Chemical Thermodynamics 2.5 - Ensemble Energy Example

[@lyddie820]

Is there a link to that graph anywhere?

[@timmy880919]

Is ensemble energy = total energy ? thank you!

[@nkyu3035]

I've got more questions! 1. one of few thing I know about E is E=hv=hc/λ[J] and if I divide both sides by hc, I get E/hc=1/λ[cm^-1]. then why does E[J] and E/hc[cm^-1] represent the same thing as energy? 2. if h is constant why does it have unit? is it usual for constant to have unit?

[@TMPChem]

Physical constants typically have units, as they often represent a specific quantity or the ratio between them. The speed of light gives us the relationship between a photon's wavelength and its frequency. The Boltzmann constant represents the connection between energy and temperature. Planck's constant tells us the scale of action at which quantum mechanics becomes important. If E is in [J] and we divide by h [J * s] we get E[1/s] = E[Hz]. Since h is a constant, Hz is another type of unit with which we can represent energy (assuming the energy is equal to a photon of that frequency). If we divide E[J] by hc [J * s * (cm / s)] = [J/cm^-1] we get E[cm^-1], where the energy is the energy of a photon with that given inverse wavelength. Since h and c are both constants, there is always a direct proportional relationship between these systems of units.

[@nkyu3035]

Thank you again. i think i'm getting a clue. but what i know about unit conversion is that for example when you have 200cm, you multiply it by 1m/100cm and as a result you get 2m. the reason i'm confused is that 1m/100cm is basically just 1. h or hc is not 1. is it because 2.9977*10^10 cm is the same thing with 1 seconds for a photon? / you've drawn 3 layers of line as E1, E2 and E3 and figured out . and i'm not sure what specific situation this is. is it like 3 orbitals and 1 electron? / i'm asking you a number of question. do you see it as shortage of background knowledge? like it's too difficult for me?

[@TMPChem]

Not necessarily any background shortage. It's just a different approach to the problem in a more technical language that isn't often taught at lower levels, but behind the abstractions really isn't much different than what you already know how to do. The bars just represent individual allowed states of the system. The vertical locations of the bars indicate the energy of these states. Different states have a different probability of being occupied, and the average energy is a weighted average of the energy of every state, where the weights are the probability of each state. Using energy units of cm^-1, Hz, J, etc. are all based on the fact that the speed of light and Planck's constant are constant, thus the frequency and/or inverse wavelength of a photon are directly proportional to the energy of that photon. We can thus use it as an energy scale to discuss units. It's often very convenient to do so in spectroscopy, where peaks are already labeled in terms of these units, and the peaks represent the energy of a photon which produced the transition from the initial to the final states which the peak is measuring.

[@nkyu3035]

appreciate the comment^^ I'll give it a more thought with your help

[@nkyu3035]

you divided K_b by hc and substituted this for K_b inside exponential. how does this make sense?

[@TMPChem]

The exponent always contains three things: energy, temperature, and the constant to convert between them. Temperature almost always has units of Kelvin. Energy has a variety of common units, depending on what's convenient for the system at hand, whether J, kJ, kJ/mol, kcal/mol, Hz, cm^-1, etc. Dividing k_b by hc just changes it from being in units of J/K to units of cm^-1/K, as for many molecular systems cm^-1 is a convenient choice of energy units.

[@nkyu3035]

so unit of E in the exponent was cm^-1? then it's same as multiplying -E/kT by hc/hc, unit of E is Joule here, resulting in changing of the unit of E from Joule to cm^-1. right?

[@TMPChem]

Yes, the three energy levels are labeled in units of (cm^-1), as is in the right most column of the final table. Therefore, in order to make the exponent unitless we must use k_b in units of cm^-1/K, which we do by dividing it by hc.

[@nkyu3035]

i feel silly o_o;; Thank you for the reply^^