
In the manufacture of ammonia by Haber’s process, the condition which would give maximum yield is: \[{N_2} + 3{H_2} \rightleftharpoons 2N{H_3} + Q\,kcal\]
A.High temperature, high pressure and high concentrations of the reactants
B.High temperature, low pressure and low concentrations of the reactants
C.Low temperature and high pressure
D.Low temperature, low pressure and low concentration of \[{H_2}\]
Answer
589.2k+ views
Hint: In this question, we have to first determine the type of changes that take place because of this reaction. These changes can be both physical and chemical in nature. Then the conditions that promote these changes must be incorporated to get the maximum yield in any given reaction.
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 for understanding the conditions that are the most favourable for this process, we need to understand the changes that are involved with this process.
The formation of ammonia using Haber’s Process involves the release of energy. This means that the formation of ammonia is an exothermic process.
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 favour the above-mentioned conditions, we need to provide the following conditions:
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.
In a reaction where the number of moles of the reactants is greater than the number of moles of 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
Hence, Option C is the correct option.
Note:
Another way to get the maximum yield is the introduction of a catalyst. In Haber’s Process, iron can be used as a catalyst. It helps to achieve an acceptable yield in an acceptable time.
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 for understanding the conditions that are the most favourable for this process, we need to understand the changes that are involved with this process.
The formation of ammonia using Haber’s Process involves the release of energy. This means that the formation of ammonia is an exothermic process.
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 favour the above-mentioned conditions, we need to provide the following conditions:
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.
In a reaction where the number of moles of the reactants is greater than the number of moles of 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
Hence, Option C is the correct option.
Note:
Another way to get the maximum yield is the introduction of a catalyst. In Haber’s Process, iron can be used as a catalyst. It helps to achieve an acceptable yield in an acceptable time.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

10 examples of friction in our daily life

