Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gibbs free energy going to zero (equilibrium)

  1. Aug 10, 2013 #1
    Hi I have a 4 part question about phase changes and I hope you guys can help me out here thanks :)

    The chemical equation for vaporization for water is H2O(l)-->H2O(g)

    1) When ΔG is 0 during a phase change we would use the formula ΔH=TΔS and put in values to get the boiling/melting point by rearranging it to T=ΔH/ΔS. However, after learning that every reaction at any temperature would slowly go into equilibrium where ΔG=0, why would the boiling point of water be 373K since equilibrium can be reached at even 298K?

    2)I read that the ΔS used represents the vapor at 1 atm which would be different from the vapor at a lower atm. But I don't quite understand why they would be different. Their explanation was that water vapor at 1 atm occupies more space than at a lower atm so the ΔS is greater. But since the temperature are different shouldn't the volume be equal to each other too? So why would the entropy change be different when the vapor pressures are different?

    3)They also mentioned that the ΔG would go to zero only when it reaches vapor pressure for that specific temperature. I agree with that but since when using ΔG°=-RTlnKeq is it assumed that the reaction is for 1 mole of liquid water to 1 mole of gaseous water at 1atm. So when ΔG=0 does it matter if more than 1 mole has or less than 1 mole had been vaporized? I would think only 1 mole can be reacted since for ΔH we would use the vaporization of 1 mole. So would the container of the gas be fixed to be such that 1 mole would be vaporized at the pressure defined (like 1 atm) and at the temperature defined (like 373K)? If so then would there be any difference if more or less than 1 mole of water vaporizes instead? It would not right since the moles double so would ΔH and ΔS so there isn't a change? Or is there another reason?

    4)But even so at 298K, Gibbs free energy can also be represented with ΔG=G products-G reactants so for G products it would water vapor at 298K which I thought was impossible as well?

    Thanks so much for the help :)
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted