# What is the activation energy barrier of uncatalyzed nitrogen fixation?

I'm interested in the activation energy ($$E_\mathrm a$$) of uncatalyzed aqueous nitrogen fixation: $$\ce{3 H2 + N2 -> 2 NH3}$$ Biology performs this reaction with the nitrogenase enzyme at room temperature. Humans have come up with ways of doing this reaction with a metal catalyst like rubidium at ~300 °C. I am not interested in the activation energy of nitrogen fixation when catalyzed by biology or by industrial chemists. I want to know the activation energy of the uncatalyzed reaction.

The most luck I have had is with a book titled "Ammonia: Principles and Industrial Practice", which states, "According to estimates[99], the homogenous reaction of nitrogen with hydrogen to form ammonia in the gas phase requires an activation energy of 230–420 kJ/mol." Reference [99] is cited as "W. G. Frankenburger, Catalysis 1954-1960 3 (1955) 171" which I haven't found a copy of yet. I would prefer to have a measured (not estimated) value that is a bit more precise.

Any help would be much appreciated!

• For an activation energy, you first need a sensible mechanism under some defined conditions. I don't think there is one. The thermal decomposition is sure to take another route. You can of course make assumptions, but I strongly doubt the number has any scientific value. Sorry. I'd be glad to get convinced of the opposite. – Karl Jul 21 at 14:01
• Side note: 400 kJ/mol is equivalent to a temperature of about 50 kK. That's right, kilo Kelvin. – Karl Jul 21 at 14:05
• @Karl I understand what you mean that a specific mechanism is needed to define an activation energy. I'm not sure what the mechanism would be. I would imagine though that at ~2000 K it would still be thermodynamically favorable for nitrogen and hydrogen to make a fair bit of ammonia, and some small fraction of the molecules that have enough energy would react via some mechanism. I am interested in the activation energy of that scenario. – nicholaswogan Jul 22 at 0:05
• I think the point is you won't get any ammonia at all. You'd need a three- or four-particle reaction, and if those 400kJ/mol are anywhere near a realistic value, at 2000K you will get just nothing. I'm sure you could do it at extreme pressures, but that's likely a different reaction then. – Karl Jul 23 at 16:40
• What are you actually interested in, i.e. why are you asking this? The number itself seems not so terribly interesting, perhaps we can string up sth. else? – Karl Jul 23 at 16:44