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### Topic: Why is TΔS part of the Gibbs Free Energy formula?  (Read 6342 times)

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#### Technicalhuman

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##### Why is TΔS part of the Gibbs Free Energy formula?
« on: July 21, 2013, 01:14:17 AM »
Hi ChemicalForums, hope I can get some help with the understanding of some concepts here.

When calculating ΔS, isn't temperature already accounted for? For example, ΔS is positive when water is heated from 25 degrees to 35 degrees. So in the ΔS formula, Sfinal - Sinitial aren't the temperatures already taken into account?

And even if it has to be taken into account, how can T be a constant? When heating up the liquid, the temperature changes too. And in other chemical reactions, temperature usually isn't constant. So what's the basis that our temperature can be taken to be constant?

#### curiouscat

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##### Re: Why is TΔS part of the Gibbs Free Energy formula?
« Reply #1 on: July 21, 2013, 03:16:44 AM »

And even if it has to be taken into account, how can T be a constant? When heating up the liquid, the temperature changes too. And in other chemical reactions, temperature usually isn't constant. So what's the basis that our temperature can be taken to be constant?

If T isn't constant you use differentials and use integration.

#### Technicalhuman

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##### Re: Why is TΔS part of the Gibbs Free Energy formula?
« Reply #2 on: July 21, 2013, 04:16:11 AM »

And even if it has to be taken into account, how can T be a constant? When heating up the liquid, the temperature changes too. And in other chemical reactions, temperature usually isn't constant. So what's the basis that our temperature can be taken to be constant?

If T isn't constant you use differentials and use integration.

Hello Curiouscat, thank you for the reply.

But what about the final and initial state thing? For example in the heating of water, the temperature increased. So would the Sfinal be for water at 35 degrees?

Lastly, why is it that we can allow T to be constant in so many cases? Because from many examples, I feel that the temperature change would be prominent.

Thank you for the help on the post too.

#### curiouscat

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##### Re: Why is TΔS part of the Gibbs Free Energy formula?
« Reply #3 on: July 21, 2013, 04:20:34 AM »

But what about the final and initial state thing? For example in the heating of water, the temperature increased. So would the Sfinal be for water at 35 degrees?

Yes. Final state S is evaluated at final T, whatever that may be.

Quote
Lastly, why is it that we can allow T to be constant in so many cases? Because from many examples, I feel that the temperature change would be prominent.

What do you mean? T either is constant or it isn't. It isn't a question of what we allow. It is a constraint imposed by the process or the problem.

i.e. Isothermal = T is constant

If you are supplying heat to water in a pot open to atmosphere T is definitely not constant.

#### Technicalhuman

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##### Re: Why is TΔS part of the Gibbs Free Energy formula?
« Reply #4 on: July 21, 2013, 06:34:10 AM »

But what about the final and initial state thing? For example in the heating of water, the temperature increased. So would the Sfinal be for water at 35 degrees?

Yes. Final state S is evaluated at final T, whatever that may be.

Quote
Lastly, why is it that we can allow T to be constant in so many cases? Because from many examples, I feel that the temperature change would be prominent.

What do you mean? T either is constant or it isn't. It isn't a question of what we allow. It is a constraint imposed by the process or the problem.

i.e. Isothermal = T is constant

If you are supplying heat to water in a pot open to atmosphere T is definitely not constant.

Hi thanks again for the reply.

I thought in many chemical reactions naturally heat is going to be absorbed or released?

For instance in this example, I'm sure T cannot be constant. So I guess we would have to use differentiation? So in what case would the temperature remain constant?

Thank you for the help Curiouscat I really appreciate it.

#### curiouscat

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##### Re: Why is TΔS part of the Gibbs Free Energy formula?
« Reply #5 on: July 21, 2013, 07:16:47 AM »

I thought in many chemical reactions naturally heat is going to be absorbed or released?

True.

Quote
For instance in this example, I'm sure T cannot be constant. So I guess we would have to use differentiation? So in what case would the temperature remain constant?

If you hold it in a constant temp. water bath, say. Or if there's a lot of solvent. i.e. high heat capacity.

#### Technicalhuman

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##### Re: Why is TΔS part of the Gibbs Free Energy formula?
« Reply #6 on: July 21, 2013, 10:35:11 AM »

I thought in many chemical reactions naturally heat is going to be absorbed or released?

True.

Quote
For instance in this example, I'm sure T cannot be constant. So I guess we would have to use differentiation? So in what case would the temperature remain constant?

If you hold it in a constant temp. water bath, say. Or if there's a lot of solvent. i.e. high heat capacity.

Hi thanks you for the great response.

So when we place it in a water bath where temperature remains relatively constant, now both the Sfinal and Sinitial would be for the same temperature?

Regarding the same topic of Gibbs Free Energy, why is the TΔS subtracted from ΔH? My chemistry course is pretty basic now and we use the magnitude of ΔG to determine if a reaction is spontaneous or not. However, the notes briefly mentioned that it is the energy capable of doing non-expansion work.

What does that mean? I know that the ΔH at constant P is the heat given out or absorbed. But I'm not too sure why misusing TΔS with ΔH would give us the energy to do non expansion work. Because that would mean TΔS does expansive work? Why would this be so?

Thanks so much for the great help provided here.

#### Yggdrasil

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##### Re: Why is TΔS part of the Gibbs Free Energy formula?
« Reply #7 on: July 21, 2013, 01:10:20 PM »
Regarding the same topic of Gibbs Free Energy, why is the TΔS subtracted from ΔH? My chemistry course is pretty basic now and we use the magnitude of ΔG to determine if a reaction is spontaneous or not.

The principle that ΔG must always decrease for a spontaneous reaction is a simple restatement of the second law of thermodynamics – that the entropy of the universe is always increasing.

Now, there are two ways to increase the entropy of the universe.  The most obvious way is to increase the entropy of the system, thus as ΔS increases, ΔG becomes more negative.

The other way of increasing the entropy of the universe is to increase the entropy of the surroundings.  This process occurs when the system transfers heat to the surroundings.  In essence, you can think of this as the energy becoming more disordered because it starts as potential energy confined in the system and gets released as heat energy into the larger surroundings.   The entropy of the surroundings is represented by the ΔH term in the expression for ΔG, and as the reaction releases more heat (ΔH is more negative), ΔG becomes more negative as well.

Of course, there are many examples when these two processes come into conflict.  Many reactions increase the entropy of the surroundings while decreasing the entropy of the system and vice versa.  Under these conditions, what side of the equation wins?  What controls the balance between these two factors is the temperature.  At lower temperatures, it becomes more favorable to transfer heat to the surroundings (specifically, this comes from the equation that ΔSsurroundings = q/T where q is the amount heat being transferred to the surroundings).

For a full derivation of the expression for ΔG from the second law see http://www.chemicalforums.com/index.php?topic=64531.msg232158#msg232158

Now how is the free energy of the reaction used to do non-expansion work?  Often, reactions that release free energy can be coupled to endergonic reactions (reactions with a positive ΔG) in order to make those reactions occur spontaneously.  This process occurs ubiquitously in biology, for example, in the coupling of ATP hydrolysis (a reaction with a very negative ΔG) to make unfavorable processes occur spontaneously (for example, the transport of ions against their concentration gradient, directional movement of molecular motors, synthesis of complex molecules from simpler molecules, etc.).

#### Technicalhuman

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##### Re: Why is TΔS part of the Gibbs Free Energy formula?
« Reply #8 on: July 22, 2013, 07:32:38 AM »
Regarding the same topic of Gibbs Free Energy, why is the TΔS subtracted from ΔH? My chemistry course is pretty basic now and we use the magnitude of ΔG to determine if a reaction is spontaneous or not.

The principle that ΔG must always decrease for a spontaneous reaction is a simple restatement of the second law of thermodynamics – that the entropy of the universe is always increasing.

Now, there are two ways to increase the entropy of the universe.  The most obvious way is to increase the entropy of the system, thus as ΔS increases, ΔG becomes more negative.

The other way of increasing the entropy of the universe is to increase the entropy of the surroundings.  This process occurs when the system transfers heat to the surroundings.  In essence, you can think of this as the energy becoming more disordered because it starts as potential energy confined in the system and gets released as heat energy into the larger surroundings.   The entropy of the surroundings is represented by the ΔH term in the expression for ΔG, and as the reaction releases more heat (ΔH is more negative), ΔG becomes more negative as well.

Of course, there are many examples when these two processes come into conflict.  Many reactions increase the entropy of the surroundings while decreasing the entropy of the system and vice versa.  Under these conditions, what side of the equation wins?  What controls the balance between these two factors is the temperature.  At lower temperatures, it becomes more favorable to transfer heat to the surroundings (specifically, this comes from the equation that ΔSsurroundings = q/T where q is the amount heat being transferred to the surroundings).

For a full derivation of the expression for ΔG from the second law see http://www.chemicalforums.com/index.php?topic=64531.msg232158#msg232158

Now how is the free energy of the reaction used to do non-expansion work?  Often, reactions that release free energy can be coupled to endergonic reactions (reactions with a positive ΔG) in order to make those reactions occur spontaneously.  This process occurs ubiquitously in biology, for example, in the coupling of ATP hydrolysis (a reaction with a very negative ΔG) to make unfavorable processes occur spontaneously (for example, the transport of ions against their concentration gradient, directional movement of molecular motors, synthesis of complex molecules from simpler molecules, etc.).

Hello Yggdrasil, thanks for the great post it gave me new insight on the topic.

After reading the explanation in detail, I realized that its entropy that is main concept here where the ΔH is just the entropy change of the surroundings. However, if we consider the entropy change of the surroundings, shouldn't temperature be including in the expression like in the -TΔS component of the equation?

Also, regarding the non-expansion work, I still don't quite understand why if we subtract the ΔSsurrounding with TΔSsystem we would get the work done for non-expansive work. Why would the TΔSsystem part be the work done for expansion work and why would the ΔSsystem represent the total energy capable of doing work?

Thank you so much for the great input.

#### Yggdrasil

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##### Re: Why is TΔS part of the Gibbs Free Energy formula?
« Reply #9 on: July 23, 2013, 07:12:55 PM »
Hello Yggdrasil, thanks for the great post it gave me new insight on the topic.

After reading the explanation in detail, I realized that its entropy that is main concept here where the ΔH is just the entropy change of the surroundings. However, if we consider the entropy change of the surroundings, shouldn't temperature be including in the expression like in the -TΔS component of the equation?

You just as easily define a "free entropy" as ΔH/T -ΔS, which shows the real relationship between the temperature and the change in entropy of the surroundings.  However, it's more convenient to multiply the expression by the temperature to get ΔH - TΔS.

Quote
Also, regarding the non-expansion work, I still don't quite understand why if we subtract the ΔSsurrounding with TΔSsystem we would get the work done for non-expansive work. Why would the TΔSsystem part be the work done for expansion work and why would the ΔSsystem represent the total energy capable of doing work?

When a system releases (or absorbs) energy, we can write the change in internal energy (U) of the system as:

ΔU = q + wpv + wnon

where wpv represents the pressure-volume (expansion) work done on the system and wnon represents the non-expansion work done on the system.

If we solve for the non-expansion work done by the system, we get:

wnon = ΔU - wpv - q

For a process occurring reversibly at constant pressure,  ΔU - wpv = ΔH.  Similarly, since ΔS = q/T for a reversible process, we can write q as TΔS.

Substituting these expressions into our equation for the non-expansion work we get:

wnon = ΔH - TΔS

Note that ΔH = ΔU - wpv and q = -TΔS only for reversible processes, so the above equation defines only the upper limit for wnon.
« Last Edit: July 23, 2013, 09:53:26 PM by Yggdrasil »