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Question: At what temperature liquid water will be in equilibrium with water vapour? \[\Delta H_{vap} = 40.73...

At what temperature liquid water will be in equilibrium with water vapour?

ΔHvap=40.73kJmol−1,ΔSvap=0.109kJK−1mol−1\Delta H_{vap} = 40.73kJmol^{- 1},\Delta S_{vap} = 0.109kJK^{- 1}mol^{- 1}

A

282.4K282.4K

B

373.6K373.6K

C

100 K

D

400 K

Answer

373.6K373.6K

Explanation

Solution

: At equilibrium, ΔG=0\Delta G = 0 for ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

or T=ΔHΔS=40.730.109=373.6KT = \frac{\Delta H}{\Delta S} = \frac{40.73}{0.109} = 373.6K

=2.303×1×8.314log⁡10010=19.14JK−1= 2.303 \times 1 \times 8.314\log{}\frac{100}{10} = 19.14JK^{- 1}