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### Topic: Why always ∆G, never absolute G?  (Read 6908 times)

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#### poobear

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##### Why always ∆G, never absolute G?
« on: April 17, 2011, 10:31:40 AM »
Hi,

Why does people always calculate the difference in Gibbs free energy, ∆G, instead of absolute values for reactants and products? In my education I always hear that absolute measurements are better than relative measurements, so the fact that everyone calculates ∆G makes me believe that the absolute G is too hard to obtain. So if this is the case, why is it so hard to obtain?

Thanks

#### rabolisk

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##### Re: Why always ∆G, never absolute G?
« Reply #1 on: April 17, 2011, 01:20:08 PM »
There is no such thing as absolute G. The :delta: G of formation of elements in their standard state is 0 by definition.

#### poobear

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##### Re: Why always ∆G, never absolute G?
« Reply #2 on: April 17, 2011, 04:37:38 PM »
There is no such thing as absolute G. The :delta: G of formation of elements in their standard state is 0 by definition.
Hmm I don't really understand. But if you take a molecule, there must be a enthalpy and entropy associated with it no? It's bond energy and number of states.

#### Jorriss

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##### Re: Why always ∆G, never absolute G?
« Reply #3 on: April 17, 2011, 05:07:13 PM »
There is no such thing as absolute G. The :delta: G of formation of elements in their standard state is 0 by definition.
Hmm I don't really understand. But if you take a molecule, there must be a enthalpy and entropy associated with it no? It's bond energy and number of states.
Enthalpy is not just bond energy. It's the energy needed to create the system plus expand it against an environment to a given volume. That energy includes potential energies, kinetic, rest energy, and every other energy associated with the system. Energy considerations are not just absolute, you need something to measure relative to.

Entropy I'm not certain, I don't understand entropy well enough to comment.

#### Mm04302011

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##### Re: Why always ∆G, never absolute G?
« Reply #4 on: June 11, 2011, 01:25:08 PM »
The absolute entropy of a system can be defined through the use of statistical mechanics (see an undergraduate physical chemistry text--McQuarrie, if you are well-versed in calculus).  Using Boltzmann statistics, one introduces, W, the weight of configuration.  This is defined as the most probable configuration of a system.  I find this to be utterly abstract for all but the most simplistic systems.  It is easier to look at things instead using, q, the partition function (the wikipedia page on this topic is fairly informative: http://en.wikipedia.org/wiki/Partition_function_(statistical_mechanics).  Using this formalism, one can derive the relationship of entropy with temperature:
$request=http://www.forkosh.dreamhost.com/mimetex.cgi?{+S(T)=+\frac{U(T)-U(0)}{T}+Nkln{q}+}&hash=4dd1cefa059d13254dd5a65f46596485$
This is the general form with which entropy can be computed.  As far as the absolute enthalpy and free energy, if one introduced the Maxwell relations, we can attain these using classical thermodynamic relations (taking internal energy and entropy to be known quantities which are obtained from earlier derivations).  Chiefly, we must find an expression for the pressure of a system as a function of the partition function.  Using a relationship gained from deriving the Maxwell relations we know that:
$request=http://www.forkosh.dreamhost.com/mimetex.cgi?{+p=kT\bigg(\frac{\partial{ln{Q}}}{\partial{V}}\bigg)_T+}&hash=147c7fc89561b4d24a76530dc7b632fd$, where Q is the more general molecular partion function. From classical thermodynamics we know that H=U+PV, and we know all of the quantities in question and can obtain the absolute enthalpy as a function of the partition function:
$request=http://www.forkosh.dreamhost.com/mimetex.cgi?{H=-\bigg(\frac{\partial{ln{Q}}}{\partial{\beta}}\bigg)+kVT\bigg(\frac{\partial{ln{Q}}}{\partial{V}}\bigg)_T+}&hash=70da6b6c85ba5d5e5ba4e5057d66d7fc$
Clearly, at this point we can put together H and S and form the aformentioned G.  Using the farmiliar relation: G=H-TS.  doing so, we get:
$request=http://www.forkosh.dreamhost.com/mimetex.cgi?{+G=+-kTln{Q}+kTV\bigg(\frac{\partial{ln{Q}}}{\partial{V}}\bigg)_T+}&hash=c92e1aa067c9862d22047e82ffa1042e$
Where I've used the more tractable expressions for A, the Helmholtz energy, and p, the pressure, and also the relationship:
G=H-TS=A+pV.
The absolute Gibbs Free energy can be obtained exactly (in theory, although the partition functions required for analysis are rarely exact except for the most ideal systems eg. monatomic ideal gas, ideal diatomic gas...).  Approximate partition functions are used in the absence of exact ones and most electronic structure packages have the ability to compute the absolute entropy, enthalpy and free energy after calculating the vibrational frequencies.  From there theoretical entropies, enthalpies and free energies of reaction or activation (using transition state structures) can be postulated.

Hope this helps!

sdm

« Last Edit: June 11, 2011, 01:39:47 PM by Mm04302011 »

#### poobear

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##### Re: Why always ∆G, never absolute G?
« Reply #5 on: June 11, 2011, 03:48:01 PM »
Mm04302011: Wow thanks!

#### Yggdrasil

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##### Re: Why always ∆G, never absolute G?
« Reply #6 on: June 12, 2011, 02:46:34 PM »
Mm04302011, in calculating the partition function, you need to know the energies of the various states your system can occupy.  Here, you implicitly need to choose some arbitrary reference energy, usually taking the energy of a completely isolated atoms to be your E = 0 reference.  So, even these "absolute" enthalpies and free energies calculated from statistical mechanics are made relative to an arbitrarily chosen reference energy.

You are correct, however, that you can calculate an absolute entropy for your system.

#### Mm04302011

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##### Re: Why always ∆G, never absolute G?
« Reply #7 on: June 12, 2011, 05:04:46 PM »
Yggdrasil,
Thank you for the response.  Had you not posted this, I would have just taken my answer as "good enough."  You bring up a valid point in suggesting that we always need a reference point.  Choosing E=0 with all atoms at infinite separation seems like a pragmatic choice.  However, partition functions and entropies, enthalpies, and free energies are available for these atoms as well.  Clearly, this nullifies the previous reference point (I know you just picked this point arbitrarily, not absolutely).  Now, we must reference our system to a new E=0..  My response is almost entirely semantics--I'm not trying to offend you.  I argue that the true reference point for any system has to be the same system at absolute zero, because the statistical thermodynamic equations for enthalpy and free energy to not invoke any reference point (aside from absolute zero).

#### fledarmus

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##### Re: Why always ∆G, never absolute G?
« Reply #8 on: June 16, 2011, 03:07:25 PM »
Actually, for matter to exist at all there must be a certain amount of energy associated with it. That amount of energy can also be calculated - E=mc2

#### nj_bartel

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##### Re: Why always ∆G, never absolute G?
« Reply #9 on: June 16, 2011, 05:03:59 PM »
If my understanding is right

E=mc^2 is a good measure of absolute G, but there is really no utility for absolute G outside of nuclear (E=mc^2) purposes.  This is because it disregards molecular differences in stability between compounds, due to those differences being insignificant relative to the nuclear energy released just based upon mass.  Chemical reactions due not consider nuclear energy/energy due to mass though - there's no utility there.  You want to look at the difference in stability between two molecules, hence the use of dG.

I suppose you could potentially true G value through the modification of E=mc^2 to include some kind of heat of formation/entropy equation, but it would really be meaningless.

But someone please confirm/deny, I'm just waxing philosophical.

#### fledarmus

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##### Re: Why always ∆G, never absolute G?
« Reply #10 on: June 17, 2011, 08:58:05 AM »
If my understanding is right

E=mc^2 is a good measure of absolute G, but there is really no utility for absolute G outside of nuclear (E=mc^2) purposes.  This is because it disregards molecular differences in stability between compounds, due to those differences being insignificant relative to the nuclear energy released just based upon mass.  Chemical reactions due not consider nuclear energy/energy due to mass though - there's no utility there.  You want to look at the difference in stability between two molecules, hence the use of dG.

I suppose you could potentially true G value through the modification of E=mc^2 to include some kind of heat of formation/entropy equation, but it would really be meaningless.

But someone please confirm/deny, I'm just waxing philosophical.

Exactly. Chemists have never really had the luxury of a ground state being the actual ground, as physicists have (potential energy = U = mgh, where h = height above the ground), so chemical potential is always measured in terms of changes in energy rather than specific energies. (Of course, the physicists never considered that you might be changing the potential energy by digging a hole underneath the object and therefore lowering the "ground state" :-)  )

#### slavinzing56

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##### Re: Why always ∆G, never absolute G?
« Reply #11 on: July 04, 2011, 04:13:45 AM »
If my understanding is right

E=mc^2 is a good measure of absolute G, but there is really no utility for absolute G outside of nuclear (E=mc^2) purposes.  This is because it disregards molecular differences in stability between compounds, due to those differences being insignificant relative to the nuclear energy released just based upon mass.  Chemical reactions due not consider nuclear energy/energy due to mass though - there's no utility there.  You want to look at the difference in stability between two molecules, hence the use of dG.

I suppose you could potentially true G value through the modification of E=mc^2 to include some kind of heat of formation/entropy equation, but it would really be meaningless.

But someone please confirm/deny, I'm just waxing philosophical.

Exactly. Chemists have never really had the luxury of a ground state being the actual ground, as physicists have (potential energy = U = mgh, where h = height above the ground), so chemical potential is always measured in terms of changes in energy rather than specific energies. (Of course, the physicists never considered that you might be changing the potential energy by digging a hole underneath the object and therefore lowering the "ground state" :-)  )

It right,
thank you very much