The slow step is the rate determining step. In the slow step in this reaction, 1 molecule of NO3 (intermediate) and 1 molecule of NO combine to form the products. Therefore, rate of this reaction depends upon 1 concentration term of NO3 and 1 concentration term of NO.
∴ Molecularity of the reaction = 1 + 1 = 2.
Rate = K [NO3] [NO]
NO3 is an intermediate which is formed rapidly by the collision of 1 molecule of NO and 1 molecule of O2.
[NO3] ∝ [NO][O2]
Rate = K1 [NO3] [NO]
= K [NO] [O2] [NO]
= K [NO]2[O2]
Order of reaction is 2 + 1 = 3.
This reaction follows second order kinetics.
So that, the rate equation for this reaction will
Rate, R = k[X]2 .............(1)
Let initial concentration is x mol L−1,
Plug the value in equation (1)
Rate, R1 = k .(a)2
Given that concentration is increasing by 3 times so new concentration will 3a mol L−1
Plug the value in equation (1) we get
Rate, R2 = k (3a)2
We have already get that R1 = ka2 plus this value we get
R2 = 9 R1
So that, the rate of formation will increase by 9 times.
Rate = k[A]2
If concentration of X is increased to three times,
Rate = k[3A]2
or Rate = 9 k A2
Thus, rate will increase 9 times.