equilibrium - Why is the Haber process carried out at such high temperatures?

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?

Answer

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.