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Question:

ΔfGo at 500 K for substance 'S' in liquid state and gaseous state are +100.7 kcal mol⁻¹ and +103 kcal mol⁻¹, respectively. Vapour pressure of liquid 'S' at 500 K is approximately equal to: (R = 2 cal K⁻¹mol⁻¹). 0.1 atm, 10 atm, 1 atm, 100 atm

10atm

100atm

1atm

0.1atm

Solution:

ΔGorxn = ΔfGo(vapour) - ΔfGo(liquid)
ΔGorxn = 103 - 100.7 = 2.3 kcal/mol
ΔGorxn = -RTlnK
2.3 kcal/mol × 1000 cal/kcal = -2 cal K⁻¹mol⁻¹ × 500 K × lnK
lnK = -2.3
K = 10 atm = Vapour pressure of liquid 'S'