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Find Overall Order of Reaction from Mechanism | Kinetics | CHEMCA

Find Overall Order of Reaction from Mechanism | Kinetics | CHEMCA

Finding the Overall Order of a Reaction From Its Mechanism

Published by Abhishek Sengar | CHEMCA India

In Chemical Kinetics, questions giving you a multi-step reaction mechanism and asking for the overall order are extremely popular in JEE and NEET.

The standard trick is to write the rate law using the Slow Step. However, there is a major trap: What happens if the slow step contains an unstable intermediate that doesn't exist in the overall balanced equation? Let's solve a classic Previous Year Question (PYQ) to find out!

Video Tutorial: The Intermediate Substitution Method

Watch Abhishek Sengar sir from CHEMCA break down the algebra required to substitute an intermediate out of a rate law using the fast-equilibrium step.

Step-by-Step Problem Breakdown

The Reaction Mechanism Given:

  • Step 1 (Fast Equilibrium): NO + Br2 ⇌ NOBr2
  • Step 2 (Slow): NOBr2 + NO → 2NOBr
  1. Write the Initial Rate Law:
    The rate of a complex reaction is always determined by its slowest step (the Rate Determining Step). From Step 2, we write:
    Rate = k [NOBr2]1 [NO]1
  2. Identify the Problem (The Trap):
    Problem: NOBr2 is an Intermediate! It is formed in Step 1 and consumed in Step 2. A proper rate law can never contain reaction intermediates; it must only contain true reactants or catalysts. We must substitute it out!
  3. Use the Fast Step to Substitute:
    Since Step 1 is a fast equilibrium, we can write an Equilibrium Constant (Kc) expression for it:
    Kc = [Products] / [Reactants]
    Kc = [NOBr2] / ([NO][Br2])
    Rearrange this to solve for the intermediate:
    [NOBr2] = Kc [NO] [Br2]
  4. Determine the Final Rate Law:
    Substitute the expression we just found back into our original rate equation:
    Rate = k × (Kc [NO] [Br2]) × [NO]
    Combine the constants (k × Kc = k') and multiply the [NO] terms:
    Final Rate = k' [NO]2 [Br2]1
The Substitution Workflow 1. Initial Rate Law (Slow Step) Rate = k [NOBr2] [NO] NOBr   is an invalid intermediate! 2. Solve Equilibrium (Fast Step) Kc = [NOBr2] / ([NO][Br2]) → [NOBr2] = Kc[NO][Br2] Substitute! 3. Final Valid Rate Law Rate = k' [NO]2 [Br2]1

Fig: Visually tracking the intermediate substitution process.

Calculating the Overall Order

The overall order of a reaction is simply the sum of the powers of the concentration terms in the final, valid rate law.

Power of [NO] = 2

Power of [Br2] = 1

Overall Order = 2 + 1 = 3

Practice Questions for JEE & NEET

Master this exact substitution logic by applying it to these highly-tested variations!

Question 1: Why can't we just look at the overall balanced chemical equation (2NO + Br2 → 2NOBr) to determine the order of the reaction?

Answer: Because Order is a strictly experimental quantity.

Reasoning:

A balanced overall equation only shows the starting materials and final products; it tells you absolutely nothing about the pathway (mechanism) the molecules take to get there. For complex (multi-step) reactions, the stoichiometric coefficients in the balanced equation rarely match the powers in the rate law. You must derive the rate law exclusively from experimental data or the slowest elementary step of the mechanism.

Question 2 (The Ozone Trap): The decomposition of ozone (2O3 → 3O2) follows this mechanism:
Step 1 (Fast eq): O3 ⇌ O2 + O
Step 2 (Slow): O + O3 → 2O2
Calculate the overall order of this reaction.

Answer: Order = 1 (Specifically: 2 with respect to O3, and -1 with respect to O2)

Step-by-Step Solution:

  • 1. Slow Step Rate Law: Rate = k [O] [O3]. (Note: atomic Oxygen 'O' is an intermediate!)
  • 2. Fast Eq to solve for [O]:
    Kc = ([O2][O]) / [O3]
    Therefore, [O] = Kc [O3] / [O2].
  • 3. Substitute:
    Rate = k × (Kc [O3] / [O2]) × [O3]
    Rate = k' [O3]2 [O2]-1.
  • 4. Calculate Order: 2 + (-1) = 1. (This proves that orders can be negative, meaning an increase in O2 actually slows down the reaction!).

Crush Physical Chemistry Numericals!

Don't let multi-step algebra intimidate you. Visit www.chemca.in today to access Abhishek Sir's complete Chemical Kinetics notes and full-length mock tests designed specifically for JEE Main & NEET.

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