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Homework Help: Gibb's free energy confusion!

  1. Jan 7, 2009 #1
    [tex]\Delta G = \Delta H - T \Delta S [/tex]

    I don't understand this equation satisfactorily.
    I have learned so far that endothermic reaction having a positive [tex]\Delta S_{system}[/tex] is spontaneous at higher temperature because [tex]T\Delta S[/tex] out-compete the enthalpy term and make the free energy change negative.
    But I don't understand why [tex]T\Delta S _{sorrounding}[/tex] should not increase with temperature as [tex]T\Delta S _{system}[/tex] does.
    Clearly [tex]\Delta H[/tex] =[tex]-T\Delta S _{sorrounding}[/tex]
    so we can rewrite the gibbs free energy equation
    [tex]\Delta G= -T\Delta S _{sorrounding} - T\Delta S _{system}[/tex]
    [tex]\Delta G = -T ( \Delta S _{sorrounding} + \Delta S _{system})[/tex]

    if [tex]\Delta S _{sorrounding}[/tex] and [tex]\Delta S _{system}[/tex] both are constant then the sign of [tex]\Delta G[/tex] is independent of temperature. If [tex]\Delta G[/tex] is positive increasing temperature just make it more positive and If [tex]\Delta G[/tex] is negative increasing temperature just make it more negative. Temperature cannot change the sign.
    I know I must be missing something. Please help!!

    A[tex]^{ }_{A}[/tex]
    Last edited: Jan 7, 2009
  2. jcsd
  3. Jan 7, 2009 #2


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    Staff: Mentor

    When you use the equation to describe isolated system you don't have to worry about the rest of the Universe.
  4. Jan 7, 2009 #3
    But I need to know why?
    Isn't Del H = -T*Del S (sorrounding)? Clearly it is a function of T, then why it would be unaltered by temperature {but T*Del S (system) is altered}?
    Last edited: Jan 7, 2009
  5. Jan 7, 2009 #4


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    Not clearly. That equation only holds if you have a very large heat bath surrounding the system with the same temperature T, so that all heat transfer is reversible. It's important to make this distinction.

    But that is a side point. The key issue is this: if you're writing [itex]\Delta H=T\Delta S^\mathrm{surr}[/itex], you're already assuming that the reaction is spontaneous; otherwise energy would not be transferred. And when you write [itex]\Delta G= -T\Delta S^\mathrm{surr} - T\Delta S^\mathrm{sys}[/itex], you're assuming that it's barely spontaneous, that [itex]\Delta G=0[/itex]. Otherwise more energy would be present from the spontaneous forward reaction.

    So you can no longer vary T and ask whether the reaction is spontaneous!
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