[Mark S 2014]

Q. In a thermally equilibrated sample, how would you expect the fraction of vibrationally excited ²H⁷⁹Br molecules to compare with the fraction of excited ¹H⁷⁹Br molecules?

I can't seem to work out this question. I thought that it may be simply something to do with the natural abundance of each isotope but I'm not convinced that's right. If anyone could show me in the right direction I would be very grateful.

Mark

[Corribus]

Do you know about partition functions?

I have written some posts at these forums on the topic.

http://www.chemicalforums.com/index.php?topic=66033.msg238397#msg238397
http://www.chemicalforums.com/index.php?topic=76141.0
http://www.chemicalforums.com/index.php?topic=77155.msg281359#msg281359

[Mark S 2014]

Hi Corribus, thanks for the quick reply.

No, sorry, I have never heard of partition functions and I don't think I am expected to know of these as they are not mentioned in either text book or lecture notes.

I'm thinking about maybe just using the boltzmann equation to work out both of the populations and simply using the results as my answer. I'll do this now and I will post back when I have finished.

Thanks again,

Mark

[Mark S 2014]

As it was thermally equilibrated I took the temperature to be 298 K, I worked out the populations using the boltzmann equation and got the HBr population to be 4.36 x10^-6 and the DBr to be 1.51 x10^-4, so in other words the upper state in the DBr is much more populated than that of HBr.

Does that sound right ? Or is this just nonsense ?

[Corribus]

First, it doesn't sound from the way the problem is worded that you are actually expected to calculate anything, which seems like it may be beyond the scope of the material. So let's not worry about numbers. Let's just worry about whether excited vibrational states of HBr or DBr are expected to be more thermally populated at a given temperature ("thermally equilibrated" isn't implying a particular temperature, so your answer should be generalized to any temperature). You concluded that the "upper state of DBr is much more populated than that of HBr". Let's see if that makes conceptual sense.

Thermal population of excited-states happens because there is enough background thermal energy to promote a statistical portion of (identical) quantum systems into one or more upper states. In the limit of absolute zero temperature, no thermal population of excite-states occurs: the probability of any given system being in the lowest energy state is exactly 1. At higher temperatures, there is a nonzero probability of any upper state being populated at a given time. The probability of an upper state being populated is correlated to how much energy is required to get there.  If the difference between the lowest state and the next lowest state is smaller, there will be a higher probability of the next lowest state being thermally populated, no matter what the temperature is. The actual temperature only impacts what the magnitude of that probability is. (I.e., the genearl trends are the same.)

So, the question to answer is this: of HBr or DBr, which do you expect to have smaller energy separation between the lowest energy (ground) vibrational state and the first excited state? If it is HBr, then HBr molecules have a higher probability of being thermally, vibrationally excited at any temperature. And likewise for DBr.

Given this information, is your conclusion correct? (Again, forgetting about numbers.) If you don't know how to go about answering this question, start with the energy formula for a quantum harmonic oscillator, and make some assumptions about the force constant and reduced mass.