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In Gibbs Free Energy, why multiply -T?

  1. Jan 28, 2015 #1
    When you get the equation for the Gibbs free energy, why do you multiply -T? Can't you multiply just T? I thought about this and came up with an answer of my own, which goes like 'Because its Gibbs 'Free' Energy, when you have that free energy ( \Delta G >0) you have the energy to do 'unfavorable' reactions, but when you don't ( \Delta G <0), you can only do 'favorable' reactions" I'm not so sure about this interpretation so I'd like to get some comments for this. Thank you in advance.
     
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  3. Jan 28, 2015 #2

    DrDu

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    I would say the sign of G is fixed by the demand to be an energy like quantity, as it's name suggests. Specifically, if S=0, G=E.
     
  4. Jan 28, 2015 #3
    Would the interpretation(?) I wrote above be suitable?
     
  5. Jan 28, 2015 #4

    DrDu

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    ##\Delta G## tells you whether an isothermal, isobaric reaction to occur is possible or not. I don't know what you mean with "favourable" vs. "unvavourable" reactions.
     
  6. Jan 28, 2015 #5
    I haven't heard that term used with Gibbs free energy as I really don't know much about this. But I've learned that when ΔG is positive, a favorable (spontaneous) reaction will occur and non-spontaneous reactions when its negative.
     
  7. Jan 28, 2015 #6

    DrDu

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    Only if ##\Delta G## is negative, any reaction will occur at all, hence it makes little sense to say that an "unvavourable reaction occurs".
     
  8. Jan 28, 2015 #7
    Sorry I meant favorable reactions at negative Gibbs Energy. Also are you saying that reactions only occur at negative Gibbs energy?
     
  9. Jan 28, 2015 #8

    DrDu

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    Yes.
     
  10. Jan 28, 2015 #9
  11. Jan 28, 2015 #10

    DrDu

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    Where do they say that unvavourable reactions actually "occur"?

    PS: There are some point in that presentation I don't like. E.g. a negative value of Delta G does not really mean that a reaction occurs spontaneously. E.g. a mixture of hydrogen and oxigen gas won't react spontaneously to form water, although it is strongly exergonic. Another example is the decomposition of an explosive. These reactions need a trigger and one has to differentiate between thermodynamical possibility and the kinetics of the reaction.
     
  12. Jan 28, 2015 #11
    You lost me, so they don't occur? But... why?
     
  13. Jan 28, 2015 #12

    DrDu

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    Maybe you are referring to the following passage: "The change in the free energy of the system that occurs during a reaction measures the balance between the two driving forces that determine whether a reaction is spontaneous."
    This is at best misleading. You will not be able to observe a reaction to occur where Delta G increases in the same way as you won't observe a stone jumping up spontaneously gaining it's kinetic energy from the heat of the flor.
     
  14. Jan 28, 2015 #13
    So what's actually happening? I think my textbook said that reactions occur at negative Gibbs energy also, are they missing something?
     
  15. Jan 28, 2015 #14
    I mean positive gibbs energy
     
  16. Jan 28, 2015 #15

    DrDu

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    Get a better text book. :-)
     
  17. Jan 28, 2015 #16
    Yeah I too thought the textbook wasn't that good, only covered the basics and didn't explain much. So I'd like to read a more scientific explanation, do you have to provide energy when Gibbs energy is positive? Please bear with me sir, I really want to understand this. :s
     
  18. Jan 28, 2015 #17
    If you give energy to something with negative Gibbs energy, then you can make reactions happen right? But not when you don't give energy.

    Like the reaction here; 6CO2+6H2O−→−−−−−−LightEnergyC6H12O6+CO2ΔG∘=+2870kJ/mol
     
  19. Jan 28, 2015 #18

    Ygggdrasil

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    Ok, so we've established that thermodynamically favorable reactions have a ΔG < 0, and that the two components that determine the free energy change are the change in enthalpy of the system (ΔH) and change in entropy of the system (ΔS). In general, will reactions that absorb heat from the surroundings (ΔH > 0) or reactions that release heat to the surroundings (ΔH < 0) be more thermodynamically favorable? Similarly, will reactions that increase the entropy of the system (ΔS > 0) or decrease the entropy of the system (ΔS < 0) be more thermodynamically favorable? What do these answers tell you about the signs of ΔH and ΔS as they relate to ΔG?
     
  20. Jan 28, 2015 #19
    I don't know where you're getting at, but I know the relations between ΔH,ΔS and ΔG. When H is negative and S is positive, the reaction is favorable and when H is positive and S is negative, the reaction is unfavorable.
     
  21. Jan 28, 2015 #20

    Ygggdrasil

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    Right. So, which equation: ΔG = ΔH +TΔS or ΔG = ΔH -TΔS is consistent with these ideas.
     
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