Question

For vaporization of water at 1 atmospheric pressure, the values of $\Delta H$ and $\Delta S$ are $40.63 \, kJ \, mol^{-1}$ and $108.8 \, JK^{-1} mol^{-1}$, respectively. The temperature when Gibbs energy change $(\Delta G)$ for this transformation will be zero, is

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

Answer 1

The Correct Option is B

To determine the temperature at which the Gibbs free energy change $(\Delta G)$ for the vaporization of water is zero, we use the relationship:

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

where:

• \(\Delta H\) is the enthalpy change, given as 40.63 \, \text{kJ} \, \text{mol}^{-1}
• \(\Delta S\) is the entropy change, given as 108.8 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1}
• T is the temperature in Kelvin

For the condition where \(\Delta G = 0\) (equilibrium state), the equation becomes:

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

Rearranging for temperature T, we have:

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

Substituting the given values (note the conversion from kJ to J for \(\Delta H\) since \(\Delta S\) is in J):

\[\Delta H = 40.63 \, \text{kJ} \, \text{mol}^{-1} = 40630 \, \text{J} \, \text{mol}^{-1}\]

\[T = \frac{40630}{108.8} \approx 373.4 \, \text{K}\]

Thus, the temperature at which \(\Delta G = 0\) is 373.4 K.

This matches the normal boiling point of water at 1 atmospheric pressure, confirming the answer. Therefore, the correct option is:

• 373.4 \, \text{K}