On a large scale, ammonia is prepared via the Haber process:

$$\ce{N2(g) + 3H2(g)->2NH3(g)} \qquad \Delta _\mathrm{f}H^\circ = -46.1~\mathrm{kJ \cdot mol^{-1}}$$

The optimum conditions for the productions of ammonia are a pressure of $200~\mathrm{atm}$ and a temperature of about $700~\mathrm{K}$.

The process obviously is exothermic and $700~\mathrm{K}$ is, by no means, a low temperature. Shouldn't the temperature be much lower for optimum production of ammonia?


3 Answers 3


As others have pointed out, it is purely kinetics, but you may still wonder, why.

For a reaction to actually occur (in both directions) and thus for an equilibrium to be reached, you need to overcome the activation energy. In the case of the Haber-Bosch process, this involves breaking the highly stable $\ce{N#N}$ triple bond. Even with the catalysts used, the energy required to break apart $\ce{N2}$ is still enormous. Therefore, a lower temperature may give a better yield of ammonia theoretically (i.e. based on equilibrium and Le Châtelier considerations) but the reaction speed would be a lot slower.

Even if you consider a batch-wise process of generating ammonia (which, as orthocresol points out, isn’t the case), it is more efficient to run two batches in half the time for two sets of $15~\%$ than to run a single batch for twice the time to get an overall yield of maybe (note: This yield is a ballpark estimate) $25~\%$ — with a batch size where $1~\%$ yield is equivalent to $100~\mathrm{kg}~\ce{NH3}$, two runs at higher temperature give $3~\mathrm{t}\ \ce{NH3}$ and one run at lower temperature $2.5~\mathrm{t}\ \ce{NH3}$ in the same time frame.

  • $\begingroup$ I think that last part may not necessarily accurately reflect the numbers. If you assume the percent yield is of the form $B e^{C/T}$, the time to run a batch is $A e^{E/T}$, and the desired final percent yield is some given number $p$, then the whole process takes time $\frac{p A e^{E/T}}{B e^{C/T}}$. The dependence here on $T$ depends on how $E$ and $C$ compare to one another. (I know my formulae are oversimplified.) $\endgroup$
    – Ian
    Jan 6, 2016 at 19:20
  • $\begingroup$ @Ian The last part most certainly does not accurately reflect numbers, since it is a ballpark estimate. If you have good numbers, do please share! $\endgroup$
    – Jan
    Jan 6, 2016 at 19:21
  • 1
    $\begingroup$ Sorry, I don't mean exact numbers, I mean the way the calculation would go if you had the exact numbers. There is a nontrivial competition between the yield per run and the time to make a run. Specifically, if $\Delta H$ is sufficiently negative compared to $E_a$, then (at least based on my heuristics above), the best thing to do would be to make a single very cold run. (Of course, once we start talking about "very cold", the very assumptions that make these heuristics make sense start to break down.) $\endgroup$
    – Ian
    Jan 6, 2016 at 19:24
  • $\begingroup$ @Ian Okay … All I know about the Haber-Bosch process is that it has been practised for about a century now and that the conditions used are probably already optimised to a tenth of a degree centigrade … So as I said, I was really just throwing numbers out there to show that going for highest batch yield is not necessarily a good thing ^^ $\endgroup$
    – Jan
    Jan 6, 2016 at 19:27
  • $\begingroup$ So let me summarise this. If we take the temperature higher, then according to Le Chatelier's principle, equilibrium will shift backwards, though the reaction rate increases. And on making the temperature lower, according to the Le Chatelier's principle, equilibrium will shift forward, but the rate will be too slow, right? $\endgroup$ May 25, 2017 at 10:17

The answer mainly has to do with kinetic considerations, as aml points out. I want to point out another thing.

In a typical industrial setting, you don't just mix the $\ce{N2}$ with the $\ce{H2}$ at a certain $T$ and $p$, collect the ammonia, and throw the unused reactants away. That would be horribly inefficient.

Greenwood & Earnshaw, Chemistry of the Elements (2nd ed) writes:

The gas leaving the catalyst beds contains about $15\%~\ce{NH3}$; this is condensed by refrigeration and the remaining gas mixed with more incoming synthesis gas and recycled.

The "synthesis gas" that they speak of is a mixture of $\ce{H2}$ and $\ce{N2}$ that is freshly produced via a series of reactions.

So, at first glance, the "yield" is absolutely rubbish. But reusing the reactants allows you to get high overall yields (albeit with more steps).


While it is true that the Haber Process would be much more efficient at a lower temperature it's carried out at a higher one because it happens much faster at the higher temperature and while the industry wants an efficient process they also need it to happen fast enough to be commercially viable.

So you are correct but it's an example of practical realities coming before theoretical ideals. The use of the word optimum is misleading in this context


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