Effect of Temperature on Spontaneity: Sign of ΔH and ΔS
Chemistry · Thermodynamics · NEET
A reaction is spontaneous when ΔG = ΔH − TΔS is negative. The signs of ΔH and ΔS decide this, and temperature (T) can flip the answer. Memory hook: "Negative ΔG goes." Both ΔH negative and ΔS positive means it goes at every temperature; both positive means it needs HIGH temperature; both negative means it needs LOW temperature.
ΔG changes with temperature. When ΔH and ΔS share a sign, the ΔG line crosses zero at T = ΔH/ΔS: above it or below it the reaction becomes spontaneous. The table gives the four NEET cases.
Your doubts, answered
When is a reaction spontaneous at ALL temperatures?
When ΔH is negative AND ΔS is positive. In ΔG = ΔH − TΔS, the first term (ΔH) is already negative, and −TΔS is also negative because ΔS is positive and T is always positive. So ΔG stays negative no matter how big or small T is. This is the exact NEET 2016 answer (option C).
When is a reaction NON-spontaneous at all temperatures?
When ΔH is positive AND ΔS is negative. Now ΔH is positive and −TΔS is also positive (because ΔS is negative). Both terms are positive, so ΔG is always positive. The reaction never goes on its own, at any temperature.
If both ΔH and ΔS are POSITIVE, is the reaction spontaneous?
Only at HIGH temperature. ΔH is positive (works against the reaction), but −TΔS is negative (helps it). At high T, the −TΔS term becomes large enough to beat ΔH, so ΔG turns negative. The switch happens at T = ΔH/ΔS. Below that T it is non-spontaneous; above it, spontaneous.
If both ΔH and ΔS are NEGATIVE, when does it go?
Only at LOW temperature. ΔH is negative (helps), but −TΔS is positive (works against, because ΔS is negative). At low T that positive term is small, so ΔH wins and ΔG is negative. At high T the −TΔS term grows and makes ΔG positive, stopping the reaction.
How do I find the exact temperature where a reaction becomes spontaneous?
Set ΔG = 0. Then ΔH − TΔS = 0, so T = ΔH/ΔS. This is the crossover temperature. Watch the direction: if ΔH and ΔS are both positive, the reaction is spontaneous ABOVE this T; if both negative, spontaneous BELOW this T. Always convert ΔH to J to match ΔS in J/K before dividing.
Why must ΔH and ΔS have the SAME units before I divide?
ΔH is usually given in kJ (or kJ/mol) and ΔS in J/K. If you divide directly you get a wrong T by 1000 times. First change ΔH to joules (multiply kJ by 1000), then T = ΔH/ΔS gives kelvin. This unit slip is the most common mistake in NEET calculation questions.
⚠️ The NEET trap ✗ Picking ΔH < 0 and ΔS < 0 as 'spontaneous at all temperatures' because ΔH is negative. ✓ Spontaneous at ALL temperatures needs ΔH < 0 AND ΔS > 0 (both terms negative in ΔG = ΔH − TΔS). ΔH < 0 with ΔS < 0 is spontaneous only at LOW temperature. 🧠 For 'all temperatures', T must cancel out — that only happens when ΔH < 0 and ΔS > 0. If ΔS is negative, temperature always matters, so it is not 'all temperatures'.
Real NEET questions
NEET 2016
The correct thermodynamic conditions for the spontaneous reaction at all temperatures is:
A · ΔH < 0 and ΔS = 0
B · ΔH > 0 and ΔS < 0
C · ΔH < 0 and ΔS > 0 ✓
D · ΔH < 0 and ΔS < 0
Solution: Use ΔG = ΔH − TΔS. For the reaction to go at EVERY temperature, ΔG must be negative for all T. This needs ΔH < 0 (first term negative) and ΔS > 0 (so −TΔS is also negative for any positive T). Both parts are negative, so ΔG < 0 always. Option D (ΔH < 0, ΔS < 0) only works at low T, so it is the trap answer.
NEET 2017
For a given reaction, ΔH = 35.5 kJ mol⁻¹ and ΔS = 83.6 J K⁻¹ mol⁻¹. The reaction is spontaneous at: (Assume ΔH and ΔS do not vary with temperature.)
A · T < 425 K
B · T > 425 K ✓
C · All temperatures
D · T > 298 K
Solution: Both ΔH and ΔS are positive, so the reaction needs high temperature. It becomes spontaneous when TΔS > ΔH, i.e. T > ΔH/ΔS. Convert ΔH to joules: 35.5 × 1000 = 35500 J. Then T > 35500 / 83.6 ≈ 425 K. So it is spontaneous at T > 425 K.
NEET 2026
For 2A(g) + B(g) → 2D(g), ΔU° = −10 kJ mol⁻¹ and ΔS° = −44 J K⁻¹ at 298 K. Identify ΔG° and the spontaneity at 298 K. (R = 8.31 J mol⁻¹ K⁻¹)
A · −1.635 kJ mol⁻¹, spontaneous
B · −0.63568 kJ mol⁻¹, spontaneous
C · +0.63568 kJ mol⁻¹, non-spontaneous ✓
D · +1.635 kJ mol⁻¹, non-spontaneous
Solution: Δn_g = 2 − (2+1) = −1. First get ΔH°: ΔH° = ΔU° + Δn_g RT = −10 + (−1)(8.31)(298)/1000 = −10 − 2.476 = −12.476 kJ. Then ΔG° = ΔH° − TΔS° = −12.476 − 298(−44)/1000 = −12.476 + 13.112 = +0.636 kJ mol⁻¹. Since ΔG° > 0, the reaction is non-spontaneous at 298 K.
Solved Thermodynamics NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula linking spontaneity and temperature?
ΔG = ΔH − TΔS. The reaction is spontaneous when ΔG < 0. Temperature T only affects the −TΔS part, so changing T can change the sign of ΔG when ΔH and ΔS have the same sign.
What is the crossover temperature?
It is the temperature where ΔG = 0, given by T = ΔH/ΔS. On one side of this temperature the reaction is spontaneous; on the other side it is not. Its direction depends on the signs of ΔH and ΔS.
Does a spontaneous reaction have to be fast?
No. Spontaneity (ΔG < 0) only tells you the reaction can happen on its own. It says nothing about speed. Rusting of iron is spontaneous but slow. Speed is a kinetics topic, not thermodynamics.
What are the four sign combinations of ΔH and ΔS for NEET?
(1) ΔH<0, ΔS>0: spontaneous at all T. (2) ΔH>0, ΔS<0: never spontaneous. (3) ΔH>0, ΔS>0: spontaneous at high T. (4) ΔH<0, ΔS<0: spontaneous at low T. Memorising this table saves time in NEET.
Why does this matter for NEET?
NEET asks 1–2 questions almost every year on the sign of ΔH and ΔS or on finding the temperature where ΔG = 0. It is a high-scoring, formula-based topic you can answer in under a minute.
