Consider the electrolysis of water reaction: $$\ce{H_2O -> H_2 + 1/2O_2}$$ At room temperature and atmospheric pressure, the change in enthalpy of this reaction is $\Delta H = \pu{+286 kJ mol-1}$. Now at constant pressure, $\Delta H = Q + W_\text{other}$, so in order for this reaction to occur, at the bare minimum, we must provide an amount of energy equal to $\pu{286 kJ mol-1}$. The question then becomes, of the $\pu{286 kJ mol-1}$ 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):

$$S_\ce{H2O} = \pu{70 J/K} \qquad S_\ce{H2} = \pu{131 J/K} \qquad S_\ce{O2} = \pu{205 J/K}$$

So if this reaction occurs, the increase in the systems entropy according to these values will be: $$\Delta S_\text{sys} = (131 + \frac{1}{2}\cdot 205-70) = \pu{+163 J/K}$$

My textbook (Schroeder's Introduction to thermal physics) then says that because $\Delta S_\text{sys} = \pu{+163 J/K}$: "the maximum amount of heat that can enter the system is therefore $T\Delta S = (\pu{298 K})(\pu{163 J/K}) = \pu{49 kJ}$". This fact is bugging me a bit. I can see that because the systems entropy is increasing by $\Delta S_\text{sys} = \pu{+163 J/K}$, we can afford for the entropy of the surroundings $(S_\text{surr})$ to decrease by a maximum of $\Delta S_\text{surr} = \pu{-163 J/K}$ without violating the requirement $\Delta S_\text{universe} \gt 0$. Now since the entropy of the surroundings is reduced when heat is provided to the system $(\Delta S_\text{surr} = -\frac{Q_\text{to sys}}{T})$, the maximum heat the surroundings can supply is simply the aforementioned $T\Delta S_\text{surr} = (\pu{298 K})(\pu{163 J/K}) = \pu{49 kJ}$. We can always have the surroundings supply an amount of heat less than $\pu{49 kJ}$ provided we increase $W_\text{other}$ but we may never have the surroundings supply heat more than $\pu{49 kJ}$.

My problem is the fact that when an amount of heat $Q \lt \pu{49 kJ}$ is supplied to the system, we have $\Delta S_\text{surr} \gt \pu{-163 J/K}$. But we also have that $\Delta S_\text{sys} \lt \pu{163 J/K}$ because $\Delta S_\text{sys} = \frac{Q_\text{to sys}}{T}$. But this change in the systems entropy is clearly less than the required change of $\Delta S_\text{sys} = \pu{+163 J/K}$ given by the tables. So surely it is the case that the amount of heat supplied by the surroundings must always be equal to $Q = \pu{49 kJ}$, for if $Q \gt \pu{49 kJ}$, then we violate the second law, but if $Q \gt \pu{49 kJ}$ then we fail to create enough entropy for the system to meet its entropic requirement of $\Delta S_\text{sys} = \pu{+163 J/K}$.

What am I missing here? Is entropy created within the system by other means in just the right amount? If so, how?

  • $\begingroup$ I would suggest to make the Q smaller , simpler. Otherwise , no one would be interested to read the Q. Also , make speakers for paragraphs $\endgroup$
    – S.M.T
    Commented Jun 14, 2021 at 7:03
  • $\begingroup$ As you truly mentioned, Q cannot be more than 49 KJ. Q=49 KJ happens when the system works under reversible conditions. Q<49 KJ happens when system is not working reversibly (i.e., the reaction rate is fast and not enough heat can be supplied by the surrounding). In such a condition, some entropy is generated because of irreversibility which adds to thermal entropy. $\endgroup$ Commented Dec 1, 2022 at 3:13
  • $\begingroup$ Do you know why $\Delta S^0\neq\frac{\Delta H^0}{T}$ for a reaction of ideal gas species? $\endgroup$ Commented Apr 30, 2023 at 10:12
  • $\begingroup$ $2^{nd}$ law will not be violated anyways. $\endgroup$ Commented Apr 30, 2023 at 18:06
  • $\begingroup$ $\Delta S_\text{sys} = \frac{Q_\text{to sys}}{T}$ Is incorrect. Entropy is a state function, so it is independent on the path that determines work and heat exchange. The surrounding entropy change is related to the heat exchange because nothing else happens in the surrounding. $\endgroup$
    – Karsten
    Commented May 31, 2023 at 3:21

2 Answers 2


The amount of heat that must be added is $Q=T\Delta S^0=49$ kJ. The amount of electrical work that must be done on the system is $-\Delta G^0=-\Delta H^0+T\Delta S^0=286-49=237\ kJ$. The real question for you is "why isn't Q equal to $\Delta H^0?"$


The answers to all your questions can be provided by using the open-system (control volume) version of the 1st and 2nd laws of thermodynamics to analyze a specific continuous flow system featuring negligible irreversibility. The control volume contains an ideal Van't Hoff equilibrium box reactor, followed by ideal isothermal flow compressors to compress the pure hydrogen and pure oxygen streams coming out of the reactor.

The reactor receives a pure feed stream of liquid water in the standard state, and discharges pure oxygen and hydrogen through semipermeable membranes; the pressures of the pure oxygen and hydrogen exiting the reactor through the membranes are the same as the partial pressures within the reactor; so they are essentially at equilibrium with the gas mixture in the reactor. The reactor is operating at steady state, so the amounts of pure oxygen and hydrogen streams exiting the reactor correspond stoichiometrically to the amounts present in the inlet water stream.

Once the pure oxygen and hydrogen streams exit the reactor (at pressures << 1bar), they are fed to isothermal continuous flow compresors which raise their pressure up to the stand state pressure of 1 bar. Since the compressors are operating isothermally, the enthalpy change of these streams in the compressors is zero; therefore, all the enthalpy change of the overall combined process takes place in the reactor.


From the open system version of the 1st law applied to the reactor, we have: $$0=Q_R-W_{SR}-\Delta H_R=0$$or $$Q_R-W_{SR}=\Delta H=+\pu{286 kJ/mol}$$where $W_{SR}$ is the shaft work done by the reactor on the surroundings.

From the open system version of the 2nd law applied to the reactor, we have $$Q_R=T\Delta S_R=\pu{298 K}\cdot \Delta S_R$$If we combine these two equations, we obtain $$\Delta G_R=-W_{SR}=\Delta H_R-T\Delta S_R=0$$where $\Delta G_R=-W_{SR}=0$ because the reactor is operating at quasi-static equilibrium. Therefore, we have $$Q_R=\Delta H=\pu{285 kJ/mol}=T\Delta S_R$$$$\Delta S_R=\frac{\Delta H_R}{298}=\pu{959 J mol-1 K-1}$$


From the open system version of the 1st law applied to the isothermal;, we have: $$0=Q_C-W_{C}-\Delta H_C=0$$Since the compressors are operating isothermally, $\Delta H_C=0$ or $$Q_C=W_{SC}$$ where $W_{SC}$ is the shaft work done by the compressors on the surroundings.

From the open system version of the 2nd law applied to the compressors, we have $$Q_C=T\Delta S_C=\pu{298 K}\cdot \Delta S_C$$

Since the Gibbs free energy change of the reactor is zero and the Gibbs free energy change of the overall process in the control volume is 237 kJ/mol, the Gibbs free energy change of the compressors is 237 kJ/mol. It therefore follows that $$-T\Delta S_C=\pu{237 kJ/mol}$$ $$\Delta S_C=\pu{-795 J mol-1 K-1}$$and$$Q_C=\pu{-237 kJ/mol}$$


$$\Delta H=\Delta H_R+\Delta H_C=(286+0) \pu{kJ/mol}=\pu{286 kJ/mol}$$ $$Q=Q_R+Q_C=(286 -237)\pu{kJ/mol} =\pu{49 kJ/mol }$$ $$W_S=W_{SR}-W_(SC)=(0-237) \pu{kJ/mol}=\pu{-237 kJ/mol}$$ $$\Delta S=\Delta S_R+\Delta S_C=(959-795) \pu{J mol-1 K-1}=\pu{164 J mol-1 K-1}$$ $$\Delta G=\Delta G_R+\Delta G_C=(0+237) \pu{kJ/mol}=\pu{237 kJ/mol}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.