Understanding the use of Le Chatelier's principle in the industrial production of ammonia

I understand why the reaction


$$\ce{N2 + 3 H2 -> 2 NH3} \qquad \Delta H = \pu{-92.2 kJ mol-1}$$


isn't carried out in the lowest possible temperature (even though it seems that it should be as it favors the right side of the reaction). As the rate of reaction would be very slow, the amount of $\ce{NH3}$ produced would be lesser than the amount that would've been produced if the reaction was carried out at the optimum temperature in a given time span.


Why isn't then the reaction carried out at the highest possible temperature since the rate of reaction would be very high and the yield would be very high consequently as well?





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> Why isn't then the reaction carried out at the highest possible temperature since the rate of reaction would be very high and the yield would be very high consequently as well?
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The rate constant would be high, but that does not mean that the net forward rate is high. If you start without product, the initial rate would be high, but would drop to zero very quickly as equilibrium is reached. The equilibrium concentration of product would be very low at high temperature, as you said, so the yield would be very low.


If you had a very cheap and highly efficient process to remove product directly from the reaction vessel, you could run the reaction at high temperature. However, the product removal is slow and expensive (cool down the reaction mixture, remove ammonia, and reheat the remaining reactant to feed back into the reactor). Because this is costly, the reaction is run at a compromise temperature where at equilibrium, about 15% of the reaction mixture is product. 


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I think the simplest way to conceptualize this would be to pretend the heat produced by the reaction is a "product" (it's not really correct to say that, but it may help you understand it). Since the reaction is exothermic, you could rewrite the reaction as: N2 + 3H2 ⟶ 2NH3 + Heat


If you raised the temperature, you'd add heat, which is a "product". As per Le Chatelier's Principle, increasing the concentration of a product shifts the reaction equilibrium left, which is why the reactants predominate at a higher temperature.


If you're familiar with Gibbs Free Energy, I think that would be a better (and more accurate) way to look at it. The forward reaction converts four moles of gas (3 H2 + 1 N2) to two moles of gas (2 NH3). This reduces the total entropy of the system.


Now, we know that: ΔG = ΔH - TΔS


We also know that since entropy is lowered by the forward reaction, ΔS < 0, which makes the term -(TΔS) positive. Lastly, we know a reaction will be spontaneous only if ΔG < 0.


If the temperature were very high, the (TΔS) term would also be very high, making ΔG positive. For the reaction to be thermodynamically possible, the temperature needs to be very low, otherwise the decrease in entropy would offset the favorability added by the loss of enthalpy. This is why there's an "optimal temperature", and why you can't just try and maximize the temperature during ammonia synthesis.


There may be some other ways to look at it, but that's how I would conceptualize it.


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