Solveeit Logo
Login

Question

Question: In the reaction \({{N}_{2}}+{{O}_{2}}\rightleftharpoons 2NO\) , the moles/litre of \({{N}_{2}},{{O}_...

In the reaction N2+O2⇌2NO{{N}_{2}}+{{O}_{2}}\rightleftharpoons 2NO , the moles/litre of N2,O2{{N}_{2}},{{O}_{2}} and NONOrespectively 0.25,0.050.25,0.05 and 1.01.0 at equilibrium, the initial concentration of N2{{N}_{2}} and O2{{O}_{2}} will be respectively.
A.0.75 mol/litre, 0.55 mole/litre.0.75\text{ }mol/litre,\text{ }0.55\text{ }mole/litre.
B.0.50 mol/litre, 0.75 mole/litre.0.50\text{ }mol/litre,\text{ }0.75\text{ }mole/litre.
C.0.25 mol/litre, 0.50 mole/litre.0.25\text{ }mol/litre,\text{ }0.50\text{ }mole/litre.
D.0.25 mol/litre, 1.0 mole/litre.0.25\text{ }mol/litre,\text{ }1.0\text{ }mole/litre.

Explanation

Solution

We know that the formation of Nitrogen oxide from the reaction of Nitrogen and oxygen is exothermic as a large amount of heat is released after the bond formation. The equilibrium between the reactions is set. We need to shift it in a direction where it will favor product formation

Complete step by step solution:
Le Chatelier’s principle which is also called the equilibrium law is used to predict the effect of some change in factors on a thermodynamic system that is in chemical equilibrium. The principle was given by French chemist Henry Louis Le Chatelier. He said that equilibrium can adjust the forward and backward reactions in such a way as to negate the changes that are affecting the equilibrium.
According to the Le Chatelier Principle, states that a change in temperature, pressure, or concentration of reactants in an equilibrium system will shift equilibrium to change into a new equilibrium to negate effect of that change. In the case of changing temperature, adding or removing heat shifts the equilibrium. The given reaction is exothermic, so in case of an exothermic reaction, increasing the temperature will shift the equilibrium in the reverse direction to remove the effect of extra heat but we want a good yield of NO.NO. Here we have reaction and let a and b be initial concentration of N2{{N}_{2}} and O2{{O}_{2}} respectively;
N2a−x +O2b−x ⇌2NO2x \underset{a-x}{\mathop{{{N}_{2}}}}\,+\underset{b-x}{\mathop{{{O}_{2}}}}\,\rightleftharpoons \underset{2x}{\mathop{2NO}}\,
According to question; we have N2=0.25,O2=0.05{{N}_{2}}=0.25,{{O}_{2}}=0.05 and NO=1.0NO=1.0
On further solving for NONO we get; 2x=1.02x=1.0; ∴x=0.5\therefore x=0.5 and thus ⇒NO=0.5\Rightarrow NO=0.5
Similarly the equation for N2{{N}_{2}} is a−x=0.25a-x=0.25; ∴a=0.75\therefore a=0.75 and thus ⇒N2=0.75\Rightarrow {{N}_{2}}=0.75
Similarly the equation for O2{{O}_{2}} is b−x=0.05b-x=0.05; ∴a=0.55\therefore a=0.55 and thus ⇒O2=0.5\Rightarrow {{O}_{2}}=0.5

Therefore, the correct answer is option A.

Note:
Remember that exothermic reactions are those reactions in which heat is evolved from the system to the surrounding after the product formation. In the case of endothermic reactions, the heat is absorbed from the surroundings to the system.