Using the given bond dissociation energies, calculate H°rxn for the following reaction:2 H – Cl(g) + Br – Br(g)  2 H – Br(g) + Cl – Cl(g) A. + 78 kJB. - 308 kJC. + 14 kJD. + 564 kJBE (kJ/mol)H – Cl: 427H – Br: 363Cl – Cl: 243Br – Br: 193 ### Calculating the Enthalpy Change (ΔH°) for a Reaction#### Problem:Using the given bond dissociation energies, calculate the standard enthalpy change (ΔH°) for the following reaction:\[ 2 \, \text{H–Cl(g)} + \text{Br–Br(g)} \rightarrow 2 \, \text{H–Br(g)} + \text{Cl–Cl(g)} \]#### Options for ΔH°:- A. +78 kJ- B. -308 kJ- C. +14 kJ- D. +564 kJ#### Bond Dissociation Energies (BE in kJ/mol):| Bond | Energy (kJ/mol) ||----------|-----------------|| H–Cl | 427 || H–Br | 363 || Cl–Cl | 243 || Br–Br | 193 |#### Explanation:To solve for ΔH°, you need to:1. Calculate the total energy required to break all the bonds in the reactants.2. Calculate the total energy released when the new bonds are formed in the products.3. Subtract the energy released from the energy required to find the ΔH° for the reaction.#### Steps:1. **Bonds Broken** (Reactants): - 2 × H–Cl bonds - 1 × Br–Br bond2. **Bonds Formed** (Products): - 2 × H–Br bonds - 1 × Cl–Cl bond3. **Calculate Energies**: - Energy to break bonds = (2 × 427) + 193 - Energy to form bonds = (2 × 363) + 2434. **Compute ΔH°**: \[ ΔH° = \text{Energy to break bonds} - \text{Energy to form bonds} \]Following the calculations will lead you to the correct value of ΔH° among the given options.

Given reaction: 2 H-Cl (g) + Br-Br (g) → 2 H-Br (g) + Cl-Cl (g) Given bond energies: H-Cl = 427 kJ/mol H-Br = 363 kJ/mol Cl-Cl = 243 kJ/mol Br-Br = 193 kJ/mol We have to calculate the ∆H∘rxn for the above reaction.

Answered: Using the given bond dissociation…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Using the given bond dissociation energies,
calculate H°rxn for the following reaction:
2 H – Cl(g) + Br – Br(g)  2 H – Br(g) + Cl – Cl(g)
A. + 78 kJ
B. - 308 kJ
C. + 14 kJ
D. + 564 kJ
BE (kJ/mol)
H – Cl: 427
H – Br: 363
Cl – Cl: 243
Br – Br: 193

### Calculating the Enthalpy Change (ΔH°) for a Reaction

#### Problem:
Using the given bond dissociation energies, calculate the standard enthalpy change (ΔH°) for the following reaction:

\[ 2 \, \text{H–Cl(g)} + \text{Br–Br(g)} \rightarrow 2 \, \text{H–Br(g)} + \text{Cl–Cl(g)} \]

#### Options for ΔH°:
- A. +78 kJ
- B. -308 kJ
- C. +14 kJ
- D. +564 kJ

#### Bond Dissociation Energies (BE in kJ/mol):

| Bond | Energy (kJ/mol) |
|----------|-----------------|
| H–Cl | 427 |
| H–Br | 363 |
| Cl–Cl | 243 |
| Br–Br | 193 |

#### Explanation:
To solve for ΔH°, you need to:
1. Calculate the total energy required to break all the bonds in the reactants.
2. Calculate the total energy released when the new bonds are formed in the products.
3. Subtract the energy released from the energy required to find the ΔH° for the reaction.

#### Steps:
1. **Bonds Broken** (Reactants):
- 2 × H–Cl bonds
- 1 × Br–Br bond

2. **Bonds Formed** (Products):
- 2 × H–Br bonds
- 1 × Cl–Cl bond

3. **Calculate Energies**:
- Energy to break bonds = (2 × 427) + 193
- Energy to form bonds = (2 × 363) + 243

4. **Compute ΔH°**:

\[ ΔH° = \text{Energy to break bonds} - \text{Energy to form bonds} \]

Following the calculations will lead you to the correct value of ΔH° among the given options.

Formula Formula Bond dissociation energy (BDE) is the energy required to break a bond, making it an endothermic process. BDE is calculated for a particular bond and therefore consists of fragments such as radicals since it undergoes homolytic bond cleavage. For the homolysis of a X-Y molecule, the energy of bond dissociation is calculated as the difference in the total enthalpy of formation for the reactants and products. X-Y → X + Y BDE = Δ H f X + Δ H f Y – Δ H f X-Y where, ΔHf is the heat of formation.

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