For vaporization of water at 1 atmospheric pressure the values
of `DeltaH` and `DeltaS` are `40.63KJmol^(-1)` and `108JK^(-1)mol^(-1)` ,
respectively. The temperature when Gibbs energy change `(DeltaG)`
for this transformation will be zero is

Options

A. 395.3 K
B. 376 K
C. 373 K
D. 380 K

Answer

To find the temperature at which the Gibbs free energy change (ΔG) for the vaporization of water is zero, we can use the relationship between ΔG, ΔH, and ΔS given by the equation:

\[ \Delta G = \Delta H - T \Delta S \]

### Step-by-Step Solution:

1. **Identify the Given Values**:
- ΔH = 40.63 kJ/mol
- ΔS = 108 J/K·mol

2. **Convert ΔH to Joules**:
Since ΔH is given in kilojoules, we need to convert it to joules for consistency with ΔS:
\[ \Delta H = 40.63 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 40630 \, \text{J/mol} \]

3. **Set ΔG to Zero**:
We want to find the temperature (T) when ΔG = 0:
\[ 0 = \Delta H - T \Delta S \]

4. **Rearrange the Equation**:
Rearranging the equation gives us:
\[ \Delta H = T \Delta S \]

5. **Solve for Temperature (T)**:
Now, we can solve for T:
\[ T = \frac{\Delta H}{\Delta S} \]
Substituting the values we have:
\[ T = \frac{40630 \, \text{J/mol}}{108 \, \text{J/K·mol}} \]

6. **Calculate T**:
Performing the division:
\[ T = 376.85 \, \text{K} \]
Rounding to three significant figures, we get:
\[ T \approx 377 \, \text{K} \]

### Final Answer:
The temperature at which the Gibbs energy change (ΔG) for the vaporization of water is zero is approximately **377 K**.