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Gibbs free energy -- mathematical expression

  1. Feb 23, 2016 #1
    I am not able to understand the mathematical expression of "change in Gibbs free energy",

    For a chemical reaction occurring at constant temperature and constant pressure,
    (ΔS)total = (ΔS)system + (ΔS)surrounding

    Considering that reaction is exothermic, ΔH be the heat supplied by system to surrounding at constant pressure and temperature,

    (ΔS)total = (ΔS)system + (-ΔH)/T

    - T * (ΔS)total = ΔH - T * (ΔS)system

    The term on left hand side is known as Change in Gibbs free energy ΔG. WHY?
    also, the temperature in above expression is Tsystem and not Tsurrounding , HOW?

    (ΔS)surrounding should be Q/Tsurrounding, where Q is heat added to surrounding.

    By intuition, ΔH amount of heat is available and T * ΔSsystem is the unavailable energy. Thus (ΔH-T*ΔSsystem) is the amount of energy that is available to be converted to work and should be Gibbs free energy right? I am not able to understand it from the mathematical equation.

    Please explain what am I missing out? and where am I wrong? Please tolerate me for any mistakes, this is my first post and I do not have chemical engineering background. Thank you.
     
  2. jcsd
  3. Feb 23, 2016 #2

    DrDu

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    The equation ##\Delta S_\mathrm{surrounding}=-\Delta H/T ## only holds if the heat is transferred under equilibrium conditions, hence ##T_\mathrm{system}=T_\mathrm{surrounding}##.
     
  4. Feb 23, 2016 #3
    Hi rajendra123,

    Your question is very puzzling to me. The Gibbs Free Energy of a system is defined as G≡H-TS. I don't know why you have even introduced the surroundings into the discussion. For a chemical reaction at constant temperature and pressure, the change in Gibbs Free Energy is defined with respect to the following two thermodynamic equilibrium states of a specific system:

    State 1: Pure reactants in stoichiometric molar proportions in separate containers, each at temperature T and pressure P

    State 2: Pure products in corresponding stoichiometric molar proportions in separate containers, each at temperature T and pressure P

    Now, how you determine the change in free energy between State 1 and State 2 might be a little complicated, and might involve devices like ideal semi-permeable membranes and a Van't Hoff equilibrium box, but I can be done. And it is done only using reversible steps.

    Chet
     
  5. Feb 23, 2016 #4

    DrDu

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    How else would you prove that a reaction is spontaneous if you don't take the entropy of the surrounding into account?
     
  6. Feb 23, 2016 #5
    Huh?? All reactions are spontaneous if you start out with pure reactants. It's just a matter of where the system equilibrates. So, I have no idea how the entropy of the surroundings can matter.

    The usual very rough rule of thumb for a reaction being considered spontaneous is if the change in Free Energy (as defined in my previous response) at 1 atm and 298 K is negative.

    Chet
     
  7. Feb 23, 2016 #6

    DrDu

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    Ok, so let me invert my question. How do you show that ##\Delta G=0## corresponds to chemical equilibrium? The argument of rajendra123 is that at ##\Delta G=0##, also the total entropy of the system and it's surrounding doesn't change. Hence we are at equilibrium.
     
  8. Feb 23, 2016 #7
    Wow. That's not how I interpreted his question. But certainly, that's a way of showing it. Now your original answer makes more sense to me.
     
  9. Feb 23, 2016 #8
    First of all thank you for your quick reply Friends.

    Do you mean that Gibbs Free Energy is calculated for ISOLATED systems only? If the way I have mentioned in the question is not right then how else could we get mathematical relation.

    Yes, but to maintain temperature constant there must be heat interaction with something (e.g. surrounding) right?
     
  10. Feb 23, 2016 #9
    YES, both of you are right, though I didn't mean that, it's also a way of showing it.
     
  11. Feb 23, 2016 #10
    If Tsystem = Tsurrounding then how can heat transfer take place?
     
  12. Feb 24, 2016 #11

    DrDu

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    If the two temperatures are equal, the heat exchange will be reversible but it will take infinite time. So you have to understand this as a limit of a very small temperature difference and a correspondingly slow heat exchange rate.
     
  13. Feb 24, 2016 #12
    No. I'm saying that the Gibbs Free Energy is a very precise quantity that we define as being equal to H-TS. It definitely doesn't only apply to isolated systems. It can be applied to any closed system.
    I don't know what relation you are referring to, but, as I said, the Gibbs Free Energy of a closed system is a very precise mathematical definition.
    Yes, but so what? I've always felt that it is a very bad idea to include the surroundings in the development because it confuses the issue and because one has much less control over what is happening in the surroundings in terms of reversibility. In my judgment, the focus should always be on the system.
     
  14. Feb 24, 2016 #13
    Okay, so to calculate ΔG we have to have such a device which will transfer heat at such a rate that Tsystem will be constant and heat transfer shall take place between infinitesimal temperature difference.
     
  15. Feb 24, 2016 #14
    I have found this link which says "The whole Gibbs relationship or function is about entropy change."
    http://2ndlaw.oxy.edu/gibbs.html
     
  16. Feb 24, 2016 #15
    Yes. And, if you're considering a chemical reaction (which can occur spontaneously, if permitted), identifying a physical system (at least conceptually) that only allows the reaction to process reversibly presents an interesting challenge.
     
  17. Feb 24, 2016 #16
    I have no idea what this article is talking about, but, whatever it is doesn't work for me.
     
  18. Feb 24, 2016 #17
    So, you mean to say, if a chemical reaction occurring in closed container releasing ΔH amount of heat somehow we convert (ΔH-Tsystem*ΔSsystem) to work i.e. without interacting with surrounding. This work is ΔG?
     
  19. Feb 24, 2016 #18
    No. I'm saying that we don't need to specify the details of how the interaction with the surroundings takes place. In this case, the heat transferred to the surroundings is ΔH, and the system may do work on the surroundings. But the details are not our concern. Our focus is on the system.

    Incidentally, ΔG is not the reversible work done on the surroundings. It is the maximum non-PV work that can be done on the surroundings.
     
  20. Feb 24, 2016 #19
    okay. yes it's the maximum non-PV work for closed system.
     
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