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To calculate the enthalpy of combustion for ethane \((C_2H_6)\) using bond enthalpies, we follow these steps:

### Balanced Combustion Reaction:
\[
2 \ C_2H_6(g) + 7 \ O_2(g) \rightarrow 4 \ CO_2(g) + 6 \ H_2O(l)
\]

### Bond Enthalpy Approach:
We need to consider:
1. **Bonds Broken** (Reactants):
   - 2 molecules of ethane \((C_2H_6)\)
   - 7 molecules of oxygen \((O_2)\)

2. **Bonds Formed** (Products):
   - 4 molecules of carbon dioxide \((CO_2)\)
   - 6 molecules of water \((H_2O)\)

### Bond Enthalpies (approximate values in \( kJ/mol \)):
- **C-H bond** (in \(C_2H_6\)): 413 \(kJ/mol\)
- **C-C bond** (in \(C_2H_6\)): 347 \(kJ/mol\)
- **O=O bond** (in \(O_2\)): 498 \(kJ/mol\)
- **C=O bond** (in \(CO_2\)): 799 \(kJ/mol\)
- **O-H bond** (in \(H_2O\)): 467 \(kJ/mol\)

### Step 1: Energy to Break Bonds (Reactants):
For 2 molecules of \(C_2H_6\), and 7 molecules of \(O_2\):
- Each \(C_2H_6\) has 6 C-H bonds and 1 C-C bond.
  \[
  \text{Total C-H bonds: } 2 \times 6 = 12 \text{ bonds}
  \]
  \[
  \text{Total C-C bonds: } 2 \times 1 = 2 \text{ bonds}
  \]
  \[
  \text{Energy to break C-H bonds: } 12 \times 413 = 4956 \ kJ
  \]
  \[
  \text{Energy to break C-C bonds: } 2 \times 347 = 694 \ kJ
  \]
  \[
  \text{Energy to break O=O bonds: } 7 \times 498 = 3486 \ kJ
  \]
  Total energy to break all bonds:
  \[
  4956 + 694 + 3486 = 9136 \ kJ
  \]

### Step 2: Energy to Form Bonds (Products):
For 4 molecules of \(CO_2\) and 6 molecules of \(H_2O\):
- Each \(CO_2\) has 2 C=O bonds.
  \[
  \text{Total C=O bonds: } 4 \times 2 = 8 \text{ bonds}
  \]
  \[
  \text{Total O-H bonds: } 6 \times 2 = 12 \text{ bonds}
  \]
  \[
  \text{Energy to form C=O bonds: } 8 \times 799 = 6392 \ kJ
  \]
  \[
  \text{Energy to form O-H bonds: } 12 \times 467 = 5604 \ kJ
  \]
  Total energy released by forming bonds:
  \[
  6392 + 5604 = 11996 \ kJ
  \]

### Step 3: Calculate the Enthalpy of Combustion:
The enthalpy change (\(\Delta H\)) is the difference between the energy required to break bonds and the energy released when new bonds form:
\[
\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed}
\]
\[
\Delta H = 9136 \ kJ - 11996 \ kJ = -2860 \ kJ
\]

Thus, the enthalpy of combustion for ethane is \(-2860 \ kJ\).

Would you like further details or have any questions?

Here are 5 related questions you might explore:
1. How does the enthalpy of combustion change for different hydrocarbons?
2. Why are bond enthalpy values different for various chemical bonds?
3. How does temperature affect the enthalpy of combustion?
4. How would the enthalpy of combustion differ if ethane combusted in a non-oxygenated environment?
5. What role does bond dissociation energy play in chemical reaction rates?

**Tip**: When calculating the enthalpy of combustion using bond enthalpies, always ensure to use the appropriate phase for reactants and products (e.g., \(H_2O(l)\) vs. \(H_2O(g)\)) as it affects the values.

Use the bond enthalpy values to calculate the enthalpy of combustion for ethane, 2 C2H6(g) + 7 O2(g) -> 4 CO2(g) + 6 H2O(l).

Learn how to calculate the enthalpy of combustion for ethane using bond enthalpy values. This detailed step-by-step solution includes bond enthalpy calculations for C-H, C-C, O=O, C=O, and O-H bonds, showing how to find the enthalpy change for the combustion reaction.

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