Chem Question.

[sketchy]

"The CO molecule absorbs a 2134cm^-1 infrared photon at room temperature, and is promoted from the lowest (v=0) to the next highest (V=1) quantized vibration energy level, where v is the vibrational quantum number. In general, no degeneracies are involved with vibrational energy states. What is the energy difference between these two vibrational states (in J) and what is the relative populations of these two vibrational states at room temperature?"

So we did this last year but I can't remember how. I'm thinking the rydberg equation, maybe? But isn't that only applicable to H? Any help would be appreciated.

:rant:

[Miluardo]

I have no fucking clue what you just said.

[Mantikore]

im not sure, but i would guess if they said there were no degeneracies involved it would just mean to calculated the energy difference between the two states, in which case you know that its a photon with a certain wavelength.

wavelength * frequency = speed of light constant..................solve for frequency.
planck constant * frequency = energy......................solve for energy

i dunno. im just guessing about what i already know from physics, which is limited

[DJ Meaty Cheeks]

It sounds like something that might be tabulated in a handbook, like the CRC Standard Handbook of Chemistry and Physics.

[Ivy Mike]

Couldn't you just figure out the energy of the photon?

[The Jitterskull]

Sounds like a physical organic chemistry question. Not my field. I can interpret IR for you though if you want.

[Zip]

Degeneracies are probably referring to equivalent scissoring/bending vibrations which don't occur in the CO molecule. We know it's IR active from the question so we can actually approximate the vibration frequency (wavenumber) v for stretching vibrations using Hooke's Law:

v = [1/(2πc)][F(m1+m2)/(m1m2)]^(1/2)

Where c is the speed of light, F is a force constant for the bond, the triple bond in CO can be approximated by F = 15e5 dyne/cm, and m1, m2 are the masses of C and O in grams. Calculate this out and you get v = 1925 1/cm which is a reasonably close approximation to the 2134 value in the question.

Since you are already given this value though, you don't need to do much more to calculate the energy which is just E = hcv where h is the Planck constant.

E = (6.6e-34 J)(3e10 cm/s)(2134 1/cm) = 4.2e-20 J

You can probably get a more reasonable value working in units of kJ/mol.

I don't remember how to do the second half of the question but hopefully this helps with understanding.

[KaBoom]

wavelength * frequency = speed of light constant..................solve for frequency.
planck constant * frequency = energy......................solve for energy



Pretty sure you'll be needing these at some point...