Is the reaction involving propane and oxygen spontaneous based on the ΔG° value?

Is the reaction involving propane and oxygen spontaneous based on the ΔG° value?

• A. Yes
• B. No
• C. Depends on temperature
• D. Not enough information

Answer: (A)

A spontaneous reaction is one that occurs naturally under given conditions without needing to be driven by an external force. For a reaction, spontaneity can be assessed using the Gibbs free energy change (ΔG°) for the reaction. If ΔG° is negative, the reaction is spontaneous at standard conditions. In the case of propane and oxygen, the combustion reaction produces carbon dioxide and water: \[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \] This reaction is highly exothermic, meaning it releases significant heat energy, which contributes to making ΔG° negative. Therefore, under standard conditions, the combustion of propane is spontaneous since the decrease in free energy indicates that the reaction can proceed without additional energy input. Notably, the spontaneity can also depend on specific conditions such as temperature or pressure, but for standard conditions typically applied, this combustion reaction is spontaneous due to the favorable free energy change. Thus, identifying the spontaneity of the reaction based solely on a negative ΔG° confirms it is indeed spontaneous, which aligns with the provided answer.

A spontaneous reaction is one that occurs naturally under given conditions without needing to be driven by an external force. For a reaction, spontaneity can be assessed using the Gibbs free energy change (ΔG°) for the reaction. If ΔG° is negative, the reaction is spontaneous at standard conditions.

In the case of propane and oxygen, the combustion reaction produces carbon dioxide and water:

[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} ]

This reaction is highly exothermic, meaning it releases significant heat energy, which contributes to making ΔG° negative. Therefore, under standard conditions, the combustion of propane is spontaneous since the decrease in free energy indicates that the reaction can proceed without additional energy input.

Notably, the spontaneity can also depend on specific conditions such as temperature or pressure, but for standard conditions typically applied, this combustion reaction is spontaneous due to the favorable free energy change. Thus, identifying the spontaneity of the reaction based solely on a negative ΔG° confirms it is indeed spontaneous, which aligns with the provided answer.