Chemistry · General chemistry II · Worked example

Rule a mechanism in or out for 2NO + Br₂

Nitric oxide and bromine form nitrosyl bromide, 2NO(g) + Br₂(g) → 2NOBr(g). Experiments give rate = k[NO]²[Br₂]. Which of these two-step mechanisms is consistent with that rate law? Mechanism A: NO + Br₂ → NOBr₂ (slow), then NOBr₂ + NO → 2NOBr (fast). Mechanism B: NO + Br₂ ⇌ NOBr₂ (fast), then NOBr₂ + NO → 2NOBr (slow).

Write what the experiment says

Start from the overall reaction and the observed rate law. The orders, 2 in NO and 1 in Br₂, are the target every mechanism must reproduce.

2 NO + {Br}_{2} \to 2 NOBr, rate = k {[NO]}^{2} [{Br}_{2}]

Check 1 for both: do the steps add up?

Both mechanisms use the same two equations; only which step is slow differs. Adding them, NOBr₂ appears on both sides and cancels. The sum is the overall reaction, so neither mechanism fails check 1. NOBr₂ is an intermediate: made in the first step, used in the second.

\begin{aligned} NO + {Br}_{2} & \to {NOBr}_{2} \\ {NOBr}_{2} + NO & \to 2 NOBr \\ sum: 2 NO + {Br}_{2} & \to 2 NOBr \end{aligned}

Mechanism A: write the slow step’s rate law

In A the first step is slow. It is elementary, so its rate law uses its own coefficients: one NO and one Br₂. It contains no intermediate, so check 3 passes immediately.

rate = k_{1} [NO] [{Br}_{2}]

Mechanism A: compare, and rule it out

A predicts first order in NO, but the experiment shows second order. The orders disagree, so Mechanism A is ruled out. The steps adding up correctly was not enough.

k_{1} {[NO]}^{1} [{Br}_{2}] \neq k {[NO]}^{2} [{Br}_{2}]

Mechanism B: the slow step contains an intermediate

In B the second step is slow, so the rate law comes from it. It contains NOBr₂, an intermediate that the experiment never measures. Check 3 sends us to the fast first step.

rate = k_{2} [{NOBr}_{2}] [NO]

Replace the intermediate using the fast equilibrium

Because the first step is fast and reversible, its forward and reverse rates are equal. Solve that equality for [NOBr₂].

k_{1} [NO] [{Br}_{2}] = k_{- 1} [{NOBr}_{2}] \Rightarrow [{NOBr}_{2}] = \frac{k_{1}}{k_{- 1}} [NO] [{Br}_{2}]

Substitute and compare

Put that expression into the slow step’s rate law and collect the constants into one k. The prediction is second order in NO and first order in Br₂, matching the experiment. Mechanism B passes all four checks.

rate = k_{2} \frac{k_{1}}{k_{- 1}} {[NO]}^{2} [{Br}_{2}] = k {[NO]}^{2} [{Br}_{2}]

State the conclusion carefully

Mechanism B is consistent with the data, so it is possible. It is not proved: a single step in which two NO molecules and one Br₂ collide at once would predict the same rate law, although three-particle collisions are rare. Detecting NOBr₂ during the reaction would be evidence for B.

Result

Mechanism A is ruled out: it predicts rate = k[NO][Br₂]. Mechanism B is consistent: its fast equilibrium turns k₂[NOBr₂][NO] into k[NO]²[Br₂]. Consistent means possible, not proved.

Your turn

For H₂ + 2ICl → I₂ + 2HCl the observed rate law is rate = k[H₂][ICl]. Is this mechanism consistent? Step 1: H₂ + ICl → HI + HCl (slow). Step 2: HI + ICl → I₂ + HCl (fast).

Show the answer and explanation

Yes. It is consistent with the rate law, though not proved.

The steps add up to the overall reaction, with HI as the intermediate. The slow first step is elementary, so it predicts rate = k[H₂][ICl], which matches. No intermediate appears, so no substitution is needed.

rate = k [H_{2}] [ICl]

Keep exploring

Open the mechanism in a Chemistry box, then change which step is slow and watch the rate-law row change its verdict.

Return to the concept →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

Tro, Chemistry: A Molecular Approach, 4th ed., §14.6 Reaction Mechanisms, pp. 648–653 (a valid mechanism: steps sum to the overall reaction and predict the observed rate law, p. 651)
OpenStax Chemistry 2e — Reaction mechanisms