December 01, 2021, 10:42:36 PM
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Topic: Novice question about Gibbs energy  (Read 3555 times)

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Offline Corribus

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Re: Novice question about Gibbs energy
« Reply #15 on: June 17, 2021, 11:20:25 PM »
It doesn't matter what the fate of the radiated heat is. Just so, if I do a chemical reaction in a flask, the Gibbs energy change is the Gibbs energy change, as are the enthalpy and entropy changes. Whether or not I use the products to perform another chemical reaction tomorrow, or just sit it on a shelf, makes no difference to those values for the reaction I did today.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline ASmith

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Re: Novice question about Gibbs energy
« Reply #16 on: June 18, 2021, 05:06:53 AM »
It doesn't matter what the fate of the radiated heat is.

We seem to agree on something.  But entropy is described as a measure of unusable heat, and the reacting particles don't know how much of the released heat will be used.  They can't calculate the ensuing increase in disorder.  So the free energy of the reaction can't depend on entropy.  It just depends on the energy released (at the time it's released) not on entropy - which is the party line that keeps getting repeating.

Offline ASmith

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Re: Novice question about Gibbs energy
« Reply #17 on: July 06, 2021, 02:53:40 PM »
If any readers still think the textbooks are correct, where are your logical arguments to defend the use of entropy in the Gibbs equation?

Offline ASmith

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Re: Novice question about Gibbs energy
« Reply #18 on: July 19, 2021, 04:07:19 PM »
I thought science textbooks were supposed to promote rationality rather than beliefs from earlier generations.  Do readers think this is no longer the case?

Offline ASmith

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Re: Novice question about Gibbs energy
« Reply #19 on: July 28, 2021, 11:37:20 AM »
If no one can defend the Gibbs equation it may be easier to attack an alternative.

Apart from the illogicality of being based on entropy the Gibbs equation has a number of drawbacks.  The feasibility of a reaction depends on whether heat will be released both in the system and its surroundings, but the two heat contributions have opposite signs.  One of these also involves multiplying the entropy change of the system by the temperature of the surroundings.  So the equation is neither intuitive nor easily remembered in relation to its simplicity.

Moreover the free energy needs to be negative for the reaction to proceed.  If there is no free energy, nothing happens.  So there must be less than no free energy before there is enough free energy.  Does this strike students as sensible?  It seems much simpler to require a release of positive energy, so the “spontaneity” test can be:
ΔE_sys/T_sys  + ΔE_surr/T_surr > 0.  Everything now has a positive sign and the two terms are consistent, i.e. they don’t mix heat and temperature from different sides of the system/surroundings boundary.

Although I’ve never actually used the equation it seems there may be situations (involving a heater for example) when it would be helpful to use more than two different temperatures.  So the test might be generalised to just Σ ΔEi/Ti > 0.  Even I can remember this.

Offline Borek

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Re: Novice question about Gibbs energy
« Reply #20 on: July 28, 2021, 06:51:20 PM »
If no one can defend the Gibbs equation

Well, it doesn't need any defense. And not because we say "it is right" pretending to be some kind of authority on the subject. It is one of these things that are tested again and again, day by day, by thousands of of physicists, chemists, biologists, as they use it to predict behavior of thousands of systems. It works perfectly every time, it is combat proven for over 150 years now.

I am always fascinated by the level of arrogance required to say "I don't get it so everyone else is wrong".

I am locking the topic, it became a useless soapbox.
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