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Question

# At 25∘C, for the process H2O(l)⇌H2O(g); ΔG∘ is 8.6 kJ. The vapour pressure of water at this temperature is nearly:

A
29 torr
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B
285 torr
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C
23.17 torr
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D
100 torr
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Solution

## The correct option is D 23.17 torrConsider the vaporization as equilibrium reaction. The the activities satisfy equilibrium equation K=a(H2O(g)a(H₂O(l)The vapour pressure is the partial pressure over gaseous water over pure water. Since liquid water water is pure , it has activity 1. a(H2O(l))=1 Assuming ideal gas phase, the activity of gaseous water is equal to ratio of partial pressure to total pressure: a(H2O(g))=p(H2O(g))po Hence K=p(HO(g))po =p(H2O(g))=po⋅K Equilibrium constant and ΔG are related as ΔG=−R⋅T⋅ln(K) K=e−ΔGR⋅T Therefore p(H2O(g))=po⋅e−ΔGR⋅T =101325Pa⋅e−8600J/mol(8.314472J/molK⋅298.15K)=3155Paconverting Pa to torr, we get 23.17 as answer Hence, te correct option is C

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