Questions: Helmholtz and Gibbs Free Energy: Maximum Work

Question 1

A reaction has ΔG° = +20 kJ/mol at 298 K. Which statement is most accurate?

• A. The reaction is thermodynamically impossible and will never proceed in the forward direction under any conditions
• B. At standard conditions the reverse reaction is spontaneous, but the forward reaction can still proceed spontaneously if the reaction quotient Q is less than K_eq
• C. The system is at equilibrium under standard conditions because ΔG° represents the equilibrium free energy
• D. The reaction is spontaneous because the positive sign indicates energy is being released to the system

Answer: (B)

ΔG° is the free energy change at standard conditions (all species at 1 bar, 298 K). A positive ΔG° means the reverse reaction is spontaneous from standard state, and K_eq < 1. But ΔG = ΔG° + RT ln(Q) — if Q < K_eq (e.g., starting with pure reactants and no products), ΔG can be negative even when ΔG° is positive, and the reaction proceeds forward. ΔG° governs the equilibrium position; actual spontaneity at non-standard conditions depends on both ΔG° and Q.

Question 2

An engineer analyzes an isothermal process occurring inside a sealed rigid vessel (constant T and V). Which thermodynamic potential determines the maximum work available from this process?

• A. Gibbs free energy G = H − TS, because it applies to all isothermal processes regardless of volume constraint
• B. Helmholtz free energy A = U − TS, because constant temperature and volume is the natural domain of A, and −ΔA equals maximum work
• C. Enthalpy H, because the process occurs at constant pressure inside the rigid container
• D. Either A or G can be used interchangeably at constant temperature

Answer: (B)

The choice of free energy depends on the constraints. At constant T and V, Helmholtz free energy A is the correct potential: dA ≤ 0 for spontaneous processes, and −ΔA gives the maximum useful work extractable. At constant T and P (the more common engineering case), Gibbs free energy G applies and −ΔG gives maximum non-PV work. Using G for a constant-volume process introduces a PV term that doesn't correspond to any physical work in the system.

Question 3

At equilibrium, ΔG = 0, not ΔG < 0; a negative ΔG indicates that the system has not yet reached equilibrium and will spontaneously move toward it.

• T. True
• F. False

Answer: True

ΔG = 0 is the condition for equilibrium at constant T and P — the system has minimized its Gibbs free energy. ΔG < 0 means the system can still lower its free energy by converting more reactants to products, so the forward reaction proceeds spontaneously. ΔG > 0 means the reverse is spontaneous. The common error is thinking 'spontaneous' and 'at equilibrium' are synonymous — they are not. Equilibrium is the state toward which spontaneous processes drive the system.

Question 4

A reaction with ΔG < 0 will proceed to completion, consuming most reactants and converting them mostly to products.

• T. True
• F. False

Answer: False

ΔG < 0 (at standard conditions, ΔG°) means products are favored at equilibrium — K_eq > 1. But this does not mean complete conversion. The reaction proceeds until ΔG = 0 (equilibrium), at which point both reactants and products are present. How far it goes depends on the magnitude of K_eq: a very large K means nearly complete conversion, but true completion (K → ∞) is only asymptotically approached. ΔG < 0 indicates direction and tendency, not completion.

Question 5

Why can Gibbs free energy determine whether a process is spontaneous without explicitly tracking the entropy change of the surroundings?

Model answer: For a process at constant T and P, the heat exchanged with the surroundings equals the enthalpy change: q_p = ΔH. This heat transfer changes the surroundings' entropy by ΔS_surr = −ΔH/T. The second law requires ΔS_total = ΔS_sys + ΔS_surr ≥ 0, which becomes ΔS_sys − ΔH/T ≥ 0, or equivalently ΔH − TΔS_sys ≤ 0. Defining G = H − TS, this is just ΔG ≤ 0. Gibbs free energy bundles both the system's entropy gain and the surroundings' entropy cost into a single system-property quantity, so you only need to track the system.

This is the practical power of free energy: it converts the second law from a statement about the entire universe into a criterion applied only to the system. No explicit accounting of the surroundings is required — their contribution is embedded in the enthalpy term ΔH.