If you were to super-heat a mole of carbon dioxide gas, at what point would it decompose into carbon and oxygen? If it decomposes into carbon monoxide and oxygen, at what point do those products decompose? I'm interested in the process from a hypothetical perspective, so energy is not a concern, nor is industrial feasibility.

I'm looking for the heat required for carbon dioxide to break down into atomic carbon and oxygen, if carbon monoxide is being formed then its not hot enough yet!


1 Answer 1


Disclaimer: I am too lazy to check whether the reaction $\ce{CO2(g) <=> CO(g) + 1/2 O2(g)}$ is more favourable.

There is no single tipping point where all carbon dioxide will suddenly be converted to carbon and oxygen. As you heat carbon dioxide, the percentage of oxygen present would gradually increase.

You're interested in the reaction

$$\ce{CO2(g) <=> C(graphite) + O2(g)}$$

for which, at $298~\mathrm{K}$, $\Delta_\mathrm{r} H^\circ = +393.51~\mathrm{kJ~mol^{-1}}$ and $\Delta_\mathrm{r} S^\circ = -2.862~\mathrm{J~K^{-1}~mol^{-1}}$ (data from Atkins, Physical Chemistry 10th ed., pp 974-7). The equilibrium constant for this reaction is remarkably simple:

$$K = \exp{\left(\frac{-\Delta_\mathrm{r}G^\circ}{RT}\right)} = \frac{a_{\ce{O2}}}{a_{\ce{CO2}}}$$

and if we assume ideality, then both activities are proportional to the partial pressures, which are in turn proportional to the number of moles of both substances:

$$K \approx \frac{n_{\ce{O2}}}{n_{\ce{CO2}}}$$

Therefore if $K = 100$, then you have a hundred times more oxygen in your system than carbon dioxide. If $K = 1$, then you have equal amounts. The pressure and volume of the system do not matter in this interpretation of the value of the equilibrium constant. Our goal is now to find out the variation of $K$ with temperature.

The variation of $\Delta_\mathrm{r} H^\circ$ with temperature is given by Kirchhoff's law, and to simplify things, let's assume that the heat capacities at constant pressure are independent of temperature over the temperature range we are interested in:

$$\begin{array}{cc} \hline \text{Compound} & C_{p,\mathrm{m}}^\circ\text{ / }\mathrm{J~K^{-1}~mol^{-1}} \\ \hline \ce{C(graphite)} & 8.527 \\ \ce{O2(g)} & 29.355 \\ \ce{CO2(g)} & 37.11 \\ \hline \end{array}$$

$$\Delta_\mathrm{r}H^\circ(T) = +393.51~\mathrm{kJ~mol^{-1}} + (0.77~\mathrm{J~K^{-1}~mol^{-1}})(T - 298~\mathrm{K})$$

One can derive a similar expression for the entropy change, which is "left as an exercise for the reader":

$$\Delta_\mathrm{r}S^\circ(T) = -2.862~\mathrm{J~K^{-1}~mol^{-1}} + (0.77~\mathrm{J~K^{-1}~mol^{-1}})\left[\ln{\left(\frac{T}{298~\mathrm{K}}\right)}\right]$$

All that remains is to calculate the Gibbs free energy change

$$\Delta_\mathrm{r}G^\circ = \Delta_\mathrm{r}H^\circ - T\Delta_\mathrm{r}S^\circ$$

and the equilibrium constant

$$K = \exp{\left(\frac{-\Delta_\mathrm{r}G^\circ}{RT}\right)}$$

You can do this easily with a spreadsheet. I've thrown together some very messy Mathematica code and I'm sure someone will come along and tell me that there's a much more concise way of doing it, but:


The first number in the curly brackets gives you the value of $T$ in Kelvins, and the second gives you the value of $K$ at that temperature $T$. Here are some of the numbers:

$$\begin{array}{cc} \hline T\text{ / K} & K \\ \hline 298 & 7.513 \times 10^{-70} \\ 10^3 & 2.073 \times 10^{-21} \\ 10^4 & 0.007895 \\ 10^5 & 0.6899 \\ 10^6 & 1.307 \\ 10^9 & 2.598 \\ 10^{12} & 4.9257 \\ \hline \end{array}$$

$K = 1$ when $T = 2.52 \times 10^5~\mathrm{K}$.

This is well above the temperature at which carbon vaporises ($4827~\mathrm{^\circ C}$ according to Wikipedia). If you redo the calculations using instead the equation

$$\ce{CO2(g) <=> C(g) + O2(g)}$$

then you get these new expressions (thermodynamic data again from Atkins):

$$\Delta_\mathrm{r}H^\circ(T) = +1110.19~\mathrm{kJ~mol^{-1}} + (13.083~\mathrm{J~K^{-1}~mol^{-1}})(T - 298~\mathrm{K})$$

$$\Delta_\mathrm{r}S^\circ(T) = +149.498~\mathrm{J~K^{-1}~mol^{-1}} + (13.083~\mathrm{J~K^{-1}~mol^{-1}})\left[\ln{\left(\frac{T}{298~\mathrm{K}}\right)}\right]$$

and much more favourable numbers:

$$\begin{array}{cc} \hline T\text{ / K} & K \\ \hline 298 & 1.637 \times 10^{-187} \\ 1000 & 1.470 \times 10^{-50} \\ 2000 & 3.417 \times 10^{-21} \\ 5000 & 0.003130 \\ 6000 & 0.351794 \\ 7000 & 10.65 \\ 10000 & 5594.31 \\ \hline \end{array}$$

This time, $K = 1$ at $T = 6276~\mathrm{K}$. You can play around with the code here:


Why does $K$ for this reaction increase so drastically with temperature? It is likely to be because of the entropy contribution: because the entropy change is much more positive, increasing the temperature makes the term $-T\Delta_\mathrm{r}S^\circ$ much more negative, lowering $\Delta_\mathrm{r}G^\circ$ and increasing $K$.

  • $\begingroup$ To all readers: Please tell me if there's an error somewhere, I wouldn't be surprised, the maths is rather fiddly. $\endgroup$ May 10, 2016 at 19:25
  • $\begingroup$ Dissociation to CO is more favorable, but it later also dissociates. $\endgroup$
    – Mithoron
    May 10, 2016 at 19:40
  • 1
    $\begingroup$ @orthocresol when it is hot enough for atomic C to form, monoatomic O rather than diatomic O predominates $\endgroup$
    – DavePhD
    May 10, 2016 at 20:00
  • $\begingroup$ @DavePhD Thanks. I thought so too, but couldn't really be bothered to calculate it, maybe I'll do the maths later. $\endgroup$ May 10, 2016 at 20:03
  • 2
    $\begingroup$ I wouldn't even mention temperatures like $10^5\;\mathrm K$ and above. Anyway, by the time we get there, all chemistry is long gone, and the only real process that matters is the ionization of atoms. $\endgroup$ May 10, 2016 at 20:09

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