---
title: "A student investigates the kinetics of the reaction represented by the balanced equation below.  \\[ 2\\text{ NO}(g) + \\text{Br}_2(g) \\rightarrow 2\\text{ NOBr}(g) \\]  To determine the experimental rate law, the student measures the initial rate of formation of \\(\\text{NOBr}(g)\\) across three trials at \\(298\\text{ K}\\), collecting the data shown in the following table.  | Trial | Initial \\([\\text{NO}]\\) (\\(\\text{M}\\)) | Initial \\([\\text{Br}_2]\\) (\\(\\text{M}\\)) | Initial Rate of Formation of \\(\\text{NOBr}\\) (\\(\\text{M}\\cdot\\text{s}^{-1}\\)) | | :—: | :—: | :—: | :—: | | \\(1\\) | \\(0.10\\) | \\(0.10\\) | \\(1.2 \\times 10^{-3}\\) | | \\(2\\) | \\(0.20\\) | \\(0.10\\) | \\(4.8 \\times 10^{-3}\\) | | \\(3\\) | \\(0.20\\) | \\(0.20\\) | \\(9.6 \\times 10^{-3}\\) |  Three different reaction mechanisms are proposed:  Mechanism I Step 1: \\(\\text{NO}(g) + \\text{Br}_2(g) \\rightarrow \\text{NOBr}_2(g)\\) (slow) Step 2: \\(\\text{NOBr}_2(g) + \\text{NO}(g) \\rightarrow 2\\text{ NOBr}(g)\\) (fast)  Mechanism II Step 1: \\(\\text{NO}(g) + \\text{Br}_2(g) \\rightleftharpoons \\text{NOBr}_2(g)\\) (fast equilibrium) Step 2: \\(\\text{NOBr}_2(g) + \\text{NO}(g) \\rightarrow 2\\text{ NOBr}(g)\\) (slow)  Mechanism III Step 1: \\(\\text{Br}_2(g) \\rightleftharpoons 2\\text{ Br}(g)\\) (fast equilibrium) Step 2: \\(\\text{NO}(g) + \\text{Br}(g) \\rightarrow \\text{NOBr}(g)\\) (slow)  Based on the experimental data, which mechanism is consistent with the observed rate law, and why?"
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url: "https://nerd-notes.com/ubq/123689/"
date_modified: "2026-09-28T12:01:59+00:00"
---

# A student investigates the kinetics of the reaction represented by the balanced equation below.

\[ 2\text{ NO}(g) + \text{Br}_2(g) \rightarrow 2\text{ NOBr}(g) \]

To determine the experimental rate law, the student measures the initial rate of formation of \(\text{NOBr}(g)\) across three trials at \(298\text{ K}\), collecting the data shown in the following table.

| Trial | Initial \([\text{NO}]\) (\(\text{M}\)) | Initial \([\text{Br}_2]\) (\(\text{M}\)) | Initial Rate of Formation of \(\text{NOBr}\) (\(\text{M}\cdot\text{s}^{-1}\)) |
| :—: | :—: | :—: | :—: |
| \(1\) | \(0.10\) | \(0.10\) | \(1.2 \times 10^{-3}\) |
| \(2\) | \(0.20\) | \(0.10\) | \(4.8 \times 10^{-3}\) |
| \(3\) | \(0.20\) | \(0.20\) | \(9.6 \times 10^{-3}\) |

Three different reaction mechanisms are proposed:

Mechanism I
Step 1: \(\text{NO}(g) + \text{Br}_2(g) \rightarrow \text{NOBr}_2(g)\) (slow)
Step 2: \(\text{NOBr}_2(g) + \text{NO}(g) \rightarrow 2\text{ NOBr}(g)\) (fast)

Mechanism II
Step 1: \(\text{NO}(g) + \text{Br}_2(g) \rightleftharpoons \text{NOBr}_2(g)\) (fast equilibrium)
Step 2: \(\text{NOBr}_2(g) + \text{NO}(g) \rightarrow 2\text{ NOBr}(g)\) (slow)

Mechanism III
Step 1: \(\text{Br}_2(g) \rightleftharpoons 2\text{ Br}(g)\) (fast equilibrium)
Step 2: \(\text{NO}(g) + \text{Br}(g) \rightarrow \text{NOBr}(g)\) (slow)

Based on the experimental data, which mechanism is consistent with the observed rate law, and why?

A student investigates the kinetics of the reaction represented by the balanced equation below.

\[ 2\text{ NO}(g) + \text{Br}_2(g) \rightarrow 2\text{ NOBr}(g) \]

To determine the experimental rate law, the student measures the initial rate of formation of \(\text{NOBr}(g)\) across three trials at \(298\text{ K}\), collecting the data shown in the following table.

| Trial | Initial \([\text{NO}]\) (\(\text{M}\)) | Initial \([\text{Br}_2]\) (\(\text{M}\)) | Initial Rate of Formation of \(\text{NOBr}\) (\(\text{M}\cdot\text{s}^{-1}\)) |
| :---: | :---: | :---: | :---: |
| \(1\) | \(0.10\) | \(0.10\) | \(1.2 \times 10^{-3}\) |
| \(2\) | \(0.20\) | \(0.10\) | \(4.8 \times 10^{-3}\) |
| \(3\) | \(0.20\) | \(0.20\) | \(9.6 \times 10^{-3}\) |

Three different reaction mechanisms are proposed:

Mechanism I
Step 1: \(\text{NO}(g) + \text{Br}_2(g) \rightarrow \text{NOBr}_2(g)\) (slow)
Step 2: \(\text{NOBr}_2(g) + \text{NO}(g) \rightarrow 2\text{ NOBr}(g)\) (fast)

Mechanism II
Step 1: \(\text{NO}(g) + \text{Br}_2(g) \rightleftharpoons \text{NOBr}_2(g)\) (fast equilibrium)
Step 2: \(\text{NOBr}_2(g) + \text{NO}(g) \rightarrow 2\text{ NOBr}(g)\) (slow)

Mechanism III
Step 1: \(\text{Br}_2(g) \rightleftharpoons 2\text{ Br}(g)\) (fast equilibrium)
Step 2: \(\text{NO}(g) + \text{Br}(g) \rightarrow \text{NOBr}(g)\) (slow)

Based on the experimental data, which mechanism is consistent with the observed rate law, and why?

- **A.** Mechanism I, because the rate-determining step involves the bimolecular collision of one \(\text{NO}\) molecule and one \(\text{Br}_2\) molecule, which directly determines the overall reaction rate.
- **B.** Mechanism I, because the elementary steps add together to yield the correct stoichiometry of the overall balanced equation.
- **C.** Mechanism II, because the slow step depends on the intermediate \(\text{NOBr}_2\), whose pre-equilibrium concentration is proportional to \([\text{NO}][\text{Br}_2]\), producing an overall rate law of \(\text{Rate} = k[\text{NO}]^2[\text{Br}_2]\).
- **D.** Mechanism III, because the rapid equilibrium dissociation of \(\text{Br}_2\) into \(\text{Br}\) atoms establishes a second-order dependence on \([\text{Br}_2]\) and a first-order dependence on \([\text{NO}]\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123689/*