Under standard conditions, a negative ∆G° indicates the reaction is spontaneous.

Under standard conditions, a negative ∆G° indicates the reaction is spontaneous.

• A. True
• B. False
• C. Only at high temperature
• D. Only if ∆H is negative

Answer: (A)

The main idea is that the sign of the standard Gibbs free energy change tells you thermodynamic favorability under standard conditions. If ∆G° is negative, the process tends to proceed toward products when starting from standard states, so it is spontaneous in that standard state. This comes from the relationship ∆G° = -RT ln K: a negative ∆G° corresponds to a equilibrium constant K greater than 1, meaning products are favored at equilibrium. So under standard conditions (1 atm for gases, 1 M for solutions, at 25°C), the reaction will proceed spontaneously in the forward direction. Remember that spontaneity refers to thermodynamics, not rate—a negative ∆G° indicates the reaction can run without input of external energy, but it doesn’t guarantee it happens quickly.

The main idea is that the sign of the standard Gibbs free energy change tells you thermodynamic favorability under standard conditions. If ∆G° is negative, the process tends to proceed toward products when starting from standard states, so it is spontaneous in that standard state. This comes from the relationship ∆G° = -RT ln K: a negative ∆G° corresponds to a equilibrium constant K greater than 1, meaning products are favored at equilibrium. So under standard conditions (1 atm for gases, 1 M for solutions, at 25°C), the reaction will proceed spontaneously in the forward direction. Remember that spontaneity refers to thermodynamics, not rate—a negative ∆G° indicates the reaction can run without input of external energy, but it doesn’t guarantee it happens quickly.