Question

Calculate the Gibbs energy of freezing (DeltaGfreezing) in units of J/mol when supercooled water freezes at -3degreeC at constant T and P. Delta H fusion = 6000 J/mol at 0degreeC. The molar heat capacity of water and ice are 75.3 J/molK and 38 J/molK, respectively, and both are independent of temperature over this range. State any assumptions you make in your calculation!

Answer 1

Explanation:

It is known that the change in Gibb's free energy varies with temperature as follows.

            $\Delta G(T) = \Delta H(T) - T \Delta S(T)$

                             = $\Delta H(T_{f}) - \Delta C_{p,m} (T - T_{f}) - T[\Delta S(T_{f}) - \Delta C_{p,m} ln (\frac{T}{T_{f}})]$

         $\Delta H(T_{f}) = -\Delta_{fus} H(T_{f})$        (assumption)

                     = $\Delta H(T_{f}) - \frac{T}{T_{f}} \Delta H(T_{f}) - \Delta C_{p, m}(T - T_{f} - T ln \frac{T}{T_{f}})$

                     = $(\frac{T}{T_{f}} - 1) \Delta_{fus} H(T_{f}) - \Delta C_{p,m}(T - T_{f} - Tln (\frac{T}{T_{f}}))$

As, T = $-3^{o}C$ = (-3 + 273) = 270 K,   $T_{f} = 0^{o}C = 0 + 273 K = 273 K$.

Therefore, calculate the change in Gibb's free energy as follows.

     $\Delta G(T) = (\frac{T}{T_{f}} - 1) \Delta_{fus} H(T_{f}) - \Delta C_{p,m}(T - T_{f} - Tln (\frac{T}{T_{f}}))$

= $(\frac{270 K}{273 K} - 1)(6000 J/mol K) - (75.3 - 38) J/mol K (270 K - 273 K - 270 K ln \frac{270 K}{273 K})$

                  = -65.93 J/mol K + 0.62 J/mol K

                  = -65.31 J/mol K

Thus, we can conclude that Gibbs energy of freezing for the given reaction is -65.31 J/mol K.