Question

A ${N_2}(g) + 3{H_2}(g)\underset {} \leftrightarrows 2N{H_3}(g) + heat$
The reaction above is carried out in a sealed vessel. Which changes would both shift the equilibrium position to favour in the formation of products?
A. Heat the vessel, increase the pressure
B. Cool the vessel, increase the pressure
C. Heat the vessel, decrease the pressure
D. Cool the vessel, decrease the pressure.

Answer 1

The above process is Haber’s process for the preparation of ammonia. The favourable conditions for ammonia are – high pressure, low temperature, and use of catalyst.

Complete step-by-step answer:
The given reaction is the very famous Haber’s Process used for the preparation of ammonia. Nitrogen can be extracted from air while the hydrogen can be produced by reacting methane with steam.
The raw materials here are nitrogen and hydrogen. The conditions of any reaction is determined by Le Chatelier’s Principle. The given reaction is exothermic in nature i.e. it liberates heat.
Hence any increase in temperature will shift the reaction in a backward direction. We need to cool the vessel to shift it in a forward direction.
High pressure will push the reaction in a forward direction. It will increase the number of moles on the product side hence increasing the yield.
Lower temperature and high pressure are favourable for this reaction.

Hence, the correct option is B.

Note: The conditions for above reaction is based on Le Chatelier’s Principle which states that if any system is in equilibrium and if any change is made in any of the conditions, then the system corresponds in a way to counteract the change.