Question

A mixture of nitrogen and hydrogen is at an initial pressure of $200atm$ . If $20\% $ of the mixture reacts by the times equilibrium is reached, the equilibrium pressure of the mixture is:
A. $160atm$
B. $180atm$
C. $170atm$
D. Data insufficient

Answer 1

Hint:According to the Le Chatelier's Principle a system at equilibrium will adjust to relieve stress when changes in the concentration of a reactant or product, the partial pressures of components, the volume of the system, and the temperature of reaction takes place. Equilibrium shift also relates to the number of moles.

Complete step by step answer:
According to the question, reaction of hydrogen and nitrogen will takes place as:
$3{H_2} + {N_2} \rightleftharpoons 2N{H_3}$
It is clear that, in the reactant side there are total $4$ moles that are three moles of hydrogen and one mole of nitrogen.
But on the product side there are a total $2$ moles of ammonia only. So, the difference between number of moles in this reaction is $2 - 4 \Rightarrow - 2mol$
Now as the question said that $20\% $ of the mixture reacted that is $40atm$ then the pressure of ammonia formed will be only $20atm$ .
That means there will be a drop of pressure by $20atm$ .
So the equilibrium pressure of the mixture will be $(200 - 20)atm \Rightarrow 180atm$

Hence, option B is correct.
Note:
When a system at equilibrium and change in pressure takes place, then the equilibrium of the system will shift to offset the change and determine a new equilibrium. The system can be shifted in two ways : either towards the reactants or towards the products. When there is an increase in pressure, the equilibrium will shift towards the side of the reaction with lesser moles of gas. When there is a decrease in pressure, the equilibrium will shift towards the side of the reaction with more moles of the gas.