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Ch.14 - Chemical Kinetics
All textbooksBrown 14th EditionCh.14 - Chemical KineticsProblem 121c
Chapter 14, Problem 121c

The mechanism for the oxidation of HBr by O2 to form 2 H2O and Br2 is shown in Exercise 14.74. (c) Draw a plausible Lewis structure for the intermediate HOOBr. To what familiar compound of hydrogen and oxygen does it appear similar?

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1
Identify the atoms involved in the intermediate HOOBr: hydrogen (H), oxygen (O), and bromine (Br).
Determine the total number of valence electrons available for the structure: H (1 valence electron), O (6 valence electrons each), and Br (7 valence electrons).
Arrange the atoms with the least electronegative atom (Br) as the central atom, and connect the atoms with single bonds: H-O-O-Br.
Distribute the remaining valence electrons to satisfy the octet rule for each atom, starting with the outer atoms (O and Br) and then the central atom.
Compare the structure of HOOBr to a familiar compound of hydrogen and oxygen, such as hydrogen peroxide (H2O2), noting the similar O-O bond.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Lewis Structures

Lewis structures are diagrams that represent the bonding between atoms in a molecule and the lone pairs of electrons that may exist. They help visualize the arrangement of electrons and the connectivity of atoms, which is crucial for understanding molecular geometry and reactivity. In the context of HOOBr, drawing its Lewis structure will reveal how the atoms are bonded and the presence of any formal charges.
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Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions involve the transfer of electrons between species, leading to changes in oxidation states. In the oxidation of HBr by O2, HBr is oxidized, and O2 is reduced, resulting in the formation of water and bromine. Understanding redox processes is essential for analyzing the reaction mechanism and the role of intermediates like HOOBr.
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Intermediates in Reaction Mechanisms

Intermediates are transient species formed during the course of a chemical reaction that are not present in the final products. They often have distinct structures and properties that can provide insight into the reaction pathway. In this case, HOOBr serves as an intermediate, and recognizing its similarity to familiar compounds, such as hydrogen peroxide (H2O2), can aid in predicting its behavior and stability.
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Related Practice
Textbook Question

The gas-phase reaction of NO with F2 to form NOF and F has an activation energy of Ea = 6.3 kJ/mol. and a frequency factor of A = 6.0 × 108 M-1 s-1. The reaction is believed to be bimolecular: NO(g) + F2(g) → NOF(g) + F(g) (b) Draw the Lewis structures for the NO and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule.

Textbook Question

The gas-phase reaction of NO with F2 to form NOF and F has an activation energy of Ea = 6.3 kJ>mol. and a frequency factor of A = 6.0 * 108 M-1 s-1. The reaction is believed to be bimolecular: NO1g2 + F21g2 ¡ NOF1g2 + F1g2 (e) Suggest a reason for the low activation energy for the reaction.

Textbook Question

The mechanism for the oxidation of HBr by O2 to form 2 H2O and Br2 is shown in Exercise 14.74. (a) Calculate the overall standard enthalpy change for the reaction process.

Textbook Question

The rates of many atmospheric reactions are accelerated by the absorption of light by one of the reactants. For example, consider the reaction between methane and chlorine to produce methyl chloride and hydrogen chloride:

Reaction 1: CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)

This reaction is very slow in the absence of light. However, Cl2(g) can absorb light to form Cl atoms:

Reaction 2: Cl2(g) + hv → 2 Cl(g)

Once the Cl atoms are generated, they can catalyze the reaction of CH4 and Cl2, according to the following proposed mechanism:

Reaction 3: CH4(g) + Cl(g) → CH3(g) + HCl(g)

Reaction 4: CH3(g) + Cl2(g) → CH3Cl(g) + Cl(g)

The enthalpy changes and activation energies for these two reactions are tabulated as follows:

Reaction ΔH° (kJ/mol) Ea (kJ/mol)

3 +4 17

4 -109 4 

(b) By using the data tabulated here, sketch a quantitative energy profile for the catalyzed reaction represented by reactions 3 and 4.