In my book I have a solved question:

$$\ce{4 Fe(s) + 3 O2(g) -> 2 Fe2O3(s)}$$ For that reaction: $$\Delta H=-1648\ \mathrm{kJ/mol},\ \Delta S=-549.4\ \mathrm{J/(K\ mol)}\ \text{at}\ 298\ \mathrm K$$

It is asked in the question that why is this reaction spontaneous despite negative entropy.

The solution they give goes like this:

$$ \Delta S_\mathrm{total} = \Delta S_\mathrm{sys} + \Delta S_\mathrm{surr}$$

$$\Delta S_\mathrm{surr} =- \Delta_\mathrm{r}H/T = -(-1648000/298) = \pu{5530J/mol}$$ $$ \Delta S_\mathrm{total} = 5530 + (-549.4) = \pu{4980.6J/K mol}$$

But my question is: shouldn't the $\Delta S_\mathrm{sys} = \Delta_\mathrm{r}H/T$?
Which would be exactly equal in magnitude to $\Delta S_\mathrm{surr}$ but opposite in sign assuming the temperature of iron (system) and surrounding is same?

Then why is it provided as -549.4 instead?

  • $\begingroup$ In given temperature. for spontaneous reaction, $\Delta G = \Delta H - T \Delta S \lt 0$. Thus in here: $$-1648 - (-0.5494) \cdot 298 = -1484.3 \lt 0$$ $\endgroup$ Commented Apr 24 at 8:07
  • $\begingroup$ Note that if there are ongoing spontaneous processes in the system, the entropy of the system grows, even if the change of enthalpy and the exchange of heat and work is zero. $\endgroup$
    – Poutnik
    Commented Apr 24 at 11:04
  • $\begingroup$ There is a reason that $\Delta H^0$ differs from $T^0\Delta S^0$. Do you know what it is? $\endgroup$ Commented Apr 24 at 14:17

3 Answers 3


During the isothermal reversible process of going from pure reactants at 298 K and 1 bar to pure products at 298 K and 1 bar, there are steps in the process where the pressures of the pure reactants and products differ from 1 bar; only the initial and final pressures for the overall process are 1 bar. So, during these specific steps, $\Delta H\neq Q$, but, rather $\Delta H=0$; at the same time, for these specific steps, $\Delta S\neq 0.$

As a result, for the overall isothermal reversible process, $\Delta H^0\neq T^0\Delta S^0$.


[OP] But my question is shouldn't the ∆Ssys =∆rH/T.

That is the first misconception. The entropy of the surrounding increases because heat is transferred from system to surroundings. (The entropy of the system changes because the system goes from one state to another - with initial and final temperature being the same because it is in contact with a large surrounding that determines the initial and final temperature).

[OP] And even in real life I would assume iron would rust without presence of temperature difference between system and surroundings.

That is the second misconception. The temperature of the system and the surroundings are equal, at 298 K (before and after the reaction). You choose the surroundings to be sufficiently large (a huge thermostat bath) that the heat transfer results in a negligible change in temperature (but a relevant change in entropy).

[OP] But according to ∆Stotal=∆Ssys+∆Ssurr & ∆S=q(rev)/T it means that in order to have a ∆Stotal>0 there must be a temperature difference between system and surroundings otherwise ∆Stotal=0.

If that argument were correct, any isothermal process would be at equilibrium. However, reactions happen even if system and surroundings are at the same temperature, or there "is no surrounding" (insulated systems).

  • $\begingroup$ It would be very helpful if you could show me how they reached the value of ∆Ssys = -549.4j/k mol mathematically. Or how they generally reach such ∆S values. From the ∆Stotal equation I would just take a guess and would say ∆rH/T but that doesn't seem to be the case (∆rH/T doesn't make intuitive sense either as it doesn't accommodate for the change of state such as o2 gas in reactants and all solid products.) $\endgroup$
    – Pankaj
    Commented Apr 24 at 4:53
  • 1
    $\begingroup$ You have to measure the entropy change or look it up from entropies of formation. It is independent of the reaction enthalpy. $\endgroup$
    – Karsten
    Commented Apr 24 at 9:45
  • $\begingroup$ If entropy change values are independent of reaction enthalpy then why do we have ∆S=∆H/T? (Thanks for helping me out) $\endgroup$
    – Pankaj
    Commented Apr 24 at 13:19
  • $\begingroup$ A more better way to ask my question would be how do I measure the entropy change of the system? $\endgroup$
    – Pankaj
    Commented Apr 24 at 13:26
  • $\begingroup$ You’re missing the subscripts. The enthalpy change of the system is related to the entropy change of the surrounding, under certain conditions $\endgroup$
    – Karsten
    Commented Apr 24 at 17:07

The following table contains the standard entropies and enthalpies of formation of the species involved in the shown oxidation reaction of iron.

species $\Delta H ^\circ ~~\pu{(kJ/mol)}$ $\Delta S ^\circ ~~\pu{(J/molK)}$
$\ce{O2(g)}$ 0 205.1
$\ce{Fe(s)}$ 0 27.3
$\ce{Fe2O3(s)}$ -824.2 87.4

(source: Silbey and Alberty, Physical Chemistry 3rd Ed, Wiley)

The data provided are standard enthalpies and entropies at 298 K and 1 bar pressure (as noted in another answer), so the notation used should be $\Delta H ^\circ$ and $\Delta S ^\circ$. This is an oversight encountered frequently. The parameters do not describe changes at the conditions of reaction equilibrium. Only at equilibrium between products and reactants is $\Delta S_\textrm{universe}=0$ so that $\Delta S_\textrm{system}=-\Delta S_\textrm{surroundings}$ and, for constant pressure, $T\Delta S_\textrm{system} = T\Delta S= \Delta H$.

The equilibrium partial pressure of oxygen is given by

$$p_\ce{O2(g)} = \exp\left(\frac{\Delta G^\circ}{RT}\right)= p^\circ \exp\left(\frac{\Delta H^\circ}{RT}-\frac{\Delta S^\circ}{R}\right)$$ which is very small at 298 K ($p^\circ = \pu{1 bar}$).


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