The enthalpy of vaporisation of water at `100^(@)C` is `40.63 KJ mol^(-1)`. The value `Delta E` for the process would be :-

Options

A. `37.53 KJ mol^(-1)`
B. `39.08 KJ mol^(-1)`
C. `42.19 KJ mol^(-1)`
D. `43.73 KJ mol^(-1)`

Explanation

To find the value of ΔE (change in internal energy) for the vaporization of water at 100°C, we can use the relationship between enthalpy change (ΔH), internal energy change (ΔE), and the change in the number of moles of gas (Δn) during the process.

### Step-by-Step Solution:

1. **Understand the Process**:
The process involves the vaporization of water (H₂O) from liquid to gas at 100°C. The given enthalpy of vaporization (ΔH) is 40.63 kJ/mol.

2. **Use the Formula**:
The relationship between ΔH and ΔE is given by the equation:
\[ \Delta H = \Delta E + \Delta n \cdot R \cdot T \]
where:
- ΔH = Enthalpy change
- ΔE = Internal energy change
- Δn = Change in the number of moles of gas
- R = Ideal gas constant (8.314 J/(mol·K))
- T = Temperature in Kelvin

3. **Identify Δn**:
In this case, we have:
- 1 mole of liquid water (H₂O) is converted to 1 mole of water vapor (gas).
- Therefore, Δn = moles of gas products - moles of liquid reactants = 1 - 0 = 1.

4. **Convert Units**:
Since ΔH is given in kJ, we need to convert R into kJ for consistency.
\[ R = 8.314 \, \text{J/(mol·K)} = 8.314 \times 10^{-3} \, \text{kJ/(mol·K)} \]

5. **Convert Temperature to Kelvin**:
The temperature in Celsius (100°C) needs to be converted to Kelvin:
\[ T = 100 + 273.15 = 373.15 \, \text{K} \]

6. **Substitute Values into the Equation**:
Now we can substitute the values into the equation:
\[ \Delta E = \Delta H - \Delta n \cdot R \cdot T \]
\[ \Delta E = 40.63 \, \text{kJ/mol} - (1) \cdot (8.314 \times 10^{-3} \, \text{kJ/(mol·K)}) \cdot (373.15 \, \text{K}) \]

7. **Calculate the Second Term**:
Calculate \( \Delta n \cdot R \cdot T \):
\[ \Delta n \cdot R \cdot T = 1 \cdot 8.314 \times 10^{-3} \cdot 373.15 \approx 3.103 \, \text{kJ/mol} \]

8. **Final Calculation**:
Now substitute this back into the equation for ΔE:
\[ \Delta E = 40.63 \, \text{kJ/mol} - 3.103 \, \text{kJ/mol} \approx 37.527 \, \text{kJ/mol} \]

9. **Round the Final Answer**:
Rounding to two decimal places, we get:
\[ \Delta E \approx 37.53 \, \text{kJ/mol} \]

### Final Answer:
\[ \Delta E \approx 37.53 \, \text{kJ/mol} \]