The oxidation of NO to NO₂ occurs via the mechanism given below.

2NO  N₂O₂

N₂O₂ + O₂ \(\stackrel{\rm k_2}{\longrightarrow}\) 2NO₂

\(\rm\frac{d\left[NO_2\right]}{dt}\) in the presence of large excess of O₂ can be written as

Concept:-

• The rate of a chemical reaction is defined as the speed at which the reactants are converted into products. The rate of a reaction depends on the composition and the temperature of the reaction mixture.
• It is the amount of chemical change occurring with time.
• According to the steady state approximation, When a reaction proceeds steadily, there will be no overall accumulation of the intermediate, and there would be a stationary concentration of the same.
• When the concentration of the intermediate is very small, then according to the steady state approximation

​

• Let us consider the following consecutive reaction,

​

• If the concentration of the reactive intermediate 'B' has a small value that practically remains constant throughout the reaction, then we can apply steady state approximation for B as,

• Therefore for the consecutive reaction,

• Again, the Rate of the reaction is given by,

Explanation:-

2NO  N2O2    ----------(i) -2nd order reaction

N2O2 + O2  2NO2 ----------(ii) -2nd order reaction.

The rate expression for N₂O₂ is from the above two equations;

So, 

or, 

or, 

or, 

• or,  -----------(iii)

Now as per the question, rate law for  is,

From equation (ii)

By putting the value of N₂O₂ from eq.(i) we get

It is given that O2 is present in excess. So K₂O₂>>K_-1

So, 

Conclusion:-

 Therefore the 2k1(NO)2 is correct.