Question

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

Short answer

The balanced equation for the combustion of octane is: \(C_8H_{18}(l) + 12.5 \space O_2(g) \rightarrow 8 \space CO_2(g) + 9 \space H_2O(l)\). For this combustion reaction, the standard Gibbs Free Energy change (∆G°) is more negative than the standard Enthalpy change (∆H°) because it is highly exothermic and produces more gas molecules, resulting in a positive entropy change (∆S°).

Step-by-step solution

(a) Writing a balanced equation

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: \(C_8H_{18}(l) + 12.5 \space O_2(g) \rightarrow 8 \space CO_2(g) + 9 \space H_2O(l)\)

(b) Predicting the relationship between ∆G° and ∆H°

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°).

Key concepts

Balancing Chemical Equations

Balancing chemical equations is an essential step in chemical reactions. It ensures that the same number of each type of atom appears on both the reactants and products sides of the equation. To balance the combustion reaction of octane ( C_8H_{18}(l) ), start by writing the equation: C_8H_{18}(l) + O_2(g) → CO_2(g) + H_2O(l) .
Begin by balancing the carbon atoms: since there are eight carbon atoms in octane, place 8 CO_2 molecules as products. Next, balance the hydrogen atoms, requiring 9 H_2O molecules for the 18 hydrogen atoms in octane.
Finally, balance the oxygen atoms. Count 25 oxygen atoms needed on the product side, resulting in 12.5 O_2 molecules since each oxygen molecule contains two atoms. The balanced equation ends as: C_8H_{18}(l) + 12.5 O_2(g) ightarrow 8 CO_2(g) + 9 H_2O(l) . Note that sometimes the actual use in practice would multiply through by 2 to avoid having a fractional coefficient.

Gibbs Free Energy

Gibbs Free Energy, denoted as ΔG^ ∘ , helps us determine the spontaneity of a reaction. The formula ΔG^ ∘ = ΔH^ ∘ - T ΔS^ ∘ links the change in free energy with enthalpy change( ΔH^ ∘ ) and entropy change( ΔS^ ∘ ) influenced by temperature (T).
A negative ΔG^ ∘ indicates a spontaneous reaction. For the combustion of octane, anticipate a more negative ΔG^ ∘ than ΔH^ ∘ . This is because ΔH^ ∘ is highly exothermic and ΔS^ ∘ adds further negative impact due to increased disorder among gas molecules.
When the number of product gas molecules surpasses the reactants, the increase in disorder leads to a positive ΔS^ ∘ . Combined with temperature, this makes T ΔS^ ∘ a negative value, driving ΔG^ ∘ to be more negative.

Enthalpy Change

Enthalpy change ( ΔH^ ∘ ) quantifies the heat absorbed or released during a reaction under constant pressure. During the combustion of octane, ΔH^ ∘ is significantly negative, manifesting that the process is exothermic, as it releases a large amount of energy.
Combustion reactions inherently release energy by breaking chemical bonds and forming more stable products, such as CO_2 and H_2O . It's crucial within thermodynamics, particularly predicting whether a process is endothermic (absorbs heat) or exothermic (releases heat).
A negative ΔH^ ∘ in a combustion reaction is anticipated since hydrocarbons heavily release thermal energy when reacting with oxygen due to strong bond formation in products.

Entropy Change

Entropy change ( ΔS^ ∘ ) measures the increase in the disorder or randomness within a system. For combustion reactions like octane's, ΔS^ ∘ appears positive.
When octane ( C_8H_{18}(l) ) combusts, it results in more gas molecules ( 8 CO_2 vs 12.5 O_2 consumed). This conversion, by amplifying the number of gas molecules, increases entropy and disorder in the system.
An understanding of entropy is pivotal, especially concerning how it influences the ΔG^ ∘ equation. A positive ΔS^ ∘ decreases ΔG^ ∘ , encouraging a reaction to be spontaneous when paired with an exothermic ΔH^ ∘ .