C8H18(l) + 252O2(g) ? 8CO2(g) + 9H2O(g)
1) Calculate ?Hrxn for the combustion of octane (C8H18), a component of gasoline, by using average bond energies. Express your answer using three significant figures.
2) Calculate ?Hrxn for the combustion of octane by using enthalpies of formation from Appendix IIB in the textbook. The standard enthalpy of formation of C8H18 is -250 kJ/mol. Express your answer using three significant figures.
What is the percent difference between the two results? Express your answer using two significant figures.
The Correct Answer and Explanation is:
To solve this, we will follow two different approaches to calculate the enthalpy change (ΔH_rxn) of the combustion of octane (C₈H₁₈):
1) Using Average Bond Energies
The combustion reaction is:
$$
\text{C}8\text{H}{18}(l) + \frac{25}{2}\text{O}_2(g) \rightarrow 8\text{CO}_2(g) + 9\text{H}_2\text{O}(g)
$$
Step 1: Bonds broken in reactants
- In 1 mol of C₈H₁₈, approximate bonds:
- C–C: 7
- C–H: 18
- 12.5 mol of O=O bonds (each O₂ molecule has a double bond)
Bond energies (approximate in kJ/mol):
- C–C: 348
- C–H: 412
- O=O: 498
Total energy to break bonds:
$$
(7 \times 348) + (18 \times 412) + (12.5 \times 498) = 2436 + 7416 + 6225 = 16,077 \, \text{kJ}
$$
Step 2: Bonds formed in products
- 8 CO₂: each has 2 C=O → 16 C=O bonds
- 9 H₂O: each has 2 O–H → 18 O–H bonds
Bond energies:
- C=O (in CO₂): 799
- O–H: 463
Total energy released:
$$
(16 \times 799) + (18 \times 463) = 12,784 + 8,334 = 21,118 \, \text{kJ}
$$
ΔH_rxn ≈ Bonds broken – Bonds formed:
$$
\Delta H_\text{rxn} \approx 16,077 – 21,118 = -5,041 \, \text{kJ/mol}
$$
2) Using Standard Enthalpies of Formation
$$
\Delta H^\circ_\text{rxn} = \sum \Delta H_f^\circ (\text{products}) – \sum \Delta H_f^\circ (\text{reactants})
$$
Values (in kJ/mol):
- ΔH_f°(C₈H₁₈) = –250
- ΔH_f°(CO₂) = –393.5
- ΔH_f°(H₂O, g) = –241.8
- ΔH_f°(O₂) = 0 (element)
$$
\Delta H_\text{rxn} = [8(-393.5) + 9(-241.8)] – [-250 + 0]
= [-3148 + (-2176.2)] – (-250)
= -5324.2 + 250 = \boxed{-5074 \, \text{kJ/mol}}
$$
3) Percent Difference
$$
\text{Percent difference} = \left| \frac{-5041 – (-5074)}{-5074} \right| \times 100 = \left| \frac{33}{5074} \right| \times 100 \approx \boxed{0.65\%}
$$
The combustion of octane (C₈H₁₈), a key gasoline component, is highly exothermic, meaning it releases a large amount of energy. We estimated this energy change (ΔH_rxn) using two methods.
The first method used average bond energies, which estimate enthalpy change by subtracting the total energy released forming bonds in products from the total energy required to break bonds in reactants. This method yielded a value of approximately –5041 kJ/mol. While useful for approximations, it often lacks precision because average bond energies don’t account for the specific molecular environments.
The second method used standard enthalpies of formation, which is more accurate for known compounds. It calculates ΔH_rxn by taking the difference in enthalpy between products and reactants based on their respective heats of formation. This gave us –5074 kJ/mol, very close to the first result.
The percent difference between the two values is about 0.65%, indicating a high level of agreement despite the simplifications in bond energy calculations. This low percent difference shows that bond energy estimations can be reasonably accurate for combustion reactions, especially when dealing with well-characterized molecules like octane.
Overall, the enthalpy of combustion provides critical insight into the energy content of fuels. The large negative ΔH_rxn confirms why octane is a powerful fuel—its combustion releases substantial energy, which powers engines. However, accurate thermodynamic data (like enthalpies of formation) should be used in precise engineering or scientific applications.
