Consider the Electrolysis of water reaction $$H_2O \rightarrow H_2+\frac{1}{2}O_2$$ At room temperature and atmospheric pressure, the change in enthalpy of this reaction is $\Delta H=+286kJ/mol$. Now at constant pressure, $\Delta H=Q +W_{other}$, so in order for this reaction to occur, at the bare minimum, we must provide an amount of energy equal to $286kJ/mol$. The question then becomes, of the $286kJ/mol$ required for this reaction to occur, how much can we supply as heat and how much can we supply as work such that the 2nd law isn't violated. Now according to standard reference tables (all values being per mol):

$$S_{H_2O}=70 J/K \,\,\,\,\,\,\,\, S_{H_2}=131 J/K\,\,\,\,\,\, S_{O_2}=205 J/K$$

So if this reaction occurs, the increase in the systems entropy according to these values will be $$\Delta S_{sys}=(131+\frac{1}{2}\cdot 205-70)=+163\,J/K$$ My textbook (Schroeder's Introduction to thermal physics) then says that because of $\Delta S_{sys}=+163\,J/K$, we have that "the maximum amount of heat that can enter the system is therefore $T\Delta S=(298K)(163J/K)=49 kJ$". This fact is bugging me a bit. I can see that because the systems entropy is increasing by $\Delta S_{sys}=+163\,J/K$, we can afford for the entropy of the surroundings to decrease by a maximum of $\Delta S_{surroundings}=-163\,J/K$ without violating the requirement $\Delta S_{universe}>0$. Now since the entropy of the surroundings is reduced when heat is provided to the system ($\Delta S_{surroundings}=-\frac{Q_{to\,system}}{T}$), the max heat the surroundings can supply is simply the aforementioned $T\Delta S_{surroundings}=(298K)(163J/K)=49 kJ$. We can always have the surroundings supply an amount of heat less than $49kJ$ provided we increase $W_{other}$ but we may never have the surroundings supply heat more than $49kJ$.

My problem is the fact that when an amount of heat $Q<49 kJ$ is supplied to the system, we have $\Delta S_{surroundings}>-163\,J/K$. But we also have that $\Delta S_{sys}<163\,J/K$ because $\Delta S_{sys}=\frac{Q_{to\,system}}{T}$. But this change in the systems entropy is clearly less than the required change of $\Delta S_{sys}=+163 J/K$ given by the tables. So surely it is the case that the amount of heat supplied by the surroundings must always be equal to $Q=49kJ$, for if $Q>49kJ$, then we violate the second law, but if $Q<49kJ$ then we fail to create enough entropy for the system to meet its entropic requirement of $\Delta S_{sys}=+163\,J/K$. What am I missing here? Is entropy created within the system by other means in just the right amount? If so, how?

Any help on this would be most appreciated!

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    $\begingroup$ Have you tried analyzing this same standard change using a Van't Hoff equilibrium box in conjunction with gas compressions and expansions? $\endgroup$ Jun 8, 2021 at 14:59
  • $\begingroup$ @ChetMiller No I haven't. To be honest I don't even know what a Van't Hoff equilibrium Box is but I'll google. Chemistry isn't my strength. $\endgroup$ Jun 8, 2021 at 15:08
  • $\begingroup$ Why can't the rest of the $\Delta H$ be accounted for by the work? That would be 237 kJ/mol. This would be consistent with the Gibbs free energy change of the reaction. $\endgroup$ Jun 8, 2021 at 19:35
  • $\begingroup$ @ChetMiller For this reaction, table values state that $\Delta S_{sys}=+163J/K$. So in order for this reaction to occur under any circumstances, we need for $\Delta S_{surrounds}>=-163J/K$. Thus the maximum heat that can enter the system is $T\Delta S_{sys}=49kJ$ to ensure $\Delta S_{uni}>=0$. Heat less than this is perfectly acceptable (satisfies $\Delta S_{uni}>=0$). We must just do enough work to make up for the reduction in heat. But if we do transfer less heat than $49kJ$, say for example $Q_o=10kJ$. Then $\Delta S_{sys}=\frac{10 kJ}{298K}=33 J/K$ but this is not enough to match the ... $\endgroup$ Jun 8, 2021 at 19:54
  • $\begingroup$ (2/n) to match the $\Delta S_{sys}=+163J/K$ mentioned in the tables. So if the heat transfer to the system is less than $T\Delta S_{sys}=49kJ$, in order to ensure that $\Delta S_{sys}=+163J/K$ (i.e. to ensure the system gets enough entropy so that the reaction confirms what the tables state about the reaction), something else has to create the necessary entropy because heat transfer is insufficient. So it must be the work that creates it? How can this be? $\endgroup$ Jun 8, 2021 at 19:59

1 Answer 1


For the situation you are looking at $$\Delta H=Q+W_{elec}$$where $W_{elec}$ is the electrical work done by the surroundings on the system. From the 2nd law, if follows that $$Q=T\Delta S-T\sigma$$ where $\sigma$ is the entropy generated within the system by irreversibility of the process (e.g., oversupplying electrical work). So, combining these we have $$\Delta H=T\Delta S-T\sigma+W_{elec}$$. Between the same two end states, if we do the electrolysis reversibly, we get $$\Delta H=T\Delta S+W_{elec,min}$$where $W_{elec,min}$ is the minimum electrical work required to do the electrolysis. So, if we combine these, we get $$W_{elec}=W_{elec,min}+T\sigma$$and, along with this goes $$Q=Q_{max}-T\sigma$$

  • $\begingroup$ Thank you so much! Everything is pretty much clear now. So the extra work that must be done when $Q<Q_{max}$ always generates the correct amount of entropy to ensure that we get the correct $\Delta S^{reaction}_{sys}$? $\endgroup$ Jun 8, 2021 at 20:50
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    $\begingroup$ Yes, with the heat transfer adjusting automatically when we make the final state is the same. The real interesting issue is the mechanism of entropy generation in the case of a chemical reaction that is being forced to proceed at finite reaction rate. For some analysis of this, see Chapter 24 of Transport Phenomena by Bird, Stewart, and Lightfoot. $\endgroup$ Jun 8, 2021 at 22:34

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