Gibb's free energy confusion!

  • #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]
 
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Answers and Replies

  • #2
Borek
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When you use the equation to describe isolated system you don't have to worry about the rest of the Universe.
 
  • #3
When you use the equation to describe isolated system you don't have to worry about the rest of the Universe.
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}?
 
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  • #4
Mapes
Science Advisor
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Clearly [tex]\Delta[/tex] H = -T[tex]\Delta[/tex]S[tex]^{ }_{sorrounding}[/tex]
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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