)`

Answer

To find the enthalpy of hydration of the Na⁺ ion, we will follow these steps:

### Step 1: Understand the given data
- Lattice energy of NaCl (ΔH_lattice) = 180 kcal/mol
- Enthalpy of solution (ΔH_solution) = 1 kcal/mol
- Hydration energies of Na⁺ and Cl⁻ are in the ratio 6:5.

### Step 2: Use the formula for enthalpy of solution
The enthalpy of solution can be expressed as:
\[ \Delta H_{\text{solution}} = \Delta H_{\text{lattice}} + \Delta H_{\text{hydration}} \]
Where:
- ΔH_hydration = ΔH_hydration(Na⁺) + ΔH_hydration(Cl⁻)

### Step 3: Rearrange the formula to find hydration energy
We can rearrange the formula to find the total hydration energy:
\[ \Delta H_{\text{hydration}} = \Delta H_{\text{solution}} - \Delta H_{\text{lattice}} \]

### Step 4: Substitute the values
Substituting the known values:
\[ \Delta H_{\text{hydration}} = 1 \text{ kcal/mol} - 180 \text{ kcal/mol} \]
\[ \Delta H_{\text{hydration}} = -179 \text{ kcal/mol} \]

### Step 5: Determine the individual hydration energies
Let the enthalpy of hydration of Na⁺ be \( x \) and that of Cl⁻ be \( y \). Given the ratio of their hydration energies:
\[ \frac{x}{y} = \frac{6}{5} \]
This means:
\[ x = \frac{6}{5}y \]

### Step 6: Express the total hydration energy in terms of \( y \)
Substituting \( x \) in the total hydration energy equation:
\[ \Delta H_{\text{hydration}} = x + y = \frac{6}{5}y + y = \frac{6y + 5y}{5} = \frac{11y}{5} \]

### Step 7: Set the total hydration energy equal to the calculated value
Now, we can set this equal to the total hydration energy we calculated:
\[ \frac{11y}{5} = -179 \text{ kcal/mol} \]

### Step 8: Solve for \( y \)
Multiplying both sides by 5:
\[ 11y = -895 \]
\[ y = -\frac{895}{11} \approx -81.36 \text{ kcal/mol} \]

### Step 9: Find \( x \) (enthalpy of hydration of Na⁺)
Now substituting \( y \) back to find \( x \):
\[ x = \frac{6}{5}y = \frac{6}{5} \times -81.36 \approx -97.63 \text{ kcal/mol} \]

### Final Answer
The enthalpy of hydration of the Na⁺ ion is approximately:
\[ \Delta H_{\text{hydration}}(Na^+) \approx -97.63 \text{ kcal/mol} \]