Rate law from mechanism — rate-determining step approach

Question

The reaction $2\text{NO}_2 + \text{F}_2 \rightarrow 2\text{NO}_2\text{F}$ proceeds by the following mechanism:

Step 1 (slow): $\text{NO}_2 + \text{F}_2 \rightarrow \text{NO}_2\text{F} + \text{F}$

Step 2 (fast): $\text{NO}_2 + \text{F} \rightarrow \text{NO}_2\text{F}$

Derive the rate law from this mechanism.

(JEE Advanced 2023, similar pattern)

Solution — Step by Step

The rate-determining step (RDS) is the slowest step in the mechanism — it acts as the bottleneck. Here, Step 1 is slow, so the overall rate is governed by Step 1.

The rate law is written directly from the molecularity of the RDS:

\text{Rate} = k_1[\text{NO}_2][\text{F}_2]

The species in our rate law — $\text{NO}_2$ and $\text{F}_2$ — are both reactants (not intermediates). The fluorine atom F produced in Step 1 is an intermediate but does not appear in our rate law. Good — no further manipulation needed.

Adding Step 1 and Step 2:

$\text{NO}_2 + \text{F}_2 + \text{NO}_2 + \text{F} \rightarrow \text{NO}_2\text{F} + \text{F} + \text{NO}_2\text{F}$

Cancel the intermediate F from both sides:

$2\text{NO}_2 + \text{F}_2 \rightarrow 2\text{NO}_2\text{F}$ — matches the overall reaction.

\boxed{\text{Rate} = k[\text{NO}_2][\text{F}_2]}

The reaction is first order in $\text{NO}_2$ and first order in $\text{F}_2$ — overall second order.

Why This Works

The rate-determining step controls the overall rate because the fast step has to “wait” for the slow step to produce intermediates. Think of it like an assembly line: the slowest station determines the throughput of the entire line.

Notice that the experimentally observed rate law ( $\text{Rate} = k[\text{NO}_2][\text{F}_2]$ ) is NOT what you would predict from the stoichiometry of the overall reaction (which would suggest $\text{Rate} = k[\text{NO}_2]^2[\text{F}_2]$ ). This mismatch between stoichiometry and rate law is a strong hint that the reaction proceeds through a multi-step mechanism.

Alternative Method

When the RDS involves an intermediate, we must use the steady-state approximation or the pre-equilibrium approximation to eliminate the intermediate from the rate law. In this problem, the intermediate (F atom) only appears in Step 2 (the fast step), not in the RDS, so we did not need these techniques. But for problems where the RDS involves an intermediate, we set $d[\text{intermediate}]/dt = 0$ and solve.

For JEE Advanced, the algorithm is: (1) write rate from the slow step, (2) if an intermediate appears in the rate law, eliminate it using the equilibrium expression from the preceding fast step, (3) verify the mechanism adds up to the overall reaction. Practice this on 5-6 mechanisms and the pattern becomes automatic.

Common Mistake

The biggest error: writing the rate law from the overall balanced equation instead of from the mechanism. The rate law can only be determined experimentally or derived from the mechanism. The stoichiometric coefficients of the overall reaction have nothing to do with the order of the reaction unless it is an elementary reaction (single step).

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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