The dissolution of ammonium nitrate in water is endothermic (ΔH > 0) yet occurs spontaneously at room temperature. What is the correct thermodynamic explanation?
AThe sign convention for ΔH must be wrong — all spontaneous processes at room temperature release heat
BThe entropy increase from dissolution (ΔS > 0) is large enough that the −TΔS term outweighs ΔH, making ΔG < 0
CDissolution is a physical rather than chemical process, so thermodynamic criteria for spontaneity do not apply
DThe reaction is not truly spontaneous; external stirring supplies the necessary energy to drive it
This is the canonical counterexample to the misconception that spontaneous processes always release heat. ΔG = ΔH − TΔS, and spontaneity requires only ΔG < 0 — not ΔH < 0. When dissolution disperses tightly ordered ions into solution, the entropy increase is large. At room temperature, TΔS exceeds ΔH, so ΔG is negative. This is exactly the entropic driving force at work: the system gains enough microstates from the ions' new translational freedom to more than compensate for the energy cost of breaking the lattice.
Question 2 Multiple Choice
A reaction has ΔH = +50 kJ/mol and ΔS = +200 J/(mol·K). At what temperatures is this reaction spontaneous?
ANever — positive ΔH means the reaction always absorbs heat, so ΔG is always positive
BOnly at very low temperatures, where entropy effects are small and ΔH dominates
CAt all temperatures, because positive ΔS always drives spontaneity
DAt temperatures above 250 K, where TΔS (= T × 0.200 kJ/K) exceeds ΔH (50 kJ)
ΔG = ΔH − TΔS. With ΔH = +50 kJ and ΔS = +0.200 kJ/K, ΔG = 50 − 0.200T. This is negative when T > 250 K. Below 250 K, the T multiplier is too small to make −TΔS overcome the unfavorable ΔH. This pattern — endothermic reactions that become spontaneous above a crossover temperature — appears throughout chemistry: protein denaturation, some dissolution processes, and many gas-phase reactions are entropy-driven in exactly this way.
Question 3 True / False
A reaction with ΔH < 0 favors spontaneity because, at the molecular level, it increases the total number of microstates available to the universe — even if the system's own entropy decreases.
TTrue
FFalse
Answer: True
This is the deep insight: enthalpy is also secretly an entropy argument, just operating on the surroundings. When a reaction releases heat (ΔH < 0), that energy flows into the surroundings and increases the number of microstates available to those surroundings (more thermal motion modes become accessible). The second law requires only that the *total* entropy of system plus surroundings increases. Exothermic reactions can therefore be spontaneous even when ΔS_system < 0 — the surroundings' entropy gain wins.
Question 4 True / False
At high temperatures, whether a reaction is spontaneous is determined primarily by its enthalpy change, because heat effects are more pronounced at higher temperatures.
TTrue
FFalse
Answer: False
This is precisely backwards. At high temperatures, the −TΔS term in ΔG = ΔH − TΔS is amplified by the large T multiplier, making entropy the dominant factor. Even modest entropy increases become thermodynamically decisive at high T. Enthalpy dominates at low temperatures, where the T multiplier is small. This is why reactions that are entropically unfavorable (ΔS < 0) become less spontaneous as temperature rises, and why endothermic reactions with ΔS > 0 become spontaneous above a characteristic temperature.
Question 5 Short Answer
Explain in molecular terms why the melting of ice is spontaneous above 273 K but freezing is spontaneous below 273 K, using the concept of competing microstate contributions.
Think about your answer, then reveal below.
Model answer: Above 273 K, the liquid state has far more accessible microstates (translational and rotational freedom of water molecules) than the crystal. The entropy gain of the system (TΔS, with large T) exceeds the energy cost of breaking hydrogen bonds (ΔH). ΔG < 0, so melting is spontaneous. Below 273 K, T is small enough that TΔS no longer outweighs ΔH — the energy released to the surroundings by forming hydrogen bonds (increasing surroundings' microstates) dominates. At exactly 273 K, ΔH = TΔS, so ΔG = 0 and the two phases coexist in equilibrium.
The crossover at 273 K is not arbitrary — it is the temperature at which the two entropy contributions (system disorder vs surroundings' thermal modes) exactly balance. The Gibbs energy packages this competition into a single number: G < 0 for the phase that wins at each temperature. This is why every substance has a characteristic melting point: it is the temperature at which ΔG_fusion = 0.