Close a finite reaction without counting heat and work twice

A favorable reaction is not a promise of full conversion, rapid approach or delivered electrical energy.

Reconstruct the evidence

At total .060 M and K3, equilibrium B is .045 M. With kf+kr=.100 s⁻¹, after10 s B=.045(1−e⁻¹)=.0284454 M. In .050 L the extent is .00142227 mol and the supplied constant-pressure reaction heat is−28.4454 J.

For conserved C=A+B, equilibrium gives B_eq=CK/(1+K). The reversible first-order law dB/dt=kf(C−B)−kr B has solution B=B_eq[1−exp(−(kf+kr)t)] for initially pure A. Rate ratio fixes equilibrium, while the sum fixes relaxation time. The90%-of-equilibrium time is ln10/.100=23.02585 s for every supplied K, but the fraction of initial A converted at that time is .90K/(1+K).

Separate state functions and kinetic paths

For this ideal isothermal/isobaric mixture, Gibbs excess above equilibrium is RT V[A ln(A/A_eq)+B ln(B/B_eq)]. Concentrations and volume use matching mol/L and L units; ratios are dimensionless. At B=0 the limit B ln B is zero, not an error or infinite total energy. Gibbs excess falls along the supplied relaxation. Its decrease bounds reversible non-expansion work between those states; the freely relaxing kinetic path does not itself deliver that bound.

The stipulated constant reaction enthalpy gives constant-pressure reaction heat −20000 V B joules when only pressure-volume work is performed. The Gibbs bound is a different thermodynamic account, not additional energy to add to the heat. An alternative work-producing path would require its own heat/work balance. A hypothetical quantitative two-electron recovery of B gives Q_max=2FVB, not actual usable voltage, current efficiency or a device specification.

Changed case: predict, then check

At C=.040 M and K1, equilibrium B=.020 M; at10 s B=.0126424 M. In .050 L the extent is .000632121 mol, so the heat under the stated no-nonexpansion-work path is−12.6424 J. At90% of the equilibrium amount only45% of initial A has converted. Changing the common rate scale changes the approach time without changing K; adding a catalyst cannot establish a new fixed-temperature equilibrium constant.

Retained investigation

Compare two K values at the same total rate coefficient and two initial totals at fixed K. Retain concentration, extent, heat, Gibbs-excess decrease and hypothetical charge ceilings with their distinct assumptions. Why is the reversible work bound neither actual output from the kinetic path nor an extra energy source to add to reaction heat?

Source and safety boundary

Primary conceptual references checked30 October2026: OpenStax equilibrium calculations, free energy, and potential/free-energy/equilibrium. Original fictional data and analysis; no physical manipulation, actual assay certification or independent learner mastery is asserted. Retain first report, critique, revision and a separately frozen new assignment.