Energy and the meaning of stability

Which state is favoured?

Ice and liquid water are familiar states whose stability changes with conditions. Materials also have alternative crystal structures and phase combinations. To compare their tendencies, specify temperature, pressure, composition and what the system can exchange with its surroundings.

Consider a closed specimen with two possible phases, A and B, of exactly the same chemical composition. A phase is a region with a particular material state or structure; here the candidates are alternative bulk structures of the same substance. Which would be preferred after equilibration at a chosen temperature and pressure? The answer can change with those conditions. It cannot be found from a picture of the atomic arrangement alone.

Closed means that matter does not cross the chosen system boundary. It can still exchange energy with surroundings that maintain its temperature and pressure. A sealed specimen in thermal contact with a large temperature-controlled environment is therefore not the same constraint as a perfectly isolated specimen. Our comparison also excludes imposed electrical fields, additional mechanical constraints and significant surface or interface contributions. Those can require additional terms or permitted-state restrictions.

Why a lower internal energy is not the whole comparison

Internal energy U includes the microscopic energies within the specimen. Enthalpy is H = U + pV, where p is pressure and V is volume; the pV term accounts for the pressure–volume contribution appropriate to this description. Enthalpy is not a synonym for temperature. Two phases at the same temperature can have different enthalpies.

Entropy S describes how the macroscopic state is associated with microscopic possibilities, including atomic arrangements and ways of distributing energy. A crystal can have entropy through atomic vibrations even when its average atomic positions look regular. Visible untidiness alone is therefore a poor definition. A higher entropy is not automatically a higher temperature, and entropy has units of energy divided by temperature.

For a closed bulk system at fixed temperature T and pressure p under the stated simple conditions, compare Gibbs free energy G = H − TS. T must be absolute temperature in kelvin. Multiplying entropy by T gives energy units, so H and TS can be combined. At these constraints the equilibrium comparison favours lower G. The entropy term is why choosing the lowest U, or even the lowest H, can give the wrong preference.

The expression is a criterion for comparing permitted states, rather than an extra store of heat. Its temperature factor means that a given positive entropy difference contributes more strongly to lowering G at a higher comparison temperature. The next model makes that competition explicit without needing calculus or a microscopic count of states.

Compare like amounts at the same conditions

Use molar quantities g = G/n, h = H/n and s = S/n, where n is amount of substance in moles. The phases must use the same molar basis, composition, temperature and pressure in each comparison. Define Δg = g_B − g_A, Δh = h_B − h_A and Δs = s_B − s_A. The symbol Δ here means B minus A at one temperature; it does not mean the change caused by heating A from one temperature to another.

Because the two candidates are compared at the same T, subtracting their expressions gives Δg = Δh − TΔs. If Δg is negative, B has the lower molar Gibbs energy. If it is positive, A is lower. Zero means equality between these candidates under the model; it does not identify a transformation rate.

All following numerical phase data are original hypothetical teaching inputs, not measurements of ice, water or an alloy. At one chosen fixed pressure, stipulate Δh = 4 kJ/mol and Δs = 10 J/(mol K) throughout 200–600 K. Treating these differences as constant is part of this limited model. Real differences can change with temperature and pressure and must be evaluated for the conditions used.

Convert entropy to 0.010 kJ/(mol K) before combining units. The comparison becomes Δg = 4 − 0.010T in kJ/mol when T is in K. At 200 K the enthalpy difference costs B 4 kJ/mol, while TΔs contributes 2 kJ/mol in B’s favour: Δg = +2 kJ/mol, so A is preferred. At 600 K the entropy contribution is 6 kJ/mol, exceeding the same 4 kJ/mol cost: Δg = −2 kJ/mol, so B is preferred.

Read the crossover as a change of preference

Set 4 − 0.010T = 0. The model’s equality temperature is T = 4/0.010 = 400 K. Below 400 K in the stated range, A has lower g; above 400 K, B does. Thus a higher-enthalpy phase can be preferred at the higher temperature when its positive entropy difference is large enough. This is a result of these supplied signs and values, not a rule that any higher-entropy phase wins under every condition.

The graph plots the difference g_B − g_A against temperature. Its zero line is an equality reference. A point above zero means B is higher than A, not that either absolute g is positive. A point below zero reverses that ordering. Each plotted temperature represents a separate comparison of both phases at the same temperature and chosen pressure. The graph is not a time trace of a specimen heating or transforming.

At 400 K the bulk model gives equal molar g for A and B. It does not select a unique phase fraction, create a measurable waiting time, or include the energy of an interface between them. For phases of different compositions, a single same-composition subtraction also cannot replace the required comparison of allowed phase combinations. The later phase-equilibrium lessons develop that broader problem.

Use the appropriate energy

At fixed temperature and pressure, Gibbs free energy is a useful quantity for comparing allowed states of a closed system. The stable equilibrium minimises it subject to the constraints. Lowest internal energy alone is not a universal rule: entropy and work or exchange conditions matter. A thermodynamic tendency also says nothing by itself about speed.

Read the retained qualitative sketch separately from the temperature graph. Its horizontal axis is a transformation coordinate: a schematic description of progress through configurations at one set of conditions. It is not temperature or elapsed time. Well B is lower than well A, while the path drawn between them rises over a barrier. The picture supplies neither the numerical molar differences in our table nor a calibrated barrier height.

A local minimum resists small changes along the pictured coordinate, yet need not be the lowest permitted state. A specimen may persist there because finding or completing a pathway takes time. The worked case below retains that distinction; Metastability and persistence examines local minima and barriers further, while Equilibrium tendencies and transformation rates develops pathways and observations of progress. Neither a negative Δg nor the shape of this uncalibrated sketch gives a lifetime.

State what has and has not been established

At 600 K, assign g_A an arbitrary relative reference of 0 kJ/mol. The calculated difference then places B at −2 kJ/mol. If a previously unconsidered permitted phase C is at −3 kJ/mol on that same basis and under the same conditions, C is lower than both. Calling B the globally stable state from the A/B calculation alone would have missed a candidate. Shifting every candidate by the same energy reference changes none of these differences.

Equilibrium minimises the appropriate potential over the allowed states, which can include phase combinations. Our simple ranking establishes only the preference among the candidates supplied with the stated constraints. It does not assess unlisted reactions, changing composition, a constrained thin film, surface-dominated particles or a specimen held at fixed volume instead of fixed pressure. A specimen open to chemical exchange likewise requires its exchanges to be included.

A useful stability statement therefore names the system, conditions, permitted candidates and comparison quantity. To predict what will actually be observed after a particular hold, add an appropriate kinetic description and evidence. To change pressure or environment, obtain a comparison appropriate to that change instead of extending the illustrative straight line beyond its assumptions.

Worked examples

Try the reasoning

1. Can a stability statement omit pressure for every material?

   Show the answer and reasoning

   No.

   Pressure can change which structure is favoured. The changed-pressure worked example demonstrates the consequence using explicitly supplied new differences; the old-pressure line alone cannot quantify it.

2. Why state whether matter can enter or leave?

   Show the answer and reasoning

   Exchange changes the permitted states and thermodynamic description.

   A sealed alloy and a reacting open surface have different constraints. Closed matter still permits energy exchange. Include exchanges and use the appropriate constraints before choosing a thermodynamic comparison.

3. A learner inserts the numerical Celsius value 26.85 into 4 − 0.010T for a specimen at 300 K and obtains about +3.73 kJ/mol. Diagnose and correct the calculation.

   Show the answer and reasoning

   The formula requires kelvin, giving Δg = +1 kJ/mol at 300 K.

   The term TΔs uses absolute temperature: 300 × 0.010 = 3 kJ/mol. Subtract it from 4. Celsius has a shifted zero, so its numerical value cannot replace the kelvin temperature in this expression even though a one-degree temperature interval has the same size.

4. At 500 K in the original hypothetical model, what is g_A − g_B? Does reversing the subtraction change the preferred phase?

   Show the answer and reasoning

   g_A − g_B = +1 kJ/mol; B remains lower.

   The supplied definition gave g_B − g_A = −1 kJ/mol. Reversing the order reverses the sign and the wording of what it compares. It does not change either phase’s ordering.

5. For a separate hypothetical same-composition pair at a fixed pressure, take Δh = 3 kJ/mol and Δs = 5 J/(mol K), both constant over 300–700 K. Find equality and compare the pair at 400 K and 700 K.

   Show the answer and reasoning

   Equality is at 600 K. A is lower at 400 K; B is lower at 700 K.

   Convert Δs to 0.005 kJ/(mol K). Then Δg = 3 − 0.005T, so equality is 3/0.005 = 600 K. At 400 K the difference is +1 kJ/mol; at 700 K it is −0.5 kJ/mol. All evaluations lie within this separate stipulated interval.

6. At 500 K in the original model, what is G_B − G_A for 2 mol of the substance? If 7 kJ/mol is added to both molar energy references, does the preference change?

   Show the answer and reasoning

   The total difference is −2 kJ for 2 mol, and the common reference shift leaves the preference unchanged.

   Multiply the molar difference −1 kJ/mol by the same 2 mol in each candidate. A common reference cancels: (g_B + 7) − (g_A + 7) = g_B − g_A. Comparing unequal amounts without a consistent basis would answer a different question.

7. At 600 K use relative molar energies A = 0, B = −2 and permitted C = −3 kJ/mol. A report calls B globally stable because it is below A. What conclusion is actually justified by the supplied ranking?

   Show the answer and reasoning

   C is lowest among A, B and C; the A/B comparison alone did not establish global stability.

   C is 1 kJ/mol below B on the same basis. A global claim requires the allowed candidate states and combinations to have been considered. This ranking still supplies no transformation time.

8. At the original model’s 400 K equality, someone predicts exactly half A and half B after ten minutes. Which missing information makes that conclusion unjustified?

   Show the answer and reasoning

   Equality alone selects neither a unique phase fraction nor a ten-minute outcome.

   The bulk molar values are equal. The model omits interface contributions and supplies no nucleation, growth or transport kinetics. A time and a fraction cannot be read from the zero of Δg.

What to carry forward

• State the system and constraints.
• A favoured state is not necessarily reached quickly.
• At the stated fixed T and p, compare G = H − TS, using the same composition and amount basis.
• Define the subtraction and convert units before interpreting its sign. The same positive enthalpy and entropy differences can produce a temperature-dependent preference.
• A ranking among supplied candidates and a global equilibrium claim have different evidential reach. Equality, barriers and transformation time require distinct reasoning.