Gibbs Free Energy of Formation: 1°C or 25°C? (And other exciting questions.)

  1. Hi there:

    When using ΔG=ΔG°+RT ln Q to calculate the energy yield of a reaction, does it matter if I use ΔG° calculated at 1°C or 25°C? Also, why are there two choices and when are they each applicable? Finally, I have also seen ΔG°' written (note the prime). What does this mean and how does it differ from ΔG° conceptually and numerically?

    I realise that these are very basic questions, so if you want to point me towards some elementary reading material I understand.

    Thanks in advance,
  2. jcsd
  3. DrDu

    DrDu 4,500
    Science Advisor

    Yes, it matters. If you use Delta G at 1 deg C, then you will also have to use a corresponding T and then equilibrium constant will be that for this temperature. The same way using Delta G at 25 deg Celsius yields Q for this very temperature.
    The change of Q with temperature can be calculated with the van't Hoff equation:'t_Hoff_equation
  4. Thanks for your reply DrDru, but I'm afraid I don't really follow. Perhaps a concrete example would help me understand. For example, consider this reaction at 50ºC (i.e., 323.15K):


    Using ΔG=ΔG°+RT ln Q, I get something like this:

    ΔG=ΔG°+R×323.15×ln (([C][D])/([A]))

    but what is the value of ΔG° that I should use? The value at 1ºC or 25ºC?
  5. DrDu

    DrDu 4,500
    Science Advisor

    Neither of the two but the value at 50 deg. Celsius.
    If you have both the values at 1 and at 50 degrees, you could linearly interpolate as a first approximation.
    But, as I said, it would be more exact to calculate the value of Delta G0 at 50 degs from the van't Hoff or Gibbs-Helmholtz equation, see:

    E.g. ##T_1=1^\circ##C, ##T_2=25^\circ##C and ##T_3=50^\circ##C,
    ##\Delta G^0(T_1)/T_1-\Delta G^0(T_2)/T_2=\Delta H ^0(1/T_1-1/T_2)##.
    Solve this for ##\Delta H^0## and then solve
    ##\Delta G^0(T_1)/T_1-\Delta G^0(T_3)/T_3=\Delta H ^0(1/T_1-1/T_3)##
    for ##\Delta G^0(T_3)##.
  6. Aha! Makes sense. Thanks for explaining that.
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