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Free energy

  1. Jan 29, 2014 #1
    If ##ΔG_{sys} = ΔH_{sys} - TΔS_{sys}; ΔS_{sys} = \frac {-ΔH_{surr}}{T} ##

    Then, doesn't this expression just simplify to:

    ##ΔG_{sys} = ΔH_{sys} + ΔH_{surr}## and isn't ##ΔH_{sys} = -ΔH_{surr}##?

    So then ##ΔG = 0##...this does not seem correct...could anyone please clarify my mistake and the formal proof of free energy if it differs from the one I have shown?

    Also, what is the explanation for why there is no free energy change during phase changes? I was merely told this without any explanation and one would be very helpful.

    Finally, just a related question:

    If you have 1 gram of ammonia and it dissolves in 50 grams of water to release 1000 J (just hypothetical). When calculating the temperature change of the water, using q = mcΔT, is m = 50g or 51g since the ammonia is now aqueous and inseparable from the water. Yet, it has slightly different properties now. I am unsure if it is treated just like a normal sample of water or not.
    Last edited: Jan 29, 2014
  2. jcsd
  3. Jan 29, 2014 #2


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    ΔS_sys = -ΔH_surr/T only for reversible processes, and it is true that ΔG = 0 for any reversible process.

    If the phase changes are occurring reversibly (e.g. melting at the melting point, boiling at the boiling point), then ΔG = 0 for the reason above.

    For the last question, if you are really worried about an exact answer, you would also have to take into account that the heat capacity of the solution changes when you dissolve ammonia in it.
  4. Jan 29, 2014 #3
    ΔG is not equal to zero for any arbitrary reversible process. For example, in the isothermal expansion of an ideal gas, ΔG=RTln(P2/P1).
  5. Jan 30, 2014 #4


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    You are right, I also should have specified any isobaric (constant pressure) reversible process. Chemists usually take the isobaric assumption for granted since most of our chemical reactions are performed at constant (atmospheric) pressure. Thanks for the correction.
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