At \(100 .^{\circ} \mathrm{C}\) and $1.00 \mathrm{atm}, \Delta H^{\circ}=40.6 \mathrm{kJ} / \mathrm{mol}\( for the vaporization of water. Estimate \)\Delta G^{\circ}$ for the vaporization of water at \(90 .^{\circ} \mathrm{C}\) and \(110 .^{\circ} \mathrm{C}\) . Assume $\Delta H^{\circ}\( and \)\Delta S^{\circ}\( at \)100 .^{\circ} \mathrm{C}$ and 1.00 \(\mathrm{atm}\) do not depend on temperature.

The relationship between the standard Gibbs free energy change (∆G°) standard enthalpy change (∆H°), and standard entropy change (∆S°) is described by the Gibbs-Helmholtz equation: \[ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \] We are given ∆H° and T at 100°C and need to find ∆G° at different temperatures.

We first need to convert the given temperature (100°C) to Kelvin: \[ T(K) = T(^\circ C) + 273.15 \] For 100°C: \[ T(100 ^\circ C) = 100 + 273.15 = 373.15 \,\mathrm{K} \] We know that at 100°C and 1.00 atm, ∆H° = 40.6 kJ/mol

At the boiling point, the system is at equilibrium. This means that the Gibbs free energy change (∆G°) is equal to zero: \[ \Delta G^{\circ}_{eq} = 0 \] We can rearrange the Gibbs-Helmholtz equation to find ∆S°: \[ \Delta S^{\circ} = \frac{\Delta H^{\circ}}{T} \] Using the values of ∆H° and temperature: \[ \Delta S^{\circ} = \frac{40.6\,\mathrm{kJ/mol}}{373.15\, \mathrm{K}} = 0.1087\, \mathrm{kJ/(mol\cdot K)} \]

Now that we have ∆H° and ∆S° at 100°C, we can use the Gibbs-Helmholtz equation to find ∆G° at the other temperatures. First, convert 90°C and 110°C to Kelvin: For 90°C: \[ T(90 ^\circ C) = 90 + 273.15 = 363.15 \,\mathrm{K} \] For 110°C: \[ T(110 ^\circ C) = 110 + 273.15 = 383.15 \,\mathrm{K} \] Now, use the Gibbs-Helmholtz equation to find ∆G° at each temperature: For 90°C: \[ \Delta G^{\circ}(90 ^\circ C) = \Delta H^{\circ} - T\Delta S^{\circ} = 40.6\,\mathrm{kJ/mol} - (363.15\,\mathrm{K} \times 0.1087\, \mathrm{kJ/(mol\cdot K)}) \approx 0.94\,\mathrm{kJ/mol} \] For 110°C: \[ \Delta G^{\circ}(110 ^\circ C) = \Delta H^{\circ} - T\Delta S^{\circ} = 40.6\,\mathrm{kJ/mol} - (383.15\,\mathrm{K} \times 0.1087\, \mathrm{kJ/(mol\cdot K)}) \approx -2.69\,\mathrm{kJ/mol} \] So, the estimated standard Gibbs free energy change for the vaporization of water is approximately 0.94 kJ/mol at 90°C and -2.69 kJ/mol at 110°C.