Optimizing Ammonia Production: Calculating Partial Pressures and Reactant Ratios

[yolo123]

An ammonia plant operates at close to 700K and a Kp of 3.5 x 10-4 employing the stoichiometric ratio of 1:3 N22. If the equilibrium partial pressure of NH3 is 50. atm, calculate the partial pressures of the reactants and the total pressure.
The suggestion is made that, since the [NH3] is cubed in the reaction quotient, the plant could produce the same amount of ammonia if the reactant ratio were 1:6 - and at a lower pressure and increased profitability. Calculate the partial pressure of each reactant and the total pressure under these new conditions assuming PNH3 is still 50. atm at equilibrium.
Is the suggestion valid?


Since when is NH3 cubed?
What do we have to do? I don't understand anything. What is the 1:6 ratio? How do we prove improved profitability?

 

[abitslow]

I agree, cubed seems wrong.

 

[epenguin]

An ammonia plant operates at close to 700K and a Kp of 3.5 x 10-4 employing the stoichiometric ratio of 1:3 N22. If the equilibrium partial pressure of NH3 is 50. atm, calculate the partial pressures of the reactants and the total pressure.
The suggestion is made that, since the [NH3] is cubed in the reaction quotient, the plant could produce the same amount of ammonia if the reactant ratio were 1:6 - and at a lower pressure and increased profitability. Calculate the partial pressure of each reactant and the total pressure under these new conditions assuming PNH3 is still 50. atm at equilibrium.
Is the suggestion valid?


Since when is NH3 cubed?
What do we have to do? I don't understand anything. What is the 1:6 ratio? How do we prove improved profitability?

From context the the 1:6 ratio should be the N₂₂ ratio, which previously was 1:3.

I'd suggest you first try to answer the understandable and solvable questions.

When answers are evaluated by humans not computers you should normally get credit if you state the question you think you are answering and give an intelligent answer.

What is more economical could depend on the relative costs of the two reactants, or it might be independent, hard to say beforehand. You need to write some equations and attempt some calculations first.