The reaction 2NO + Br _ 2 → 2NOBr takes place through the mechanism...

On applying Law of mass action over slowest step,

$R = k [ {NOBr}_{2} ] [ NO ]$

The slowest step is the rate determining step.

On applying Law of mass action on equilibrium of Step I

$k = \dfrac{\left[{NOBr}_{2}\right]}{\left[NO\right]\left[{Br}_{2}\right]}$

\left[{NOBr}_{2}\right] = k\left[NO\right]\left[{Br}_{2}\right]$</p><p>Hence, overall rate equation will be</p><p>$R = k{\left[NO\right]}^{2}\left[{Br}_{2}\right]$</p><p>$= k ' {\left[NO\right]}^{2}\left[{Br}_{2}\right]

The overall order is equal to sum of the powers of the concentration terms in the rate law.

The overall order of the reaction is 3.