Chemistry · Thermodynamics · NEET
A reaction is spontaneous (it happens on its own) only when ΔG is NEGATIVE. When ΔG is positive the reaction is non-spontaneous in the forward direction. When ΔG is exactly zero the system is at equilibrium. So just remember: ΔG < 0 means Go, ΔG > 0 means No, ΔG = 0 means balanced. This one sign is the most tested idea in NEET thermodynamics.
The minus sign is there because entropy (disorder) FAVOURS a reaction. When ΔS is positive (disorder increases), the term −TΔS becomes negative and pulls ΔG down toward the spontaneous side. So a positive ΔS actually helps the reaction go. Nature likes lower energy (ΔH negative) AND more disorder (ΔS positive) — the minus sign lets both work together to make ΔG negative.
ΔH is the heat term (energy released or absorbed). TΔS is the disorder term (temperature times entropy change). ΔG is the NET result after balancing both. Think of it as a tug-of-war: ΔH pulls one way, TΔS pulls the other, and ΔG tells you who wins. Only ΔG decides spontaneity — not ΔH alone. A reaction can release heat (ΔH negative) but still not be spontaneous if the disorder term is against it.
A reaction is spontaneous at EVERY temperature only when ΔH is negative AND ΔS is positive. Look at ΔG = ΔH − TΔS: if ΔH is negative and ΔS is positive, then −TΔS is also negative, so ΔG stays negative no matter what T is. Both terms push in the same direction. This exact combination was asked directly in NEET 2016 (answer: ΔH < 0 and ΔS > 0).
This is the most common calculation slip. ΔH is usually given in kJ/mol but ΔS is given in J/K/mol. You MUST convert them to the same unit first. The safest way: change ΔS from J to kJ by dividing by 1000, or change ΔH from kJ to J by multiplying by 1000. Also remember to use temperature T in Kelvin. Forgetting this 1000 factor gives a wildly wrong answer.
No, and this is a trap. A negative ΔH (exothermic, heat released) only FAVOURS spontaneity, it does not guarantee it. If ΔS is very negative, the +T|ΔS| part can make ΔG positive, so the reaction is non-spontaneous even though it releases heat. Always compute ΔG = ΔH − TΔS fully before deciding. Only the sign of ΔG is the final judge.
The correct thermodynamic conditions for the spontaneous reaction at all temperatures is:
For a given reaction, ΔH = 35.5 kJ/mol and ΔS = 83.6 J/K/mol. The reaction is spontaneous at: (ΔH and ΔS do not vary with temperature.)
For 2A(g) + B(g) → 2D(g), ΔU° = −10 kJ/mol and ΔS° = −44 J/K at 298 K. Find ΔG° and the spontaneity at 298 K. (R = 8.31 J/mol/K)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
ΔG = ΔH − TΔS. Here ΔG is the Gibbs energy change, ΔH is the enthalpy change, T is the temperature in Kelvin, and ΔS is the entropy change. G is a state function and an extensive property, so ΔG depends only on the initial and final states.
ΔG < 0 means the reaction is spontaneous (goes forward on its own). ΔG > 0 means it is non-spontaneous in the forward direction. ΔG = 0 means the system is at equilibrium.
Because ΔG combines BOTH energy (ΔH) and disorder (TΔS). A reaction can be exothermic yet non-spontaneous if the entropy term opposes it. Only ΔG gives the final answer, which is why NEET always tests ΔG, not ΔH alone.
Temperature T must be in Kelvin. It multiplies ΔS, so at high T the entropy term becomes large and can flip the sign of ΔG. This is why some reactions are spontaneous only above or below a certain temperature.
Mixing units. ΔH is in kJ/mol but ΔS is in J/K/mol. You must convert them to match (divide ΔS by 1000 or multiply ΔH by 1000) before subtracting. Forgetting this changes the answer by a factor of 1000.
