Question

Octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is a liquid hydrocarbon at room temperature that is the primary constituent of gasoline. (a) Write a balanced equation for the combustion of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ}\).

Short answer

The balanced chemical equation for the combustion of octane is: \(\mathrm{C}_{8} \mathrm{H}_{18} + \frac{25}{2} \mathrm{O}_{2} \rightarrow 8 \mathrm{CO}_{2} + 9 \mathrm{H}_{2} \mathrm{O}\) We predict that \(\Delta G^{\circ}\) for this reaction is more negative than \(\Delta H^{\circ}\) without using any thermochemical data.

Step-by-step solution

Balance the chemical equation for octane combustion

First, we'll start by writing the unbalanced equation for the reaction. The combustion of octane (C8H18) forms carbon dioxide (CO2) and liquid water (H2O): \(\mathrm{C}_{8} \mathrm{H}_{18} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O}\) Next, we'll balance the equation by ensuring that the same number of each type of atom is present on both sides of the equation. We have 8 carbon atoms, 18 hydrogen atoms, and 2 oxygen atoms on the left side. To balance the carbon, we'll need 8 CO₂ molecules on the right side: \(\mathrm{C}_{8} \mathrm{H}_{18} \rightarrow 8 \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O}\) Now, we need to balance the hydrogen atoms. We have 18 hydrogen atoms in the octane molecule, so we need 9 H₂O molecules on the right side: \(\mathrm{C}_{8} \mathrm{H}_{18} \rightarrow 8 \mathrm{CO}_{2} + 9 \mathrm{H}_{2} \mathrm{O}\) Finally, we need to balance the oxygen atoms. We have 8 × 2 = 16 oxygen atoms in the CO₂ molecules and 9 oxygen atoms in the H₂O molecules, for a total of 25 oxygen atoms. For this, we need to add 25/2 O₂ molecules on the left side of the equation: \(\mathrm{C}_{8} \mathrm{H}_{18} + \frac{25}{2} \mathrm{O}_{2} \rightarrow 8 \mathrm{CO}_{2} + 9 \mathrm{H}_{2} \mathrm{O}\) That gives us the balanced equation for the combustion of octane:

Balanced Chemical Equation

\( \mathrm{C}_{8} \mathrm{H}_{18} + \frac{25}{2} \mathrm{O}_{2} \rightarrow 8 \mathrm{CO}_{2} + 9 \mathrm{H}_{2} \mathrm{O}\)

Predict the relationship between ∆G° and ∆H°

The change in Gibbs free energy (∆G°) is related to the changes in enthalpy (∆H°) and entropy (∆S°) according to the equation: \(\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\) Since we do not have any thermochemical data, we will have to analyze the relationship between ∆H° and ∆G° qualitatively. Combustion reactions are usually exothermic, meaning they release heat, so ∆H° is expected to be negative. When it comes to entropy (∆S°), the reaction results in more gas molecules being formed than were present initially (8 CO₂ on the right side versus 25/2 O₂ on the left side). This increase in the number of gas molecules leads to an increase in entropy, making ∆S° positive. Since T (temperature) is always positive, the term -T∆S° should be negative. Therefore, the value of ∆G° is expected to be more negative than ∆H°, as the negative value of ∆H° will be further decreased by the negative value of -T∆S°.

Conclusion

In summary, the balanced chemical equation for the combustion of octane is: \(\mathrm{C}_{8} \mathrm{H}_{18} + \frac{25}{2} \mathrm{O}_{2} \rightarrow 8 \mathrm{CO}_{2} + 9 \mathrm{H}_{2} \mathrm{O}\) Additionally, without using any thermochemical data, we predict that \(\Delta G^{\circ}\) for this reaction is more negative than \(\Delta H^{\circ}\).