Chemistry · Thermodynamics · NEET
No. Giving out heat (ΔH negative) makes a reaction more likely to be spontaneous, but it is not a guarantee. Spontaneity is decided by ΔG = ΔH − TΔS. So even a reaction with negative ΔH can be non-spontaneous if the entropy term TΔS works against it strongly enough. Enthalpy is one factor, not the full answer.
Yes. NCERT gives two clear examples: ½N2(g) + O2(g) → NO2(g) with ΔH = +33.2 kJ/mol, and C(graphite) + 2S(l) → CS2(l) with ΔH = +128.5 kJ/mol. Both absorb heat (ΔH positive) yet occur on their own. This proves that a decrease in enthalpy is not required for a reaction to be spontaneous.
Because it fails in real cases. People thought reactions go 'downhill' in energy like water falling down a hill. That guess works for many exothermic reactions, but it breaks for endothermic spontaneous reactions such as NO2 formation. If enthalpy were the true test, no endothermic reaction could ever be spontaneous — but many are. So enthalpy is only a contributing factor.
The Gibbs free energy change: ΔG = ΆH − TΔS. A process is spontaneous when ΔG is negative. This combines two drives: lower enthalpy (ΔH negative) AND higher entropy/disorder (ΔS positive). NCERT introduces entropy (S) as the missing second factor after showing enthalpy alone is not enough.
No. Spontaneous means 'has the potential to proceed without outside help.' It says nothing about speed. H2 + O2 can sit for years without visibly reacting, yet the reaction is still called spontaneous. So do not confuse spontaneity (direction) with rate (how fast).
When ΔH is negative AND ΔS is positive, ΔG = ΔH − TΔS is negative at every temperature, so the reaction is spontaneous at all temperatures. This exact combination (ΔH < 0 and ΔS > 0) was the answer to a 2016 NEET question. If ΔS is negative, spontaneity depends on temperature.
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: (Assume ΔH and ΔS do not vary with temperature.)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No. A negative ΔH (exothermic) is only a contributing factor. The true criterion is ΔG = ΔH − TΔS being negative, which also depends on entropy and temperature.
½N2(g) + O2(g) → NO2(g) with ΔH = +33.2 kJ/mol, and C(graphite) + 2S(l) → CS2(l) with ΔH = +128.5 kJ/mol. Both absorb heat but are spontaneous (from NCERT).
No. Spontaneous only means the reaction can proceed on its own without external help. H2 + O2 is spontaneous but extremely slow at room temperature.
ΔG < 0, where ΔG = ΔH − TΔS. This combines the enthalpy factor (ΔH) and the entropy factor (ΔS) at a given temperature T.
When ΔH < 0 and ΔS > 0. Then ΔG = ΔH − TΔS is negative at every temperature. This was the answer to a 2016 NEET question.
