While reading about Ellingham diagram in my textbook, I noticed that for:

$\ce{C_(_s_) + O2_(_g_) -> CO2_(_g_)}$,

the standard Gibbs free energy ($\ce{\Delta_fG^\circ}$) doesn't seem to vary with temperature (I even compared the line's inclination with a horizontal line and they coincided completely!).

Going through the Wikipedia page for Ellingham diagram, I was able to find the following information:

The formation free energy of carbon dioxide ($\ce{CO2}$) is almost independent of temperature, while that of carbon monoxide (CO) has negative slope and crosses the $\ce{CO2}$ line near $\mathrm{700 ^\circ C}$.

I agree with this more as I believe that $\ce{\Delta_fG^\circ}$ should vary with the temperature. Upon comparing the $\ce{CO_2}$ line (in the Ellingham diagram in Wikipedia) with a horizontal line, this time the $\ce{CO_2}$ line is inclined slightly upwards.

I still have a doubt as to why the inclination of the $\ce{CO2}$ line is so less. One reason that I am able to think of is:

As there are equal number of gases on each side of the reaction, the entropy of the reaction is going to be less (as there is no net increase of decrease in the number of gases). As $\ce{\Delta G^\circ = \Delta H^\circ -T\Delta S^\circ}$, if $\ce{\Delta S^\circ}$ (whose negative value is also the slope of the lines) is really very less as compared to $\ce{\Delta H^\circ}$, then we can conclude that $\ce{\Delta G^\circ}$ is going to be almost constant. But I seem to have a doubt (more specific to thermodynamics). We know that the entropy of reaction is "less", but "less" as compared to what?, i.e., how are we able to tell that this magnitude of entropy is lesser than the magnitude of entropies of all the metal oxidation reactions?

I think there must be a better way to explain the less magnitude of entropy of formation of $\ce{CO2}$. Could anyone please provide a concrete method / reason to prove this?

Also, are there any other factors (other than the entropy factor) governing this?

  • 1
    $\begingroup$ Your reasoning is quite sound. $\endgroup$ May 28, 2021 at 17:31

1 Answer 1


Let's subdivide the CO2 standard entropy change reversible process into 3 parts:

  1. Isothermal expansion of one mole of oxygen from 1 bar to the equilibrium partial pressure of oxygen in a reactor operating at equilibrium at a total pressure of 1 bar (a Van't Hoff equilibrium box)

  2. Injection of one mole of oxygen at its equilibrium partial pressure into the equilibrium box reactor through a semipermeable membrane while removal of one mole of CO2 through a semipermeable membrane at its equilibrium partial pressure in the reactor

  3. Isothermal compression of one mole of CO2 from its equilibrium partial pressure in the equilibrium box reactor to a pressure of 1 bar.

The overall changes (standard changes) in G, H, and S for this 3-step process are -394359 J/mole, -393509 J/mole, and 2.8508 J/mole-K. Now, let's break this down to the changes in the three steps. Based on the standard change in G for the overall process, the equilibrium partial pressures of oxygen and CO2 in the equilibrium box are 8.08E-70 bars and essentially 1 bar.

In step 2, since the reaction is operating at equilibrium, the change in G is zero, while the changes in H and S are $\Delta H_2=\Delta H^0=-393509$ J/mole and $\Delta S_2=\Delta H_2/298.15 = -1319.8$ J/mole-K.

In steps 1 and 3, the expansion of oxygen and the compression of the CO2 are isothermal, so $\Delta H_1=\Delta H_3=0$. In step 1, $\Delta G_1=R(298.15)\ln{8.08\times 10^{-70}}=-394359$ J/mole and $\Delta S_1=-R\ln{8.08\times 10^{-70}}=1322.7$ J/mole-K. In step 3, the initial and final CO2 pressures are essentially 1 bar, so $\Delta G_3=\Delta S_3=0$

So, it seems that, merely by coincidence, the changes in entropy in step 2 compared to the sum in steps 1 and 3 are of nearly equal magnitude and opposite in sign (-1319.8 vs +1322.7). This means that their difference is only 2.85, which represents the overall entropy change. It is coincidental because the mechanistic changes in step 2 compared to steps 1 and 3 are very different.

  • $\begingroup$ In the Ellingham diagram in Wikipedia, the slope of CO2 line is slightly positive. As ΔS is coming out to be positive here, does that mean that the slope of CO2 line should have been negative instead? $\endgroup$
    – Pal
    May 29, 2021 at 6:32
  • 1
    $\begingroup$ I don't see a positive slope on the Wiki graph. At 298, the slope should be slightly negative, and I think that the standard change in the heat capacities is close to zero, so that shouldn't modify the sign as temperature increases. $\endgroup$ May 29, 2021 at 10:53

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.