In Gibbs Free Energy, why multiply -T?

  • #1
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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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  • #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.
 
  • #3
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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.
Would the interpretation(?) I wrote above be suitable?
 
  • #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.
 
  • #5
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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.
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.
 
  • #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".
 
  • #7
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Sorry I meant favorable reactions at negative Gibbs Energy. Also are you saying that reactions only occur at negative Gibbs energy?
 
  • #8
DrDu
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Yes.
 
  • #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.
 
  • #11
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You lost me, so they don't occur? But... why?
 
  • #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.
 
  • #13
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So what's actually happening? I think my textbook said that reactions occur at negative Gibbs energy also, are they missing something?
 
  • #14
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I mean positive gibbs energy
 
  • #15
DrDu
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Get a better text book. :-)
 
  • #16
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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
 
  • #17
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Get a better text book. :)
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
 
  • #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?
 
  • #19
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What do these answers tell you about the signs of ΔH and ΔS as they relate to ΔG?
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.
 
  • #20
Ygggdrasil
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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.
Right. So, which equation: ΔG = ΔH +TΔS or ΔG = ΔH -TΔS is consistent with these ideas.
 
  • #21
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A positive standard Gibbs free energy change does not mean that the reaction will not occur at all. It simply means that the conversion at which the reactants will be at equilibrium with the products will be small.

Chet
 
  • #22
DrDu
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A positive standard Gibbs free energy change does not mean that the reaction will not occur at all. It simply means that the conversion at which the reactants will be at equilibrium with the products will be small.

Chet
I have difficulties to figure out the meaning of your last sentence.
 
  • #23
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Chet's just pointing out that Gibb's free energy has nothing to do with whether a reaction "goes" or does "not go." It limits the degree of "completion" of a reaction, allowing quantitative calculation of quantities/concentrations of products.
 
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  • #24
DrDu
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Chet's just pointing out that Gibb's free energy has nothing to do with whether a reaction "goes" or does "not go." It limits the degree of "completion" of a reaction, allowing quantitative calculation of quantities/concentrations of products.
I don't think so. The Delta G value always refers to specific concentrations of the reactants, which may even be held constant, e.g. in a continuously stirred reactor. Now if Delta G>0, you will never observe an increase of the ammount of moles of the products. So you really have the information that the reaction "does not go" in this direction.
You are right in that from Delta G <0 you can't infer that the reaction really will "go", as, there may be high barriers of activation or more favourable alternative reactions.
I also want to mention that the negative of Delta G is known as chemical affinity and one can show that the reaction speed is a linear function of the affinity, sufficiently close to equilibrium.
 
  • #25
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you will never observe an increase of the ammount of moles of the products.
Correct. You will also never observe zero concentrations of products, no matter how positive the free energy for the reaction is once you have mixed the reactants.
 
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