Q.For oxidation of iron, , entropy change is –549.4 JK mol at 298 K. In spite of the negative entropy change of this reaction, why is the reaction spontaneous? ( for this reaction is J mol)
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Start your 14-day free trial to unlock the full solution →The reaction is spontaneous despite a large negative entropy change because the enormous negative enthalpy change ( J/mol) dominates the Gibbs free energy equation, making negative at 298 K. The calculated J/mol confirms spontaneity.
The key insight here is that spontaneity is not decided by entropy alone — it's decided by Gibbs free energy, which combines both enthalpy and entropy. A negative entropy change means the system becomes more ordered (gas molecules turning into solid), which by itself would oppose spontaneity. But the reaction releases a massive amount of heat — that's the driving force.
Let's see why this works step by step.
- The Gibbs free energy criterion For any process at constant temperature and pressure, spontaneity is determined by the sign of :
If , the reaction is spontaneous (thermodynamically favourable). If , it is non-spontaneous. If , the system is at equilibrium.
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What we are given
- J mol (negative, so the reaction is exothermic)
- J K mol (negative, so the products are more ordered than the reactants)
- Temperature K
Notice the units: enthalpy is in J/mol, entropy in J/K·mol — they are compatible.
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Plug into the equation
Be careful with signs: subtracting a negative is the same as adding the magnitude.
- Calculate the term
So:
That's approximately J mol.
- Interpret the result is large and negative — about kJ/mol. This tells us the reaction is highly spontaneous at 298 K. The enthalpy term ( kJ) is so overwhelmingly negative that it swamps the unfavourable entropy term ( kJ, because is negative but we subtract it, making it a positive contribution to ). …
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