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Offline canonicalpartitionstudent

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Molecular Physical Chemistry
« on: August 12, 2014, 11:07:25 AM »
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."

Offline Corribus

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Re: Molecular Physical Chemistry
« Reply #1 on: August 12, 2014, 11:28:32 AM »
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?
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline canonicalpartitionstudent

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Re: Molecular Physical Chemistry
« Reply #2 on: August 12, 2014, 11:41:31 AM »
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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Re: Molecular Physical Chemistry
« Reply #3 on: August 12, 2014, 12:03:45 PM »
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.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline canonicalpartitionstudent

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Re: Molecular Physical Chemistry
« Reply #4 on: August 12, 2014, 12:40:01 PM »
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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Re: Molecular Physical Chemistry
« Reply #5 on: August 12, 2014, 12:46:13 PM »
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?
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline canonicalpartitionstudent

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Re: Molecular Physical Chemistry
« Reply #6 on: August 12, 2014, 12:59:08 PM »
Fermi-Dirac statistics?

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Re: Molecular Physical Chemistry
« Reply #7 on: August 12, 2014, 01:55:45 PM »
Have you learned about the partition function yet?

Click here.
« Last Edit: August 12, 2014, 03:31:06 PM by Corribus »
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline canonicalpartitionstudent

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Re: Molecular Physical Chemistry
« Reply #8 on: August 12, 2014, 02:24:38 PM »
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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Re: Molecular Physical Chemistry
« Reply #9 on: August 12, 2014, 03:48:00 PM »
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!
« Last Edit: August 12, 2014, 04:06:42 PM by Corribus »
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline canonicalpartitionstudent

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Re: Molecular Physical Chemistry
« Reply #10 on: August 12, 2014, 04:52:08 PM »
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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Re: Molecular Physical Chemistry
« Reply #11 on: August 12, 2014, 05:06:02 PM »
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.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline canonicalpartitionstudent

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Re: Molecular Physical Chemistry
« Reply #12 on: August 12, 2014, 06:30:37 PM »
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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Re: Molecular Physical Chemistry
« Reply #13 on: August 12, 2014, 06:49:12 PM »
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.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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Re: Molecular Physical Chemistry
« Reply #14 on: August 12, 2014, 07:00:59 PM »
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.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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