For the oxidation of iron,

$$\ce{4 Fe(s) + 3 H2O(g) -> 2 Fe2O3(s) + 3 H2(g)}$$

the entropy change is $ \pu{–549.4 J K^{-1} mol^{-1}}$ at $\pu{298 K}.$ In spite of the negative entropy change of this reaction, why is the reaction spontaneous? $(\Delta_\mathrm{r}H^\circ$ for this reaction is $\pu{–1648 kJ mol^{–1}})$

In this question the solution given is that we consider the total entropy change for the reaction, i.e. the entropy change for a system plus the entropy change of surroundings, and that comes out to be positive.

But shouldn't the entropy change for surroundings (calculated by $q_\mathrm{rev}/T)$ be the negative of the entropy change for the system (heat absorbed by the surroundings is same as the heat emitted by system)? Thus, shouldn't the total entropy change be zero?

  • $\begingroup$ your logic is on the right track, but the heat emitted is the $\Delta H$ term, not the $\Delta S$. $\endgroup$
    – Andrew
    Feb 15 '21 at 18:08
  • $\begingroup$ This number is the entropy change of the system, not including the surroundings. Please specify for us your understanding of the initial state and final state for this change. $\endgroup$ Feb 15 '21 at 23:17
  • $\begingroup$ Does this answer your question? Is ΔS of a system related to temperature and change in enthalpy? $\endgroup$ Jul 27 '21 at 20:09

The initial state of your system consists of 4 moles of $\ce{Fe(s)}$ and 3 moles of water vapor at $\pu{298 K}$ and $\ce{1 bar}$ in separate containers. The final state of your system consists of 2 moles of $\ce{Fe2O3(s)}$ and 3 moles of $\ce{H2(g)}$ at $\pu{298 K}$ and $\pu{1 bar}$ in separate containers.

The $\Delta H^\circ$ and $\Delta S^\circ$ correspond to a process for taking this system reversibly from the initial state to the final state. They don't include the changes in enthalpy and entropy of the surroundings in executing this process.

Do you have an idea of what such a process would be?

  • $\begingroup$ Yes I understand your point but what I am asking is that if the heat lost by system is the same magnitude as the heat gained by surroundings then shouldn't the entropy change be same for both of them, just different in the sign? $\endgroup$ Feb 16 '21 at 6:05
  • $\begingroup$ @ADITYAMITTAL11E Heat gain of surrounding is not only of the same magnitude, but is the same as the heat loss of the system. But entropy change is not limited to heat exchange between the system and surrounding. Imagine an isolated system with zero heat transfer, where is placed a brick with thermal gradient. The entropy of system would grow, even if the heat exchange is zero. Similarly, other cases of entropy change are other spontaneous processes like chemical reactions. $\endgroup$
    – Poutnik
    Feb 16 '21 at 7:11
  • $\begingroup$ @Poutnik If heat exchange is not the sole contributor to change in entropy then why is change in entropy calculated by Qrev/T $\endgroup$ Feb 16 '21 at 7:33
  • $\begingroup$ @ADITYAMITTAL11E The change in entropy is calculated as dS=dQrev/T IF AND ONLY IF the only change leading to a change of entropy is reversible heat exchange. If there is ongoing any irreversible spontaneous process in the system, it is not valid, respectively, it may be valid as the heat transfer contribution to the system entropy change. $\endgroup$
    – Poutnik
    Feb 16 '21 at 7:59
  • $\begingroup$ @Poutnik I think the OP is asking about what would happen if the change is carried out reversibly. I think he is wondering why $\Delta S$. is not equal to $\Delta H/T$. OP, is that a correct interpretation of what you are asking? $\endgroup$ Feb 16 '21 at 12:28

Under constant pressure and temperature, we have

$$-\frac{\Delta G}{T} = \Delta S_{tot} = \Delta S_{sys} + \Delta S_{surr}$$

hence, a negative $\Delta G$ tells us that the total entropy gets increased (= spontaneous process). A common way for textbooks to justify the definition of $G$ is by specifying $\Delta S_{surr} = -\frac{\Delta H_{sys}}{T}$, which is the entropy change by reversible heat transfer between the system and the surroundings. The part that is probably irritating you is that $\Delta S_{surr}$ is not neccesarily equal to $-\Delta S_{sys}$. We can show why this is the case by rationalizing the entropy changes.

Entropy $S$ correlates with the number of accessible microstates (= positions in the phase space, which consists of all position and momenta coordinates for all particles). Let's consider a simple dimerization:

$$\ce{2 A <=> A_2}$$

We change from two independent particles to a single particle. Hence, such a reaction decreases our accessible portion of the position subspace and entropy will decrease. This change is our $\Delta_r S$. Now besides changes in entropy, we can also have changes in the total potential energy. Let's say the bond formed by the dimerization releases a certain amount of energy $\Delta_r H$. What happens to this energy? Well, we convert potential energy to kinetic energy (= heating up the system). Such a process will increase entropy, since we gain access to higher momenta: $\Delta_{pot\to kin} S$. So the total change in entropy for the system would be $\Delta_r S + \Delta_{pot\to kin} S$. We have one problem though. We initially specified a constant temperature experiment. We can transfer the heat $\Delta_r H$ to the surroundings of course. The entropy change for the system will then just be $\Delta_r S$, while we have $\Delta_{pot\to kin} S=-\frac{\Delta_r H}{T}$ for the surroundings.

Disclaimer: Our considerations for entropy are very simplified here. The actually change in entropy for a dimerization is more complex in reality, since it also includes internal degrees of freedom like vibrations.


$\Delta_r \mathrm{H}$ and $\Delta_r \mathrm{S}$ are independent of each other

Processes can be exothermic or endothermic, and they can show a decrease or increase in entropy. You can't predict $\Delta_r \mathrm{H}$ from $\Delta_r \mathrm{S}$ or vice versa. You can combine the two to calculated the change in Gibbs energy, and this tells you about the direction the process will go in the absence of non-PV work.

$\Delta_r \mathrm{H}$ and $\Delta \mathrm{S}_\mathrm{surr}$ are linked under certain circumstances

In the absence of non-PV work, all energy exchange between reaction and surrounding will be through PV work and through heat exchange. With these restrictions and when there are no differences in the temperature before and after the reaction, $\Delta_r \mathrm{H}$ will be equal to the heat transfer to the surroundings. If the surroundings are a giant reservoir, its temperature will not change much (for the ideal case, not at all), and the heat exchange will almost be reversible (for the ideal case, reversible). In that case, $$\Delta \mathrm{S}_\mathrm{surr} = - \frac{\Delta_r \mathrm{H}}{T}$$

$\Delta_r \mathrm{S}$ and $\Delta \mathrm{S}_\mathrm{surr}$ are independent of each other

Because $\Delta_r \mathrm{S}$ and $\Delta_r \mathrm{H}$ are independent of each other, $\Delta_r \mathrm{S}$ and $\Delta \mathrm{S}_\mathrm{surr}$ are, too. The one thing that connects them is that processes for which the sum of the entropy changes is smaller than zero don't happen because that would go against the second law of thermodynamics.

Thus, shouldn't the total entropy change be zero?

No, it shouldn't. If it is, you have reached equilibrium and there would not be any changes in any state functions at all. The reaction in question, however, is described as spontaneous. When a spontaneous reaction proceeds, the total entropy will increase.


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