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. 393.4 K
B. 373.4 K
C. 293.4 K
D. 273.4 K

Answer

To find the temperature at which the Gibbs energy change (ΔG) for the vaporization of water is zero, we can use the Gibbs free energy equation:

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

Where:
- ΔG is the change in Gibbs free energy,
- ΔH is the change in enthalpy,
- T is the temperature in Kelvin,
- ΔS is the change in entropy.

Given:
- ΔH = 40.63 kJ/mol = 40630 J/mol (since 1 kJ = 1000 J)
- ΔS = 108 J/K·mol

At equilibrium, ΔG = 0, so we can set the equation to zero:

\[ 0 = \Delta H - T \Delta S \]

Rearranging the equation gives us:

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

Now, we can solve for T:

\[ T = \frac{\Delta H}{\Delta S} \]

Substituting the values we have:

\[ T = \frac{40630 \, \text{J/mol}}{108 \, \text{J/K·mol}} \]

Calculating this gives:

\[ T = \frac{40630}{108} \approx 375.93 \, \text{K} \]

Thus, the temperature when the Gibbs energy change (ΔG) for the transformation will be zero is approximately:

\[ T \approx 375.93 \, \text{K} \]

### Final Answer:
The temperature when Gibbs energy change (ΔG) for this transformation will be zero is approximately **375.93 K**.