Chemistry · Lesson
Non-equilibrium processes and limits of equilibrium models
Track entropy transport separately from internal generation. Recognize a driven steady state, and state the local assumptions needed before using a transport law.
Back to the Chemistry mapBy the end, you can
- Balance subsystem entropy changes against the composite production.
- Distinguish steady macroscopic observations from thermodynamic equilibrium.
- Check signs and units in a scoped Fourier-conduction model.
Entropy can cross a boundary as well as be produced
A system entropy balance contains entropy brought in or carried out by heat and material, plus internal entropy generation. In rate form, dS/dt = net inward entropy flow + Ṡ_gen. The second law requires nonnegative generation for the appropriate thermodynamic description, while the system’s own entropy change can be negative when it exports entropy.
Consider a stipulated 100 J of heat transferred directly from a large 500 K reservoir to a large 300 K reservoir, without extracting work. The reservoir temperatures remain fixed. The hot reservoir changes by−100/500 = −0.200 J K⁻¹ and the cold one by+100/300 ≈+0.333 J K⁻¹. The isolated composite gains approximately 0.133 J K⁻¹. The hot subsystem’s entropy decrease does not contradict that total balance.
| System and example | Boundary or transfer term | Generation or combined result | Net entropy change |
|---|---|---|---|
| Hot 500 K reservoir;100 J leaves | −100/500 = −0.200 J K⁻¹ | Ideal reservoir remains at fixed T | −0.200 J K⁻¹ |
| Cold 300 K reservoir;100 J enters | +100/300 ≈+0.333 J K⁻¹ | Ideal reservoir remains at fixed T | +0.333 J K⁻¹ |
| Isolated composite of the direct transfer | No external entropy flow | Approximately+0.133 J K⁻¹ produced | Approximately+0.133 J K⁻¹ |
| Separate driven steady system | Net outward rate 0.100 J s⁻¹ K⁻¹ | Internal production rate 0.100 J s⁻¹ K⁻¹ | dS/dt = 0, with continuing production |
Separate original entropy-budget examples. The first three rows describe one 100 J direct heat transfer between fixed-temperature reservoirs; values are entropy changes in J K⁻¹. The last row is a different driven steady-rate example, with rates in J s⁻¹ K⁻¹. Keep the system boundary and change-versus-rate basis visible.
A steady state can keep producing entropy
Suppose a driven system produces 0.100 J s⁻¹ K⁻¹ internally while exporting the same net entropy rate. Its dS/dt is zero, but generation continues. Maintained temperature, concentration or electrical differences can support fluxes even though selected macroscopic quantities remain stationary.
Equilibrium requires the absence of the relevant thermodynamic driving forces under the permitted exchanges, rather than only a flat time series. Constraints or metastability can delay equilibration. An equilibrium Gibbs landscape does not specify reaction rates, transport coefficients, nucleation times or whether an experiment reaches the stable state during observation.
A local temperature field requires its own scale assumptions
A slowly varying medium may be close enough to equilibrium within small regions to assign a local temperature T and local thermodynamic properties. This local-equilibrium approximation requires internal relaxation scales to be sufficiently short compared with the variation being modeled. It does not give the entire nonuniform body one global equilibrium temperature.
For heat-only local conduction, the entropy-production density is σ = j_q·∇(1/T), where j_q is heat flux in W m⁻². In a scalar isotropic Fourier model, j_q = −κ∇T, with nonnegative thermal conductivity κ. Then σ = κ|∇T|²/T² ≥0, in W m⁻³ K⁻¹. Material transport, reactions or other dissipative channels add terms to a broader entropy balance.
A constitutive law adds information beyond the second law
At one invented local state, T = 3.00 × 10² K, κ = 2.00 W m⁻¹ K⁻¹ and dT/dx = −1.00 × 10² K m⁻¹ give j_q,x = +2.00 × 10² W m⁻². The positive heat flux points toward decreasing temperature. Multiplying by ∂(1/T)/∂x = −(dT/dx)/T² gives σ ≈ 0.222 W m⁻³ K⁻¹.
These values are supplied inputs to a local linear model, not universal material properties. Strongly nonlocal, rapidly varying or far-from-equilibrium transport can require other descriptions. Neither a universal minimum nor a universal maximum entropy-production principle follows from the second law alone.
Worked examples
Check both subsystem and composite entropy
Take the specified direct 100 J heat transfer from a fixed 500 K reservoir to a fixed 300 K reservoir, without work extraction. Find all entropy changes.
- Hot reservoir: ΔS_hot = −q/T_hot = −0.200 J K⁻¹.
- Cold reservoir: ΔS_cold = +q/T_cold = +0.333333… J K⁻¹.
- Add them: ΔS_total = q(1/T_cold −1/T_hot).
- The total is positive even though the hot reservoir’s own change is negative.
ΔS_total ≈+0.133 J K⁻¹. This two-temperature transfer is irreversible under the stated boundary and direct-transfer assumptions.
Check a local flux and its entropy-production units
For the stated local Fourier model, use T = 3.00 × 10² K, κ = 2.00 W m⁻¹ K⁻¹ and dT/dx = −1.00 × 10² K m⁻¹. Find j_q,x and σ.
- j_q,x = −κ(dT/dx) = +200 W m⁻².
- ∂(1/T)/∂x = −(dT/dx)/T² = 100/90000 K⁻¹ m⁻¹.
- Multiply the flux and inverse-temperature gradient:σ = 200 × 100/90000.
- The units are W m⁻² ×K⁻¹ m⁻¹ = W m⁻³ K⁻¹.
j_q,x = +2.00 × 10² W m⁻² and σ ≈ 0.222 W m⁻³ K⁻¹ in this local-equilibrium, isotropic linear model.
Try it yourself
Internal generation and net outward entropy flow both equal 0.100 J s⁻¹ K⁻¹. What is dS/dt?
Show the answer and reasoning
Zero.
The outward flow balances production; generation itself remains positive.
Can the hot subsystem lose entropy in the stated spontaneous direct heat transfer?
Show the answer and reasoning
Yes.
The isolated composite entropy increases while entropy leaves the hot subsystem.
With κ ≥0, what sign does κ|∇T|²/T² have at positive T?
Show the answer and reasoning
Nonnegative.
This is the sign of the scoped heat-conduction production density.
Does an equilibrium potential alone give the time needed to cross a nucleation barrier?
Show the answer and reasoning
No.
Kinetic mechanisms and their parameters provide additional information.
What to carry forward
- Keep entropy flow distinct from generation.
- A system entropy decrease can accompany positive composite production.
- Stationary averages do not establish equilibrium.
- Local and linear transport laws require their stated scale and constitutive assumptions.
References
Teaching resources and reference material supporting this lesson’s explanations and examples.