Question
Using the Gibbs energy equation ΔG = ΔH − TΔS, predict the spontaneity of a reaction for all four possible sign combinations of ΔH and ΔS. Under what conditions is a reaction always spontaneous, never spontaneous, or temperature-dependent?
Solution — Step by Step
A reaction is spontaneous when ΔG < 0. This is the one condition the entire chapter revolves around — get this wrong and nothing else works.
From the equation ΔG = ΔH − TΔS, the sign of ΔG depends on the signs of ΔH and ΔS, plus the value of T (which is always positive in Kelvin).
ΔG = (−) − T(+) = negative − positive = always negative
Both terms push ΔG in the negative direction. This reaction is spontaneous at all temperatures. Exothermic reactions that increase disorder — like combustion of fuels — fall here. JEE loves asking you to identify these.
ΔG = (+) − T(−) = positive + positive = always positive
Both terms push ΔG positive. This reaction is non-spontaneous at all temperatures. You cannot make it happen on its own, no matter how hot or cold the system is. These are the “thermodynamic dead ends.”
ΔG = (−) − T(−) = negative + positive
At low T, the TΔS term is small, so the negative ΔH dominates → ΔG < 0, spontaneous. At high T, the +TΔS term grows large and overwhelms ΔH → ΔG becomes positive, non-spontaneous.
Spontaneous only at low temperatures. Think of reactions like water freezing — happens only when T is low enough.
ΔG = (+) − T(+) = positive − positive
At low T, the positive ΔH dominates → ΔG > 0, non-spontaneous. At high T, the TΔS term grows and eventually exceeds ΔH → ΔG becomes negative, spontaneous.
Spontaneous only at high temperatures. The classic example: melting of ice above 0°C, dissolution of many salts.
Why This Works
The Gibbs energy equation combines two natural tendencies of a system: the drive to release energy (negative ΔH) and the drive to increase disorder (positive ΔS). A system wants both, but it’s a competition weighted by temperature.
The factor T acts as a multiplier on ΔS. At high temperatures, entropy wins. At low temperatures, enthalpy wins. This is why temperature is the deciding variable in Cases 3 and 4 — it shifts the balance between the two competing drives.
| ΔH | ΔS | Spontaneity |
|---|---|---|
| − | + | Always spontaneous |
| + | − | Never spontaneous |
| − | − | Only at low T |
| + | + | Only at high T |
This table has appeared directly in CBSE board papers and as MCQ options in NEET — memorise the pattern, not just the formula.
Alternative Method
Instead of checking signs algebraically, find the crossover temperature where ΔG = 0:
At this temperature, the reaction is at equilibrium (neither spontaneous nor non-spontaneous). For Cases 3 and 4, this gives you the exact temperature where spontaneity flips.
For Case 3 (−ΔH, −ΔS): spontaneous when T < T_crossover
For Case 4 (+ΔH, +ΔS): spontaneous when T > T_crossover
This approach is faster in numerical problems where you’re given actual values of ΔH and ΔS and asked to find the temperature range.
In JEE Main 2023 Shift 2, a question gave ΔH = +40 kJ/mol and ΔS = +200 J/mol·K and asked above what temperature the reaction becomes spontaneous. Direct application: T = 40000/200 = 200 K. Units trap is real — ΔH in J, not kJ.
Common Mistake
The units trap kills marks every year. ΔH is usually given in kJ/mol but ΔS is given in J/mol·K. Students plug directly into ΔG = ΔH − TΔS without converting, getting an answer off by a factor of 1000.
Always convert ΔH to J/mol (multiply by 1000) before calculating ΔG or T_crossover. Or convert ΔS to kJ/mol·K (divide by 1000). Pick one — just be consistent.