Materials Science · Radiation, vacuum and reactive environments

Reactive gases, oxidation and aggressive media

Separate reaction driving force from the transport that controls attack. Follow a protective oxide through growth, chemical removal and continuing substrate consumption, then compare mass records with local penetration and actual chemical evidence.

Useful background: Durability in real environments; Electrical transport and electronic materials; Heat transport, expansion and thermal response. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Use reaction stoichiometry and dimensionless activities to interpret a change in chemical driving force without mistaking equilibrium for a rate.
Explain how coupled transport through an oxide can slow growth, and how cracking or a volatile product can defeat that barrier.
Derive and interrogate a bounded oxidation–removal balance, separating resident scale, gross growth, material consumption and specimen mass change.
Use electrochemical charge and spatial evidence to distinguish an area-average loss from local attack and qualify transfer to a changed environment.

A hot gas or liquid can change a surface

Oxidation, dissolution, chemical attack or species penetration can alter structure and properties. Reactions depend on chemical potential, temperature and transport. A protective layer may slow one pathway but fail under another exposure.

An oxide-coated alloy can react slowly in one gas and lose material rapidly after its chemistry changes. The key question is not simply whether oxygen is present. Which reaction is favoured, which reactants reach the reaction front, which products remain, and which species leave? The same surface layer can be a transport barrier in one environment and a reactant consumed by another.

Chemical potential is a partial molar Gibbs free energy: μ_i = (∂G/∂n_i) at fixed temperature, pressure and amounts of the other species. It has units J/mol. For an ideal gas, μ = μ° + RT ln(p/p°), where T is absolute temperature, R the gas constant and p° the standard pressure. A real gas instead requires a fugacity-based activity; a concentrated liquid generally cannot replace activity by concentration without justification.

For a reaction, Δ_rG = Δ_rG° + RT ln Q, with Q formed from dimensionless activities raised to the balanced stoichiometric powers. Negative Δ_rG favours the forward reaction at the specified state. It neither gives a reaction time nor guarantees a protective product. Melting point is another property: a solid can react, dissolve or form a volatile product well below melting.

Use the reaction, not just the gas label

Consider the illustrative oxidation 2M(s) + O₂(g) → 2MO(s). If pure M and MO each have unit activity and the gas is ideal, Q = 1/(p_O₂/p°), so Δ_rG = Δ_rG° − RT ln(p_O₂/p°). Increasing oxygen partial pressure makes this particular forward oxidation more favourable at fixed T. The equilibrium boundary is p_eq/p° = exp[Δ_rG°/(RT)].

This is a boundary for the two stipulated solid phases, not a complete phase diagram or a safe operating pressure. Alloy activities, additional oxides, dissolved oxygen and other gas reactions can change the result. A gas mixture containing water and hydrogen can impose an oxygen chemical potential through their reaction equilibrium; total pressure or an informal name such as steam does not supply every needed activity.

In a supplied teaching case at 1000 K, Δ_rG° = −200 kJ per mole of the written reaction. With R = 8.314 J/(mol·K) and p° = 1 bar, p_eq/p° ≈ 3.57 × 10⁻¹¹. At p_O₂/p° = 10⁻⁶ the forward Δ_rG is about −85.1 kJ/mol; at 10⁻¹² it is +29.7 kJ/mol. These are invented inputs for M/MO, not thermochemical data or an exposure prescription for a real material.

A protective scale slows the route to further reaction

An external oxide scale can separate gas from metal, but continued growth requires transport. Metal cations may move outward, oxygen-containing species inward, and electrons or electron holes carry compensating charge. Growth can occur at different interfaces depending on the mobile species. Grain boundaries, pores and cracks can provide routes unlike transport through an intact lattice.

For a compact planar scale with a stable composition, fixed boundary driving conditions and approximately steady transport across its thickness x, a flux proportional to a concentration difference divided by x motivates dx/dt = a/x. Here a is a supplied effective thickness-growth coefficient with units length²/time; it bundles transport, driving conditions and stoichiometric conversion. It is not simply a tracer diffusion coefficient. Integrating from a positive initial thickness x₀ gives x² − x₀² = 2at.

This parabolic thickness law predicts a declining instantaneous thickening rate as the path length grows. A reaction-controlled interval with nearly unchanged reaction area and surface conditions can instead approach linear growth. Very thin films, changing phase assemblages, gas-supply limitation and damaged scales require other models. A straight x²-versus-time segment is consistent with the simple transport model, not unique proof of it.

Protection also requires a sufficiently continuous, adherent and chemically stable barrier. Growth stresses and thermal expansion mismatch can crack or detach a scale; detachment is called spallation. The exposed regions can regrow while other regions retain old oxide. A small final scale mass can therefore reflect repeated loss, not little reaction. Internal reaction or penetration below a surface also need not be represented by the measured outer thickness.

Water vapour can consume an otherwise protective oxide

A researched example is chromia, Cr₂O₃, in suitable high-temperature oxygen/water-vapour environments. One volatile product is chromium oxyhydroxide, CrO₂(OH)₂(g), formed through ½Cr₂O₃(s) + H₂O(g) + ¾O₂(g) ⇌ CrO₂(OH)₂(g). Chromium, hydrogen and oxygen all balance: one Cr, two H and four O atoms on each side. This is chemical formation of a gaseous product, not just release of a pre-existing adsorbed contaminant.

For this reaction at a fixed temperature, the equilibrium product activity is p_v/p° = K a_Cr₂O₃^(1/2)(p_H₂O/p°)(p_O₂/p°)^(3/4) in the ideal-gas approximation. Thus water, oxygen and the oxide activity all enter. If water partial pressure rises fourfold while oxygen partial pressure falls sixteenfold, the conditional product pressure changes by 4 × (1/16)^(3/4) = 1/2, with K and oxide activity held fixed. Water alone is not an adequate description of that changed mixture.

An equilibrium pressure at the oxide surface is not a recession rate. If gas transport limits removal, a schematic molar flux is J_v = h_m(c_v,s − c_v,∞), where h_m is a mass-transfer coefficient in m/s and the concentrations are in mol/m³. Gas flow, geometry and the bulk product concentration affect this step. Surface kinetics or gas-phase reactions can instead limit it. The equation is a stated boundary-layer model, not a universal flow correlation.

Measured pressure dependence was compared with the proposed oxyhydroxide reaction. Such dependence supports an identity within a model; it is not the same as an unambiguous mass spectrum. Absolute thermochemical data and competing volatile species must also be assessed. No historical rate value or present material limit is adopted here.

Chromia chemistry and silica chemistry are not interchangeable. For example, silica can form Si(OH)₄(g) through SiO₂(s) + 2H₂O(g) ⇌ Si(OH)₄(g), with a different pressure dependence. Removing a scale-forming constituent from an alloy can also deplete the supply available to maintain its barrier. This illustrates why a contaminant addition can change a pathway rather than invalidate an earlier result under different conditions.

A constant scale thickness can hide continuing consumption

Now use a separate invented planar model, not a fit to the chromium or silica experiments. Keep a uniform, continuous oxide of fixed density and composition. Let diffusion-controlled formation add oxide-equivalent thickness at a/x and chemical removal subtract it at a constant b > 0, both per unit time. With a fixed exposed area, no spallation, no internal oxidation and unchanged kinetic coefficients, dx/dt = a/x − b, starting at x₀ > 0. The positive starting thickness avoids extrapolating the singular a/x expression to an atomically bare surface.

A steady thickness x_* = a/b follows by setting dx/dt = 0. It is stable within this model: below x_* formation exceeds removal; above it removal exceeds formation. At x_* the scale still forms and leaves at rate b. A stationary thickness is therefore not a chemically inert interface or an infinite-life state. Loss of the underlying reactant, a composition change or a crack would end the stipulated regime.

For b = 0, x = √(x₀² + 2at). For b > 0 and x₀ < x < a/b, separating variables gives t = (a/b²) ln[(a − bx₀)/(a − bx)] − (x − x₀)/b. The logarithm contains a dimensionless ratio. This implicit expression approaches infinite time as x approaches a/b; the limit is not reached at a finite plotted time.

Track two cumulative quantities as well: B = bt is removed oxide-equivalent thickness and G = ∫₀ᵗ(a/x)dt is gross formed thickness. The balance gives G = x − x₀ + B. With oxide density ρ_ox, metal mass fraction w_M and oxygen mass fraction w_O = 1 − w_M, metal consumed per area is ρ_ox w_M G. If only oxygen incorporation adds mass and loss of oxide-equivalent material removes specimen mass, net specimen change per area is Δm/A = ρ_ox(w_O G − B). Initial oxide is included in the reference specimen mass. Gas atoms joining the volatile product directly from the gas are not counted as mass previously held by the specimen.

These identities require the stated fixed stoichiometry, density and area. They cannot be applied unchanged to a carbon-containing substrate that also emits CO, to dissolved salt products, or to a multi-oxide scale. The purpose is to distinguish inventories, not to make a thin oxide a proxy for small substrate loss.

Original invented planar oxide model. Initial scale thickness is 1.00 micrometre and a = 1.00 square micrometre per hour. Solid blue has no removal, b = 0; dashed violet has b = 0.250 micrometre per hour. Top panel is resident scale thickness x, not metal lost. Its green dotted line at 4.00 micrometres is the removal model’s asymptotic steady thickness a/b, not a measured protective limit. Bottom panel is gross cumulative formed oxide-equivalent thickness G, excluding the initial scale and including material later removed. Both time axes span 0 to 24 hours. At 24 hours blue has x = 7.00 and G = 6.00 micrometres; violet has x = 3.65535 and G = 8.65535 micrometres. The thinner resident scale accompanies more gross formation. Coefficients, density and phase state are held fixed, with no spallation or internal oxidation. Curves are original model solutions, not chromia measurements.

Original mass accounting at 24 hours for the same invented model, not measured alloy data. Oxide density is 5000 kilograms per cubic metre, with metal mass fraction 0.800 and oxygen fraction 0.200. One micrometre of oxide-equivalent thickness has mass 0.500 milligram per square centimetre. Solid blue bars have no removal; hollow violet bars with dashed outlines have b = 0.250 micrometre per hour. Upper bars show positive mass of underlying metal consumed: 2.400 and 3.46214 milligrams per square centimetre, respectively. Lower signed bars show net specimen mass change from its initial oxide-containing state: plus 0.600 and minus 2.13447 milligrams per square centimetre. All bars use the same horizontal mass-per-area scale from minus 3 to plus 4. Positive metal consumed is a magnitude, not a specimen mass gain. Oxygen incorporation adds mass; oxide-equivalent removal subtracts mass. Thus the small resident-scale thickness or a net mass change cannot replace a metal-consumption measurement. Fixed phase composition, area and supplied pathways are necessary assumptions.

In a liquid, separate the current balance from the attack depth

Electrochemical corrosion provides a contrasting pathway. For iron dissolution, Fe(s) → Fe²⁺(aq) + 2e⁻ is an anodic half-reaction. In an acidic oxygen-containing electrolyte a possible cathodic reaction is O₂ + 4H⁺ + 4e⁻ → 2H₂O. Electrons move through the metal while ions carry charge in the electrolyte; anodic and cathodic sites need not coincide. Solution chemistry and film state affect the kinetics, so these balanced equations do not by themselves supply a corrosion current.

At freely corroding steady conditions, the total anodic and cathodic currents balance. A zero net external current is not zero dissolution. If the supplied anodic dissolution-current density i_a is referred to the whole exposed area, constant in time, and all that charge corresponds to M → M^n+ + ne⁻, Faraday’s law gives lost mass per area m/A = i_a t M_m/(nF). Here M_m is molar mass and F the charge per mole of electrons. Side reactions or a changing valence/current require a revised allocation.

Dividing by metal density gives an area-average equivalent penetration. If the same dissolution is confined to a supplied area fraction f_a, a uniform-depth approximation within that patch gives local depth equal to the area-average depth divided by f_a. This is a geometrical counterexample to interpreting a mean as a maximum, not a pit-growth prediction. Real local chemistry, spreading current, repassivation and three-dimensional shape can all change that distribution.

An outer film can coexist with attack beneath it, a depleted near-surface zone or a deep local defect. Cross-sectional and chemical evidence should therefore resolve the location and species relevant to the proposed mechanism. Average mass change, current and a visually intact surface answer different questions.

Specify species and conditions

Pressure, concentration, flow and contaminants can change reaction rates or products. A nominal environment name is insufficient when trace species govern damage. Include joint and coating materials because they create additional reaction paths.

To distinguish intact-scale transport from chemical removal, combine a time-resolved mass record with retained-scale thickness/composition and evidence for emitted or collected species. To test spallation, inspect detached products and exposed regions. To assess penetration, use depth-resolved sections rather than only a plan-view surface image. Each observation tests a different term in the balance.

A rate comparison should retain specimen composition and history, surface/scale state, local temperature, activities or gas partial pressures, total pressure, flow/transport geometry, duration and the measured endpoint. In a liquid add electrolyte state, potential/current interpretation and the area basis. These quantities determine whether a fitted coefficient or mechanism can transfer.

No single parabolic fit, small mass change or high melting point establishes universal resistance. Report the demonstrated pathway and conditions, the uncertainty and the unobserved alternatives. The next lesson treats combined exposures and sequence explicitly; a separate benign-gas result does not answer that coupled question.

Worked examples

Change oxygen activity without inventing a rate

For the invented M/MO case at 1000 K, use Δ_rG° = −200 kJ per mole of the written reaction 2M + O₂ → 2MO. Compare p_O₂/p° = 10⁻⁶ and 10⁻¹², with unit solid activities and R = 8.314 J/(mol·K).

The reaction consumes one O₂, so Q = (p_O₂/p°)⁻¹ and Δ_rG = −200000 − 8314 ln(p_O₂/p°), in J per mole of the written reaction. A pressure with units must not be placed directly inside a logarithm.
At 10⁻⁶, ln(p_O₂/p°) = −13.8155, giving Δ_rG ≈ −85137.8 J/mol = −85.1 kJ/mol. The stipulated forward oxidation is favoured.
At 10⁻¹², ln(p_O₂/p°) = −27.6310, giving +29724.3 J/mol = +29.7 kJ/mol. The forward reaction is not favoured relative to those stipulated pure phases.
The zero occurs at p_eq/p° = exp(−200000/8314) ≈ 3.57 × 10⁻¹¹. Neither sign computes a kinetic rate, proves which other phases exist, or establishes a real component limit.

The signs differ: approximately −85.1 and +29.7 kJ/mol. The thermodynamic comparison supplies a direction for a specified reaction, not an attack speed.

Less retained oxide, more underlying metal consumed

In the invented scale model use x₀ = 1.00 μm, a = 1.00 μm²/h, and compare b = 0 with b = 0.250 μm/h after 24.0 h. For the fictitious fixed-composition oxide take ρ_ox = 5000 kg/m³ and w_M = 0.800. Find retained thickness, gross growth, metal consumption per area and net specimen mass change. Assume the stated mass pathways only.

With no removal, x = √(1² + 2 × 1 × 24) = 7.00 μm. B = 0 and G = x − x₀ = 6.00 μm. With removal, x_* = a/b = 4.00 μm; the solution starting at 1 μm must remain below that asymptotic limit.
For removal insert the supplied coefficients into the implicit relation: t = 16 ln[0.750/(1.00 − 0.250x)] − 4(x − 1.00), using x numerically in μm and t in h. At x = 3.65 μm it gives 23.775 h, while 3.66 μm gives 24.199 h. Repeatedly narrowing that bracket at t = 24.0 h gives x ≈ 3.65535 μm. The logarithm’s ratio has matching units in its numerator and denominator.
For the removal case B = 0.250 × 24 = 6.00 μm and G = 3.65535 − 1 + 6 = 8.65535 μm. Gross formation exceeds retained thickening because material has also left.
A 1 μm oxide-equivalent layer at 5000 kg/m³ has mass 0.00500 kg/m² = 0.500 mg/cm². Metal consumption is therefore 0.800 × 0.500 × G in mg/cm²: 2.40 without removal and about 3.46214 with removal. The latter is about 44.3% greater, despite its thinner resident scale.
With w_O = 0.200, Δm/A = 0.500(0.200G − B) mg/cm². This is +0.600 mg/cm² without removal and about −2.13447 mg/cm² with removal. These net changes do not equal metal consumed; oxygen gain and oxide loss enter differently.
This is a deliberately bounded comparison with the same a. A real changed gas may also change a, scale composition, density and continuity. The numbers are not chromia data or a recommended exposure.

Resident scale is 7.00 versus 3.66 μm, but consumed-metal mass is 2.40 versus 3.46 mg/cm². Net specimen change is +0.600 versus −2.13 mg/cm²; none of these quantities can silently replace the others.

Identify what a changed gas result means

An alloy survives one oxidising gas but fails in another gas with a small contaminant addition. Is the original result disproved?

Retain the first result under its original chemistry, scale and temperature. In the changed gas, ask whether the added species alters reactant activity, reacts with the existing scale, or changes transport.
The chromia/water-vapour reaction supplies one researched candidate pathway, not an automatic diagnosis for every contaminant. Look for changed reaction products or volatile species, constituent depletion, a damaged scale and the corresponding mass/thickness history.
A matched comparison that changes the relevant species while retaining other conditions can test that explanation. A small final scale thickness or an endpoint property loss alone cannot distinguish volatilisation from spallation or a different reaction.
In the ideal steady-thickness balance, growth and removal can remain nonzero. Hence a stable-looking surface does not prove negligible material consumption.

No. Resistance is conditional on the actual exposure.

Convert a dissolution charge into an area-average loss

An invented iron record supplies an anodic dissolution current density of 20.0 μA/cm², referred to the whole exposed area, for 24.0 h. Assume all anodic charge forms Fe²⁺, with M_m = 55.845 g/mol, F = 96485 C/mol and density 7.87 g/cm³. Compare uniform loss with the same loss confined uniformly to 1.00% of the area.

The charge per whole exposed area is 20.0 × 10⁻⁶ C/(s·cm²) × 86400 s = 1.728 C/cm². Two moles of electrons accompany one mole of iron dissolved.
m/A = 1.728 × 55.845/(2 × 96485) = 0.000500079 g/cm² = 0.500079 mg/cm². Dividing by 7.87 g/cm³ gives 6.35424 × 10⁻⁵ cm = 0.635424 μm of area-average equivalent loss.
Confinement to f_a = 0.0100 gives a uniform patch depth of 0.635424/0.0100 = 63.5424 μm, with the same total mass. It also implies a local anodic density 100 times the supplied whole-area average.
This redistribution is supplied, not inferred from the current alone. No particular pit shape, initiation time, current-stability mechanism or safe remaining section has been established.

The same charge corresponds to 0.500 mg/cm² and 0.635 μm of area-average loss, or about 63.5 μm in the stipulated 1% patch. Local evidence is needed to decide which distribution exists.

Try the reasoning

Does a high melting point establish resistance to an oxygen/water-vapour mixture? Identify a chemical route that can operate while the solid remains below melting.
Show the answer and reasoning
No. A protective solid oxide can react to form a volatile hydroxide or oxyhydroxide, or another reaction can consume the solid without melting.
Thermal phase stability, reaction free energy and transport kinetics are separate. A high melting point supplies neither the relevant reaction quotient nor the rate or continuity of the barrier.
A specimen has almost no net mass change, but its surface layer is thinner and a scale-forming element is depleted below it. Why inspect the products and depth profile before calling it resistant?
Show the answer and reasoning
Opposing gain and loss can conceal continuing reaction, and the retained surface does not describe the consumed inventory or local penetration.
Oxidation adds oxygen while volatile or detached products remove specimen material. Product chemistry and spatially resolved observations test these alternatives; mass cancellation alone does not select a mechanism.
At fixed temperature and oxide activity, what happens to the equilibrium CrO₂(OH)₂ product pressure if p_H₂O is multiplied by four and p_O₂ by 1/16? May the same factor be assigned automatically to metal recession?
Show the answer and reasoning
The conditional equilibrium product pressure halves. A recession factor does not follow without a removal/transport model.
The factor is 4 × (1/16)^(3/4) = 0.5. Mass-transfer coefficients, bulk product concentration, surface kinetics and scale formation can also change.
In a new invented run a = 2.00 μm²/h and b = 0.500 μm/h. The resident scale is already at its steady thickness. Find that thickness and the additional gross oxide-equivalent formation in the next 10.0 h. Is the substrate unchanged?
Show the answer and reasoning
x_* = 4.00 μm and G increases by 5.00 μm. The scale is stationary while underlying reactant continues to be consumed.
At x_* = a/b, a/x_* = b = 0.500 μm/h. Formation and removal each accumulate 5.00 μm over ten hours; their difference is zero. Constant thickness is not zero chemical throughput.
With no removal and a = 1.00 μm²/h, compare two initially intact model scales, x₀ = 1.00 and 3.00 μm. Find x after 4.00 h and gross formed thickness G in each. Does simply resetting the clock erase the initial condition?
Show the answer and reasoning
Final thicknesses are 3.00 and √17 ≈ 4.12 μm; gross growth is 2.00 and about 1.12 μm. The initial scale remains part of the problem.
Use x² = x₀² + 2at, then subtract x₀. A thicker starting layer creates a longer transport path in this model. A real old scale can also have changed chemistry, defects or adhesion, so thickness alone may not transfer a.
Keep the same dissolution charge, metal and valence as the worked iron example, but supply a uniformly attacked patch fraction of 2.00% rather than 1.00%. Find the patch depth. Does a zero external ammeter reading prove no such corrosion current?
Show the answer and reasoning
The stipulated patch depth is about 31.8 μm. Zero external net current does not imply zero balanced internal anodic and cathodic currents.
0.635424 μm/0.0200 = 31.7712 μm. This is a charge-and-area conversion, not a model for how the active patch forms or remains stable.
An x²-versus-time segment is straight, but detached oxide was not collected and scale composition was not measured. Does that graph alone establish an intact diffusion-controlled scale and a low metal-consumption rate?
Show the answer and reasoning
No. It is consistent with a bounded parabolic description, but unmeasured loss and changing scale states prevent that unique inference.
Check retained thickness/composition, specimen mass, emitted or detached products and the relevant depth profile against one common mass and transport account. A fit cannot supply measurements excluded from the record.

What to carry forward

A reaction needs specified activities, stoichiometry and temperature; a favourable free energy is not an attack rate.
An intact scale can slow transport, but its continuity, adherence and chemical stability must persist.
Chemical removal can balance scale growth while the underlying material continues to be consumed.
Net specimen mass, retained scale and metal loss are different inventories; average loss also does not bound local depth.
Transfer resistance claims through the actual chemistry, transport, scale state, time and measured endpoint.

References

Teaching resources and research supporting this lesson’s explanations and examples.