Question

Use the table below to calculate the change in enthalpy (bond energy) for the reaction above.
Bond Type Bond Energy kJ/mol

O - H 464
C - H 411
O = O 494
C = O 799

Answer 1

To calculate the change in enthalpy (\( \Delta H \)) for a chemical reaction using bond energies, you can follow these steps:

1. **Identify the bonds broken and formed in the reaction.**
2. **Use the provided bond energies to calculate the total energy of bonds broken and the total energy of bonds formed.**
3. **Calculate the change in enthalpy using the formula:**
\[
\Delta H = (\text{Energy of bonds broken}) - (\text{Energy of bonds formed})
\]

### Example Calculation

For a hypothetical reaction:
\[ \text{Reactants} \rightarrow \text{Products} \]

Suppose the reaction involves breaking and forming the following bonds:

- **Bonds broken (assuming)**:
- 2 O-H bonds (from water or some alcohol)
- 1 C-H bond (from an alkane)

- **Bonds formed (assuming)**:
- 1 O=O bond (from molecular oxygen)
- 1 C=O bond (from a carbonyl compound)

### Step-by-Step Calculation

**Step 1: Determine the bonds involved.**
- **Bonds broken**:
- \( 2 \, \text{O-H} \)
- \( 1 \, \text{C-H} \)

**Step 2: Calculate energy for bonds broken.**
- Energy for \( 2 \, \text{O-H} \):
\[
2 \times 464 \, \text{kJ/mol} = 928 \, \text{kJ/mol}
\]
- Energy for \( 1 \, \text{C-H} \):
\[
1 \times 411 \, \text{kJ/mol} = 411 \, \text{kJ/mol}
\]

- **Total energy of bonds broken**:
\[
928 \, \text{kJ/mol} + 411 \, \text{kJ/mol} = 1339 \, \text{kJ/mol}
\]

**Step 3: Calculate energy for bonds formed.**
- **Bonds formed**:
- \( 1 \, \text{O=O} \)
- \( 1 \, \text{C=O} \)

- Energy for \( 1 \, \text{O=O} \):
\[
1 \times 494 \, \text{kJ/mol} = 494 \, \text{kJ/mol}
\]
- Energy for \( 1 \, \text{C=O} \):
\[
1 \times 799 \, \text{kJ/mol} = 799 \, \text{kJ/mol}
\]

- **Total energy of bonds formed**:
\[
494 \, \text{kJ/mol} + 799 \, \text{kJ/mol} = 1293 \, \text{kJ/mol}
\]

**Step 4: Calculate \( \Delta H \) for the reaction.**
\[
\Delta H = (\text{Energy of bonds broken}) - (\text{Energy of bonds formed})
\]
\[
\Delta H = 1339 \, \text{kJ/mol} - 1293 \, \text{kJ/mol} = 46 \, \text{kJ/mol}
\]

### Conclusion
The change in enthalpy (\( \Delta H \)) for this hypothetical reaction is \( +46 \, \text{kJ/mol} \), indicating that the reaction is endothermic. Adjust your bonds broken and formed as per the actual reaction you are analyzing.