
The gas phase reaction $$2N{O_{2(g)}} \to {N_2}{O_{4(g)}}$$ is an exothermic reaction. The decomposition of $${N_2}{O_{4(g)}}$$, in equilibrium mixture of $$N{O_{2(g)}}$$ and $${N_2}{O_{4(g)}}$$, can be increased by:
(A) Increasing the pressure
(B) Addition of inert gas at constant pressure
(C) Lowering the temperature
(D) Addition of inert gas at constant volume
Answer
600k+ views
Hint:
- Le Chetelier’s principle can decide the flow of the reaction equilibrium. Gas phase reactions are sensitive to pressure changes as well as temperature changes.
Complete answer:
The reaction given can be written as below where Delta represents heat evolved.
$$2N{O_{2(g)}} \to {N_2}{O_{4(g)}} + \Delta $$
Now the decomposition reaction of $${N_2}{O_{4(g)}}$$ can be written as below.
$${N_2}{O_{4(g)}} \rightleftharpoons 2N{O_{2(g)}}$$
As the earlier reversed reaction was exothermic, this reaction is endothermic.
Now, let’s see all the options to find the true one.
- When we increase the pressure of the reaction, according to Le Chetelier’s principle, equilibrium will shift towards the side which has less number of molecules. So, it will shift towards the left. So, option (A) is incorrect.
- When we add an inert gas to the reaction chamber at constant pressure, the equilibrium will then shift towards the side having more number of molecules.
- When we lower the temperature, as this reaction is an endothermic reaction, according to Le Chatelier’s principle, the equilibrium will shift towards the left side. So, this is not the correct sentence.
- When we add inert gas in the reaction chamber at constant volume, the pressure will increase and then equilibrium will shift towards the side having less number of molecules. So, this option is also not true.
So, option (B) Addition of inert gas at constant pressure is the correct answer.
Additional Information:
-Le Chatelier’s Principle refers to the change in the reaction equilibrium when any of the parameters respective to reaction is changed. This principle is also referred to as the law of equilibrium.
Note:
Reaction given in the question is not the reaction they have asked about, actually they have asked about its reverse reaction. As we reverse the reaction, its exothermic or endothermic property also changes as both reactions are the exact opposite of each other.
- Le Chetelier’s principle can decide the flow of the reaction equilibrium. Gas phase reactions are sensitive to pressure changes as well as temperature changes.
Complete answer:
The reaction given can be written as below where Delta represents heat evolved.
$$2N{O_{2(g)}} \to {N_2}{O_{4(g)}} + \Delta $$
Now the decomposition reaction of $${N_2}{O_{4(g)}}$$ can be written as below.
$${N_2}{O_{4(g)}} \rightleftharpoons 2N{O_{2(g)}}$$
As the earlier reversed reaction was exothermic, this reaction is endothermic.
Now, let’s see all the options to find the true one.
- When we increase the pressure of the reaction, according to Le Chetelier’s principle, equilibrium will shift towards the side which has less number of molecules. So, it will shift towards the left. So, option (A) is incorrect.
- When we add an inert gas to the reaction chamber at constant pressure, the equilibrium will then shift towards the side having more number of molecules.
- When we lower the temperature, as this reaction is an endothermic reaction, according to Le Chatelier’s principle, the equilibrium will shift towards the left side. So, this is not the correct sentence.
- When we add inert gas in the reaction chamber at constant volume, the pressure will increase and then equilibrium will shift towards the side having less number of molecules. So, this option is also not true.
So, option (B) Addition of inert gas at constant pressure is the correct answer.
Additional Information:
-Le Chatelier’s Principle refers to the change in the reaction equilibrium when any of the parameters respective to reaction is changed. This principle is also referred to as the law of equilibrium.
Note:
Reaction given in the question is not the reaction they have asked about, actually they have asked about its reverse reaction. As we reverse the reaction, its exothermic or endothermic property also changes as both reactions are the exact opposite of each other.
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