Octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is a liquid hydrocarbon at room temperature that is a 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 \(\Delta H^{\circ} .\)

To write a balanced equation for the combustion of octane, first, write the general equation: C8H18(l) + O2(g) → CO2(g) + H2O(l) Now, balance the equation: - Balance carbon atoms: There are 8 carbon atoms in C8H18, so 8 CO2 molecules are needed on the product side. - Balance hydrogen atoms: There are 18 hydrogen atoms in C8H18, so 9 H2O molecules are needed on the product side. - Finally, balance oxygen atoms: There are 16 oxygen atoms in 8 CO2 and 9 oxygen atoms in 9 H2O, adding up to 25 oxygen atoms on the right side. Since O2 has two oxygen atoms, we need 25/2, or 12.5, O2 molecules. The balanced equation is: \(C_8H_{18}(l) + 12.5 \space O_2(g) \rightarrow 8 \space CO_2(g) + 9 \space H_2O(l)\)

To predict the relationship between the standard Gibbs Free Energy change (∆G°) and the standard Enthalpy change (∆H°) for this reaction, consider the equation: \(\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ\) For a combustion reaction, which is highly exothermic, the ∆H° value will be largely negative. The reaction also leads to an increase in entropy (∆S°) as the number of gas molecules in the products is higher than in the reactants (8 CO2 gas molecules formed and only 12.5 O2 gas molecules consumed). Thus, the ∆S° value will be positive. Because the temperature (T) is always positive, a positive ∆S° leads to a negative T∆S° term, which implies that the Gibbs Free Energy change (∆G°) would be more negative than the Enthalpy change (∆H°).