Entropy and thermodynamic stability

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Main Question or Discussion Point

Hi Everyone, I have been researching for days, and am still very confused, so thought I might turn here for assistance.

I know the minimum free energy represents the state of maximum stability for a system. Free energy is decreased when enthalpy decreases and entropy increases. I understand why the stability would increase as enthalpy decreases, due to the decline in potential energy.

What I would really like to know is: how does entropy affect the stability (thermodynamic stability)?

Thanks very much, I would really appreciate any assistance regarding this issue.
 

Answers and Replies

  • #2
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First an equation for a system in some state, A, moving towards another state B of lower free energy.

[tex]\Delta[/tex]G = [tex]\Delta[/tex]H - T[tex]\Delta[/tex]S

Since you have mentioned all the variables in this equation I take it you understand which is which.

There are four possibilities. to achieve a move to lower free energy ([tex]\Delta[/tex]G -ve)

1) [tex]\Delta[/tex]H is zero & [tex]\Delta[/tex]S is positive

Change is solely due to entropy since no heat is evolved or absorbed.
Example: mixing of two gasses or liquids at constant volume

2) [tex]\Delta[/tex]H is negative & [tex]\Delta[/tex]S is positive

When an exothermic reaction is accompanied by a high increase in entropy this give a high negative value for [tex]\Delta[/tex]G

This occurs for instance in a chemical reaction where a gas is evolved
Example: adding water to calcium carbide to liberate acetylene.

3) [tex]\Delta[/tex]H is positive & [tex]\Delta[/tex]S is positive

For and endothermic reaction to have -ve [tex]\Delta[/tex]G T[tex]\Delta[/tex]S must be greater than [tex]\Delta[/tex]H
At normal temperatures this is unusual, but as T increases the reaction may become favourable.

Example: heat calcium carbonate strongly (increase T) it dissociates into calcium oxide and carbon dioxide (gas).

4) [tex]\Delta[/tex]H is negative & [tex]\Delta[/tex]S is negative

Here we must have an exothermic reaction where the heat evolved outweighs the decrease in entropy.

Example: burn hydrogen in oxygen to yield water. The entropy decreases because there are fewer gaseous molecules in the products than in the reactants.

go well
 
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  • #3
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Thanks for your reply. I understand from the mathematical side of the equation why entropy becomes dominant at higher temperature, but do not understand the chemical theory side of this. If all systems tend to a state of maximum stability/minimum free energy, why does increasing the entropy contribute to this?

I really appreciate your time in answering my questions.
 
  • #4
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Delta G is a kind of "real world lowest energy". It is not exactly stability or probability but a compromise. Look at a crystal: One free from defects would have the lowest delta H and delta S. It just doesn't happen, in real life defects creep in, always. That is called entropy so delta G is lowest energy with a compromise for what is likely to happen. I hope you see my arms waving :-)
 
  • #5
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why does increasing the entropy contribute to this?
I really thought I'd answered this.

If a system tends to a state of minimum free energy then it will follow a reaction where free energy is evolved ie [tex]\Delta[/tex]G is negative.

I listed all four cases of this.

Looking more closely when the enthalpy term in the equation absorbs energy the only thing left to create a negative [tex]\Delta[/tex]G is the entropy term.

My example (case3) was a chemical situation where this happens and a reason. The reason is most often where there are more free molecules at the end than the beginning such as when a gas is evolved froma solid.
 
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  • #6
DrDu
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I understand why the stability would increase as enthalpy decreases, due to the decline in potential energy.
Decline in potential energy is not a universal principle. The enthalpy term describes an increase of the entropy of the surrounding while the T Delta S term describes the entropy change of the system itself. So both terms together yield the total change of entropy ( times temperature) which is the only quantity of importance in terms of stability.
 
  • #7
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Okay, I'm sorry for being so thick, but it's still not sinking in! Basically, is there one particular reason why a system tends to highest entropy (eg. energy spread over more micro states)? Thanks for persevering with me everyone.
 
  • #8
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I don't wonder you are confused.
But then so am I so we are equal.

:biggrin:

You started off talking about Free Energy.

This is a classical thermodynamic function. So I provided a classical answer. And in Chemistry terms since this is the Chemistry forum.

You also linked it to some undefined concept of 'maximum stability' on which Dr Du offered a comment.

Now you are talking about an undefined form of energy 'distributed over microstates', which is not a classical function at all.

No offence but I think you would be better served properly establishing a few solid functions and developing from there.

What did you not understand about my offering?
 
  • #9
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I just don't understand why entropy is good! :cry:

I really appreciate all your help, but I just seem to be missing some crucial point, and can't quite put my finger on it.
 
  • #10
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Well it's good to start from what you know and work towards what you don't.

:wink:
 
  • #11
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I fell like I don't know anything at the moment! I am doing an assignment, and have based it around the concept that the driving force of a reaction is to attain a state of maximum stability (which is equivalent to minimum free energy). I don't know if this is strictly correct, but it all seems to make sense. The problem I am having now is relating the favourability of an entropy increase to maximizing the stability of the system.
 
  • #12
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I am worried about the concept of 'maximum stability'.

That implies that there is some scale of stability, to which I can put numbers so that I can say the state A is x times as stable as state B etc.

I don't know of such a measure.

I think your idea of stability needs brushing up, what do you understand by the term?
 
  • #13
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I am just confusing myself at the moment.

Basically, I am trying to discuss the ka for acids, which I known is dependent on their strength.

When I first started researching, I found a bunch of factors that affect the strength of the acid. The most prominent one was the stability of the conjugate base. If it was more stable, then dissociation was more likely to occur.

Then I found out more about free energy. I found that if the free energy decreases, then the reaction is more likely to occur. I also read that minimum free energy is the most stable state.

So I guess I've kind of come to see stability and free energy as one and the same, which is probably wrong!

Have you got any idea where I am coming from, or how to de-confuse this situation?
 
  • #14
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Are you, by any chance, studying the chapter on chemical equilibrium in Atkins?

He deals with this issue there.
 
  • #15
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No, I'm not. I've just been picking information from everywhere, and I think that is where I'm getting confused.

Do you have any idea if there is a link between the strength of an acid (due to stability factors such as resonance and size) and the dissociation in order to lower free energy? They both seem to act to maximize stability.
 
  • #16
5,439
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Yes I've been thinking about your question while I was driving to the airport.

I'll post my thoughts shortly,

I don't know what library facilities you have acces to but they should have a copy of Atkins Physical Chemistry. There is a graph of deltaG showing the minimum you describe - although he does not claim 'maximums stability'.

Most chemistry texts (including Atkins who handles it in a more modern way) deal with the subject in terms of activity potentials rather than mechanical thermodynamic concepts.
 
  • #17
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I don't really have access to the book. However, I really appreciate all your effort, and look forward to hearing your thoughts!
 
  • #18
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OK going from what we know to what we don't.

Consider first a simple process - melting of a solid.

This requires substantial heat input, but there is little or no mechanical work done since there is (almost) no change in volume.

So where does the heat go and why does something melt at all?

The answer is that in changing from a solid to a liquid the particles gain degrees of freedom or become more disordered. That is the entropy increases.

In fact the entropy of fusion (or melting) equals the latent heat divided by the melting temperature and is often very high compared to the specific heat of either the solid or liquid.

In a similar way consider mixing and solubility.

It is often said (wrongly) that solubility is due to molecular forces.
Any two gases, however are miscible (=soluble) in all proportions. The mixing is not due to intermolecular forces it is due to molecular motion.
The mixing constitutes an increase in entropy since there is only one configuration where the gases are separated, but many possible configurations where they are mixed.

Gases are subject to substantial space between the molecules.
If we consider much closer spacing, as in a solution, we still get the motion, but the proximity allows the molecular forces to modify the mixing either to increase or decrease the tendency.
This is because the molecular interaction supplies or consumes some of the internal energy of mixing.

Dissociation inherently leads to increased entropy since there are more products after the dissociation than in the reactant(s), similar to the gas mixing above.

However, similar to the solution situation above, the molecules (or ions) are in sufficient proximity for this to be modified by molecular forces.

This is why you measure a different Ka for different acids.

I hope this rather rambling, non mathematical, discussion will lead towards helping you connect Ka and entropy.

You will need to obtain actual measurements (or use published ones) and consider what molecular forces might be acting in the case of individual acid.

go well
 
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  • #19
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I think I still need to process everything you just said! I'll go away and see what I come up with, and do you mind if I come back with any more questions?

Thankyou so much for all your assistance and dedication though, it is really greatly appreciated!
 

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