16. Given the bond dissociation energies below (in kcal/mol), calculate the overall AH° for the following reaction: (СН3)3СH + Br2 (СН3)3СBr + HBr (СН)3С—Н (СН)3С-Br 91 65 Br-Br 46 H-Br 88 CH3-Br 70

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**Problem 16:**

Given the bond dissociation energies below (in kcal/mol), calculate the overall ΔH° for the following reaction:

\[
(\text{CH}_3)_3\text{CH} + \text{Br}_2 \rightarrow (\text{CH}_3)_3\text{CBr} + \text{HBr}
\]

**Bond Dissociation Energies:**

- \((\text{CH}_3)_3\text{C} - \text{H}:\) 91 kcal/mol
- \((\text{CH}_3)_3\text{C} - \text{Br}:\) 65 kcal/mol
- \(\text{Br} - \text{Br}:\) 46 kcal/mol
- \(\text{H} - \text{Br}:\) 88 kcal/mol
- \(\text{CH}_3 - \text{Br}:\) 70 kcal/mol

**Explanation:**

To find the overall ΔH° of the reaction, sum the energies of the bonds broken and subtract the energies of the bonds formed.

1. **Bonds Broken:**
   - \((\text{CH}_3)_3\text{C} - \text{H}\): 91 kcal/mol
   - \(\text{Br} - \text{Br}\): 46 kcal/mol

2. **Bonds Formed:**
   - \((\text{CH}_3)_3\text{C} - \text{Br}\): 65 kcal/mol
   - \(\text{H} - \text{Br}\): 88 kcal/mol

3. **Calculation:**

   \[
   \Delta H° = (\text{Bonds Broken}) - (\text{Bonds Formed})
   \]
   \[
   \Delta H° = (91 + 46) - (65 + 88) = 137 - 153 = -16 \text{ kcal/mol}
   \]

Therefore, the overall ΔH° for the reaction is \(-16\) kcal/mol, indicating that the reaction is exothermic.
Transcribed Image Text:**Problem 16:** Given the bond dissociation energies below (in kcal/mol), calculate the overall ΔH° for the following reaction: \[ (\text{CH}_3)_3\text{CH} + \text{Br}_2 \rightarrow (\text{CH}_3)_3\text{CBr} + \text{HBr} \] **Bond Dissociation Energies:** - \((\text{CH}_3)_3\text{C} - \text{H}:\) 91 kcal/mol - \((\text{CH}_3)_3\text{C} - \text{Br}:\) 65 kcal/mol - \(\text{Br} - \text{Br}:\) 46 kcal/mol - \(\text{H} - \text{Br}:\) 88 kcal/mol - \(\text{CH}_3 - \text{Br}:\) 70 kcal/mol **Explanation:** To find the overall ΔH° of the reaction, sum the energies of the bonds broken and subtract the energies of the bonds formed. 1. **Bonds Broken:** - \((\text{CH}_3)_3\text{C} - \text{H}\): 91 kcal/mol - \(\text{Br} - \text{Br}\): 46 kcal/mol 2. **Bonds Formed:** - \((\text{CH}_3)_3\text{C} - \text{Br}\): 65 kcal/mol - \(\text{H} - \text{Br}\): 88 kcal/mol 3. **Calculation:** \[ \Delta H° = (\text{Bonds Broken}) - (\text{Bonds Formed}) \] \[ \Delta H° = (91 + 46) - (65 + 88) = 137 - 153 = -16 \text{ kcal/mol} \] Therefore, the overall ΔH° for the reaction is \(-16\) kcal/mol, indicating that the reaction is exothermic.
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