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Question: Calculate the free energy per mole when liquid water boils against 1 atm pressure \((\Delta H_{vap} ...

Calculate the free energy per mole when liquid water boils against 1 atm pressure (ΔHvap=2.0723kJ/g)(\Delta H_{vap} = 2.0723kJ/g)

A

0

B

0.2

C

0.3

D

0.4

Answer

0

Explanation

Solution

ΔH=2.0723×18kJ/mole=37.3kJ/mole\Delta H = 2.0723 \times 18kJ/mole\mathbf{= 37.3kJ/mole};

T=373K\mathbf{T = 373K}; ΔGvap\mathbf{\Delta}\mathbf{G}_{\mathbf{vap}}=?

ΔS=ΔHvapT=37.3kJ373=0.1kJK−1mol−1\Delta S = \frac{\Delta H_{vap}}{T} = \frac{37.3kJ}{373} = 0.1kJK^{- 1}mol^{- 1}

ΔGvap=ΔHvap−TΔSvap=37.3−373×0.1\mathbf{\Delta}\mathbf{G}_{\mathbf{vap}}\mathbf{= \Delta}\mathbf{H}_{\mathbf{vap}}\mathbf{- T\Delta}\mathbf{S}_{\mathbf{vap}}\mathbf{= 37.3}\mathbf{-}\mathbf{373}\mathbf{\times}\mathbf{0.1}= 0