What is the relationship between temperature and Gibbs free energy?

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Discussion Overview

The discussion revolves around the relationship between temperature and Gibbs free energy, particularly focusing on the implications of the Gibbs free energy equation, ΔG = ΔH - TΔS. Participants explore the conditions under which reactions are spontaneous and how temperature affects the signs and values of ΔG, ΔH, and ΔS in different contexts.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant expresses confusion about the Gibbs free energy equation, particularly regarding how temperature influences ΔG when considering the surrounding system's entropy change (ΔSsurr).
  • Another participant asserts that when analyzing an isolated system, the broader universe's conditions do not need to be considered.
  • A follow-up question challenges the assumption that ΔH is independent of temperature, questioning why ΔH would not change with temperature if it is defined as a function of T.
  • A later reply emphasizes that the equation ΔH = -TΔSsurr is valid only under specific conditions, such as having a large heat bath at the same temperature, and highlights the importance of distinguishing between spontaneous and non-spontaneous reactions in this context.
  • It is noted that assuming ΔG = 0 implies the reaction is barely spontaneous, which complicates the analysis of how varying temperature affects spontaneity.

Areas of Agreement / Disagreement

Participants express differing views on the implications of temperature on ΔG and the conditions under which the Gibbs free energy equation applies. There is no consensus on the relationship between temperature and the signs of ΔG, ΔH, and ΔS, nor on the assumptions made in the discussion.

Contextual Notes

Participants highlight limitations in their assumptions regarding the system's surroundings and the conditions under which the Gibbs free energy equation is applied. The discussion reflects uncertainty about the implications of temperature changes on spontaneity and the definitions of the variables involved.

Ahmed Abdullah
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\Delta G = \Delta H - T \Delta S

I don't understand this equation satisfactorily.
I have learned so far that endothermic reaction having a positive \Delta S_{system} is spontaneous at higher temperature because T\Delta S out-compete the enthalpy term and make the free energy change negative.
But I don't understand why T\Delta S _{sorrounding} should not increase with temperature as T\Delta S _{system} does.
Clearly \Delta H =-T\Delta S _{sorrounding}
so we can rewrite the gibbs free energy equation
\Delta G= -T\Delta S _{sorrounding} - T\Delta S _{system}
\Delta G = -T ( \Delta S _{sorrounding} + \Delta S _{system})

if \Delta S _{sorrounding} and \Delta S _{system} both are constant then the sign of \Delta G is independent of temperature. If \Delta G is positive increasing temperature just make it more positive and If \Delta G is negative increasing temperature just make it more negative. Temperature cannot change the sign.
I know I must be missing something. Please help!

A^{ }_{A}
 
Last edited:
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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.
 
Borek said:
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}?
 
Last edited:
Ahmed Abdullah said:
Clearly \Delta H = -T\DeltaS^{ }_{sorrounding}

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 \Delta H=T\Delta S^\mathrm{surr}, you're already assuming that the reaction is spontaneous; otherwise energy would not be transferred. And when you write \Delta G= -T\Delta S^\mathrm{surr} - T\Delta S^\mathrm{sys}, you're assuming that it's barely spontaneous, that \Delta G=0. 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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