Question: n the manufacture of ammonia by Haber’s process, \({N_{2(g)}} + 3{H_{2(g)}}\overset {} \leftrightarr...

Question

n the manufacture of ammonia by Haber’s process, ${N_{2(g)}} + 3{H_{2(g)}}\overset {} \leftrightarrows 2N{H_{3(g)}} + 92.3kJ$ . Which of the following conditions is unfavorable?
A) Reducing the temperature
B) Removing ammonia as it is formed
C) Increasing the temperature
D) Increasing the pressure

Answer 1

Solution

The hint is given in the question itself as there is an evolution of heat which means this reaction is an exothermic reaction. Hence, one can apply the concept that which factor can affect the exothermic reaction which can lead to an answer

Complete step by step answer:

First of all we will learn about Haber's process, this method is one of the most yield producing and successful industrial procedures which is adopted for the production of ammonia.
The Haber’s process for the manufacture of ammonia involves the reaction of nitrogen and hydrogen. During this reaction the $+ 92.3kJ$ is used which shows the reaction is an exothermic type of reaction in which there is a release of energy. In Haber's process the number of moles of the reactants are greater than the number of moles of the product.
According to Le Chatelier’s principle, the production of ammonia is done at conditions of high pressure and low temperature. During the commercial production of ammonia, ammonia is continuously removed as a product and separated out as it is produced in the reaction. Removing the product ammonia gives more nitrogen and hydrogen to react with each other according to Le Chatelier’s principle.
Therefore, after analyzing all the facts, In the manufacture of ammonia by Haber’s process increasing the temperature is the only unfavorable condition

So, the correct answer is Option C.

Note:
The Haber’s process is generally being carried out at the pressure which is in between $200atm$ and $400atm$ temperature of ${500^0}c$ . When the temperature is increased in an equilibrium exothermic reaction then the reaction equilibrium will shift to the left side which is the reactant side and making the reaction unfavorable.