If ΔG° < 0, what does it indicate about spontaneity at standard conditions?

If ΔG° < 0, what does it indicate about spontaneity at standard conditions?

• A. The reaction is spontaneous
• B. The reaction is non-spontaneous
• C. The reaction is at equilibrium
• D. The reaction requires a catalyst

Answer: (A)

A negative standard Gibbs free energy change means the reaction tends to proceed on its own under standard conditions. The system releases free energy as it moves toward products, so no external energy input is required for the forward direction. Under standard conditions, ΔG° relates to the equilibrium constant by ΔG° = -RT ln K°. A negative ΔG° means K° is greater than 1, so products are favored at equilibrium. Keep in mind that spontaneity refers to thermodynamics, not speed— a catalyst can make the reaction go faster, but it doesn’t change whether the process is spontaneous. Conversely, a positive ΔG° would indicate non-spontaneity under standard conditions, and ΔG° = 0 corresponds to equilibrium.

A negative standard Gibbs free energy change means the reaction tends to proceed on its own under standard conditions. The system releases free energy as it moves toward products, so no external energy input is required for the forward direction. Under standard conditions, ΔG° relates to the equilibrium constant by ΔG° = -RT ln K°. A negative ΔG° means K° is greater than 1, so products are favored at equilibrium. Keep in mind that spontaneity refers to thermodynamics, not speed— a catalyst can make the reaction go faster, but it doesn’t change whether the process is spontaneous. Conversely, a positive ΔG° would indicate non-spontaneity under standard conditions, and ΔG° = 0 corresponds to equilibrium.