Question

In which of the following equilibria, the increase of pressure over the equilibrium will favor the backward reaction?
A. Decomposition equilibrium of $HI$
B. Formation equilibrium of $S{{O}_{3}}$
C. Decomposition equilibrium of $N{{H}_{3}}$
D. Formation equilibrium of $PC{{l}_{5}}$

Answer 1

According to the rule of Le Chatelier’s Principle: The effect of changing pressure on the position of an equilibrium. Le Chatelier's principle is a perception about chemical equilibria of reactions. It expresses that change in the temperature, pressure, volume, or concentration of a system will bring about predictable and opposing changes in the system so as to accomplish another equilibrium state.

Complete step by step solution: You can increase/decrease the pressure by decreasing/increasing the volume of the gases OR increasing/decreasing the concentration of gases in a similar volume.
For reactions including gases at equilibrium, an increase in pressure causes the equilibrium position to move towards the side with the most modest number of vaporous atoms as shown by the decent image condition for that reaction.
A decrease in pressure of a chemical reaction system including gases, causes the equilibrium position to move towards the side with the larger number of gaseous atoms as demonstrated by the balanced symbol equation for that reaction.
The decomposition reaction of ammonia, the number of moles of product generated is greater than of reaction involved
$2N{{H}_{3}}\rightleftharpoons {{N}_{2}}+3{{H}_{2}}$
Increase in pressure favors the backward reaction and Decrease in pressure favors the forward reaction. As per Le-Chatelier's principle, the increment of pressure on a chemical equilibrium shift in that direction quantity of gaseous molecules diminishes and subsequently in this reaction, increment in pressure favors the backward reaction.

Hence, the correct option is C.

Note: Thus, decomposition of ammonia to hydrogen (and nitrogen) requires an energy input, and since the disintegration reaction is extremely moderate even at generally high temperatures, a catalyst is likewise required. Since the reaction is an equilibrium reaction, it is not possible to totally break down ammonia into hydrogen.