Questions: Legendre Transformations and Thermodynamic Potentials

Question 1

A chemist runs a reaction at constant temperature and pressure and observes that it is spontaneous even though it is endothermic (ΔH > 0). Which explanation is correct?

• A. The reaction is spontaneous because it must be releasing heat — the chemist's ΔH measurement must be wrong
• B. ΔU < 0 is the true spontaneity criterion and must be satisfied here
• C. The entropy increase is large enough that TΔS > ΔH, making ΔG = ΔH − TΔS negative
• D. Volume work (−PΔV) drives the reaction since pressure is held constant

Answer: (C)

At constant T and P, the spontaneity criterion is ΔG < 0, not ΔH < 0 or ΔU < 0. Since G = H − TS, ΔG = ΔH − TΔS. When ΔS is sufficiently positive — the reaction substantially increases disorder — TΔS can exceed ΔH, making ΔG negative even for an endothermic process. Dissolving ammonium nitrate in water is a real example: endothermic but spontaneous because of the large entropy increase. Option B is wrong because ΔU < 0 is only the equilibrium criterion at constant S and V, not at constant T and P.

Question 2

Why do chemists characterize heats of reaction as ΔH rather than ΔU?

• A. Enthalpy is easier to measure calorimetrically than internal energy
• B. H's natural variables are S and P — at constant pressure, ΔH equals the heat absorbed, so it directly captures what a calorimeter measures
• C. Internal energy is only defined for ideal gases, while enthalpy applies to all states of matter
• D. Internal energy is not conserved in chemical reactions, making it unsuitable as a thermodynamic potential

Answer: (B)

The Legendre transformation replaces V with P to produce H = U + PV, with dH = TdS + VdP. At constant pressure (dP = 0), dH = TdS = dq — the heat exchanged reversibly. The natural variables S and P match the experimental constraint (constant P in an open calorimeter), making H the right potential and ΔH the direct measure of heat flow. Option A has some practical truth but is not the thermodynamic reason. Options C and D are false.

Question 3

Legendre transformations lose thermodynamic information — converting from U(S,V) to G(T,P) means you can no longer recover entropy or volume from G.

• T. True
• F. False

Answer: False

Legendre transformations preserve all thermodynamic information — nothing is lost, only re-expressed in different natural variables. From G(T,P), entropy is recovered as S = −(∂G/∂T)_P and volume as V = (∂G/∂P)_T. The transformation is a mathematically exact change of variables, not an approximation. This is why each potential is equally valid as a complete thermodynamic description — they are all equivalent representations of the same physics.

Question 4

At constant temperature and pressure, a spontaneous process usually decreases the system's enthalpy.

• T. True
• F. False

Answer: False

Spontaneity at constant T and P is governed by ΔG = ΔH − TΔS < 0, not by ΔH < 0 alone. Entropy-driven reactions — endothermic processes where disorder increases substantially — can be spontaneous even with ΔH > 0, because TΔS outweighs ΔH. Enthalpy decrease favors spontaneity, but it competes with entropy: at high temperatures, the TΔS term dominates. ΔH < 0 is neither necessary nor sufficient for spontaneity at constant T and P.

Question 5

Why do all four thermodynamic potentials (U, H, A, G) contain the same information, yet the choice of which to use matters enormously in practice?

Model answer: They contain the same information because each is a Legendre transform of the others — mathematically equivalent re-expressions of the same physics in different natural variables. But each potential is minimized at equilibrium only under its own natural constraints. Using the wrong potential for your experimental conditions forces you to track extra work terms explicitly rather than having them automatically absorbed into the potential's structure.

At constant T and P (most laboratory reactions), G is minimized at equilibrium — ΔG < 0 is the spontaneity criterion, and ΔG = 0 defines phase equilibrium. If you insist on working with U, you must separately account for heat exchange and PV work at every step — the bookkeeping G's structure handles automatically. The Legendre transform is precisely the operation that moves the work term inside the potential, making it invisible when the constraint is satisfied. Matching the potential to the constraint converts a complex balance of terms into a single inequality.