Haber-Bosch process ammonia yield?
I don't understand why the yield of NH3 is highest at low temperature and high pressure.
As far as I know, according to the RGT rule, such reactions are very slow at low temperatures.
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I think it has an amino group with an electronegativity difference of 0.8, and an OH group in the carboxyl group, which is also polar? Furthermore, the main chain is monoatomic.
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You are perfectly right, but there is a conflict between two rules:
Displacement of equilibrium (Lavoisier) against RGT rule.
A lower temperature would increase the yield, but as a result the speed becomes unprofitably low. We must therefore take a compromise.
m.f.G.
present
In equilibrium reactions, only one reaction at high temperatures can be favored. In this case, the formation of NH3 is exothermic, i.e. lower temperatures are cheaper. This collides with the high activation energy, which is why a catalyst is necessary.
Generally, the equilibrium at T is low and p is highest at NH3, but at the same time it also takes forever until this equilibrium is actually established.
So if you have a lot of patience, you get the most product from the starting materials. However, if the time is considered to be constant, higher yields are also obtained at higher temperatures.