100.℃ 1.00 atm water ΔH° = 40.6 kJ/mol ΔG°

At 100.℃ and 1.00 atm, ΔH° = 40.6 kJ/mol for the vaporization of water.

Estimate ΔG° for the vaporization of water at 90.℃ and 110.℃.

Assume ΔH° and ΔS° at 100.℃ and 1.00 atm do not depend on temperature.

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ΔG° = ΔH° – TΔS°

( 참고 https://ywpop.tistory.com/7438 )

물의 끓는점에서,

ΔG° = 0

0 = ΔH° – TΔS°

TΔS° = ΔH°

ΔS° = ΔH° / T

= (40.6 kJ/mol) / (273.15 + 100. K)

= 0.109 kJ/mol•K

90.℃에서,

ΔG° = ΔH° – TΔS°

= (40.6 kJ/mol) – (273.15 + 90. K) (0.109 kJ/mol•K)

= (40.6) – (273.15 + 90.) (0.109)

= 1.02 kJ/mol

---> ΔG° > 0 이므로, 비자발적.

110.℃에서,

ΔG° = ΔH° – TΔS°

= (40.6) – (273.15 + 110.) (0.109)

= –1.16 kJ/mol

---> ΔG° < 0 이므로, 자발적.

답: 1.02 kJ/mol, –1.16 kJ/mol

[키워드] ΔG° = ΔH° - TΔS° 기준, ΔG = ΔH - TΔS 기준