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 \(\Delta H^{\circ}\).

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:

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

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°.

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 \(\Delta H^{\circ}\).