For the reaction given below, the following mechanism has been given:
Reaction: 2NO + Br2 → 2NOBr
Mechanism:
NO + Br2 fast⇌ NOBr2
NOBr2 + NO slow→ 2NOBr
Hence, the rate law is:

The rate law is determined by the slow step of the reaction mechanism. The slow step is:
NOBr2 + NO → 2NOBr
The rate of this step is given by:
Rate = k[NOBr2][NO]
However, NOBr2 is an intermediate and its concentration cannot appear in the final rate law. We need to express [NOBr2] in terms of the reactants, NO and Br2.
The fast equilibrium step is:
NO + Br2 ⇌ NOBr2
The equilibrium constant for this step is:
K = [NOBr2]/([NO][Br2])
Solving for [NOBr2]:
[NOBr2] = K[NO][Br2]
Substitute this expression for [NOBr2] into the rate law:
Rate = k[NOBr2][NO] = k(K[NO][Br2])[NO] = kK[NO]2[Br2]
Since k and K are both constants, we can combine them into a new rate constant, k':
Rate = k'[NO]2[Br2]
Therefore, the rate law is k[NO]2[Br2]