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Problems · Problem 5.11

Q.For oxidation of iron, 4Fe(s)+3O2(g)→2Fe2O3(s)4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s), entropy change is –549.4 JK−1^{-1} mol−1^{-1} at 298 K. In spite of the negative entropy change of this reaction, why is the reaction spontaneous? (ΔrH⊖\Delta_r H^\ominus for this reaction is −1648×103-1648 \times 10^3 J mol−1^{-1})

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The reaction is spontaneous despite a large negative entropy change because the enormous negative enthalpy change (ΔH=−1648×103\Delta H = -1648 \times 10^3 J/mol) dominates the Gibbs free energy equation, making ΔG\Delta G negative at 298 K. The calculated ΔG⊖=−1484.3×103\Delta G^\ominus = -1484.3 \times 10^3 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.

  1. The Gibbs free energy criterion For any process at constant temperature and pressure, spontaneity is determined by the sign of ΔG\Delta G:

ΔG=ΔH−TΔS\Delta G = \Delta H - T \Delta S

If ΔG<0\Delta G < 0, the reaction is spontaneous (thermodynamically favourable). If ΔG>0\Delta G > 0, it is non-spontaneous. If ΔG=0\Delta G = 0, the system is at equilibrium.

  1. What we are given

    • ΔrH⊖=−1648×103\Delta_r H^\ominus = -1648 \times 10^3 J mol−1^{-1} (negative, so the reaction is exothermic)
    • ΔrS⊖=−549.4\Delta_r S^\ominus = -549.4 J K−1^{-1} mol−1^{-1} (negative, so the products are more ordered than the reactants)
    • Temperature T=298T = 298 K

    Notice the units: enthalpy is in J/mol, entropy in J/K·mol — they are compatible.

  2. Plug into the equation

ΔG⊖=ΔH⊖−TΔS⊖\Delta G^\ominus = \Delta H^\ominus - T \Delta S^\ominus

ΔG⊖=(−1648×103)−(298)×(−549.4)\Delta G^\ominus = (-1648 \times 10^3) - (298) \times (-549.4)

Be careful with signs: subtracting a negative is the same as adding the magnitude.

  1. Calculate the TΔST\Delta S term

TΔS=298×(−549.4)=−163,721.2 J mol−1T \Delta S = 298 \times (-549.4) = -163,721.2 \text{ J mol}^{-1}

So:

ΔG⊖=−1648×103−(−163,721.2)\Delta G^\ominus = -1648 \times 10^3 - (-163,721.2)

ΔG⊖=−1648×103+163,721.2\Delta G^\ominus = -1648 \times 10^3 + 163,721.2

ΔG⊖=−1,484,278.8 J mol−1\Delta G^\ominus = -1,484,278.8 \text{ J mol}^{-1}

That's approximately −1484.3×103-1484.3 \times 10^3 J mol−1^{-1}.

  1. Interpret the result ΔG⊖\Delta G^\ominus is large and negative — about −1484-1484 kJ/mol. This tells us the reaction is highly spontaneous at 298 K. The enthalpy term (−1648-1648 kJ) is so overwhelmingly negative that it swamps the unfavourable entropy term (+163.7+163.7 kJ, because TΔST\Delta S is negative but we subtract it, making it a positive contribution to ΔG\Delta G). …

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