The reaction 2 NO+Br₂ → 2 NOBr
takes places through the mechanism given below:
NO+Br₂ ⇔ NOBr₂ (fast)
NOBr₂+NO → 2 NOBr (slow)
The overall order of the reaction is ___________.

Solution

Chemistry ma’am · 3 steps

The slow step sets the rate, but NOBr₂ is an intermediate, so swap it out using the fast equilibrium.

1. Slow step decides. The rate of the whole reaction equals the rate of its slowest step. So rate = k₂[NOBr₂][NO].
2. Remove the intermediate. NOBr₂ is made and used up inside the mechanism, so it cannot appear in the final rate law. The fast first step stays at equilibrium, so K = [NOBr₂]/[NO][Br₂], which gives [NOBr₂] = K[NO][Br₂].

   Rate = k₂ K[NO][Br₂][NO] = k[NO]²[Br₂]

3. So. The order is 2 in NO and 1 in Br₂, so the overall order is 2 + 1 = 3.

The common slip: 2. Adding the exponents of the slow step as written, [NOBr₂]¹[NO]¹, gives 2. That keeps the intermediate in the rate law.

The idea to keep. Write the rate law from the slow step, then replace every intermediate using the fast equilibrium before it. Only reactants may remain in the final rate law.

Answer from the NTA final answer key. Solution by PrepWiser.