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Gibbs Free Energy & Spontaneity

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Question

Calculate the temperature at which liquid water will be in equilibrium with water vapour.
$Δ H_{v a p} = 40.73 k J {m o l}^{- 1}$ and $Δ S_{v a p} = 0.109 k J {m o l}^{- 1} K^{- 1}$

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Solution

Given $Δ H_{v a p} = 40.73 k J {m o l}^{- 1}; Δ S_{v a p} = 0.109 k J {m o l}^{- 1} K^{- 1}$
and $Δ G = 0$
Applying $Δ G = Δ H - T Δ S$
$0 = 40.73 - T \times 0.109$
$T = \frac{40.73}{0.109} = 373.6 K$

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

Δ H_{v a p} = 40.73 k J m o l^{- 1}

Δ S_{v a p} = 0.109 k J K^{- 1} m o l^{- 1}

Δ H_{v a p} = 30 kJ/mol

Δ S_{v a p} = 75 {J mol}^{- 1} K^{- 1}

. Find the temperature of vapour, at one atmosphere:

Δ H_{v a p} = 30 k J / m o l

Δ S_{v a p} = 75 J m o l^{- 1} K^{- 1}

. Find the temperature of vapour, at one atmosphere.

Δ H_{v a p} = 30 k J / m o l e

Δ S_{v a p} = 75

^{- 1} K^{- 1}

. Find temperature of vapour, at one atmosphere.

x

is the temperature (in

^{o} C

) at which liquid

H_{2} O

will be in equilibrium with its vapours (

Δ H a n d Δ S

for vaporization are

40.7 k J {m o l}^{- 1}

0.11 k J {m o l}^{- 1} K^{- 1}

respectively) , then

x

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