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Topic: Enthalpy in the Boltzmann distribution ?  (Read 8054 times)

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Offline GeLe5000

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Enthalpy in the Boltzmann distribution ?
« on: March 16, 2016, 06:02:01 AM »
Good morning.

In "Atkin's Physical Chemistry", there's a molecular interpretation of the response of chemical equilibrium to temperature.

On a figure, the Boltzmann distributions of molecules A and B are shown.
The lower level of molecules B distribution is higher than the lower level of molecules A distribution. Thus a transformation A ----> B is considered. There's more energy in B than in A in the case of an endothermic process, less energy in B than A if the process is exothermic.
Two curves (High and Low Temperature) show that an increase in temperature leads to an increase in the population of B at the expense of A in the case of an endothermic process; the opposite happens in an exothermic reaction.

I can't understand why the lower levels of A and B are placed at different levels since the energy considered in the Boltzmann distribution is linked to translational, rotational, vibrational and electronic motions.
The difference between A and B is a difference in Enthalpy. It's a matter of chemical energy present in chemical links. No relation with motions.

Offline GeLe5000

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Re: Enthalpy in the Boltzmann distribution ?
« Reply #1 on: March 16, 2016, 09:28:31 AM »
I reply to myself.

During my nap I've understood that it's the total energy that must be considered : Internal energy (enthalpy) + motion energy.
But I don't feel that this game with Boltzmann distributions placed at different levels is really an explanation.

Thank you for reading.

Offline mjc123

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Re: Enthalpy in the Boltzmann distribution ?
« Reply #2 on: March 16, 2016, 09:47:05 AM »
If you have any system with different energy states that it may occupy (including that of molecules A and B with different ground state energies, in chemical equilibrium), there will be a Boltzmann distribution between the available states. It's not just vibration, rotation etc.
Exercise: Using the statistical definition of entropy Smol = R ln W, show that the Boltzmann distribution between A and B is equivalent to the equation Keq = e-ΔG/RT

Offline GeLe5000

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Re: Enthalpy in the Boltzmann distribution ?
« Reply #3 on: March 16, 2016, 12:26:14 PM »
An exercise ! Wonderful. Thank you very much.
But I need time to understand the question. I know the Boltzmann formula and the equation for the equilibrium constant, but the relation between them...




Offline Irlanur

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Re: Enthalpy in the Boltzmann distribution ?
« Reply #4 on: March 17, 2016, 04:35:36 AM »
Quote
Using the statistical definition of entropy Smol = R ln W, show that the Boltzmann distribution between A and B is equivalent to the equation Keq = e-ΔG/RT

Is this really as straight-forward as you make it sound?

Offline mjc123

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Re: Enthalpy in the Boltzmann distribution ?
« Reply #5 on: March 17, 2016, 05:33:20 AM »
Well, possibly not. It would be hard to quantify the entropy change of a chemical reaction like A  :rarrow: B using S = R ln W. But working the other way round, it is possible to express the pre-exponential term of the Boltzmann distribution in the form eΔS/R. Perhaps I should have said "analogous" rather than "equivalent".

Offline GeLe5000

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Re: Enthalpy in the Boltzmann distribution ?
« Reply #6 on: March 17, 2016, 11:03:26 AM »
I've tried something.

S° = R ln W
W is the number of microstates available in the system at the temperature T

∆S° = R ∆(ln W), where ∆(ln W) is the change in the (logarithm of the) number of available microstates when the reaction has reached equilibrium

Keq = e-∆G°/RT

 ΔG° = ΔH° - T∆S°

ΔG° = ΔH° - T R Δ(ln W) = - RT ln K

ln K = - ∆H°/RT + ∆(ln W)

Equilibrium constant as a function of temperature :

d ln K / dT = d(- ∆H°/RT + ∆(ln W)) / dT = d(- ∆H°/RT) / dT + d(∆(ln W)) / dT

If the reaction is endothermic, ΔH° > 0  and - ∆H°/RT < 0
We know that d ln K / dT is > 0
Thus, d(∆(ln W)) / dT must be > 0

Logic since in a warmer system there are more available microstates.

If the reaction is exothermic, ΔH° < 0  and - ∆H°/RT > 0
We know that d ln K / dT is < 0
Thus, d(∆(ln W)) / dT must be < 0

It means that when the system loses heat in an exothermic reaction, the decrease in the number of available microstates is less important when the system is warmer.


I'm not sure that my interpretations are correct.

Offline mjc123

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Re: Enthalpy in the Boltzmann distribution ?
« Reply #7 on: March 17, 2016, 01:09:31 PM »
Quote
d ln K / dT = d(- ∆H°/RT + ∆(ln W)) / dT = d(- ∆H°/RT) / dT + d(∆(ln W)) / dT

If the reaction is endothermic, ΔH° > 0  and - ∆H°/RT < 0
We know that d ln K / dT is > 0
Thus, d(∆(ln W)) / dT must be > 0

Logic since in a warmer system there are more available microstates.

d(- ∆H°/RT) / dT  = ∆H°/RT2, so is > 0 when ΔH° > 0.
d(∆(ln W)) / dT = 0. Changing temperature doesn't change the number of microstates, only their occupancy.

Here's how I was thinking:
Boltzmann: N1/N0 = n1/n0*e-ΔE/RT
where n1 and n0 are the number of microstates in levels 1 and 0.
n1/n0 = e^ln(n1/n0) = e^Δ(ln W) = e^ΔS/R
K = N1/N0 = e^-(ΔE-TΔS)/RT
which is analogous to the equilibrium constant expression. My point was, as I said below, that
"If you have any system with different energy states that it may occupy (including that of molecules A and B with different ground state energies, in chemical equilibrium), there will be a Boltzmann distribution between the available states." I hope this lets you see the analogy, and that Atkins was justified - these aren't two completely different phenomena.

Offline GeLe5000

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Re: Enthalpy in the Boltzmann distribution ?
« Reply #8 on: March 17, 2016, 01:38:21 PM »
I meant "more posssible microstates", not "more available microstates". It's a mistake. I hope that  "more possible microstates" is right for you, because that's what I've learned.
Anyway, I was aware that my game with the equations was not the solution to your exercise.
I have to go back to Boltzmann in order to understand your demonstration.

Thank you.



Offline mjc123

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Re: Enthalpy in the Boltzmann distribution ?
« Reply #9 on: March 18, 2016, 05:19:58 AM »
Quote
I meant "more posssible microstates", not "more available microstates".
I'm not sure what you mean by this. Increasing temperature does not create any more microstates, though more of them may become thermally accessible.

Offline GeLe5000

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Re: Enthalpy in the Boltzmann distribution ?
« Reply #10 on: March 18, 2016, 07:35:25 AM »
Right ! Double mistake. I didn't mean "possible" but "accessible". I should have opened my books again to find the right word.

And, by the way, I don't understand that you wrote : "d(∆(ln W)) / dT = 0. Changing temperature doesn't change the number of microstates, only their occupancy"

If the temperature is higher, the Entropy increases. Since S° = R ln W, isn't it obvious that d(∆(ln W)) / dT > 0 in an endothermic reaction ? You say that changing the temperature changes the occupancy of the microstates. "occupancy" and "accessibility" are linked, aren't they ?


Offline mjc123

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Re: Enthalpy in the Boltzmann distribution ?
« Reply #11 on: March 18, 2016, 12:37:45 PM »
OK, perhaps I'm confusing myself a bit here, and not being clear about the meaning of W. As Irlanur suggests, maybe it's not as straightforward as it sounds.
Let's consider the case of a simple diatomic rotor. The energy levels are EJ = BJ(J+1) and the multiplicities are nJ = 2J+1. Let's assume for simplicity that levels above J = 1 are not thermally accessible, to a significant extent, at temperatures of interest.
Consider the states a molecule may occupy. There is just one way it can occupy level 0, with mJ = 0. There are 3 ways it can occupy level 1, with mJ = 1, 0 or -1. So W0 = 1 and W1 = 3, and ΔS01 = k ln 3.
For a mole of molecules ΔS = R ln 3. This is the difference between the molar entropies of the two states - i.e. between the state where all molecules are in level 0, and the one where all molecules are in level 1. Likewise ΔS° for a reaction such as A  ::equil:: B is S°(B) - S°(A), i.e. the difference between all-B and all-A. We are not yet considering an equilibrium distribution of the molecules between the states.
This value of ΔS (to a first approximation, for the case of the chemical reaction) does not change with temperature. You can see that W0 = 1 and W1 = 3 irrespective of temperature or occupancy. What changes with temperature is the entropy of the system as the occupancy of the levels changes. If at T1, 1% of the molecules are in level 1, and at T2 the fraction is 5%, then W(system) is greater at T2 than at T1 because there are more ways of producing this arrangement.
In the chemical reaction, the entropy of the system at equilibrium is S = xAA + xBB. The entropy of the system changes with temperature because xA and xB change.
I hope that hasn't made things too complicated!

{Note: d ln K / dT = d(- ∆H°/RT + ΔS°/R) / dT = ΔH°/RT2 - 1/RT dΔH°/dT + 1/R dΔS°/dT
But dΔH°/dT = ΔCp and dΔS°/dT = ΔCp/T, so the last two terms cancel out. Thus any variation of ΔS° with T doesn't affect K.}

Offline Irlanur

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Re: Enthalpy in the Boltzmann distribution ?
« Reply #12 on: March 20, 2016, 03:50:18 PM »
I think it would be easier to just use the partition function here...

Offline GeLe5000

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Re: Enthalpy in the Boltzmann distribution ?
« Reply #13 on: March 21, 2016, 11:16:53 AM »
I'd like to put the problem in another way.

For example,

N2 + 3 H2 -----> 2 NH3

is an exothermic reaction. An increase in temperature will increase the concentrations of the reactants N2 and H2 (Equilibrium constant = 90 at 500 K,  0,04 at 800 K).

When Atkins says that this effect has a molecular basis that stems from the Boltzmann distribution, he gives a molecular description of the state of the system at equilibrium.
Since it is known that (N2 + 3H2) is richer in energy than 2NH3, the reactants are placed higher on the Boltzmann distribution. Thus, the comparison of the curves at low and high temperatures shows that the increase in Temperature gives an increase in the population of the reactants.

I hope that you have seen this figure before because I can't manage to insert it here. (I can't even underlie a word. I don't know why)

It's only a description, not an explanation of the fact at high Temperature more NH3 decomposes into N2 and H2.

It's only in Kinetics that explanations can be given (collision theory), not in Thermodynamics. And there's no relation between Kinetics and Thermodynamics.


Am I right or not ?

Offline mjc123

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Re: Enthalpy in the Boltzmann distribution ?
« Reply #14 on: March 21, 2016, 01:51:39 PM »
At some level, almost any "explanation" of something can be regarded as simply a "description" of it in other terms. A physical "law" is just a description of how we observe the universe to behave. For example, Newton's explanation of the motion of the planets by the inverse square law of gravity could be regarded as simply describing their motion by saying "they behave as they would do if acted on by an inverse square law force." It doesn't explain what the force is, how it acts at a distance, why it follows an inverse square law etc. Nevertheless, this systematisation of the observed data gives insight into the phenomena and allows us to make correct predictions. It may therefore be regarded as an "explanation", something that advances our understanding.
I don't at all agree that kinetics explains while thermodynamics only describes. Kinetics gives a dynamic, time-dependent description of how a system gets from one state to another. But does it explain why this happens, any more than thermodynamics does? Alternatively, both may be considered as giving an "explanation" in the sense I have just defined, i.e. they systematise knowledge and enable predictions.
Nor do I agree that there is no relation between kinetics and thermodynamics. Both may be considered as descriptions of how a system behaves depending on what energy it has. Consider, for example, the energy profile of a simple one-step reaction. Kinetics and thermodynamics are related by e.g. ΔH = Ea,f - Ea,b. Kinetics, I suppose, may be regarded from one point of view as "the thermodynamics of transition states", i.e. energy maxima, species of only transient existence; while Thermodynamics is "the thermodynamics of (more or less) stable species", i.e. energy minima, species that exist for finite length of time.

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