Respuesta :
Explanation:
It is known that the change in Gibb's free energy varies with temperature as follows.
[tex]\Delta G(T) = \Delta H(T) - T \Delta S(T)[/tex]
= [tex]\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}})][/tex]
[tex]\Delta H(T_{f}) = -\Delta_{fus} H(T_{f})[/tex] (assumption)
= [tex]\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}})[/tex]
= [tex](\frac{T}{T_{f}} - 1) \Delta_{fus} H(T_{f}) - \Delta C_{p,m}(T - T_{f} - Tln (\frac{T}{T_{f}}))[/tex]
As, T = [tex]-3^{o}C[/tex] = (-3 + 273) = 270 K, [tex]T_{f} = 0^{o}C = 0 + 273 K = 273 K[/tex].
Therefore, calculate the change in Gibb's free energy as follows.
[tex]\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}}))[/tex]
= [tex](\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})[/tex]
= -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.
