Let's say we have a mixture of oxygen and hydrogen gas, initially at room temperature and pressure. This mixture can be used to perform work: we can burn it, using the resulting heat as a source in a Carnot cycle. However, this requires dropping some waste heat to a cold reservoir. If the mixture reaches a temperature $T_H$ when burned and our cold reservoir's temperature is $T_C$, then the maximum efficiency of the Carnot cycle will be:

$$ \eta = 1 - \frac{T_C}{T_H} $$

Standard thermodynamical logic only proves this is a maximum when we start with the mixture having already reacted. What if we try other ways of extracting work from it? I don't know much about how hydrogen fuel cells work, but apparently the principle is rather different and electric current gets produced directly through a gradient of chemical potential, so the fuel never heats up very much.

Would it be theoretically possible to get arbitrarily close to converting all of the chemical potential energy into work? If not, is the limit precisely that of the Carnot cycle?

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    $\begingroup$ Not to be too facetious, but isn't all efficiency limited by thermodynamics? $\endgroup$
    – DKNguyen
    Feb 4 at 21:50
  • $\begingroup$ ... the second law. $\endgroup$
    – Kotlopou
    Feb 5 at 8:56
  • 2
    $\begingroup$ I'm not sure I understand any of the theoretical lawyering in any of the answers. If 2 moles of H2 are combined with 1 mole of O2, how many Joules could you expect to get from a reasonably efficient combustion-driven generator, and how many from a reasonably efficient fuel cell? Are they significantly different? How much less efficient than ideal are combustion-based generators, and fuel cells? $\endgroup$
    – Neil_UK
    Feb 5 at 13:44
  • $\begingroup$ Only a reaction running infinitely slow can work without producing entropy, and be reversible. I say the fuel cell process contains reaction steps that are definitely irreversible. Ergo the reaction cannot run infinitely slow, and cannot run at 100% efficiency. One molecule of water is created, or it isn't. If it is, the speed is definitely finite already. $\endgroup$
    – Karl
    Feb 6 at 22:37
  • $\begingroup$ Carnot engines have limited efficiency because they are trying to use relatively-high-entropy heat energy, and the entropy has to go somewhere. There is already quite a lot of entropy in this universe, but luckily you can increase its capacity by adding more energy to a colder place, but that has to come out of the energy you want to extract. When you start with low-entropy energy, such as rotational motion, DC or AC electric current, or chemical potential, this thermodynamic problem doesn't occur. $\endgroup$
    – user253751
    Feb 6 at 22:42

6 Answers 6


The efficiency with which chemical potential energy can be converted to work is no different from any other efficiency with which potential energy is convertible to work: potential energy does not convert to work, only work converts to work.

When an energetic quantity, such as entropy, $\delta S$, volume, $\delta V$, electric charge $\delta e$ , heavy mass $\delta m$, chemical species , $\delta N$, etc., does work it always passes through two (2) potential levels: a high and a low. These potentials are temperature (thermal), pressure (mechanical), electric potential $\psi$, gravitational potential $\phi$, chemical potential, $\mu$, etc.

Each quantity transport is associated with an amount of work that is equal to the amount of quantity multiplied by the potential difference through which it is transported. These are: $\delta S \Delta T$, $-\delta V \Delta p$, $\delta S \Delta T$, $\delta e \Delta \psi$, $\delta m \Delta \phi$, $\delta N \Delta \mu$, etc.

In a complete formal analogy with d'Alembert's principle of the work of constraint forces being zero, the total work of these transports is zero for a reversible process: $$\delta S \Delta T-\delta V \Delta p +\delta e \Delta \psi+\delta m \Delta \phi+\delta N \Delta \mu +.. =0 \tag{1}$$ This (1) shows a balance among all the work of quantity transports, and shows how they may convert from one to another. Just as d'Alembert's principle refers to a dynamic equilibrium so is this work principle refers to a reversible process in which every step is a state of equilibrium.

Carnot's efficiency is just a trivial consequence of (1). If you denote the useful extractable work by $-\Delta W_x=-\delta V \Delta p+\Delta e \delta \psi+\delta m \Delta \phi+\delta N \Delta \mu +.. =0$ and the incoming heat by $Q=\delta S T_h$ and $\Delta T = T_h-T_{\ell}$ then $$\eta_H=\frac{\Delta W_x}{Q}=\frac{\Delta T}{T_h} \tag{2} = 1-\frac{T_{\ell}}{T_h}$$ $$\delta S \Delta T = \Delta W_x.$$

Now take your case of the balance between chemical work and electric work and assuming no other interaction. Having no other interaction just means that the associated potential dops are zero, $$\Delta T, \Delta \phi =0, \Delta p=0,..$$ and what remains is $$\delta e \Delta \psi + \delta N \Delta \mu = 0\tag{3}$$ If you now assume that $\Delta \mu = \mu_h - \mu_{\ell}$ and consider the invested chemical energy to be $E_H=\delta N \mu_h$ then you can write the efficiency of the conversion as $$\eta_C = \frac{-\delta e \Delta \psi}{E_H}=\frac{\delta N \Delta \mu}{\delta N \mu_h}=1-\frac{\mu_{\ell}}{\mu_h} \tag{4}.$$

The reason why thermal or pressure work was not involved because we assumed that the relevant quantity is transported, if any, through a zero potential difference.


The cell operates at constant pressure and temperature and the maximum recoverable work is fixed by the thermodynamics $W \leq -\Delta G$. The second law efficiency of the fuel cell is therefore $\frac{W}{-\Delta G}$ and it is always less than 1. In practice, it is of the order of $0.6$. This type of efficiency is very different from the Carnot efficiency which does not concern itself with the way in which the high temperature is obtained. For a heat engine, we can use this type of efficiency by considering that we have a chemical reaction and measure the work obtained for a certain quantity of fuel.

For the fuel cell, $\Delta S < 0$ and therefore $-\Delta G=-\Delta H+T\Delta S<-\Delta H$: The maximum recoverable work is less than the heat released by the reaction.

What's funny is that we could believe that by using hydrogen and oxygen to heat to a very high temperature, we could in this way obtain a Carnot efficiency close to 1 and therefore completely convert thermal energy into work, beyond what thermodynamics allows. This is an illusion because in this reasoning we forget that by increasing the temperature too much, the chemical reaction ends up being reversed.


The Carnot limit arises from the impossibility of destroying entropy.

Heat engines are based fundamentally on receiving entropy (and energy) from a source—the so-called hot reservoir—through heat transfer and cyclically dumping it strategically somewhere else—the so-called cold reservoir—also through heat transfer such that less energy is needed to achieve the same magnitude of entropy transfer.

What’s the trick to doing this? As the names suggest, an essential aspect is that the cold reservoir is cooler than the hot reservoir. (Heat transfer Q at temperature T shifts entropy Q/T.)

In other words, the same entropy is transferred with lower heating Q at a lower temperature. The energy difference is then available to be output as work.

Since all real reservoirs lie above absolute zero, some energy is always lost to the cold reservoir, even with perfect engineering of the engine to reduce dissipative processes (e.g., friction) to a level arbitrarily close to zero.

Non-heat engines aren’t bound by this particular restriction. Well, that is to say, they still aren’t allowed to destroy entropy, but this doesn’t incur the same fundamental restriction because they don’t deal in heat transfer.

A water wheel, for example can lower water at temperature T to a lower level also at temperature T and still recover essentially every bit of available potential energy, assuming perfect engineering as described above. The entropy balance is trivially satisfied because any entropy brought by the input water leaves with the output water. The same reasoning applies to a non-heat engine based on the transfer of charge carriers, for instance.

Another answer connects the Carnot limit to cyclic operation. This can’t be the whole story, as cyclic operation of a non-heat engine still wouldn’t involve this limit. Rather, the prohibition on destroying entropy provides a fuller understanding of the constraints on various engine types.

  • $\begingroup$ What do you mean by not destroying entropy? Do you mean that entropy can only increase (be created), and that in a reversible process the total entropy is conserved? $\endgroup$
    – Buck Thorn
    Feb 5 at 9:49
  • $\begingroup$ Yes, exactly that. $\endgroup$ Feb 5 at 18:02

The Carnot cycle efficiency applies only to a heat engine cycle.

Converting chemical potential energy completely into electrical work is a process, not a cycle, and therefore doesn’t violate the second law of thermodynamics.

Even in the case of heat, the second law doesn’t preclude the possibility of converting heat entirely to work in a process. That’s what happens in the reversible isothermal expansion process in the Carnot cycle.

In any event, to quote from the following link which compares the Carnot cycle to a fuel cell: https://www.sciencedirect.com/science/article/abs/pii/S0360319902000162

"The comparison dispels the misconception that an ideal fuel cell is potentially more efficient than an ideal heat engine."

Hope this helps.

  • $\begingroup$ This is true, and yet we often calculate the efficiency of engines with thermodynamics, because it's implied that they should work with arbitrary amounts of fuel while remaining the same size. That is the kind of setup I was thinking about. Wikipedia says that fuel cells provide energy as long as fuel is supplied to them, so they are cyclical in the same sense a steam engine is. $\endgroup$
    – Kotlopou
    Feb 4 at 20:39
  • $\begingroup$ @Kotlopou give me the link to the Wikipedia article $\endgroup$
    – Bob D
    Feb 4 at 20:56
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    $\begingroup$ @Kotlopou See update to my answer. $\endgroup$
    – Bob D
    Feb 4 at 21:41

Yes, the general idea of fuel cell technology is to avoid heat altogether and convert chemical energy directly into electrical energy, in the form of work, with no entropy changes. If one can do this then 100% efficiency is in principle available.

  • 2
    $\begingroup$ You're talking theory. In practice, fuel cells heat up to break covalent bounds and are far away from theoretical limits (citation needed, it is just from what I remember having heard sometime somewhere ). $\endgroup$ Feb 4 at 19:07


It turns out that the question has missed a subtle point: the temperature $T_C$ doesn't just depend on the difference in the chemical potentials of the reactants and products $\Delta \mu$, but also on the temperature of the reactants. This was pointed out by Vincent Fraticelli's answer.

Thus, we can reach a higher efficiency by first heating up the reactants to a temperature $T_H$, then allowing the reaction to happen, and using its heat as input to a Carnot cycle. We can make $T_H$ as high as we want, but at some temperature $T_{max}$ the reaction starts to go in the opposite direction. Per the Maximum Work Theorem, the maximal efficiency is reached when the reaction happens reversibly, which is when $\Delta G = 0$, because it happens in the hot reservoir at constant temperature and pressure.

This was pointed out by the link in Bob D's answer, in which this maximum temperature is called the combustion temperature.

To find the maximum efficiency, we set $\Delta G = 0$ and find its corresponding $T_H = T_{max}$. I will use $\Delta_H$ and $\Delta_L$ for the change in a quantity between the reactant and the products for a reaction happening at the high and low temperature, respectively. First we find $T_H$:

$$ 0 = \Delta_H G = \Delta_H H - T_H \Delta_H S $$

$$ T_H = \frac{\Delta_H H}{\Delta_H S} $$

Now the heat entering and escaping the Carnot cycle (which we assume to be positive to not get lost in sign conventions):

$$ Q_H = \Delta_H H $$

$$ Q_L = T_L \Delta_L S $$

And now the efficiency:

$$ \eta_{Carnot} = \frac{Q_H - Q_L}{Q_H} = \frac{\Delta_H H - T_L \Delta_L S}{\Delta_H H} $$

If we assume $\Delta_H H \approx \Delta_L H$ and $\Delta_H S \approx \Delta_L S$, then this becomes:

$$ \eta_{Carnot} \approx \frac{\Delta_L H - T_L \Delta_L S}{\Delta_L H} = \frac{\Delta_L G}{\Delta_L H} $$

That assumption apparently holds approximately, but not exactly.

Fuel cell

Now to the fuel cell. As pointed out by Vincent Fraticelli again, the fuel cell operates at a constant temperature and pressure, so work can be calculated from the change in the Gibbs potential: $W = \Delta G$. The total energy trapped in the reactants (i.e. the heat output of the reaction) is $Q = \Delta H$, the change in enthalpy during the reaction. Thus, the best possible efficiency for a given reaction is:

$$ \eta_{cell} = \frac{W}{Q} = \frac{\Delta G}{\Delta H} $$

These changes are measured at the low, ambient temperature at which the fuel cell operates.


According to the linked article, the efficiencies are close, but not exactly equal. For hydrogen-oxygen burning, the limit is around $94\%$. For some fuels, such as methanol, there apparently is no theoretical maximum, because $\Delta G$ is always positive. This is caused by $\Delta S$ of the reaction being positive. In those cases, it would be possible to convert the energy stored in the fuel entirely into work and additionally convert some heat from the environment into work, giving something like an efficiency $\eta > 1$.

One remaining question

I don't understand why we assume the chemical reaction takes place at constant pressure. This means the fuel might expand, "wasting" some energy on volume-change work that we never capture. I would expect instead to do the chemical reaction in a fixed volume, meaning that all of its heat can be used to power the Carnot cycle.


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