Question

Then factor for \[N{H_3}\] in the reaction: \[{N_2} + 3{H_2} \to 2N{H_3}\]
A. 2
B. 3
C. 1
D. 4

Answer 1

Hint: Chemical reactions are the chemical changes in which reactants transform into products by bond-making or bond-breaking or both between different atoms or molecules. When such a reaction is expressed in terms of symbols and formula, it gives us a chemical equation. \[N{H_3}\]which is called ammonia is a Lewis base.

Complete step by step answer:
The Haber process is used for producing ammonia. The reaction involved in this process can be given as follows: \[{N_2} + 3{H_2} \rightleftharpoons 2N{H_3} + Q\,kcal\] .
Now, the n factor is the number of \[{H^ + }\] accepted by base ammonia. In this case the n factor value 1.

So, the correct option is C.

Additional information:
Now for understanding the conditions that are the most favorable for the Haber process, we need to understand the changes that are involved with this process.
1.The formation of ammonia using Haber’s Process involves the release of energy. This means that the formation of ammonia is an exothermic process.
2.In Haber’s Process, 4 moles of reactants form 2 moles of product. Hence the number of moles of the product formed is less than the number of reactants.
Now to favor the above-mentioned conditions, we need to provide the following conditions:
1.For an exothermic process to proceed at a faster pace, the surrounding temperature must be low. This is in accordance with the laws of thermodynamics.
2.In a reaction where the number of moles of the reactants is greater than the number of moles of the product, the most favourable condition to complete the reaction is increasing the pressure of the entire setup.
Hence, in the manufacture of ammonia by Haber’s process, the condition which would give maximum yield is low temperature and high pressure.

Note: Electron density increases on the nitrogen atom due to the +Inductive effect and therefore makes it more basic whereas the +R effect involves the lone pair of Nitrogen in conjugation making it not available for further reaction and thereby reduces its basicity.
In the aqueous phase, if R is a methyl group then the order of basicity is secondary amine> primary amine > tertiary amine but if R is any group other than methyl group, then the order is secondary > tertiary > primary.