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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: canonicalpartitionstudent on August 12, 2014, 11:07:25 AM
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Hi everyone,
Could someone help me with this question for Molecular Physical Chemistry?
"Suppose a system is made of 1 mole of independent, distinguishable quantum objects that can be in 3 different states each (see Table 3 in attachment). These states, called A, B and C, have energies and degeneracies as listed in the table below. Plot the composition, energy and entropy of the system as a function of temperature. Discuss your result in view of the temperature dependence of chemical equilibria."
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You have to show some initial work to get help, as outlined in the forum policies. This is as much for our benefit as it is yours.
What part of the problem are you having trouble with? If it's simply getting started, then let's start with something basic. Forget any calculations and lets move right to the discussion part. How will you expect the system to change as a function of temperature?
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To be honest, I have no idea.. Otherwise I had certainly showed initial work.
Entropy will increase with higher temperatures, energy will increase as well, the composition will change until a new equilibrium has been reached?
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Think about state population. You have three states, A, B, and C, with ascending energies. Ignore the degeneracies for a moment. Let's say you have N "quantum objects", and the number of objects in each state at a given time we'll call NA, NB, and NC, such that ΣNi = N.
How do you expect NA, NB, and NC to change as the temperature rises, starting at absolute zero? Answer generally, not with numbers. No calculations required here.
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With increasing temperature, Na will decrease, Nc will increase? The objects are excited due to the higher temperature and hop into a higher energy state?
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Right, great. At absolute zero, everything will be in the A state. As you increase temperature, B and then C will be gradually populated. Entropy generally increases, too, because you have more states available for population, which is where the "disorder" concept comes from.
Now: do you know any important equations than can help you quantify the relative number of A, B, and C as a function of temperature?
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Fermi-Dirac statistics?
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Have you learned about the partition function yet?
Click here (http://en.wikipedia.org/wiki/Partition_function_(statistical_mechanics)).
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Yes!
Which formula should I use to begin?
This one?
q_el= ∑g e^(-βε_(el,i) )?
This formula can be related to temperature as:
ln q_el = ln(6) + ln(4) -4β + ln(2) - 6β ?
Here I need a push in the right direction.
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Basically, the probability of finding a quantum system in the nth microstate having energy En and degeneracy gn is given by the expression:
[tex]P_n=\frac{g_n e^{-\beta E_n}}{Q}[/tex]
Where Q is the Partition Function
[tex]Q = \sum_i g_i e^{-\beta E_i}[/tex]
And beta is the inverse temperature
[tex]\beta = (k_B T)^{-1}[/tex]
Are you able to do this evaluation to determine the composition of your system as a function of temperature?
EDIT: Btw, Units are important, so be careful!
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I tried to solve Q. The outcome for Q (for 5 to 5000K, taken as example) is 6:
--> Q = 6 + 4*EXP(-beta*4000)+2*EXP(-beta*6000)
With this I should be able to calculate Pn.
For P with n = A: outcome is 1.
--> P = 6/6
For P with n = B (and n = C): outcome is 0.
--> 4*EXP(-beta*4000)/6 = 0
I made a mistake. Do you know where?
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Let's be more specific, and it helps to use units. I kept everything in kJ/mol to match the energy values given, which requires converting Boltzmann's constant into this unit. It looks like you may be forgetting the "mole" part of that unit, because your numbers are indicating that even at very high temperatures, all of your population is predicted to be in the ground state.
If it helps you to see if you're doing your calculations right: at 5000 K, my value of beta is 2.41x10-2 moles/kJ.
My value of Q is 11.4 and my probabilities for A, B, and C are 0.53, 0.32, and 0.15 respectively, which add up to 1, as expected (this is always a good internal check).
I suggest setting up a spreadsheet in excel and plotting all of these values (in addition to internal energy and entropy, when you get there) for temperatures spanning 0.00001 K to ~ 5000 K with intervals of 100 K. Also include a really high temperature (10000000000 K) so you can check the infinite temperature limit.
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Thank you, I forgot to divide β by Na.
I followed your steps and I tried to calculate the entropy and the internal energy.
For the entropy, I used following equation:
S = - kb ∑ p ln p
And for the internal energy:
U = - kb T ln Q + TS
I obtain these values for 5000K:
β=0,024054472
Q=11,36427396
P(A)=0,527970377
P(B)=0,319691727
P(C)=0,152337896
S=0,008218395 kJ.K-1.mol-1
U=-59,94847077 kJ.mol-1
Observation:
With increasing temperature, entropy increases while internal energy U becomes more negative.
For the infinite temperature limit I obtain:
S=0,008409282 kJ.K-1.mol-1
U=-122513796,6 kJ.mol-1
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Good, looks like you're calculating the composition correctly. Do you notice anything about the probability values - and Q - in the limit of high temperature? Here's a hint: Q starts at 6 (T = 0) and approaches 12 (T = infinity). PA, PB, and PC approach values of 0.500, 0.333, and 0.167, respectively. This is telling you something about the impact of temperature on the population of available states.
For internal energy, it looks like you're doing something wrong. For one thing, I know what the value is, but even without that: at T = 0, all of the quantum objects are a state with E = 0, so the total energy must be zero. At higher temperatures, some objects are, on average, in states with E > 0. So the energy of your system a T > 0 should have an average E > 0. E(T > 0) > E (T = 0). The energy should be the statistically weighted average of all particles in the system - so, for example, in a two state system that has half population of the ground state and half population of the excited state, the internal energy will be halfway between the energy of the upper state and the lower state. From this information alone, you should be able to tell what it will be for you specific system here.
Actually, you should be calculating your energy (call it U) value first, then your entropy value, which can conveniently be calculated from U.
In terms of Q, the average internal energy, U, for a body of N objects, can be expressed basically as
[tex]U = - \frac {N}{Q} \frac {dQ}{d\beta}[/tex]
(Warning, that's based off memory. You may want to consult your physical chemistry book to make sure.) Really, it's a partial derivative but effectively the same for your purposes here.
For entropy, there are a couple of different ways to express it. I can't remember that equation off the top of my head, or the value I got, but I can check for you tomorrow morning.
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Oh and don't forget you're working in moles here, so N is basically the number of moles, not the number of molecules. Sorry.
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With increasing T, population of state A decreases due to the exciting of the objects to state B and then C. Q doubles, PA is reduced to half. The loss in population in state A is equal to the increase in state B and C. Population of B is (in the limit of high temperature) twice the population of state C.
In other words, with increasing T, objects will be excited to state B and to a lesser extent, to state C.
State A, the ground state, stays half populated in the limit of high temperature.
For the internal energy:
I found the same equation, with N equal to 1.
U is then equal to:
U = -1/Q.-Σi.Ei .e-β.Ei
For S, I obtain:
S = U/T + kb.ln(Q)
Correct?
Thank you very much.
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I calculated U and S and I obtain these values:
For 2000 K:
β=0,060136181
Q=10,53901093
PA=0,569313386
PB=0,298395927
PC=0,132290687
S=0,019928888 kJ.K-1.mol-1
U=0,695267987 kJ.mol-1
For 5000 K:
S=0,02036343 kJ.K-1.mol-1
U=0,776705416 kJ.mol-1
In the infinite temperature limit (10000000000 K):
S=0,020660662 kJ.K-1.mol-1
U=0,833333305 kJ.mol-1
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For the populations it looks like you're doing a good job. Aside from the trends you correctly describe, here are a few other things to notice that might help you understand the meaning of Q.
At low temperature (as T :rarrow: 0), Q :rarrow: 6. And as T :rarrow: ∞, Q :rarrow: 12. Notice that the ground state has a degeneracy of 6, and the total number of states is 12. The partition function is basically a description of how the available particles are partitioned among all the available states, weighted statistically by the temperature. At low temperatures, only the lowest energy states are available, and because of degeneracy, there are 6 of them. Q = 6. At the highest temperatures, the energies between the states become inconsequential (compared to background energy), and all states become effectively equally populated. Another way of looking at this is that the difference in energy between all the available states becomes so small compared to the amount of energy in the system, that they are all effectively degenerate. Since there are a total of 12 states available in the system, the effective degeneracy is 12, which is also the Q value. In a way, the Q value is a representation of the average number of states populated at a given temperature. This is why we determine the probability of populating a certain state as the ratio with Q in the denominator.
IMPORTANT: Many students make the mistake of assuming that at high temperature, only the upper states wills be populated, but this is not the case. At high temperature, all states in the system have equal likelihood of being populated.
You'll also notice that at high temperature, PA = 0.5, PB = 0.333 and PB = 0.167. These are values of 1/2, 1/3, and 1/6, respectively. If at high temperature all the available states have an equal likelihood of population, then the probability of an energy level being populated should be equal to the number of states in that energy level (the degeneracy) divided by the total number of microstates available in the system. States A, B, and C have 6, 4, and 2-fold degeneracy, respectively, given a total of 12 microstates available. So the probability of A, B, and C being populated when there is no "bias" do to the energy (i.e., at high temperature) are easily determined to be 6/12, 4/12, and 2/12... 0.5, 0.333, and 0.167!
For the internal energy, looks like you're still doing something wrong with the calculation, so let's work through it step by step.
If
[tex]U = - \frac {N}{Q} \frac {dQ}{d\beta}[/tex]
And
[tex]Q = 6 + 4e^{-4\beta} + 2e^{-6\beta}[/tex]
And N = 1 mole, then
[tex]U = \frac {16 e^{-4\beta} + 12 e^{-6\beta}}{6 + 4e^{-4\beta} + 2e^{-6\beta}}[/tex]
Is this what you got?
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Thank you for your explanation! Very interesting!
For the calculation of the internal energy at 5000K, I found these results:
S=0,020646648
U=2,192794286
I used an equation found in my course (see attachment). This is the same but I forgot the degeneracy to obtain 16 and 12.
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Great, you now have what I have.
Let me show you something else interesting. I know the equations look complicated and are sometimes hard to decipher, which makes the business of statistical mechanics seem very abstract. But actually, the concepts are rather simple and many things can be calculated, at least at the limits, without using equations. When you do it this way, the meanings of the equations start to become apparent.
Already we've looked at the populations, so let's look at the internal energy value.
The internal energy you're calculating is just the weighted average energy of the system per occupied state. Each microstate has a particular energy value associated with it. So the total internal energy is just equal to the sum of energy of each state multiplied by how many particles are in that state, divided by the total number of occupied states. This is basically what that equation is saying. At low temperature, everything is in the lowest energy state, which happens to have an energy of zero. So the average internal energy of the system per occupied state is also equal to zero: every particle in the system has an energy of zero, so the average energy of all particles is zero. At high temperature, we've already shown that the particles are equally distributed among all possibible states. There are 6 states with energy = 0, 4 states with energy = 4, and 2 states with energy = 6. 6 * 0 + 4 * 4 + 2 * 6 = 28. So there's a total of 28 kJ/mol of energy in the system at infinite temperature, on average, divided by 12 total populated states = 2.33 kJ/mol of average energy per populated state.
So in sum, all the internal energy is expressing is: what is the average energy per occupied state?
Anyway, this was a great problem that illustrates a lot of fundamental concepts in statistical mechanics. I'll have to remember it for the next time (if ever) I have to teach physical chemistry. Thanks for sharing it.
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Great, claryfying information (!), and as you said, it's very logic!!
There was also another question:
"Discuss your result in view of the temperature dependence of chemical equilibria”
What should I say?
I'm really glad you helped me, thank you very much for your time and patience.
Besides this exercise I have 2 other questions, which I solved for 90%. If you are interested I can share it also.
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Well, that part of the question is kind of vague. I'd probably start by saying something about how temperature impacts chemical equilibria, and in particular relate it to entropy. For example, you've shown explicitly here that raising the temperature results in an increase in entropy, so raising the temperature in a system already in equilibrium would tend to shift the equilibrium towards products with higher entropy.
But yeah, I'm not sure exactly what the question is after, here.