Equilibrium Oxygen Partial Pressures in Metal-Oxide Systems

Hello! Welcome to the fifth lesson in our module on the chemical engineering fundamentals of metallurgy.

In our previous lesson, we established that Gibbs free energy ($\Delta G$) is the key criterion for determining if a chemical reaction is thermodynamically feasible. We learned that for a metal M to be reduced from its oxide MO by a reductant R, the Gibbs free energy of formation of the reductant's oxide (RO) must be more negative than that of MO.

This tells us if a reaction can happen, but it doesn't tell us under what conditions. In metallurgy, controlling the chemical environment is as important as controlling the temperature. A reaction's direction can be reversed simply by changing the concentration of gases in the furnace.

Today's lesson addresses this directly. Our learning outcome is to: Calculate equilibrium oxygen partial pressures for metal-oxide systems at different temperatures.

We will connect the abstract concept of $\Delta G$ to a concrete, measurable, and controllable industrial parameter: the partial pressure of oxygen. Understanding this relationship is crucial for controlling a metallurgical process to ensure you are reducing the ore, not re-oxidizing the metal you just produced.

1. From Spontaneity to Equilibrium

In the last lesson, we focused on the standard Gibbs free energy change, $\Delta G^{\circ }$, which applies when all reactants and products are in their standard states (e.g., 1 bar pressure for gases). Real-world processes, however, rarely operate under these specific conditions.

The actual Gibbs free energy change, $\Delta G$, which determines spontaneity under any conditions, is related to $\Delta G^{\circ }$ by the following equation:

$$\Delta G=\Delta G^{\circ }+RT\ln Q$$

Here, $R$ is the ideal gas constant (8.314 J/mol·K), $T$ is the absolute temperature, and $Q$ is the reaction quotient. $Q$ has the same form as the equilibrium constant but uses the current concentrations or partial pressures, not the equilibrium ones.

A system is at equilibrium when the forward and reverse reaction rates are equal, and there's no more driving force for a net change. At this point, $\Delta G=0$. This leads to the most important equation linking thermodynamics and chemical equilibrium:

$$0=\Delta G^{\circ }+RT\ln K$$ $$\Delta G^{\circ }=-RT\ln K$$

Here, $K$ is the equilibrium constant, which is the value of the reaction quotient $Q$ when the system has reached equilibrium. This equation is powerful because it connects a fundamental thermodynamic property of a reaction ($\Delta G^{\circ }$) to the specific mixture of products and reactants ($K$) that will exist once the reaction settles.

The following reading explains this fundamental relationship and then applies it directly to metal oxidation.

2. The Equilibrium Oxygen Partial Pressure

As the reading explained, for a typical metal oxidation reaction, like the formation of a generic metal oxide MO:

$$2\text{M(s)}+{\text{O}}_2\text{(g)}\rightleftharpoons 2\text{MO(s)}$$

The equilibrium constant, $K_p$, is expressed in terms of the activities of the products and reactants. By convention, the activity of a pure solid or liquid is taken to be 1. For gases, we use their partial pressures (in bar or atm). Therefore, the expression for $K_p$ simplifies dramatically:

$$K_p=\frac{a_{\text{MO(s)}}^2}{a_{\text{M(s)}}^2\cdot p_{{\text{O}}_2}}=\frac{1^2}{1^2\cdot p_{{\text{O}}_2}}=\frac{1}{p_{{\text{O}}_2}}$$

Now, we can substitute this directly into our key equation from the previous section:

$$\Delta G^{\circ }=-RT\ln \left(\frac{1}{p_{{\text{O}}_2}}\right)$$

Using the logarithm property $\ln (1/x)=-\ln (x)$, this becomes:

$$\Delta G^{\circ }=RT\ln (p_{{\text{O}}_2})$$

Rearranging to solve for the equilibrium oxygen partial pressure, we get the central formula for this lesson:

$$p_{{\text{O}}_2}=\exp \left(\frac{\Delta G^{\circ }}{RT}\right)$$

This equation gives us the exact partial pressure of oxygen at which the metal, its oxide, and the surrounding gas are in perfect equilibrium at a given temperature $T$.

• If the actual oxygen partial pressure in your reactor is higher than this calculated value, the atmosphere is oxidizing, and the metal will turn into oxide.
• If the actual oxygen partial pressure is lower than this value, the atmosphere is reducing. To reach equilibrium, the oxide will decompose to form metal. This is the condition required for metal production.

3. A Worked Example: Reducing Titanium Dioxide

Let's apply this to a reaction relevant to our goal of understanding titanium production. Imagine we wanted to reduce titanium dioxide (TiO₂) using hydrogen gas at a very high temperature, say 1600 °C (1873 K).

The reduction reaction would be: ${\text{TiO}}_2\text{(s)}+2{\text{H}}_2\text{(g)}\to \text{Ti(s)}+2{\text{H}}_2\text{O(g)}$

But first, let's just consider the stability of TiO₂ itself. What is the equilibrium oxygen partial pressure for the reaction $\text{Ti(s)}+{\text{O}}_2\text{(g)}\rightleftharpoons {\text{TiO}}_2\text{(s)}$ at 1873 K?

At this temperature, the standard Gibbs free energy of formation for TiO₂ is approximately $\Delta G_{1873}^{\circ }=-136,000\text{ cal/mol}$ or about $-569,000\text{ J/mol}$.

Let's do the calculation. We need to be careful with units. We'll use $R=8.314$ J/mol·K.

$$p_{{\text{O}}_2}=\exp \left(\frac{-569,000\text{ J/mol}}{(8.314\text{ J/molcdotpK})\cdot (1873\text{ K})}\right)$$ $$p_{{\text{O}}_2}=\exp \left(\frac{-569,000}{15,570}\right)=\exp (-36.5)$$ $$p_{{\text{O}}_2}\approx 1.7\times 10^{-16}\text{ atm}$$

This is an incredibly low pressure. It tells us that titanium is extremely reactive towards oxygen. To prevent titanium metal from oxidizing at 1600 °C, you would need a vacuum better than what can be achieved in most industrial equipment. Conversely, to reduce TiO₂ back to titanium metal, you need an environment that is even more powerfully reducing—one that can maintain an effective oxygen pressure below this minuscule value. This calculation single-handedly explains why titanium production is so challenging and must be done in a carefully controlled inert atmosphere.

The following resource provides a similar calculation, though it uses calories for energy and a different value for R (1.987 cal/mol·K). The result is the same order of magnitude.

Test your understanding!

The formation of aluminum oxide (Al₂O₃) from aluminum is given by the reaction:

$$\frac{4}{3}\text{Al(l)}+{\text{O}}_2\text{(g)}\rightleftharpoons \frac{2}{3}{\text{Al}}_2{\text{O}}_3\text{(s)}$$

At 1000 °C (1273 K), the standard Gibbs free energy of formation for this reaction (per mole of O₂) is approximately $\Delta G^{\circ }=-845,000\text{ J/mol}$.

Calculate the equilibrium oxygen partial pressure required to prevent the oxidation of liquid aluminum at this temperature. Is aluminum more or less reactive than titanium?

Show answer

We use the same formula:

$$p_{{\text{O}}_2}=\exp \left(\frac{\Delta G^{\circ }}{RT}\right)$$ $$p_{{\text{O}}_2}=\exp \left(\frac{-845,000\text{ J/mol}}{(8.314\text{ J/molcdotpK})\cdot (1273\text{ K})}\right)$$ $$p_{{\text{O}}_2}=\exp \left(\frac{-845,000}{10,582}\right)=\exp (-79.85)$$ $$p_{{\text{O}}_2}\approx 1.6\times 10^{-35}\text{ atm}$$

This equilibrium oxygen partial pressure is many orders of magnitude lower than that for TiO₂. This indicates that Al₂O₃ is significantly more stable than TiO₂, and aluminum is even more reactive towards oxygen than titanium. This is a key reason why producing aluminum via the Hall-Héroult electrolytic process requires so much energy.

4. Equilibrium in Reactive Gas Mixtures

In many industrial furnaces, the atmosphere isn't just an inert gas with trace oxygen. Often, a reactive gas mixture like carbon monoxide/carbon dioxide (CO/CO₂) or hydrogen/water vapor (H₂/H₂O) is used. These mixtures "buffer" the oxygen partial pressure at a specific level determined by their ratio.

For example, consider the reaction: $2\text{CO(g)}+{\text{O}}_2\text{(g)}\rightleftharpoons 2{\text{CO}}_2\text{(g)}$.

This reaction also has an equilibrium constant and an associated $\Delta G^{\circ }$. In an atmosphere containing CO, CO₂, and O₂, all three gases will adjust their partial pressures to satisfy this equilibrium. This means the ratio of $p_{{\text{CO}}_2}/p_{\text{CO}}$ sets a specific equilibrium $p_{{\text{O}}_2}$.

When this atmosphere is used to reduce a metal oxide MO, two equilibria must be satisfied simultaneously:

1. The metal-oxide equilibrium: $2\text{M}+{\text{O}}_2\rightleftharpoons 2\text{MO}$
2. The gas mixture equilibrium: $2\text{CO}+{\text{O}}_2\rightleftharpoons 2{\text{CO}}_2$

Since there can only be one value of $p_{{\text{O}}_2}$ in the system, we can set them equal to each other. This allows us to calculate the exact $p_{{\text{CO}}_2}/p_{\text{CO}}$ ratio needed to achieve a low enough equilibrium $p_{{\text{O}}_2}$ to reduce the metal oxide.

This is the principle behind using CO gas as the primary reductant in an iron blast furnace. By controlling the gas composition, operators can control the "reducing power" of the furnace atmosphere.

Conclusion

You can now connect the Gibbs free energy of a reaction to a tangible, physical condition—the equilibrium oxygen partial pressure.

• Key Takeaways:
  • The fundamental link between thermodynamics and equilibrium is $\Delta G^{\circ }=-RT\ln K$.
  • For a metal oxidation reaction, this simplifies to $\Delta G^{\circ }=RT\ln (p_{{\text{O}}_2})$, allowing for the direct calculation of the equilibrium oxygen partial pressure.
  • This pressure represents the threshold between an oxidizing and a reducing atmosphere for a specific metal at a specific temperature.
  • The extremely low equilibrium $p_{{\text{O}}_2}$ for oxides like TiO₂ and Al₂O₃ quantifies why their reduction is so difficult and energy-intensive.
  • Reactive gas mixtures like CO/CO₂ establish their own equilibrium oxygen partial pressure, which can be harnessed for reduction processes.

We have performed these crucial calculations algebraically. However, comparing many different metals over a wide range of temperatures would be tedious. What if there were a graphical way to see all of this information at once?

In our next lesson, we will be introduced to Ellingham diagrams. These diagrams are elegant plots of $\Delta G^{\circ }$ vs. Temperature for various oxidation reactions. As we will see, they not only allow us to instantly compare the stability of different oxides but also contain clever graphical tools (nomographic scales) to read equilibrium oxygen partial pressures and required gas ratios directly from the chart, making them an indispensable tool for metallurgists.