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: 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

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)

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°.

Key concepts

Balanced Chemical Equation

A balanced chemical equation ensures that the number of each type of atom on the reactant side equals the number on the product side. When burning octane (\(\mathrm{C}_{8}\mathrm{H}_{18}(l)\)), you need to produce carbon dioxide and water.
Here's a simple way to balance it:

• Count the carbon atoms in octane. You have 8, so you need 8 CO₂.
• Next, count the hydrogen atoms. There are 18, so you need 9 H₂O, each H₂O having 2 hydrogen atoms.
• Finally, balance the oxygens. You'll end up needing 25 oxygen atoms, which comes from 12.5 O₂, because 12.5 x 2 = 25.

So, the balanced equation is:\[\mathrm{C}_{8}\mathrm{H}_{18}(l) + 12.5\ \mathrm{O}_2(g) \rightarrow 8\ \mathrm{CO}_2(g) + 9\ \mathrm{H}_2\mathrm{O}(l)\]Balancing equations ensures that mass and charge are conserved, making them fundamental to chemical reactions.

Gibbs Free Energy

Gibbs Free Energy (\(\Delta G^{\circ}\)) tells us about the spontaneity of a reaction. It combines enthalpy (\(\Delta H^{\circ}\)), temperature (T), and entropy (\(\Delta S^{\circ}\)) in one equation:\[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\]A negative \(\Delta G^{\circ}\) means that the reaction can occur spontaneously.
In combustion, octane releases a lot of energy as heat, making it an exothermic reaction (\(\Delta H^{\circ} < 0\)). There’s also an increase in entropy because the reaction produces more gas molecules. The outcome is a more negative \(\Delta G^{\circ}\) than \(\Delta H^{\circ}\), reinforcing that the reaction is spontaneous.

Entropy

Entropy (\(\Delta S\)) is a measure of disorder or randomness in a system. When octane burns, the number of gas molecules increases from 12.5 (from O₂) to 17 (8 CO₂ and 9 H₂O).
This transition results in higher entropy, as gases tend to have more disorder than liquids or solids.
During combustion:

• You start with fewer gas molecules as reactants and end up with more as products.
• This increase in randomness contributes positively to \(\Delta S\).

The greater the increase in entropy, the more it can drive a reaction to be spontaneous. This aligns well with the increase in entropy seen during the combustion of octane.