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:

Question

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