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Topic: G = H - TS. Effect of S ?  (Read 4671 times)

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Offline GeLe5000

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G = H - TS. Effect of S ?
« on: April 23, 2016, 03:29:46 PM »
I know that chemists are only interested in differences in free Enthalpies in order to know if a reaction is exergonic or endergonic.

Nevertheless, G = H - TS is the molar Free Enthalpy of a molecule.

Writing G° = H° - TS°, it is possible to calculate the molar standard Free Enthalpy of a pure substance.

I understand that a high Enthalpy H° will lead to a high Free Enthalpy G° since H° is the sum of the bond energies in the molecule. The molecule has a higher energetic content and is more stable since more energy is required to break it into its atoms.

But I don't understand why a high standard Entropy S° will decrease the energy content and (perhaps) the stability of the molecule.

Can someone give me a lead ? Thank you.

Offline mjc123

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Re: G = H - TS. Effect of S ?
« Reply #1 on: April 24, 2016, 04:42:44 PM »
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I understand that a high Enthalpy H° will lead to a high Free Enthalpy G° since H° is the sum of the bond energies in the molecule. The molecule has a higher energetic content and is more stable since more energy is required to break it into its atoms.
This is nonsense, but a common misconception. Lower energy means higher stability and vice versa. H° is not the sum of the bond energies (though ΔH may often be approximated by the difference in the sum of bond energies between reactants & products). If "more energy is required to break it into its atoms" then the molecule has a lower energy content, not higher.
The universe likes to maximise entropy, so a high S° means lower free energy (not "lower energy content") and higher stability.

Offline GeLe5000

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Re: G = H - TS. Effect of S ?
« Reply #2 on: April 25, 2016, 09:56:37 AM »
Thank you.

Thus I'm allowed to say that a compound has a low molar standard Enthalpy when its bond enthalpies are high.

Isn't it possible to speak of the effect of S° on G° in terms of vibrational, rotational and electronic motions ?

Offline mjc123

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Re: G = H - TS. Effect of S ?
« Reply #3 on: April 26, 2016, 04:44:10 PM »
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Thus I'm allowed to say that a compound has a low molar standard Enthalpy when its bond enthalpies are high.
I don't think this is a very helpful way of looking at it. There are other things that contribute to the enthalpy apart from bond energies. Don't get fixated on bond energy.
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Isn't it possible to speak of the effect of S° on G° in terms of vibrational, rotational and electronic motions ?
These contribute to both H and S through their contribution to Cp. They are not peculiar to entropy.

Offline GeLe5000

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Re: G = H - TS. Effect of S ?
« Reply #4 on: April 27, 2016, 08:23:41 AM »
I'm sorry my question is now general chemistry.
But it's a necessary step, otherwise I'll never understand.

The bond Enthalpy of H2 is + 436 kJ.

In a book I read that that 436 kJ = 2 x H°m(H) – H°m(H2)

Thus the standard Enthalpy of an atom must be zero and H°m(H2) must be negative = - 436 kJ.

My conclusion : a standard Enthalpy is always negative. Is it right ?

In G° = H° - TS°, H° being always negative and S° always positive, is it possible that G° be sometimes positive, sometimes negative ?
If it's always negative, then we have always H° > TS° (absolute values).

Offline GeLe5000

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Re: G = H - TS. Effect of S ?
« Reply #5 on: April 27, 2016, 09:25:17 AM »
I have another important question.

In a H atom, when the electron comes closer to the proton its energy decreases. But it means that it becomes more and more negative. In the last orbital, it's - 2,18 E-18 J. There's no other orbital closer and the electron can't reach the positive charge. And if it could, its energy would be infinitely negative.

I compare with the energy content of any substance. When we cool a body it's difficult to reach the zero temperature (0 K); Actually, it should be impossible, if we consider that 0 K corresponds to an infinitely negative value of energy. It's impossible to extract an infinite amount of energy from a body.
And on the graph showing the Boltzmann distribution, ei,..., e3, e2, e1, e0 are energy levels more and more negative down to e0 which is infinitely negative and arbitrarily set to 0.

Is it right ?

Offline GeLe5000

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Re: G = H - TS. Effect of S ?
« Reply #6 on: April 27, 2016, 11:13:26 AM »
Correction to my post about the negative values of H° and G°.

Since H° is always negative and S° always positive, G° is always negative.



Offline mjc123

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Re: G = H - TS. Effect of S ?
« Reply #7 on: April 28, 2016, 04:54:26 PM »
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My conclusion : a standard Enthalpy is always negative. Is it right ?
No. See this recent thread http://www.chemicalforums.com/index.php?topic=85501.0 on why there is not an absolute reference point for energy. You say that "436 kJ = 2 x H°m(H) – H°m(H2)" But this could be the difference between two very large positive numbers. It all depends how you define your zero. In fact we generally define the standard enthalpy of formation (note, not absolute enthalpy) of an element as zero, so 436 kJ/mol = 2 x ΔHf(H).
We are almost always concerned with enthalpy differences, not absolute enthalpies (which as I have said are not defined). You may have noticed that thermodynamic data tables give values for ΔH°f and ΔG°f of compounds, rather than H° and G°. (Entropy is a bit different; they do give S°.)

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if we consider that 0 K corresponds to an infinitely negative value of energy.
No it doesn't. It corresponds to zero energy - or rather, zero energy above H°(0K), however we define this.
H(T) = H(0) + int Cp(T)dT (plus any phase transition enthalpies) If H has a finite value at finite T, it can't be -∞ at 0K.
Likewise E0 is not infinitely negative. (Electron on proton is not E0 for the H atom; it is not an allowed energy state at all.)

Offline GeLe5000

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Re: G = H - TS. Effect of S ?
« Reply #8 on: April 29, 2016, 08:10:04 AM »
Then do you agree with this (sorry if it's only Philosophy) ?

It is not possible to know the Internal Energy of a molecule because Energy doesn't really exist. It's just a mathematical function.

What is a mathematical function ? It is Information.

Thus, Energy = Information.

We can't measure Energy, just like there's no device to measure "pain".

But if a Biochemist is right when he says that (biological) Information is a kind of Energy, then, according to the definition of Thermodynamics as the Science that studies energy interconversions, Information should be measured in Joules.

Offline Corribus

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Re: G = H - TS. Effect of S ?
« Reply #9 on: April 29, 2016, 09:37:40 AM »
Then do you agree with this (sorry if it's only Philosophy) ?

It is not possible to know the Internal Energy of a molecule because Energy doesn't really exist. It's just a mathematical function.

What is a mathematical function ? It is Information.

Thus, Energy = Information.

We can't measure Energy, just like there's no device to measure "pain".
Maybe mjc123 feels differently, but I don't agree with any of that. Quanta of energy can be measured quite precisely. Energy is not a matter of perception, so the analogy to pain is no good. Most of the time the utility of energy relates to its practical meaning - a measure of the capability of a system to change, or perform work. In this sense the concept of energy is most useful as a quantity relative to something else either in time or space. Therefore we most typically speak of - and measure - energy change or energy difference. Even in the case of, say, spectroscopy, we are measuring absolute energy quantities but they represent the difference in energy between discreet molecular or atomic potential energy states.

We might think of energy as similar in a way to the concept of altitude. Altitude certainly exists, and can be measured with extreme precision, but almost always relative to a point of reference. In some cases this can be specified uniquely but most of the time the point of reference is commonly understood: sea level.
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 mjc123

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Re: G = H - TS. Effect of S ?
« Reply #10 on: April 29, 2016, 02:20:48 PM »
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Maybe mjc123 feels differently
I agree with Corribus.

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