ΔG° for the vaporization of water

Question

At 100.°C and 1.00 atm, ΔH° = 40.6 kJ/mol for the vaporization of water. Estimate ΔG° for the vaporization of water at 87°C and 113°C. Assume ΔH° and ΔS° at 100.°C and 1.00 atm do not depend on temperature.

Answer 1

to do this ., we need dS @ At 100.°C and 1.00 atm, which luckily is its "normal" boiling point

at a "normal boiling point" you ahvce an equilibrium:

H2O (l)  <--> H2O (g)

at an equilibrium, the  equation dG = dH -TdS simplifies to:

0 = dH -TdS, because dG = zero at equilibrium.

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so let's find dS, ... if 0 = dH -TdS ,... then:

dS = dH / T

dS = 40.6kJ / 373Kelvin

dS = 0.1088 kJ/K

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now , find dS @ 87 Celsius (aka 360Kelvin):

dG = dH - TdS

dG = 40.6 kJ - (360K) (0.1088kJ/K)

dG = 40.6 kJ - 39.2

your answer is : dG = + 1.4 (not spontaneous)

now , find dS @ 113 Celsius (aka 360Kelvin):

dG = dH - TdS

dG = 40.6 kJ - (386K) (0.1088kJ/K)

dG = 40.6 kJ - 42.0

your answer is : dG = - 1.4 (spontaneous)