A reaction has ΔH = +80 kJ/mol and ΔS = +200 J/(mol·K). At 25°C (298 K), ΔG is positive and the reaction is non-spontaneous. At what temperature does it become spontaneous?
AIt is never spontaneous because ΔH is positive
BAbove approximately 400 K, where TΔS exceeds ΔH
CBelow 298 K, where the entropy contribution decreases
DOnly at absolute zero, where entropy has no effect
Spontaneity switches when ΔG = 0, which occurs at T = ΔH/ΔS = 80,000 J / 200 J·K⁻¹ = 400 K. Above this temperature, TΔS > ΔH and ΔG becomes negative. This is the mixed-sign case (positive ΔH, positive ΔS) where temperature is the deciding factor — the entropy contribution grows with T until it overwhelms the unfavorable enthalpy. Option A is a common misconception: endothermic reactions CAN be spontaneous at high enough temperatures.
Question 2 Multiple Choice
Ice melts spontaneously at 25°C but does not melt spontaneously at −25°C. Which explanation correctly applies the Gibbs equation?
AAt 25°C, ΔH becomes negative because of the warmer surroundings
BAt −25°C, entropy decreases when ice melts, making ΔS negative
CAt 25°C, TΔS exceeds ΔH, making ΔG negative; at −25°C, ΔH dominates and ΔG is positive
DSpontaneity depends only on enthalpy; at 25°C there is more energy available to break bonds
Melting ice has positive ΔH (energy absorbed to break hydrogen bonds) and positive ΔS (liquid is more disordered than solid). At low temperatures, ΔG = ΔH − TΔS is positive because TΔS is small. At high temperatures, TΔS grows until it exceeds ΔH and ΔG turns negative. The crossover is the melting point. Option D is incorrect because ΔH does not change with temperature in this simplified treatment — it is T that acts as the switch through the TΔS term.
Question 3 True / False
A reaction that releases heat (negative ΔH) is typically spontaneous under standard conditions.
TTrue
FFalse
Answer: False
Exothermic reactions are spontaneous at all temperatures only when ΔS is also positive. If ΔH is negative but ΔS is also negative, ΔG = ΔH − TΔS becomes positive at high temperatures (where the −TΔS term, which is positive when ΔS is negative, becomes large enough to outweigh ΔH). Spontaneity requires ΔG < 0, which depends on the combination of both ΔH and ΔS, weighted by temperature.
Question 4 True / False
For a reaction with both negative ΔH and negative ΔS, increasing temperature makes the reaction less likely to be spontaneous.
TTrue
FFalse
Answer: True
When both ΔH < 0 and ΔS < 0, ΔG = ΔH − TΔS = (negative) − T(negative) = (negative) + T|ΔS|. As T increases, the positive T|ΔS| term grows, eventually making ΔG positive and the reaction non-spontaneous. The entropy term, which opposes spontaneity here, becomes more influential at higher temperatures. This is why some reactions that are favorable at low temperatures become unfavorable when heated.
Question 5 Short Answer
Why does temperature act as a 'switch' for spontaneity in reactions where ΔH and ΔS have opposite signs, but not in reactions where they have the same sign?
Think about your answer, then reveal below.
Model answer: When ΔH and ΔS have opposite signs, the two terms in ΔG = ΔH − TΔS work against each other — one favors spontaneity while the other opposes it. Temperature controls the weight given to the entropy term (TΔS), so at some critical temperature the balance tips: ΔG switches sign. When ΔH and ΔS have the same sign, both terms favor the same outcome (or both oppose it), so no temperature can reverse the sign of ΔG — the reaction is spontaneous at all temperatures or non-spontaneous at all temperatures.
The equation ΔG = ΔH − TΔS makes this mechanical: if ΔH < 0 and ΔS > 0, both terms make ΔG negative regardless of T — always spontaneous. If ΔH > 0 and ΔS < 0, both terms make ΔG positive — never spontaneous. The mixed cases (+/− or −/+) pit the two thermodynamic drives against each other, and temperature determines which wins.