Why do phase trxns behave differently than chemical rxns?

[kayan]

If we have solid water (ice) at -5 C and atm P, then according to thermodynamics, the process that takes that ice (at -5 C & atm P) from solid to liquid is non-spontaneous, which means its ΔG > 0. Hence, it is impossible for someone to observe ice melt to liquid at -5 C, and assuming we started with a solid block of ice, this means there is no liquid water in the system.
However, if I have an arbitrary chemical reaction that is happening, with a ΔG > 0 (endergonic), there will always be an equilibrium established (given enough time) that will have some reactants AND some products present (the amount of each present depends on the K(eq) value or equivalently the ΔG).

My misunderstanding comes from why these two cases are different? Why, in the case of the phase transition, is it thermodynamically impossible to have both reactants and products present (ice and liquid), whereas in the chemical reaction, even if ΔG > 0 like before, its expected that both reactants and products will be present in the system? Hopefully everything here makes sense.

 

[Khashishi]

I'm sort of guessing, but I think it's because in chemical reactions where the reagent and products can mix, there is an entropy of mixing which can cause a local minimum in the Gibbs free energy at a constant T and P taken as a function of the extent of reaction. This allows something like https://en.wikipedia.org/wiki/File:Diag_eq.svg
With phase change between liquid and ice, there is no mixing between the liquid and ice, so there is no additional entropy by having both liquid and ice together.

 

[Chestermiller]

As Kashishi is alluding to, you are talking about the difference between multicomponent systems (mixtures/solutions) and single component systems. Even in phase transitions, for multicomponent phase equilibrium, all the components are present in all the phases, and at different concentrations. Once you get into chemical reactions, there is added complexity, because, even in going from the pure reactants to the pure products, there is an entropy change, an enthalpy change, and a free energy change (so it's not only mixing involved).

Chet

 

[kayan]

As of now, I don't think it has anything to deal with single vs multicomponent systems. I have talked to number of people about this issue recently, and what we seem to have concluded is just that the ΔG for the ice melting (at -4 C) is so largely positive that you will never macroscopically observe it melt to liquid water (which also means the K(eq) lies solely to the reactants side). However, on the microscopic scale, there is always a chance (and the reality) that a few H2O molecules get enough energy to be in the liquid state, even at -4 C.

The main reason why I don't think this is a single/multi issue is because I'm trying to develop insights into the sign and magnitude of ΔG, regardless of the components. Hence, so long as you can write a ΔG for any arbitrary process, the same conclusions should follow from the sign and magnitude of the G (kinetics is a separate issue not involved here).

 

[Chestermiller]

Liquid water can exist at -4C if you increase the pressure enough. See this phase diagram.

And, at a phase transition between liquid water and ice, liquid water and ice exist in equilibrium simultaneously.

 

[kayan]

I'm familiar with this, but I was assuming everyone understood me to mean -4 C and atm pressure. This entire conversation becomes trivial if we allow pressure to vary to let water thermodynamically favor liquid at -4 C and some P.

 

[Chestermiller]

I'm familiar with this, but I was assuming everyone understood me to mean -4 C and atm pressure. This entire conversation becomes trivial if we allow pressure to vary to let water thermodynamically favor liquid at -4 C and some P.

Sorry. I did not understand that. But this brings me back to multicomponent vs single component.

Chet

 

[Khashishi]

Have you read https://en.wikipedia.org/wiki/Chemical_equilibrium ?
The second paragraph under the thermodynamics section mentions the entropy of mixing.

 

[Chestermiller]

Have you read https://en.wikipedia.org/wiki/Chemical_equilibrium ?
The second paragraph under the thermodynamics section mentions the entropy of mixing.

Yes. I totally agree with you that mixing is a big part of the story, just not the entire story. But I just wish we could convince kayan.

Chet

 

[Chestermiller]

As of now, I don't think it has anything to deal with single vs multicomponent systems.

What does the phase rule tell you about the effect of single vs multicomponent with regard to phase equilibrium and chemical equilibrium? (The phase rule focuses specifically on this effect).

Chet

 

[kayan]

Yes. I totally agree with you that mixing is a big part of the story, just not the entire story. But I just wish we could convince kayan.
Chet

You don't need to convince me of this. I'm sure this plays a role, but it's not addressing the question I posed. I only posed a question concerning the magnitude and sign of ΔG. Sure, the entropy of mixing will have a role on the value of ΔG, but that is outside the scope of my question. My question assumes that you start from a given ΔG, hence, we should (EDIT: not) have to worry about the details therein unless you want to argue magnitudes in the ΔG values.

My question boiled down to, given two different processes with the same sign of ΔG > 0, why does one have (at EQ) a possible abundance of both reactants and products (as is common for many chemical reactions), whereas the other quite strictly has only reactants (except for a few molecules at the micro level), which is common for phase transitions? I'm looking for an argument from the ΔG standpoint, because in the end, that is final factor.

 

[Chestermiller]

I'm sorry, but I'm still not clear on what you are asking. Would you please consider focusing on a specific system or systems so that we have something more concrete to work with? If yes, then thanks.

Chet

 

[Tom_K]

As of now, I don't think it has anything to deal with single vs multicomponent systems. I have talked to number of people about this issue recently, and what we seem to have concluded is just that the ΔG for the ice melting (at -4 C) is so largely positive that you will never macroscopically observe it melt to liquid water (which also means the K(eq) lies solely to the reactants side). However, on the microscopic scale, there is always a chance (and the reality) that a few H2O molecules get enough energy to be in the liquid state, even at -4 C.

The main reason why I don't think this is a single/multi issue is because I'm trying to develop insights into the sign and magnitude of ΔG, regardless of the components. Hence, so long as you can write a ΔG for any arbitrary process, the same conclusions should follow from the sign and magnitude of the G (kinetics is a separate issue not involved here).

As I see it a chemical reaction requires that the separate components come into contact with one another (mixing) for the reaction to take place and this places another condition upon the process which the phase change does not have. Unless the mixing state is somehow incorporated into the ΔG factor for the chemical reaction, the ΔG being of the same magnitude is not a sufficient condition upon which to base a comparison between the two different processes of a phase change and a chemical reaction.

 

[kayan]

Are you suggesting that at a given T & P that the criteria for spontaneity is something more than ΔG < 0? My knowledge of thermodynamics relies on the fact that at a const T, P the only criteria for a process being spontaneous thermodynamically is ΔG < 0.

 

[Khashishi]

If you want to be more correct, it's not ΔG that you should use, but $dG/d\xi$, where $\xi$ is the amount of reactant that is consumed, or the amount of product created. ΔG suggests some kind of discrete reaction step. Other than that, I think G is all you need to determine the spontaneity, but you need the full G, which includes mixing effects.

 

[kayan]

I see your point but I'm not sure that dG/dξ is anymore correct than ΔG, since, for any process that has reached equilibrium, the total G's (regardless if the process was discrete or continuous) from beginning to end or from reactant to product must be equal, otherwise the system could lower its overall G by going one way or the other. In any case, I think we understand each other and this is besides the main question.

 

[Chestermiller]

From the past few posts, and, looking back on your original post, I think I finally understand what you are referring to with regard to chemical reactions. There is quite a bit of ambiguity in the literature between the ΔG of a reaction, and the ΔG that occurs when reactants and products are not initially at equilibrium and the system is allowed to spontaneously equilibrate. If we are talking about the ΔG of a reaction at T and P, then we are talking about the following:

State 1: Pure reactants (in separate containers) at T and P
State 2: Pure products (in separate containers) at T and P

This can either be positive or negative.

If we are talking about a system when reactants and products are not initially at equilibrium at T and P (either in separate containters or already mixed) and the system is allowed to equilibrate at T and P, then ΔG for the change between these two states is always negative. It will be negative even if we start with pure reactants in separate containers and allow the system to equilibrate or if we start with pure products in separate containers and allow the system to equilibrate.

Chet

 

[Tom_K]

I see your point but I'm not sure that dG/dξ is anymore correct than ΔG, since, for any process that has reached equilibrium, the total G's (regardless if the process was discrete or continuous) from beginning to end or from reactant to product must be equal, otherwise the system could lower its overall G by going one way or the other. In any case, I think we understand each other and this is besides the main question.

If I understand your main question, it concerns the difference between a chemical reaction and a phase change, as regards the amount of reactants and products. I am not sure that focusing on ΔG will lead to an answer.

A chemical reaction that has reached equilibrium is in a state of Dynamic equilibrium.

Dynamic Equilibrium

At dynamic equilibrium, reactants are converted to products and products are converted to reactants at an equal and constant rate. Reactions do not necessarily—and most often do not—end up with equal concentrations. Equilibrium is the state of equal, opposite rates, not equal concentrations.

There are both reactants and products present, or likely to be present after this state of dynamic equilibrium is reached.

A thermodynamic equilibrium that results from a phase change, such as water changing into ice, also results in a dynamic equilibrium, at least on the microscopic level, but at the macroscopic level it is more like a static equilibrium.

Static Equilibrium

Static equilibrium, also called mechanical equilibrium, occurs when all particles in the reaction are at rest and there is no motion between reactants and products. Static equilibrium can also be seen as a steady-state system in a physics-based view. Dynamic forces are not acting on the potential energies of the reverse and forward reactions.You might say this is a state of quasi-static equilibrium: dynamic microscopically but static macroscopically.

I think that might be the difference you are looking for.

 

[kayan]

Chet
Yes, this is getting to addressing my original question. OK, just to be clear, let me go through this scenario with a chemical reaction.

N2(g) + 3 H2(g)

2 NH3(g)
[PLAIN]http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch21/graphics/delta.gifG[/I][SUP]o[/SUP] = -32.96 kJ
Alright, let's assume we are at standard conditions & partial pressures for this reaction, hence we know its ΔG. Now, since the rxn is downhill as written, let's consider the reverse reaction. So let's start by putting the amount of moles of pure NH3(g) into a container (that corresponds to its std pressure) at standard conditions and we ask the question, what will happen given an infinite amount of time? According to what I've been taught, even though the rxn that takes NH3 to N2 and H2 has ΔG>0, it will still happen. Further, it will occur to the extent until the Q = K_eq = exp[-32.96kJ/(2mol*Rg*298K)] = 1.3E-3 = P(N2) * P(H2)³/P(NH3)². Hence, there will still be (macroscopic?) amounts of N2 and H2 formed from a reaction that started out with ΔG > 0.

Compare this to H2O(s)

H2O(l) at atm P and -10 C. Now, I actually haven't calculated the ΔG for this reaction at these conditions, but one thing is for sure: ΔG > 0 because there is no way that this reaction is spontaneous with these conditions. However, macroscopically, you will never observe liquid water. We know this fact so well that my thermo teacher said that it would break the 2nd Law for this reaction to happen under these conditions.
See the place for confusion? Have I made any inaccurate conclusions?

 

[kayan]

If I understand your main question, it concerns the difference between a chemical reaction and a phase change, as regards the amount of reactants and products. I am not sure that focusing on ΔG will lead to an answer.
Dynamic Equilibrium
...

Static Equilibrium
...

I think that might be the difference you are looking for.

You may be right that this is what I'm looking for, because almost by definition, for a chemical reaction, we are talking about microscopic particles, but for a block of ice, it's large.
(EDIT: but since we know everything is made of atoms, is there really such a thing as static EQ? I think not...and this is related to my question)

However, I'm not sure I agree with the first statement about focusing on ΔG...For the conditions I'm talking about ΔG is the key criteria. Not only can you not disregard it, but it tells you everything you need to know about a system's thermo EQ if you know it. No way you can understand something thermodynamically without understanding how this relates to the process.

 

[Chestermiller]

Chet
Yes, this is getting to addressing my original question. OK, just to be clear, let me go through this scenario with a chemical reaction.

N2(g) + 3 H2(g)

2 NH3(g)
[PLAIN]http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch21/graphics/delta.gifG[/I][SUP]o[/SUP] = -32.96 kJ
Alright, let's assume we are at standard conditions & partial pressures for this reaction, hence we know its ΔG.

Yes. This is the free energy change between the following two states:

State 1: Pure reactants at 298 K and 1 atm.
State 2: Pure products at 298 K and 1 atm.

So there are no partial pressures involved. Each species in each state is at 1 atm. That's what the superscript 0 means.

Now, since the rxn is downhill as written, let's consider the reverse reaction. So let's start by putting the amount of moles of pure NH3(g) into a container (that corresponds to its std pressure) at standard conditions and we ask the question, what will happen given an infinite amount of time? According to what I've been taught, even though the rxn that takes NH3 to N2 and H2 has ΔG>0, it will still happen.

Yes. That ΔG referred to here is for the reaction going from pure NH3 to pure N2 and H2.

Further, it will occur to the extent until the Q = K_eq = exp[-32.96kJ/(2mol*Rg*298K)] = 1.3E-3 = P(N2) * P(H2)³/P(NH3)². Hence, there will still be (macroscopic?) amounts of N2 and H2 formed from a reaction that started out with ΔG > 0.

No. If you calculate the ΔG between the following two states, it will be negative:

State 1: NH3 at 1 atm and 298 K
State 2: Corresponding equilibrium mixture of N2, H2, and NH3 at a total pressure of 1 atm. and 298 K.

ΔG for this change is negative here because the partial pressures in the final mixture are not each 1 atm.

As I said in my previous post, we are talking about 2 different ΔG's here, corresponding to two different initial and final equilibrium states.

Compare this to H2O(s)

H2O(l) at atm P and -10 C. Now, I actually haven't calculated the ΔG for this reaction at these conditions, but one thing is for sure: ΔG > 0 because there is no way that this reaction is spontaneous with these conditions. However, macroscopically, you will never observe liquid water. We know this fact so well that my thermo teacher said that it would break the 2nd Law for this reaction to happen under these conditions.
See the place for confusion? Have I made any inaccurate conclusions?

I'm having trouble with this one because liquid H2O cannot be in a thermodynamic equilibrium state at -10C and 1 atm. pressure.

Chet

 

[kayan]

I'm having trouble with this one because liquid H2O cannot be in a thermodynamic equilibrium state at -10C and 1 atm. pressure.

But you can still calculate a G value for it. Even though the EQ state of water is ice at -10C, this doesn't mean that there is not a G value for liquid water at -10C. For example, as I'm sure you're aware, if you start with liquid water above 0C and carefully cool below the freezing point, you can retain the liquid state in a metastable phase. Once that carefully sub-cooled liquid reaches -10C, it will still have a G value, however, it will certainly be > the ice G value to reflect the fact that ice is the preferred state.

 

[Chestermiller]

But you can still calculate a G value for it. Even though the EQ state of water is ice at -10C, this doesn't mean that there is not a G value for liquid water at -10C. For example, as I'm sure you're aware, if you start with liquid water above 0C and carefully cool below the freezing point, you can retain the liquid state in a metastable phase. Once that carefully sub-cooled liquid reaches -10C, it will still have a G value, however, it will certainly be > the ice G value to reflect the fact that ice is the preferred state.

OK. No problem.

I think that the first thing that needs to be done is to solidify your understanding of the different ΔG's involved when you have chemical reaction. Are you feeling more comfortable with what I wrote in my previous post? It might be helpful for you to actually calculate the ΔG for the NH3 dissociation process that you described, to confirm for yourself that it is actually negative. Are you familiar with the van't Hoff equilibrium box? This might be very helpful for you to read about too. The van't Hoff equilibrium box enables you to properly get ΔG in going from pure reactants at arbitrary pressures to pure products at arbitrary pressures using an equilibrium mixture present within the box.

Chet

 

[kayan]

You're definitely right that I'm rusty on my ΔG interpretation (its been a few years since I did these calcs in a good thermo class), so I'll take some time and do some thinking, then hopefully I won't forget to come back for any lingering questions. Appreciate the help.