Question

The vibrational constant for the (non-degenerate) inversion or “umbrella” motion of NH3 is 950 cm-1. Calculate the percentage of NH3 molecules in a sample at 373 K that will be in the v = 1 state of the inversion mode.

Answer 1

2.50% of the ammonia molecules are in the v=1 state of the inversion mode at 373K.

To find the percentage of ammonia molecules first I solved for the vibrational partition function, q vibrational.

I did that by dividing one by one minus "e" raised to the negative vibrational constant (950/cm) divided by kB (0.6950 cm^-1/K) times temperature (373K) which equalled 1.026.

Finding that answer gives you the total potential energy of one particle for all the positions.

Then I found the fraction of the sample in the v=1 state from the canonical distribution by raising "e" to the negative vibrational quantum number (v=1) times the vibrational constant (950/cm) divided by kB times temperature (373K) and dividing that by q vibrational to get the fraction of the sample in the v=1 state. Lastly, I multiplied the fraction by 100 to get the answer into a percentage