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Gibbs free energy problem

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  1. Mar 16, 2019 at 7:15 AM #1
    • 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?
     
  2. jcsd
  3. Mar 16, 2019 at 8:15 AM #2

    Ygggdrasil

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    What is the equilibrium vapor pressure of water at 100°C?
     
  4. Mar 16, 2019 at 9:14 AM #3
    1 bar? So it's really not an equilibrium..?
     
  5. Mar 16, 2019 at 10:08 AM #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.
     
  6. Mar 16, 2019 at 10:23 AM #5
    So it's wrong to call it an equilibrium?
     
  7. Mar 16, 2019 at 10:34 AM #6

    Ygggdrasil

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    The reaction is at equilibrium only at a specific P(H2O).
     
  8. Mar 16, 2019 at 10:37 AM #7
    Thank you!
    isn't it wrong to write the equation with a "=" sign between the reactants and products then?
     
  9. Mar 16, 2019 at 3:14 PM #8

    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.
     
  10. Mar 17, 2019 at 7:00 AM #9
    I think it would have been better if they had used an arrow. ---->
     
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