One mole of H 2 O (liquid) is super-cooled to -2.25 C at 1 bar pressure. The equilibrium freezing temperature at this pressure is 0.0 C. The transformation H 2 O ( l i q u i d ) H 2 O ( s o l i d ) is suddenly observed to occur. Show that the t

The spontaneity of a process at constant temperature and pressure can be determined using the Gibbs free energy change: {eq}Delta G = Delta H - T Delta S {/eq} where {eq}Delta H {/eq} is the enthalpy change, {eq}T {/eq} is the temperature, and {eq}Delta S {/eq} is the entropy change. If {eq}Delta G < 0 {/eq}, the process is spontaneous. At the equilibrium freezing temperature of 0.0 {eq}^{circ} {/eq}C, the Gibbs free energy change for the transformation {eq}H_2O(liquid) rightarrow H_2O(solid) {/eq} is: {eq}Delta G_{eq} = Delta H_{fusion} {/eq} At the actual temperature of -2.25 {eq}^{circ} {/eq}C, the Gibbs free energy change for the transformation is: {eq}Delta G = Delta H_{fusion} - T Delta S {/eq} To find {eq}Delta S {/eq}, we can use the heat capacities to calculate the entropy change of the transformation: {eq}Delta S = \frac{Delta H_{fusion}}{T_{fus}} + int_{T_{fus}}^{T} \frac{C_p(T)}{T}dT {/eq} where {eq}T_{fus} {/eq} is the melting temperature, and {eq}C_p(T) {/eq} is the heat capacity at constant pressure. At 1 bar pressure, the melting temperature of water is 0.0 {eq}^{circ} {/eq}C, so we have: {eq}Delta S = \frac{Delta H_{fusion}}{T_{fus}} + int_{0}^{-2.25} \frac{C_p(T)}{T}dT {/eq} Using the given values, we get: {eq}Delta S = 38.174 J/K\cdot mol {/eq} Substituting into the equation for {eq}Delta G {/eq}, we get: {eq}Delta G = 6.008 kJ/mol - (-2.25 ^{circ}{rm C} + 273.15 {rm K})\cdot 38.174 J/K\cdot mol = -0.814 kJ/mol {/eq} Since {eq}Delta G < 0 {/eq}, the transformation {eq}H_2O(liquid) rightarrow H_2O(solid) {/eq} is spontaneous at this state point.