Question

The balanced equation for methane is CH4 + 2 O2 ---> CO2 + 2 H2O The bond energies for the reactants and products are provided in the chart below:

Bond C-H has Bond Energy 413 kJ/mol
Bond O-H has Bond Energy 467 kJ/mol
Bond O=O has Bond Energy 498 kJ/mol
Bond C=) has Bond Energy 799 kJ/mol

Using the bond energies in the chart, calculate the enthalpy (total energy change) for the combustion of ethanol.
A. 2648 kJ/mol
B. -818 kJ/mol
C. -1484 kJ/mol
D. 3466 kJ/mol

Answer 1

To calculate the enthalpy change for the combustion of methane, we will use the bond energies provided and apply the following method:

1. **Identify the bonds in the reactants and products.**
2. **Calculate the total energy required to break the bonds in the reactants.**
3. **Calculate the total energy released when new bonds are formed in the products.**
4. **Determine the enthalpy change (ΔH) for the reaction.**

### The Balanced Equation
The balanced equation for the combustion of methane is:
\[ \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \]

### Step 1: Identify Bonds
**Reactants:**
- **CH4** has 4 C-H bonds
- **2 O2** has 2 O=O bonds

**Products:**
- **CO2** has 2 C=O bonds
- **2 H2O** has 4 O-H bonds (2 for each water molecule)

### Step 2: Calculate Energy to Break Bonds in Reactants
Breaking the bonds:
- C-H bonds: \(4 \times 413 \, \text{kJ/mol} = 1652 \, \text{kJ/mol}\)
- O=O bonds: \(2 \times 498 \, \text{kJ/mol} = 996 \, \text{kJ/mol}\)

Total energy to break bonds in reactants:
\[
\text{Energy to break} = 1652 \, \text{kJ/mol} + 996 \, \text{kJ/mol} = 2648 \, \text{kJ/mol}
\]

### Step 3: Calculate Energy Released in Products
Forming the bonds:
- C=O bonds: \(2 \times 799 \, \text{kJ/mol} = 1598 \, \text{kJ/mol}\)
- O-H bonds: \(4 \times 467 \, \text{kJ/mol} = 1868 \, \text{kJ/mol}\)

Total energy released when new bonds are formed:
\[
\text{Energy released} = 1598 \, \text{kJ/mol} + 1868 \, \text{kJ/mol} = 3466 \, \text{kJ/mol}
\]

### Step 4: Calculate Enthalpy Change (ΔH)
\[
\Delta H = \text{Energy to break bonds} - \text{Energy released}
\]
\[
\Delta H = 2648 \, \text{kJ/mol} - 3466 \, \text{kJ/mol} = -818 \, \text{kJ/mol}
\]

### Final Answer
Thus, the enthalpy change (ΔH) for the combustion of methane is:
**B. -818 kJ/mol**