4
$\begingroup$

My textbook1 quotes the following as the limitation of Ellingham diagram:

The interpretation of $\Delta G_\mathrm{f}^\circ$(standard Gibbs free energy of formation) is based on $K$ ($\Delta G_\mathrm{f}^\circ = -RT \ln K$).Thus it is presumed that the reactants and products are in equilibrium.

Definition of Standard Gibbs free energy of formation:

The standard Gibbs free energy of formation of a compound is the change of Gibbs free energy that accompanies the formation of $\pu{1 mol}$ of a substance in its standard state from its constituent elements in their standard states (the most stable form of the element at $\pu{1 bar}$ of pressure and the specified temperature, usually $\pu{298.15 K}$ or $\pu{25 ^\circ C}$).

But in Ellingham diagram reactants are normalized to consume one mole of oxygen having partial pressure of $\pu{1 atm}$. Thus products formed are not one mole. So how can it be a graph of Standard Gibbs free energy of formation and how this limitation even applies?


1. NCERT Chemistry Textbook for Class XII - Part I, Chapter 6, page 152

$\endgroup$
0
7
$\begingroup$

The Ellingham diagram doesn't actually use molar Gibbs energies of formation $\Delta G_\mathrm{f}^\circ$ per se; it is more accurate to say that it uses molar Gibbs energies of reaction $\Delta G_\mathrm{r}^\circ$. The difference is that the formation energy is only relevant to one specific chemical equation, for example:

$$\ce{Ca + 1/2O2 -> CaO} \qquad \qquad \Delta G_\mathrm{r}^\circ = \Delta G_\mathrm{f}^\circ(\ce{CaO})$$

in which the stoichiometric coefficient of CaO is equal to 1.** On the other hand, for any (balanced) equation with any stoichiometric coefficients, it is valid to define a Gibbs energy of reaction:

$$\ce{2Ca + O2 -> 2CaO} \qquad \qquad \Delta G_\mathrm{r}^\circ = 2\times \Delta G_\mathrm{f}^\circ(\ce{CaO})$$

which is related to the energy of formation, but is not the same thing, as evidenced by the factor of 2.

In the Ellingham diagram, every reaction has the same stoichiometric coefficient for $\ce{O2}$, which is typically 1. This is needed to make sure that different reactions are comparable. Let's say, for example, you want to see whether the reaction

$$\ce{C + 2CaO -> CO2 + 2Ca}$$

is feasible. This is done by checking the sign of $\Delta G_\mathrm{r}^\circ$: if it is negative, then the reaction is feasible, and vice versa. The point is that this $\Delta G_\mathrm{r}^\circ$ can be calculated by subtracting two reactions together:

$$\begin{align} \ce{C + O2 &-> CO2} & \Delta G_\mathrm{r}^\circ &= c_1 = \Delta G_\mathrm{f}^\circ(\ce{CO2}) \\ \ce{2Ca + O2 &-> 2CaO} & \Delta G_\mathrm{r}^\circ &= c_2 = 2 \times \Delta G_\mathrm{f}^\circ(\ce{CaO}) \\ \hline \ce{C + 2CaO &-> 2Ca + CO2} & \Delta G_\mathrm{r}^\circ &= c_1 - c_2 \\ \end{align}$$

but these two equations add up nicely only if the coefficients of $\ce{O2}$ in both equations are the same. What the Ellingham diagram does is to plot the Gibbs energies of reaction, $c_1$ and $c_2$: if $c_1 < c_2$, then the reaction is feasible. It doesn't plot the Gibbs energies of formation, because comparing those wouldn't tell us anything about the sign of $c_1 - c_2$.

As a final remark, note also that the equation

$$\Delta G_\mathrm{r}^\circ = -RT \ln K$$

holds true for any reaction, whether or not it actually corresponds to a formation reaction.


** Having a stoichiometric coefficient equal to $x$ does not mean the same thing as $x$ moles of the compound are produced in the reaction. The coefficient is purely a mathematical expression which tells us the stoichiometric relationship between different species in the reaction. It does not correspond to a real-life reaction, where a defined quantity of reactant is added to a defined quantity of product. To illustrate this, let's say you go to a lab and mix 0.4 mol of HCl to 0.4 mol of NaOH. You're asked to write a balanced equation for this. You can write

$$\ce{0.4 HCl + 0.4 NaOH -> 0.4 NaCl + 0.4 H2O,}$$

and that would be correct, but it is hardly the only correct possibility: the more conventional

$$\ce{HCl + NaOH -> NaCl + H2O}$$

is equally correct, even though the stoichiometric coefficients (1 in all cases) do not match up with the actual amount of substance used in the reaction (0.4 mol). Note also that the units are different: stoichiometric coefficients are dimensionless, but amount of substance is measured in moles. The difference is subtle, but one well worth pondering about, as mixing these two up can lead to a lot of misconceptions in thermodynamics.

$\endgroup$
3
$\begingroup$

We may interpret the energy variable as free energy of formation per mole of $\ce{O2}$. Thus, for instance, silicon would be favored to react with a limited amount of $\ce{O2}$ versus iron, because silica has a more negative free energy of formation per mole of $\ce{O2}$ than iron oxides; even though $\ce{Fe3O4}$ might be more negative per mole of compound because $\ce{Fe3O4}$ uses two moles of $\ce{O2}$ per mole of complound versus $\ce{SiO2}$ using one. This, of course, is one of the main drivers behind silicon going into slag when we smelt and purify iron.

$\endgroup$
2
$\begingroup$

For the Ellingham diagram:

(1) We write down the reaction/reactions that we are interested in. The $\Delta G_f$ (formation) per mole of product is calculated using the individual free energies of any intermediate reaction, and using Hess' law to sum them up.

(2) This energy is then normalised per mole of $O_2$, using stoichiometry.

(3) The normalised energy is plotted on the diagram.

Coming to your question - how is the Ellingham diagram representative if the products are not necessarily 1 mole?

There are two ways to resolve this:

(1) The easier, slightly less accurate : You consider $O_2$ to be the main product, such that you are writing a reduction reaction. Then just reverse the reaction to make it an oxidation reaction (this can be done for redox reactions). Consider this just a convention. You could equally make a diagram of the reduction reactions; it would convey the same information if you interpret it correctly.

(2) The more abstract, more accurate reason: As shown by @orthocresol, conversion from individual free energies of formation to final free energy of the reaction requires a combination of several reactions. Stoichiometry demands that different reactions be normalised. Since we are dealing with redox reactions that necessarily contain oxygen here, it makes the most sense to normalise the net reaction with respect to oxygen.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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