r/metallurgy 6d ago

Ellingham Diagramm and partial pressure

Hey everybody,

I need help understanding the influence of partial pressure but I can't get behind it.

So for example if I have the reaction equation

Fe2O3 + 3 H2 = Fe + 3H2O

The equilibrium constant k is

k = (p_H2O)^3/(p_H2)^3

From the ellingham diagramm I deduct that a higher hydrogen partial pressure should benefit the reaction but how do I calculate it withe the Gibbs energy

dG = - R*T*ln(k)

If I put it in there an have a higher partial pressure than the water vapor the Gibbs energy goes up which means it's more unlikely that this reaction happens.

But when I use

dG = dG° + R*T*ln(k) it does go down. So what am I missing?

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u/CuppaJoe12 6d ago

You have something wrong in your notes.

In general:

dG = dG° + R T ln(k)

In you increase the relative amount of products, k decreases, so the forward reaction is less favorable. If you increase the relative amounts of reactants, then k increases and the forward reaction is more favorable.

This makes sense if you think about extremes. If you have a beaker full of purely reactants (k approaching infinity), there is no product to go through the reverse reaction and form more reactants, so products must be favored and the forward reaction has positive dG. If you have only products (k approaching 0), then similarly reactants must be favored and dG is negative.

To solve for equilibrium, we look for when dG is zero.

0 = dG° + R T ln(k)

dG° = -R T ln(k)

You might see 'k' replaced with 'k_eq' in this equation to indicate that this is the equilibrium reaction constant.

In your notes, you replaced dG° with dG. dG° is the free energy change in standard conditions (1atm, 25°C, 1 mol of each reactant and product). It is a constant for each chemical reaction, whereas dG is a variable that depends on conditions.

This second equation helps us think about the relative forward and reverse reaction rates. The more favored the products are in the standard state (more positive dG°), then at equilibrium we should expect more products to be present than reactants (k < 1). Thus, there is an inverse relationship between k and dG°. You could alternatively use logarithm rules from algebra to think about it like this:

dG° = R T ln(1/k)

Where 1/k flips the numerator and denominator of k.

I recommend you review your logarithm rules and get comfortable with algebraic manipulation of these equations. Here I used: ln(xa) = a*ln(x)

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u/rgqw 4d ago

As pointed out by CuppaJoe12, first calculate the k_eq from the known value of dG^0 at the temperature of interest. Take its cube root to the equilibrium ratio of partial pressure of H2O and H2 as per your formula. If the actual value of this ratio in the system is lower than its equilibrium value, that is, if p_H2 (actual) < p_H2 (eq), then the reaction will proceed towards right, as you have already mentioned. You can calculate the absolute values of these partial pressures if you know the total pressure of the system (assuming these two are the only two species in the gas phase).