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JEE Main 2023: Reaction Mechanism & Order

JEE Main 2023: Reaction Mechanism & Order | chemca
Home › Class XII › Physical Chemistry › Chemical Kinetics › JEE Main 2023 Mechanism Problem
Daily Challenge JEE Main 2023 (12 Apr, Shift 1)

Determining Order from Mechanism

Derive the rate law using the Pre-equilibrium Approximation method.

Question:

The reaction $2NO + Br_2 \rightarrow 2NOBr$ takes place through the mechanism given below:

Step 1:  $NO + Br_2 \rightleftharpoons NOBr_2$    (fast)
Step 2:  $NOBr_2 + NO \rightarrow 2NOBr$    (slow)

The overall order of the reaction is ________.

Detailed Solution

Step 1: Identify the Rate-Determining Step (RDS)

The slowest step in a complex reaction mechanism dictates the overall rate of the reaction. Here, Step 2 is the slow step.

$\text{Rate} = k[NOBr_2][NO]$

The Golden Rule of Rate Laws

The final rate law expression cannot contain reaction intermediates. An intermediate is formed and consumed during the reaction, so its concentration isn't easily measurable.

In our rate equation, $NOBr_2$ is an intermediate. We must eliminate it by substituting it with stable reactants using the fast equilibrium step!

Step 2: Express Intermediate in terms of Reactants

Look at the fast equilibrium step: $NO + Br_2 \rightleftharpoons NOBr_2$

We can write the equilibrium constant ($K_{eq}$) for this step:

$$ K_{eq} = \frac{[NOBr_2]}{[NO][Br_2]} $$

Rearranging this to solve for the concentration of the intermediate ($[NOBr_2]$):

$$ [NOBr_2] = K_{eq} [NO][Br_2] $$

Step 3: Substitute and Find Overall Order

Now, substitute the expression for $[NOBr_2]$ back into our original rate equation from Step 1:

$$ \text{Rate} = k \left( K_{eq} [NO][Br_2] \right) [NO] $$ $$ \text{Rate} = (k \cdot K_{eq}) [NO]^2 [Br_2]^1 $$ $$ \text{Rate} = k' [NO]^2 [Br_2]^1 $$

The overall order of the reaction is the sum of the powers of the concentration terms in the final rate law:

Overall Order = $2 + 1$
Answer = 3

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