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Topic: Novice question about Gibbs energy  (Read 3554 times)

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Offline ASmith

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Novice question about Gibbs energy
« 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.)

Offline ASmith

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Re: Novice question about Gibbs energy
« Reply #1 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?

Offline Orcio_Dojek

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Re: Novice question about Gibbs energy
« Reply #2 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.


Offline ASmith

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Re: Novice question about Gibbs energy
« Reply #3 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.

Offline Orcio_Dojek

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Re: Novice question about Gibbs energy
« Reply #4 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.

Offline ASmith

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Re: Novice question about Gibbs energy
« Reply #5 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.

Offline Orcio_Dojek

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Re: Novice question about Gibbs energy
« Reply #6 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.
« Last Edit: June 12, 2021, 03:26:22 AM by Orcio_Dojek »

Offline ASmith

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Re: Novice question about Gibbs energy
« Reply #7 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?

Offline Borek

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Re: Novice question about Gibbs energy
« Reply #8 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 [itex]dS \geq \frac{dQ}T[/itex].
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Offline ASmith

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Re: Novice question about Gibbs energy
« Reply #9 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?

Offline ASmith

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Re: Novice question about Gibbs energy
« Reply #10 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.

Offline Corribus

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Re: Novice question about Gibbs energy
« Reply #11 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.
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 #12 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.

Offline Corribus

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Re: Novice question about Gibbs energy
« Reply #13 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.
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 #14 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?

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