Gibbs free energy problem

  • #1
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Please post this type of questions in HW section using the template.
Calculate deltaG for the reaction:

H2O(l) = H2O(g). 100 degrees celsius, water is clean. P(H2O) = 0.1 bar.

Given that it is an equilibrium, I'd think that deltaG would be zero. But the answer is in fact negative. How is that possible?
 

Answers and Replies

  • #2
Ygggdrasil
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What is the equilibrium vapor pressure of water at 100°C?
 
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1 bar? So it's really not an equilibrium..?
 
  • #4
Ygggdrasil
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At equilibrium, the partial pressure of water would be 1 atm. Because P(H2O) is less than that, the forward reaction is favored.
 
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So it's wrong to call it an equilibrium?
 
  • #6
Ygggdrasil
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The reaction is at equilibrium only at a specific P(H2O).
 
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Thank you!
isn't it wrong to write the equation with a "=" sign between the reactants and products then?
 
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Ygggdrasil
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H2O(g) ##\rightleftharpoons## H2O(l) is the correct notation.

The reaction is reversible, with vapor molecules continuously condensing into the liquid phase and liquid molecules continuously evaporating into the gas phase. When P(H2O) is less than 1 atm, the rate of evaporation exceeds the rate of condensation. When P(H2O) is greater than 1 atm, the rate of condensation exceeds the rate of evaporation. Only when P(H2O) = 1 atm will evaporation and condensation occur at the same rate, so there will be no net exchange of material between the liquid and gaseous phases.
 
  • #9
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Thank you!
isn't it wrong to write the equation with a "=" sign between the reactants and products then?
I think it would have been better if they had used an arrow. ---->
 

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