From eV to kJ/mol

What is the conversion factor from eV to kJ/mol? To go from eV to J, you have to multiply with the unit charge (formally speaking you insert the e into eV). To go to J/mol you multiply with Avogadro's number. What is the product of the unit charge and Avogadro's number - the Faraday constant. And if you were treated to Chemistry Olympiad at school then, hopefully, you are still able to blurt out its value even when woken up in the middle of the night: 96 485 C/mol. If we divide this by 1000, we get the desired result 1 eV is 96.485 kJ/mol. Or more broadly speaking, you just have to multiply with a factor of 100 if you want to go from eV to kJ/mol. I have never realised that connection before.

Marcus-Levich-Hush equation

Here is an interesting piece of math:

Or in other words show that in the Marcus-Levich-Hush theory, electron transfer stops when you approach zero temperature. Mathematica tells me that this is the case but I could not really show it. The problem has the following type:

It is not an apparent l'Hospital application and I do not really know what else to do. Anyway, I believe Mathematica and it is also the thing you'd expect physically.

In the full quantum picture you actually have a non-zero transfer rate even when approaching zero temperature that comes from nuclear tunneling (as was derived here and is reviewed here).

Interestingely it is quite easy to find derivations of this quantum formula. But I did not find much about the semi-classical formula, that is shown above. Only that it is some kind of application of Fermi's golden rule which apparently gives this prefactor for the rate equation.

Energy

It's been quite a while since I've written anything here. I don't know how exciting the topic is that I picked for coming back. But it's something to think about once. In physical chemistry (which is the common favorite of everyone doing chemistry) you often run across terms of the following form:



Now the question is how do we estimate that if we don't have a calculator at hand? And more important how do we do it with all the different units that are around?

We start out with a handy table that gives us the values of kT (at room temperature) in the different units that we are interested in. (Non-energy units are converted using Planck's constant and the speed of light in the usual way.)

298	K
2.478	kJ/mol
0.592	kcal/mol
0.026	eV
0.00094	Hartree
4.11E-21	J
207.1	cm-1
6.21E12	Hz
48281	nm

Now we can estimate things because we know that:



and



For example we can say that a hydrogen stretch at 4000 cm^-1 will not be active at room temperature because it's energy is much higher than 207 cm^-1 which corresponds to kT. Or we can say that a reaction with a Gibbs free reaction energy below -25 kJ/mol will be almost quantitative.
But to be certain we need the following two formulae which will help to estimate the order of magnitude (ln is the natural logarithm, lg the decadic logarithm)[1]:



and



Now we can do pretty much everything we ever wanted to do.

In the first two examples we see that we are in the order of 10^-4 and the guess was correct.
If we stick to normal modes, we can for example be wondering to what extent a 600 cm^-1 normal mode is in its first excited state. Using the first formula we get:



That's below10% but well above 1% (the exact value is 5.5%). And we know that a mode at 600 cm^-1 is still excited a little bit.

But of course you can also do it for more complex things. For example coalescence in NMR. For an isomerising molecule with activation barrier ΔG coalescence is reached if the following condition is fulfilled:



Δν is the frequency difference between the peaks, e.g. 0.5 ppm of 600MHz which is 300 Hz. kT/h is the frequency term in the Eyring equation, it actually corresponds to my value from above, 6.21E12 Hz, but I think for a different reason [2]. If you divide by it you get:



Apply the logarithm formula and put in the value of kT (or actually RT) and you get:



And that's about the actual value.

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[1] The advantage is that you have to remember only to digits 2.3 and you almost get four digit accuracy compared to the exact value 2.302585...

[2] I would say that the value is chosen because it is a typical frequency of a skeletal mode but it is coincidence that this is close to kT.

Science sometimes does work

When it comes to labs, I don't think I am much of a synthetic chemist. In fact I don't believe it is possible to get more than 50% "theoretical yield" or that someone is actually performing syntheses with more than two steps. However, now is my physical chemistry lab and it is much easier to make me happy there.

What we did is, suck out all the air out of a flask filled with our analyte substance. After closing the valve we measure the pressure in there which is the vapour pressure. Do that at different temperatures and you can see what the phase boundary looks like. First thing that amazed me is that we actually got a nice looking graph.



So what do we do with this? We just need the laws of thermodynamics and the ideal gas law. We rearrange them a little bit according to Mr. Clausius and Mr. Clapeyron and get their equation (I am not showing the proof even though it is kind of cool):



By using the chain rule we can also say:



The T² cancels out and we get:



That means that we get the enthalpy of vaporisation just from the change of vapour pressure with temperature. Who would have thought that? Let's plot the logarithm of the pressure against the reciprocal value of the temperature:



Make a regression, derive, get the enthalpy of vaporisation:



Why is it getting lower? Even that can be explained. At the critical point there is no difference between the phases. Hence there is no enthalpy of vaporisation. You can expect the enthalpy to become smaller when you raise the temperature because you are getting closer to the critical point.

We can even do more. Extrapolate the curve to standard pressure and get the boiling point at 39°C. Now Mr. Trouton tells us that at standard pressure the entropy of vaporisation for every liquid is about 88 J / (mol K). (It seems that this is thermodynamically derived.)

Can we get the entropy just from this curve? Yes we can. We know that Δ_vG = Δ_vH - TΔ_vS. Since we are at equilibrium Δ_vG = 0. Then Δ_vS = Δ_vH / T (we actually used this equation already the other way around for deriving the Clausius-Clapeyron equation).

Divide the enthalpy at the boiling point by the temperature and you get 75,8 J / (mol K). That is almost what Mr. Trouton said.

It took me some time to appreciate thermodynamics. But now I think it is amazing how all those values are related and how much you can do with it. Actually synthetic chemistry is cool, too (on paper at least).