Equilibrium tendencies and transformation rates

A prediction of the end does not specify the journey

In an invented record, two samples of one alloy enter the same 500 K hold in the same initial state. One is removed after 2 min and contains 8% of a second phase by volume; another is removed after 8 min and contains 19%. A separate equilibrium assessment gives 20% at 500 K. Neither a phase label nor the final temperature alone explains which sample will be obtained. We need the route and its timing.

A phase diagram describes equilibrium states under its stated conditions. Producing those states may require atoms to diffuse over substantial distances. Rapid cooling can interrupt that rearrangement and preserve a different structure.

Equilibrium describes the state favoured under specified constraints. For a closed composition at fixed temperature and pressure, the preceding stability lesson uses the minimum Gibbs free energy among the allowed alternatives. Kinetics asks how the state changes: which atomic rearrangements occur, how new regions appear, how they grow, and how quickly these processes proceed. The equilibrium comparison does not supply the clock.

Here the same overall composition can be present as one supersaturated solid solution or as a matrix containing precipitates. A solid solution holds different chemical species within one solid phase. Supersaturated means that more of the solute remains dissolved than would be present in the equilibrium matrix at the new conditions. A precipitate is a region of a distinct phase formed from that solution; it is not a grain merely rotated into another orientation.

Mechanisms set the timescale

Nucleation creates a region of a new phase; growth expands it. Both can have kinetic barriers. Temperature changes driving forces and atomic mobility, sometimes in competing ways. A treatment should therefore be assessed with transformation evidence and time, rather than by reading equilibrium alone.

Take a representative binary alloy, containing a host and a solute. At a suitable preparation temperature, its solute is dissolved in phase α, called alpha. Transfer it sufficiently rapidly to 500 K and the initially retained α contains excess dissolved solute. Under the specified closed-system conditions the equilibrium state contains α plus a solute-rich phase β, called beta. The matrix is the surrounding, continuous α phase.

Nucleation establishes small new β regions able to continue growing under the local conditions. Many smaller embryos can disappear instead. Creating an α–β interface has a cost, so a favourable final phase combination does not mean that every tiny fluctuation grows. Surfaces, existing interfaces and defects can alter the available nucleation routes. The metastability lesson develops barriers in detail; here their role is that formation of suitable starting regions can consume time.

Growth then advances the boundary of an existing β region. In this example the growing precipitate needs extra solute. Atoms redistribute through the solid by repeated local movements, called diffusion, and are incorporated across the α–β interface. The adjacent matrix can be depleted relative to more distant matrix, providing a net solute supply toward the particle. The specimen need not melt, and the whole precipitate need not travel through the matrix to collect atoms.

The closed specimen gains no new solute overall: solute enrichment of β accompanies depletion of α. As transformation proceeds, the available supersaturation and diffusion fields change. Particle growth need not continue at a constant speed. Nucleation and growth can also overlap; they are different processes, not necessarily two separate clock intervals.

This is a diffusion-involving precipitation route. It is not a rule that every material transformation requires long-range solute transport. Some structural changes can occur with local rearrangements or interface motion and need their own kinetic description. The definitions explain what evidence to seek; they do not provide a universal precipitation temperature or time.

Measure phase amount before calculating a rate

The table gives an original invented isothermal record: the hold temperature is constant at 500 K. Separate matched specimens begin in the same homogeneous α state. Transfer to the hold and the final cooling used for examination are stipulated to cause negligible further transformation in this exercise. The hold clock starts when the specimen reaches 500 K. These conditions permit the recorded structures to be assigned to the hold, rather than to an unknown cooling interval.

Let f_β be β volume divided by the total representative specimen volume. It is a dimensionless fraction: f_β = 0.08 means 8% β by volume. The remaining phase here is α, so it occupies 92%. These are supplied volume-fraction results; a count of dark pixels on an arbitrary polished section would not automatically establish that bulk quantity.

The separately supplied equilibrium β fraction at 500 K is f_β,eq = 0.20. Define q = f_β/f_β,eq as the normalised β amount for this record, which starts with no β. At 2 min, q = 0.08/0.20 = 0.40: 40% of the equilibrium β amount, while only 8% of the specimen is β. At 8 min the corresponding values are 95% and 19%. Do not call either number the percentage of atoms that have stopped moving.

An average rate over a chosen interval is the change in phase fraction divided by elapsed time: (f_β,2 − f_β,1)/(t_2 − t_1). Between 1 and 2 min it is (0.08 − 0.02)/1 = 0.06 min⁻¹, or 6 percentage points of β volume per minute. Between 4 and 8 min it is (0.19 − 0.16)/4 = 0.0075 min⁻¹, or 0.75 percentage points per minute. The same record has different interval averages.

A phase fraction is an amount; a rate describes its change per time. “8% β” is neither “8% per minute” nor a particle growth speed in nm/s. The measured amount combines the sizes and numbers of regions. To separate nucleation from growth, also examine how region counts and dimensions change with suitable resolution.

The graph joins the observations as a reading aid, not a kinetic law between them. The supplied β fraction increases monotonically in this example, but a particular value at an unmeasured intermediate time is not established by the straight line. Nor does matching the single equilibrium phase fraction at 16 min prove complete equilibrium: phase compositions and the spatial arrangement still need their own evidence.

Follow the state through the whole thermal path

Compare two specific paths. Both begin with the same prepared α solution, use the same rapid transfer to 500 K and finish with the same rapid cooling for examination. Path S holds for 2 min; path L holds for 8 min. After the shorter hold, β occupies 8% of the total volume within an α matrix; after the longer, it occupies 19%. The β regions and the remaining dissolved-solute distribution differ, although the bulk chemical composition is unchanged.

For the chosen precipitation route, the first rapid transfer retains a supersaturated solution rather than immediately producing the equilibrium phase amounts. The hold gives time for new regions, solute transport and interface advance. The final rapid cooling preserves the observed hold state only because this example explicitly supplies that condition. A real quench, storage delay or thick specimen can continue to change internally and must be checked.

A record for an isothermal hold cannot simply be read as a continuously cooling trajectory. During continuous cooling both the driving tendency and the available atomic movements change, while the structure produced at one temperature becomes the starting state at the next. Adding minutes spent at different temperatures does not turn them into equivalent minutes at 500 K. Appropriate path-dependent evidence or a justified kinetic model is required.

Even matching β amounts would not establish identical microstructure. Many small particles and fewer large particles can have the same total phase volume. Their arrangement and interfaces can matter to useful properties. The previous temperature–time lesson connects changing particle populations to strengthening; this lesson’s phase-fraction record alone does not predict hardness or strength.

Compare temperatures using the same endpoint

Temperature changes more than a clock speed. In many diffusion-involving transformations, warmer conditions permit faster atomic redistribution. Near a phase-coexistence boundary, however, the free-energy benefit of forming a new phase can be small. Cooling farther below that boundary can increase the benefit, while reducing the mobility needed to establish and grow new regions. Both nucleation and growth require atomic movements; “colder always nucleates faster” is not a general rule.

Use a second original invented record for matched specimens from the same initial α state. Each is transferred to and held at one listed temperature. The endpoint is the first time β reaches the same physical volume fraction, 16%. A separate thermodynamic assessment allows at least that amount at each listed temperature. Transfer effects are negligible here, and the recorded fraction increases monotonically up to the endpoint. At 500 K the 4 min result is the same point already used above.

Among these three temperatures, 500 K reaches the endpoint fastest: 4 min, compared with 12 min at 600 K and 24 min at 400 K. This nonmonotonic order is consistent with competing driving-force and mobility effects. The times alone do not uniquely identify which microscopic step controls each specimen, locate the optimum at an untested temperature, or give a diffusion coefficient.

For this common start and endpoint, the average rates from the start are 0.16/4 = 0.04 min⁻¹ at 500 K and 0.16/12 ≈ 0.0133 min⁻¹ at 600 K. The first is three times the second for this particular interval. That is a defensible fixed-endpoint comparison; it does not show that every atomic jump or every later transformation interval is three times faster.

Keep equilibrium, progress and mechanism evidence distinct

Equilibrium information supplies the permitted phase combination and amounts under its conditions; it does not establish a processing schedule. A kinetic record supplies progress under its actual starting state and path. A microstructure measurement tests which regions and interfaces developed. Use these together rather than treating any one as a substitute for the others.

For real measurements, retain temperature uniformity, specimen size, initial solute distribution, defects, phase-identification method, sampled volume, resolution and uncertainty. A flat phase-fraction trace might reflect a measurement limit or a slow process, not equilibrium. Conversely, continuing atomic movements need not change the net phase amount. The metastability lesson treats persistence and finite observation intervals more fully.

The calculations here are readings and averages of supplied data. They are not a rate law for another alloy, a prediction of an unmeasured cooling path, or a heat-treatment recipe. Later processing lessons develop quantitative diffusion and transformation models. At this stage, the useful skill is to connect the proposed pathway to a specific structural endpoint and the time evidence that actually supports it.

Worked examples

Try the reasoning

1. Does a phase diagram prescribe an exact cooling rate?

   Show the answer and reasoning

   No.

   Rates require kinetic information. The equilibrium 20% β input does not tell whether a 2 min or 8 min hold produces it. The separate time record is needed, and isothermal data do not supply a continuous-cooling trajectory.

2. What does diffusion add to a transformation explanation?

   Show the answer and reasoning

   A mechanism and timescale for redistribution of atoms.

   Equilibrium composition alone cannot tell how fast atoms move. For the chosen solute-rich β particles, transport through α and incorporation across the interface feed growth while depleting the matrix. This route is not a claim that every transformation needs long-range diffusion.

3. A different alloy has an independently supplied equilibrium β volume fraction of 12% at its hold temperature. A measured specimen contains 9% β. Calculate q for the same zero-β starting convention, and identify how much specimen volume is actually β.

   Show the answer and reasoning

   q = 9/12 = 0.75, or 75% of the equilibrium β amount. The actual β volume fraction is still 9%.

   Normalisation changes the reference amount, not the specimen’s phase volumes. The other 91% is not β. Neither 75% nor 9% alone supplies a transformation rate.

4. In a separate isothermal record, β volume fraction rises from 8% to 14% during a 3 min interval. Find its average rate. Does that result establish the next interval’s rate or a particle speed?

   Show the answer and reasoning

   The average is 2 percentage points/min = 0.02 min⁻¹. It establishes neither the next interval’s rate nor a particle speed.

   The fraction change is 0.14 − 0.08 = 0.06; divide by 3 min. A measured phase amount can change through growth, formation of additional regions, or both. A rate in min⁻¹ is not a length per time.

5. The monotonic 500 K record has 8% β at 2 min and 16% at 4 min. Must the 3 min value be exactly 12% because the graph joins the points with a straight segment?

   Show the answer and reasoning

   No. Monotonicity alone bounds it from 8% to 16%. The separate first-arrival statement at 4 min further places the 3 min value below 16%; neither fact establishes 12%.

   The straight connection is a reading aid, not an adopted linear rate model. For example, 10% or 14% at 3 min could both fit the endpoint and first-arrival statements. A more resolved time record or a justified kinetic model is needed for a specific intermediate prediction.

6. Two batches have the same supplied bulk equilibrium phase amounts at 500 K, but different densities of effective nucleation sites. Their β amount after a short common hold differs. Does that discrepancy alone contradict the phase diagram?

   Show the answer and reasoning

   No. The same equilibrium target need not be approached at the same rate from different starting structures.

   Nucleation routes and the subsequent growth population can differ. Keep the initial defects and solute distribution with the kinetic record. The statement supplies the common bulk equilibrium assessment; it does not infer it merely from equal composition.

7. Using the three-temperature record, compare the mean rate from zero β to 16% at 400 K with that at 500 K. Does the ratio establish a constant rate multiplier for every future time?

   Show the answer and reasoning

   The averages are 0.16/24 ≈ 0.00667 min⁻¹ and 0.16/4 = 0.04 min⁻¹. The 500 K average is six times the 400 K average for this endpoint, with no general future multiplier established.

   These use the same initial amount and endpoint, so the interval comparison is meaningful. Changing temperature can also change later equilibrium amounts and the controlling processes; one endpoint is not a complete rate law.

8. At 16 min the invented record has 20% β, matching the supplied equilibrium β fraction. Must every part of the structure now be at equilibrium, with all atomic motion stopped?

   Show the answer and reasoning

   No. Agreement in one phase-amount measure is not a full equilibrium demonstration, and equilibrium does not require atoms to stop moving.

   Check phase compositions and other relevant structural degrees of freedom under the same constraints. Local motions can continue without a net macroscopic transformation. The phase-fraction endpoint does not determine a final particle size distribution or every useful property.

What to carry forward

• Equilibrium identifies tendencies; kinetics describes pathways and rates.
• Cooling history can preserve nonequilibrium structure.
• Trace nucleation, solute transport and interface advance as different parts of a chosen pathway.
• Keep the phase-fraction basis, initial state, temperature path and observation times with a measured rate.
• A normalised phase amount is not the actual fraction of the whole specimen, and one interval average is not a universal kinetic law.
• Compare temperatures at a common physical endpoint; isothermal observations do not by themselves prescribe continuous cooling.