# Percentage yield of Haber process

The yield of a single pass of the Haber process is around $$\pu{10\%}$$ to $$\pu{15\%}$$ of ammonia when operating at $$\pu{200 °C}$$ and $$\pu{450 atm}.$$ However, most graphs show that at this condition will yield about $$\pu{35\%}$$ instead (see graph below). Where did the value of $$\pu{35\%}$$ come from? Why does it differ so drastically from the $$\pu{10\%}$$ to $$\pu{15\%}$$ range?

Chemical reactions may be performed in sequences (equally known as batches), or continuously. The two differ e.g., by residence time of the reagents and products in this container.

The former form is like what you are used to, e.g., in the kitchen. You charge your pot with potatoes, close it, cook them, open it; done. In a flow reactor (the later form), while the reaction is ongoing, reagents enter and products leave the reactor. If the reagent's time of residence is less than the time necessary to establish the thermodynamic equilibrium $$\ce{N2 + 3 H2 <=> 2 NH3}$$, then the chemical yield obtained will be below of what could be obtained.

Adjusting the time of residence depends on multiple parameters. Some of them chemical kinetics of the reaction(s) in question, other ones e.g., about cost of production (e.g., management of pressure and temperature prior/after the reactor, and «time is money»). In the case of the ammonia synthesis, the reactor is a (massive) tube, in other cases it may be a stirred vessel or a cascade of them. Each approach has its pros and cons. Welcome to an important aspect of chemical engineering.

– Tham
Nov 11, 2021 at 12:34

It is the question of equilibrium versus nonequilibrium yield and production rate. Maximizing the ammonia output flow is not the same as maximizing the reaction yield.

Approaching the equilibrium yield would lead to just minimal production. OTOH, the top production rate would be near zero reaction yield and unlimited input flow, what is not technically applicable.

$$\ce{NH3}$$ output flow is proportional to the reaction yield and the $$\ce{H2 + N2}$$ stoichiometric input flow.

If there are 2 variants of production, the one with a lower yield but higher production rate is the way to go. 6 times higher input flow of $$\ce{N2}$$ + $$\ce{H2}$$ with the 1/3 reaction yield gives 2 times higher output $$\ce{NH3}$$ flow.

$$\ce{N2}$$ + $$\ce{H2}$$, that are left, are recycled for another catalyst pass.

• ... and maximizing the profit is not the same as maximizing the product output either. Nov 12, 2021 at 8:47
• @fraxinus Sure, it is implicitly included in my answer. As for the component flow rates, there is the flow with the minimum value of cost/output ratio. Nov 12, 2021 at 8:57