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

Expert verified
The balanced chemical equation for the combustion of octane is: C8H18(l) + 12.5O2(g) -> 8CO2(g) + 9H2O(l) ∆G° for this reaction is more negative than ∆H°, indicating the reaction is spontaneous at room temperature.

Step by step solution

01

Write the balanced chemical equation for the combustion of octane

To balance the chemical equation, we need to make sure that the number of carbon, hydrogen, and oxygen atoms in the reactants equals the number of those same atoms in the products. Reactants: C8H18(l) + O2(g) Products: CO2(g) + H2O(l) Start by balancing the carbons: C8H18(l) + O2(g) -> 8CO2(g) + H2O(l) Next, balance the hydrogens: C8H18(l) + O2(g) -> 8CO2(g) + 9H2O(l) Lastly, balance the oxygens: C8H18(l) + 12.5O2(g) -> 8CO2(g) + 9H2O(l) The balanced chemical equation for the combustion of octane is: C8H18(l) + 12.5O2(g) -> 8CO2(g) + 9H2O(l)
02

Predict if ∆G° is more negative or less negative than ∆H°

We know that the relationship between Gibbs free energy (∆G°), enthalpy (∆H°), and entropy (∆S°) is given by the equation: ∆G° = ∆H° - T∆S° In this case, we are not given any specific values for ∆H°, ∆S°, or T, so we cannot calculate the exact value of ∆G°. However, we can still make predictions based on general trends. During the combustion of octane, a large amount of heat is released, which means that the reaction is exothermic (∆H° < 0). Moreover, the reaction has more gas molecules as products than reactants (8 moles of CO2 on the product side versus 12.5 moles of O2 on the reactant side), which leads to an increase in entropy (∆S° > 0). Since both ∆H° and ∆S° have negative and positive values, respectively, their product (-T∆S°) will be negative. Therefore, ∆G° will be more negative than ∆H° in this reaction, indicating the reaction is spontaneous at room temperature. In conclusion, the balanced chemical equation for the given reaction is: C8H18(l) + 12.5O2(g) -> 8CO2(g) + 9H2O(l) And ∆G° for this reaction is more negative than ∆H°.

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