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Topic: The Clausius Inequality, ΔG, and Thermal Equilibrium  (Read 2957 times)

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Offline orthoformate

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The Clausius Inequality, ΔG, and Thermal Equilibrium
« on: July 07, 2016, 02:57:23 PM »
Hello,

I have learned that Gibbs Free Energy can be derived from The Clausius Inequality.

dSsurr+dSsystem≥0 for processes that are spontaneous or in equilibrium.

dSsurr=-q/T

dSsystem-q/T≥0

TdSsystem-q≥0

assuming constant pressure q=H and assuming H does not vary with temperature

0≥ H-TdSsys or ΔG=ΔH-TΔS

Based on the Clausius inequality derivation the T in ΔG=ΔH-TΔS is the Temperature of the surroundings.

Do the temperature of the surroundings and the temperature of the system have to be in equilibrium for this definition to hold?

For example. A reaction with a ΔH: -500Kj/mol occurs in a reaction flask at 298K. The heat generated in transferred to the surroundings keeping the temperature constant. The temperature of the surroundings is 315K (it's really hot in the lab) and does not change with the addition of this heat.

ΔSsys:-1.68 Kj/K*mol
ΔSsurr:1.59 Kj/K*mol

ΔSuniverse:-0.09 Kj/K*mol

ΔG=-500Kj-315K(-500kJ/298K)= 29.2Kj

Am I doing this correctly?

Offline mjc123

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Re: The Clausius Inequality, ΔG, and Thermal Equilibrium
« Reply #1 on: July 08, 2016, 11:29:22 AM »
Congratulations, you have just discovered why heat does not flow from a cold body to a hot one! Did you really think heat generated in the reaction is transferred to the (warmer) surroundings? How?
Why doesn't the flask warm up to lab temp? Is it in a constant temperature bath at 298K? Then that is the "temperature of surroundings".
Your calculation of entropy is wrong. There will be an entropy change of reaction; the difference between the entropy of products and reactants. In the isothermal case, this is equal to ΔSsys. ΔH is lost to/gained from the surroundings, and makes no contribution to Ssys. It comes from the chemical potential energy of the reagents, and goes to the surroundings; it doesn't contribute to the thermal energy of the system.
T in ΔG = ΔH - TΔS is the temperature of the system. The whole point, in a sense, is to express "ΔSuniverse > 0" in terms of quantities of the system, so it is equivalent to "ΔGsystem < 0". But that equivalency only holds for isothermal processes in thermal equilibrium with the surroundings. In other cases ΔG is not so useful. See this thread (which I notice you contributed to) for an example where ΔG > 0 for a spontaneous non-isothermal process: http://www.chemicalforums.com/index.php?topic=81854.msg297962#msg297962

Offline orthoformate

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Re: The Clausius Inequality, ΔG, and Thermal Equilibrium
« Reply #2 on: July 08, 2016, 10:55:46 PM »
Hi mjc123,

Thanks for the response, and pointing me back to that old thread. It is reassuring to know that thermal equilibrium is a requirement for the validity of a Gibbs free energy calculation. That makes sense to me.

In retrospect, the question I posed is ridiculous. The reaction vessel would of course just heat up. I am just trying to develop a deeper understanding of Entropy.

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