Question

Bromine-containing species play a role in environmental chemistry. For example, they are evolved in volcanic eruptions. (a) The following molecules are important in bromine environmental chemistry: HBr, BrO, and HOBr. Which is an odd-electron molecule? (b) Use bond dissociation enthalpies to estimate \(\Delta_{\mathrm{r}} H\) for three reactions of bromine: $$\begin{array}{c} \mathrm{Br}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{Br}(\mathrm{g}) \\ 2 \mathrm{Br}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{BrO}(\mathrm{g}) \end{array}$$ \(\mathrm{BrO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{HOBr}(\mathrm{g})+\mathrm{OH}(\mathrm{g})\) (c) Using bond dissociation enthalpies, estimate the standard enthalpy of formation of HOBr(g) from \(\mathrm{H}_{2}(\mathrm{g}), \mathrm{O}_{2}(\mathrm{g}),\) and \(\mathrm{Br}_{2}(\mathrm{g})\) (d) Are the reactions in parts (b) and (c) exothermic or endothermic?

Short answer

(a) BrO is odd-electron. (b) Δ_rH: Reaction 1 = 193 kJ/mol, Reaction 2 = -62 kJ/mol, Reaction 3 = 50 kJ/mol. (c) Δ_fH(HOBr) = 50 kJ/mol. (d) Reaction 1 & 3 endothermic, Reaction 2 exothermic.

Step-by-step solution

Identify the Odd-Electron Molecule

Bromine-containing molecules include HBr, BrO, and HOBr. To determine which is an odd-electron molecule, consider the structure and electron count for each species. HBr and HOBr have complete electron pairs, whereas BrO has an unpaired electron due to the electron configuration of bromine (Br) and oxygen (O). Hence, BrO is the odd-electron molecule.

Reaction 1 - Calculate Δ_rH for Br2(g) → 2 Br(g)

Use bond dissociation enthalpy data to estimate the reaction enthalpy. The enthalpy change, Δ_rH, for breaking a Br—Br bond is equal to the bond dissociation energy of Br2, which is approximately 193 kJ/mol. Therefore, for the reaction \( \text{Br}_2(g) \rightarrow 2 \text{Br}(g) \), the estimated enthalpy change is \( \Delta_rH = 193 \text{ kJ/mol} \).

Reaction 2 - Estimate Δ_rH for 2 Br(g) + O2(g) → 2 BrO(g)

Use bond dissociation enthalpies to estimate the reaction enthalpy. The formation of 2 BrO molecules involves forming two Br—O bonds. Given that the bond dissociation enthalpy for Br—O is approximately 218 kJ/mol, for 2 BrO molecules it is \( 2 \times 218 = 436 \text{ kJ/mol} \). Thus, \( \Delta_rH = 2[BrO] - 2[Br] - [O2] = 436 - 0 - 498 = -62 \text{ kJ/mol} \), given that the O=O bond dissociation energy is approximately 498 kJ/mol.

Reaction 3 - Calculate Δ_rH for BrO(g) + H2O(g) → HOBr(g) + OH(g)

Estimate the enthalpies for forming HOBr and OH while breaking Br—O in BrO and O—H in water. The bond energies are: O—H ≈ 463 kJ/mol, Br—O for BrO ≈ 218 kJ/mol, Br—O in HOBr ≈ 268 kJ/mol, and O—H in OH ≈ 463 kJ/mol. Calculate the change: \( \Delta_rH = 268 + 463 - (218 + 463) = 50 \text{ kJ/mol} \).

Estimation - Standard Enthalpy of Formation of HOBr(g)

Construct the formation reaction: H2(g) + 1/2 O2(g) + 1/2 Br2(g) → HOBr(g). The enthalpy change is approximated by bond dissociation energies: Formation requires creating one Br—O and one O—H bond in HOBr. Thus, using reaction enthalpies already calculated, \( \Delta_fH[HOBr] = -218 + 268 = 50 \text{ kJ/mol} \).

Determine Thermochemistry - Exothermic or Endothermic

Exothermic reactions release heat (negative Δ_rH), while endothermic reactions absorb heat (positive Δ_rH). For these reactions: Reaction 1 (193 kJ/mol) and Reaction 3 (50 kJ/mol) are endothermic, Reaction 2 (-62 kJ/mol) is exothermic. The formation of HOBr with Δ_fH = 50 kJ/mol is endothermic.

Key concepts

Odd-Electron Molecule

In chemistry, understanding which molecules have unpaired electrons is crucial for predicting their reactivity and behavior. Molecules with an odd number of electrons have at least one unpaired electron, making them odd-electron or radical species. These are often more reactive due to the presence of unpaired electrons.
In bromine environmental chemistry, we're looking at HBr, BrO, and HOBr. To find the odd-electron molecule, consider their electron structures. HBr and HOBr complete their electron pairs, meaning all their electrons are paired up in bonds or lone pairs. BrO, on the other hand, has an unpaired electron, making it a radical. This unpaired electron is significant because it makes BrO highly reactive and important in chemical processes involved in the environment, such as during volcanic eruptions.

Bond Dissociation Enthalpy

Bond dissociation enthalpy (BDE) is a key concept in understanding chemical reactions. It represents the energy needed to break a bond in a molecule, leading to the separation of atoms. The higher the BDE, the stronger the bond.
In the exercise, BDEs are used to estimate reaction enthalpies for bromine-related reactions. For example:

• The dissociation of \( \text{Br}_2 \): The energy required is 193 kJ/mol, showing a strong Br—Br bond.
• Formation of BrO from Br: Although for forming BrO, bonds like Br—O are formed with an enthalpy of 218 kJ/mol per bond, these contribute to an overall reaction calculation that informs us about the energy dynamics of the process.

BDEs help in predicting whether a particular reaction will be endothermic or exothermic, adding insights into the feasibility and conditions needed for chemical processes.

Standard Enthalpy of Formation

The standard enthalpy of formation, represented as \( \Delta_fH \, \Delta_fH \), is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. It's a fundamental concept in thermochemistry.
For HOBr(g), the formation reaction from \( \text{H}_2(g) \, \text{O}_2(g) \, \text{and Br}_2(g) \) involves determining the energy changes involved in converting these elemental states into HOBr:

• We need to create one Br—O bond and one O—H bond during the formation.
• Using BDE values and calculated reaction enthalpies, \( \Delta_fH \) gives us an estimate of +50 kJ/mol.

This information is crucial for understanding the energy requirements and the chemical feasibility of synthesizing HOBr, particularly relevant for its environmental roles.

Exothermic and Endothermic Reactions

Understanding whether a reaction is exothermic or endothermic helps predict its heat absorption or emission.
Exothermic reactions release heat, resulting in negative enthalpy change \( (\Delta_rH) \), whereas endothermic reactions absorb heat, giving a positive \( \Delta_rH \). Analyzing the exercise:

• For \( \text{Br}_2(g) \rightarrow 2 \text{Br}(g) \, \Delta_rH \) is +193 kJ/mol, indicating the reaction is endothermic—it requires energy input to break bonds.
• Conversely, the formation of BrO from Br \( (2\text{Br}(g) + \text{O}_2(g) \rightarrow 2\text{BrO}(g)\,-62\text{ kJ/mol}) \) is exothermic, releasing energy as new bonds form efficiently.

Each reaction's energy exchange—with an endothermic step needing energy input and an exothermic step releasing energy—is crucial in identifying how different reactions progress, especially in environmental contexts like those involving bromine compounds.