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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: ASmith on June 09, 2021, 05:46:01 PM

Title: Novice question about Gibbs energy
Post by: ASmith on June 09, 2021, 05:46:01 PM
Why is the heat flow from a system divided by T equated to the entropy change in the surroundings?  Some of this heat could be stored in other systems and so would not appear as entropy.  (To be pedantic it's the magnitudes of H_sys/T and S_surr that are equated as there is a difference in sign.)
Title: Re: Novice question about Gibbs energy
Post by: ASmith on June 10, 2021, 04:39:09 AM
As no one has replied yet perhaps I should explain a bit more.  I recently read a chemistry textbook (which I found more interesting than expected) and it says ΔS_univ = ΔS_sys +ΔS_surr = ΔS_sys -ΔH_sys/T.  This is where the equality is assumed.  After multiplying by T we get TΔS_sys -ΔH = G = Gibbs free energy.

Suppose an internal combustion engine is ticking over and doing no work.  All of the heat is radiated into the surroundings and can be said to contribute to entropy.  But suppose some of the heat from the exhaust is used (via the Seebeck effect for example) to store energy in a battery.  This heat wouldn't contribute to the entropy of the surroundings.  So the equality isn't correct.  Where am I going wrong?
Title: Re: Novice question about Gibbs energy
Post by: Orcio_87 on June 10, 2021, 08:20:34 AM
Quote
Suppose an internal combustion engine is ticking over and doing no work.  All of the heat is radiated into the surroundings and can be said to contribute to entropy.  But suppose some of the heat from the exhaust is used (via the Seebeck effect for example) to store energy in a battery.  This heat wouldn't contribute to the entropy of the surroundings.  So the equality isn't correct.  Where am I going wrong?
ΔS_univ = ΔS_sys +ΔS_surr

Enthropy of the system (engine) will not change, as all the heat will be radiated out.

Some of the heat can be used to store energy in a battery, but never 100 %, so enthropy of the surrounding will still rise.

Title: Re: Novice question about Gibbs energy
Post by: ASmith on June 10, 2021, 03:57:29 PM
Thanks for replying. (By enthropy I assume you mean entropy, not enthalpy.)  The entropy of the surroundings seems to rise by different amounts in the two different situations, but the heat emitted by burning a given amount of fuel will be the same, i.e. ΔH.  The problem is getting the equality to work in different situations.

Suppose when no energy is stored, the surroundings are at 25 deg.C and the heat flow, ΔH, increases the entropy of the surroundings by ΔS.  In the second case suppose 20% of the fuel's energy is stored in a battery using an alternator.

If the engine (i.e. the system) is kept at the same temperature in both cases and the Gibbs energy stays the same then TΔS is constant (from G = ΔH - TΔS.)   But only 80% of the heat now contributes to the entropy of the surroundings.  So to keep TΔS the same, the surroundings need to be 25% hotter.  By my reckoning this means the temperature needs to rise to about 100 deg.C.

Assuming this doesn't happen, and also assuming that the Gibbs equation gives the right answers, it seems obvious that it's the conversion of chemical energy into enthalpy, ΔH, which is important in determining the viability of reactions as far as the surroundings are concerned.  This is the quantity used in the equation.  The change of entropy per se is irrelevant.
Title: Re: Novice question about Gibbs energy
Post by: Orcio_87 on June 11, 2021, 09:41:59 AM
Engine (or a car on the whole) loses enthalpy, as it burns out the fuel (quantity of fuel is source of its potential energy -> enthalpy).

Its entropy will not change, as - even if it generates out the heat it is still the same engine (car) as before.

As of surroundings....

Some of the heat can be stored back in a battery, rest is dispersed in the air.

Heat stored in the battery is calculated as increase of the enthalpy, while dispersed in the air - as increase of the entropy of the surrounding.

After all....

ΔG = ΔG_car + ΔG_surroundings

ΔG = (ΔH - TΔS)_car + (ΔH - TΔS)_surroundings

If some of the heat is stored as an energy in a battery term ΔH_surr will increase at cost of TΔS_surr.

But ΔG for whole process should not change.
Title: Re: Novice question about Gibbs energy
Post by: ASmith on June 11, 2021, 02:21:27 PM
Yes, ΔG doesn't change and neither does ΔH, the heat lost from the system, so TΔS must be the same in both situations.  But ΔS does change.  Moreover this change can't be compensated by a change in T.  So the traditional interpretation doesn't make sense.
Title: Re: Novice question about Gibbs energy
Post by: Orcio_87 on June 12, 2021, 02:52:21 AM
Quote
ΔG doesn't change and neither does ΔH, the heat lost from the system, so TΔS must be the same in both situations.
Are you 100 % sure ? I think that for 100 % conversion of heat (fuel) into energy of battery overall ΔH = 0 and TΔS = 0, while for 0 % conversion - ΔH < 0 and TΔS > 0. Conversion of heat into energy of a battery reduces growth of entropy.

If you 100 % sure I can't help you any further.
Title: Re: Novice question about Gibbs energy
Post by: ASmith on June 12, 2021, 04:49:01 AM
"Conversion of heat into energy of a battery reduces growth of entropy."

Exactly.  The entropy varies, it depends on how ΔH is used, but ΔH is constant.  Therefore ΔH can't be equated with TΔS.  ΔH = TΔS is false.

The answer seems fairly simple.  The chemical bonds in the fuel neither know nor care how the energy released may be used.  It's energy that matters (as well as temperature) not entropy.

Does anyone else wish to defend the use of entropy instead of energy?
Title: Re: Novice question about Gibbs energy
Post by: Borek on June 12, 2021, 06:07:36 AM
Why is the heat flow from a system divided by T equated to the entropy change in the surroundings?

Only for reversible processes. General form is $dS \geq \frac{dQ}T$.
Title: Re: Novice question about Gibbs energy
Post by: ASmith on June 12, 2021, 07:14:23 AM
I thought all reactions were reversible - depending on concentrations etc.  - but I don't see what this has to do with the point I'm making.  How can the physics at the instant of a reaction depend on the fate of the energy that's released?

To try to make this clearer suppose there is negative Gibbs energy for a mole of products from reactions in the Sun.  The energy could all be radiated into space and hence be said to contribute to the entropy of the surroundings.  The energy though could fall on, for example, a solar panel on a spacecraft and some of it stored in a battery.  So not all of this energy would increase the entropy of the surroundings.  Hence ΔS is variable but ΔH and ΔG aren't, so the equality is false.  Or do you think the Gibbs energy varies to take account of what will happen to the radiated energy at some time in the future?
Title: Re: Novice question about Gibbs energy
Post by: ASmith on June 12, 2021, 09:08:44 AM
With reference to the two different cases in the previous solar example, please will someone tell me where they think I'm going wrong and why:

1)  There are two different values of entropy which I'll call ΔS1 and ΔS2.
2)  G is invariant
3)  H is invariant
4)  T is invariant at the time the solar reactions occur.
5)  If ΔG = ΔH - TΔS1 then ΔG can't equal ΔH - TΔS2 as ΔS1 <> ΔS2.
6)  The Gibbs equation is false.
Title: Re: Novice question about Gibbs energy
Post by: Corribus on June 16, 2021, 06:44:41 PM
ASmith, you are using a lot of terms loosely, which makes it hard to follow what you're asking. Also (if I'm reading you right), in the last question you are defining two completely different processes that take place over very different spatial areas and have different physical/chemical steps, but you are defining the system and surroundings as identical in both cases-- then concluding that the Gibbs law is violated when your thought experiment fails.

A process can involve as many steps as you want to define it. if you define one process as A-->B and another as A-->B-->C, why would you expect that the Gibbs energy change for A-->B is the same as A-->B-->C? The system is not the same in both cases, and nor are the surroundings.
Title: Re: Novice question about Gibbs energy
Post by: ASmith on June 17, 2021, 07:28:00 AM
...  you are defining the system and surroundings as identical in both cases-- then concluding that the Gibbs law is violated when your thought experiment fails.

...  why would you expect that the Gibbs energy change for A-->B is the same as A-->B-->C?

In both situations I'm defining the Sun as the system, and its surroundings as the surroundings.

I'm not saying " A-->B is the same as A-->B-->C".  I'm saying A-->B differs from A-->C as regards the entropy of the two different situations, but the Gibbs energy of the solar reactions is the same.
Title: Re: Novice question about Gibbs energy
Post by: Corribus on June 17, 2021, 08:12:17 AM
Why do you assume the Gibbs energy for the two processes are the same? If you are including another chemical or physical change far away from the sun, that can't be ignored when determining the Gibbs energies for the two processes.
Title: Re: Novice question about Gibbs energy
Post by: ASmith on June 17, 2021, 03:36:50 PM
When the reaction occurs the fate of the radiated heat is unknown to the radiating solar matter.  So how can the free energy of the reaction vary to suit an unknown quantity?
Title: Re: Novice question about Gibbs energy
Post by: Corribus 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.
Title: Re: Novice question about Gibbs energy
Post by: ASmith 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.
Title: Re: Novice question about Gibbs energy
Post by: ASmith 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?
Title: Re: Novice question about Gibbs energy
Post by: ASmith 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?
Title: Re: Novice question about Gibbs energy
Post by: ASmith 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.
Title: Re: Novice question about Gibbs energy
Post by: Borek 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.