Consider the melting of ice (solid water) to liquid water at a pressure of \(101.3 \mathrm{kPa}\). (a) Is this process endothermic or exothermic? (b) In what temperature range is it a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) At what temperature are the two phases in equilibrium?

To determine whether the process is endothermic or exothermic, we need to look at the change in enthalpy, denoted as \(\Delta H\). For melting of ice to water, the process requires energy input (usually as heat) to break down the hydrogen bonds between the water molecules in the solid state and create a liquid state. In an endothermic process, energy is absorbed, so \(\Delta H > 0\). Therefore, the melting of ice to liquid water at a pressure of \(101.3\,\mathrm{kPa}\) is an endothermic process.

To determine the temperature range for which the melting of ice is a spontaneous process, we need to examine the change in Gibbs free energy, or \(\Delta G\). The Gibbs free energy change can be calculated as follows: \(\Delta G = \Delta H - T\Delta S\) Where \(\Delta S\) is the change in entropy, and \(T\) is the temperature. A process is spontaneous if \(\Delta G < 0\). Since the process is endothermic from part (a), we know that \(\Delta H > 0\). Also, for the melting of ice, the entropy change is positive, as the solid ice transitions into a more disordered liquid water state, hence \(\Delta S > 0\). Therefore, if the product \(T\Delta S\) is larger than \(\Delta H\), then \(\Delta G\) will be negative, and the process will be spontaneous. This occurs at higher temperatures. So, the process is spontaneous in the temperature range where \(\Delta G < 0\), which means \(T\Delta S > \Delta H\) for higher temperatures.

Now, we need to find the temperature range where the melting process is nonspontaneous. A process is considered nonspontaneous when \(\Delta G > 0\). Since the process is endothermic and the melting of ice increases the entropy, \(\Delta H > 0\) and \(\Delta S >0\). If the product \(T\Delta S\) is less than \(\Delta H\), then \(\Delta G\) will be positive, and the process will be nonspontaneous. This occurs at lower temperatures. Thus, the process is nonspontaneous in the temperature range where \(\Delta G > 0\), which means \(T\Delta S < \Delta H\) for lower temperatures.

At equilibrium, the Gibbs free energy change is zero, or \(\Delta G=0\). We can use the equation: \(\Delta G = \Delta H - T\Delta S\) Substituting \(\Delta G = 0\), we get: \(T\Delta S = \Delta H\) Since the Standard enthalpy of fusion for ice at \(1\,\mathrm{atm}\) is \(\Delta H = 6.01\,\mathrm{kJ/mol}\) and the Standard entropy of fusion for ice at \(1\,\mathrm{atm}\) is \(\Delta S = 21.97\,\mathrm{J/mol\cdot K}\), we can calculate the equilibrium temperature: \(T = \frac{\Delta H}{\Delta S} = \frac{6.01\times10^{3}\,\mathrm{J/mol}}{21.97\,\mathrm{J/mol\cdot K}} = 273.15\,\mathrm{K}\) So the equilibrium temperature between ice and liquid water at a pressure of \(101.3\,\mathrm{kPa}\) is \(273.15\,\mathrm{K}\).