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Gibbs free energy change at constant pressure is zero?

  1. Feb 19, 2012 #1
    gibbs free energy change at constant pressure is zero??

    IS gibbs free energy change at constant pressure zero?
    ΔS = q / T. At constant pressure q = ΔH so ΔS = ΔH / T
    So ΔG = ΔH - ΔH/T . T
    = 0

    I think i am wrong but where? My friend tells me that q involved in enetropy is different from the one we consider in ΔH. I dont understand this. IS heat due to entropy different from the one involved in enthalpy change
    I'm all messed up
     
  2. jcsd
  3. Feb 20, 2012 #2
    Re: gibbs free energy change at constant pressure is zero??

    In my opinion you unwittingly made two assumptions:

    1) heat is exchanged reversibly (in the definition of ΔS there is reversible heat)
    2) temperature is constant (otherwise ΔG = ΔH - TΔS does not hold, generally ΔG = ΔH - Δ(TS) )

    So you have an equilibrium situation with constant T and P in which case G is indeed constant.
     
  4. Feb 20, 2012 #3

    DrDu

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    Re: gibbs free energy change at constant pressure is zero??

    When a chemical reaction is involved
    then
    [itex] dG=Vdp-SdT+\Delta G d\xi [/itex]
    where [itex]\xi[/itex] is the reaction coordinate. So even if p and T are hold constant, G will change as a result of the chemical reaction taking place.
    The problem with using [itex]\Delta S=q/T[/itex] is that it presuposes equilibrium. However, when there is no chemical equilibrium, the formula is not applicable.
     
  5. Feb 20, 2012 #4

    DrDu

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    Re: gibbs free energy change at constant pressure is zero??

    Not quite,
    [itex]dU=TdS-pdV+\sum_i \mu_i dn_i[/itex]
    [itex]dH=TdS+Vdp+\sum_i \mu_i dn_i[/itex]
    [itex]dG=-SdT+Vdp+\sum_i \mu_i dn_i[/itex]
    so [itex]\mu_i=\partial G/\partial{n_i}|_{T,p}=\partial H/\partial {n_i}|_{S,p} [/itex] etc.

    For a reaction [itex] \sum_i \nu_i X_i=0[/itex] the reaction coordinate is defined as
    [itex] dn_i/\nu_i=d\xi [/itex] and [itex]\Delta G=\sum_i \mu_i \nu_i [/itex].
    Hence [itex]\Delta G=\partial G/\partial \xi |_{p,T}=\partial H/\partial \xi|_{p,S}[/itex]
    Now by definition
    [itex]\Delta H\equiv \partial H /\partial \xi|_{p,T}=\partial H /\partial \xi|_{p,S}+\partial H/\partial S|_{p,\xi}\partial S/\partial \xi|_{p,T}=\Delta G+T\Delta S[/itex] as [itex]\partial S/\partial \xi|_{p,T}\equiv \Delta S[/itex]. So ΔG = ΔH - TΔS holds always.
     
  6. Feb 20, 2012 #5
    Re: gibbs free energy change at constant pressure is zero??

    You confuse Δ for the total change of a state variable (what I use in my posts) with the derivative (slope) of the variable with respect to the reaction extent, which is unfortunately denoted by Δ as well, nonetheless, there should be subscript r (like reaction) in that case - to distinguish from mere change.

    My arguments are independent on the process happening in the system, be it chemical reaction or anything else. If Δ is just change (what is it's original meaning), then ΔG = ΔH - TΔS cannot hold always which is seen immediately from the mere fact that with varying temperature we wouldn't know what value from the range should be used for T:-)

    To sum up: we are not in disagreement, we just interpret the question differently. I believe the asker was refering to simple change by Δ, not to the above-mentioned differential quantity, am I right?
     
  7. Feb 20, 2012 #6

    DrDu

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    Re: gibbs free energy change at constant pressure is zero??

    Asym,

    of course you are right. My impression is that this kind of questions mostly arise because people don't know the difference of the simple [itex] \Delta G[/itex] and the partial free energy change [itex]\Delta_r G[/itex], with chemists having mostly the second one in mind.
     
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