Question

Given that delta hvap is 58.2 kj/mol and the boiling point is 83.4 c 1atm if one mole of this substance is vaporized at 1atm calculate delta ssurr

Answer 1

Change in Gibb's free energy of system (ΔG) = ΔH - TΔS.........(Eq. 1)
Now, if magnitude of ΔG <0, then reaction is spontaneous.
         if magnitude of ΔG > 0, then reaction is non-spontaneous. 
         At equilibrium, ΔG = 0
When at boiling point, liquid state is in equilibrium with vapour state. Hence, it present case ΔG = 0

∴ Eq 1 becomes, ΔH = TΔS
here, ΔH = 58.2 kj/mol (Given),
∴ At T = 83.4 oC = 356.4 K, ΔS = 0.1633 kj/mol.K