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Why do phase trxns behave differently than chemical rxns?

  1. Jun 15, 2015 #1
    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.
     
  2. jcsd
  3. Jun 16, 2015 #2
    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.
     
  4. Jun 16, 2015 #3
    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
     
  5. Jun 17, 2015 #4
    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).
     
  6. Jun 17, 2015 #5
    Liquid water can exist at -4C if you increase the pressure enough. See this phase diagram. Water.PNG

    And, at a phase transition between liquid water and ice, liquid water and ice exist in equilibrium simultaneously.
     
  7. Jun 17, 2015 #6
    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.
     
  8. Jun 17, 2015 #7
    Sorry. I did not understand that. But this brings me back to multicomponent vs single component.

    Chet
     
  9. Jun 17, 2015 #8
  10. Jun 17, 2015 #9
    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
     
  11. Jun 18, 2015 #10
    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
     
  12. Jun 18, 2015 #11
    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.
     
    Last edited: Jun 18, 2015
  13. Jun 18, 2015 #12
    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
     
  14. Jun 19, 2015 #13

    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.
     
  15. Jun 19, 2015 #14
    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.
     
  16. Jun 19, 2015 #15
    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.
     
  17. Jun 19, 2015 #16
    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.
     
  18. Jun 19, 2015 #17
    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
     
    Last edited: Jun 19, 2015
  19. Jun 19, 2015 #18
    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.
     
    Last edited by a moderator: May 7, 2017
  20. Jun 19, 2015 #19
    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) equilibr.gif 2 NH3(g)
    [PLAIN]http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch21/graphics/delta.gifG[/I][SUP]o[/SUP] [Broken] = -32.96 kJ
    Alright, lets assume we are at standard conditions & partial pressures for this reaction, hence we know its ΔG. Now, since the rxn is downhill as written, lets consider the reverse reaction. So lets 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 = Keq = 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) equilibr.gif 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?
     

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    Last edited by a moderator: May 7, 2017
  21. Jun 19, 2015 #20
    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.
     
    Last edited by a moderator: May 7, 2017
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