Problem 43 - Chemistry The Molecular Science

Question

Explain briefly why the exothermic combustion of propane is product-favored. $$ \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$

Step-by-Step Solution

Answer

The reaction is product-favored due to its exothermic nature and potential for increased entropy, making it energetically favorable.

Step 1: Identify Products and Reactants

In the combustion of propane, the reactants are propane \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})\) and oxygen \(\mathrm{O}_{2}(\mathrm{~g})\). The products are carbon dioxide \(\mathrm{CO}_{2}(\mathrm{~g})\) and water vapor \(\mathrm{H}_{2}\mathrm{O(g)}\).

Step 2: Understand Exothermic Reactions

Exothermic reactions release energy in the form of heat. In the combustion reaction, chemical energy stored in \(\mathrm{C}_{3} \mathrm{H}_{8}\) and \(\mathrm{O}_{2}\) is converted and released as heat, leading to a lower energy state in the products (\(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\)).

Step 3: Consider Energy Favorability

Since energy is released, the chemical system moves to a more stable, lower energy state, which is thermodynamically more favorable. This release of energy is a key reason why the reaction is product-favored.

Step 4: Evaluate Entropy Changes

Entropy, a measure of disorder, typically increases as reactions go from higher-order molecules to products like gases. Here, both reactants and products are gases, but the increase in total number of gas molecules from 6 to 7 can contribute favorably to entropy.

Step 5: Relate to Spontaneity

Combining the exothermic nature (enthalpy change \(\Delta H < 0\)) and increase in entropy (\(\Delta S \geq 0\)), the Gibbs Free Energy \(\Delta G\) is negative, demonstrating the reaction is spontaneous and product-favored.

Key Concepts

Exothermic reaction

Exothermic reactions are fascinating chemical processes that release energy. In the context of propane combustion, the energy is emitted primarily as heat. When propane \(\mathrm{C}_{3} \mathrm{H}_{8}\) combines with oxygen \(\mathrm{O}_{2}\), a lot of energy stored in their chemical bonds is released into the environment. This energy release occurs because the products, carbon dioxide \(\mathrm{CO}_{2}\) and water vapor \(\mathrm{H}_{2}\mathrm{O}\), possess lower energy compared to the reactants.
This decrease in energy means that the products are more stable. In everyday terms, it's like rolling a ball downhill; it wants to reach the lowest energy state naturally. Therefore, exothermic reactions tend to proceed on their own, making them spontaneous under the right conditions. Overall, the heat energy released makes the surroundings warmer, which is a common characteristic of such reactions.

Thermodynamic favorability

Thermodynamic favorability refers to whether a reaction will proceed on its own based on energy and entropy considerations. In the combustion of propane, the reaction is thermodynamically favorable because it releases energy.
The system transitions to a more stable, lower energy state as energy is released in the form of heat. Scientists describe this as being 'energy efficient' because less energy is retained in the form of chemical bonds.

• Lower energy states are typically more stable.
• Stability is associated with less bond energy needed to maintain the structure of product molecules.

Moreover, when a reaction is thermodynamically favorable, it is often because the products are more stable than the reactants. This stability is a key driver in why many reactions occur in the first place, as systems naturally aim to reach more favorable energy conditions.

Entropy change

Entropy is a measure of disorder or randomness in a system. In chemical reactions, especially those involving gases, entropy plays a significant role. During the combustion of propane, the entropy of the system slightly increases because you end up counting more gas molecules in the product side.
Initially, there are 6 molecules of gaseous reactants (1 propane and 5 oxygen), and it changes to 7 molecules of gaseous products (3 carbon dioxide and 4 water vapor).

• With more molecules, there is a greater dispersion of energy and hence more disorder.
• This increase in entropy makes the reaction more likely to occur spontaneously.

It's important to remember that overall changes in entropy, along with changes in enthalpy, contribute to the spontaneity of a reaction as seen in Gibbs Free Energy calculations.

Gibbs Free Energy

Gibbs Free Energy, symbolized as \(\Delta G\), is a vital concept in thermodynamics that combines enthalpy and entropy into a single value to predict reaction spontaneity.
For the combustion of propane, \(\Delta G\) shows spontaneity in energy terms because both the enthalpy \(\Delta H\) and entropy \(\Delta S\) are favorable.

• The reaction is exothermic, so \(\Delta H < 0\) indicating heat release.
• Entropy slightly increases, \(\Delta S \geq 0\), due to a greater number of gas molecules in the products.

These factors make \(\Delta G\) negative, as calculated with the formula \(\Delta G = \Delta H - T\Delta S\), where \(T\) is temperature.
A negative \(\Delta G\) means that the reaction is not only possible but also spontaneous under constant temperature and pressure. It's the conclusive measure scientists use to assess whether a process will naturally occur or require additional energy input.