A reaction has ΔH = +80 kJ/mol and ΔS = +200 J/(mol·K). At what temperatures is this reaction spontaneous?
ANever — endothermic reactions cannot be spontaneous because they increase the system's energy
BOnly at very low temperatures, where entropy changes are negligible
CAbove T = 400 K, where the TΔS term exceeds ΔH
DAt all temperatures, because positive ΔS always guarantees spontaneity
ΔG = ΔH − TΔS. For spontaneity, ΔG < 0. With ΔH = +80,000 J/mol and ΔS = +200 J/(mol·K), ΔG < 0 requires T > 80,000/200 = 400 K. Above this temperature, the entropy term TΔS dominates and ΔG goes negative. Option A embodies the key misconception: endothermic reactions can be spontaneous when entropy increases enough. The TS term in G is precisely what allows entropy to overcome unfavorable enthalpy at high temperatures.
Question 2 Multiple Choice
At the melting point T_m of ice, solid ice and liquid water coexist at equilibrium under constant pressure. What can be said about their Gibbs free energies?
AG_liquid < G_solid, which is why melting is spontaneous
BG_solid < G_liquid, which is why the solid phase remains present
CG_solid = G_liquid at T_m; neither phase is preferred
DG is undefined at a phase transition because entropy is discontinuous
Phase equilibrium is the condition G_solid = G_liquid. If G_liquid were lower, all the ice would melt; if G_solid were lower, all the water would freeze. Coexistence requires equality. Below T_m the solid has lower G and is stable; above T_m the liquid wins. The transition point is exactly where the two G curves cross. This is the Gibbs criterion for phase equilibrium, and it makes phase transitions transparent: you just find where two G surfaces intersect.
Question 3 True / False
A reaction releases heat (ΔH < 0). This guarantees the reaction is spontaneous at most temperatures.
TTrue
FFalse
Answer: False
ΔG = ΔH − TΔS. Even with ΔH < 0, if ΔS is also negative (the reaction decreases disorder), then at high temperatures TΔS becomes a large positive number, making ΔG = ΔH − TΔS positive. A classic example: crystallization releases heat (ΔH < 0) but greatly reduces entropy (ΔS < 0), so it becomes non-spontaneous above a certain temperature. Spontaneity requires the full ΔG criterion; enthalpy alone is not sufficient.
Question 4 True / False
The Gibbs free energy G is the appropriate thermodynamic potential for processes occurring at constant temperature and pressure.
TTrue
FFalse
Answer: True
Different fixed conditions call for different natural potentials. At constant T and V, use Helmholtz free energy F = U − TS. At constant T and P — the usual condition for chemistry open to the atmosphere and for most biological processes — use G = H − TS = U + PV − TS. At constant T and P, spontaneous processes decrease G, and equilibrium occurs at minimum G. This is why G dominates chemistry and biochemistry: lab reactions and living cells both operate under approximately constant pressure.
Question 5 Short Answer
Why can an endothermic reaction (ΔH > 0) still proceed spontaneously? What determines whether it will?
Think about your answer, then reveal below.
Model answer: Spontaneity is governed by ΔG = ΔH − TΔS, not by ΔH alone. If the reaction increases entropy (ΔS > 0), then at sufficiently high temperature the term TΔS becomes large, making ΔG = ΔH − TΔS negative even when ΔH is positive. The crossover temperature is T = ΔH/ΔS; above this temperature, entropy wins the competition with enthalpy and the reaction proceeds spontaneously.
This captures the central conceptual insight of Gibbs free energy: it collapses the two-law requirement — minimize energy (enthalpy) and maximize entropy — into a single temperature-weighted competition. At low temperatures enthalpy dominates; at high temperatures entropy dominates. G quantifies exactly which effect wins at any given T and P.