Test when a pre-equilibrium substitution is defensible
Derive an intermediate substitution, compare exact held-input relaxation and distinguish transient error from persistent consumption.
Learning outcome: Use a fast reversible-step approximation with explicit assumptions and diagnose when intermediate drain or short observation time invalidates it.
Starting recap
A reversible step has forward and reverse fluxes. A nearly steady intermediate need not be at equilibrium if another step consumes it.
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Try
R forms I reversibly while I reacts with S. Does waiting make forward and reverse rates equal when I continues draining into product?
Open the experimentPredict
Predict whether waiting longer repairs an approximation when drain equals reversal. Compare this with observing a reversal-dominant packet too early.
Experiment
Compare zero, one and ten relaxation times. Change the drain/reversal scenario and inspect supply, return and product fluxes separately. Capture an approximation within the stated tolerance and one that cannot pass even at late time.
Test the pre-equilibrium approximation against intermediate balance
Worked teaching model · synthetic values · Original open-system teaching model R ⇌ I, I + S → P. External reservoirs hold free R = 0.100 mol/L and S fixed, with product removed and I initially zero. kf = 0.010 s⁻¹ and k2 = 1 L mol⁻¹ s⁻¹ are stipulated. The exact linear intermediate balance applies only under these held-input assumptions; it is not a closed-batch simulation or measured chemical equilibrium.
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R ⇌ I, followed by I + S → P
R and I are hypothetical forms with the same composition; P contains the combined R and S material. External reservoirs hold free [R] = 0.100 mol/L and [S] = 0.1 mol/L. Product is removed. This is an open, held-input model; a closed batch would deplete reactants and require different equations.
Forward constant kf = 0.010 s⁻¹, reverse kr = 100 s⁻¹, and second-step k₂ = 1 L mol⁻¹ s⁻¹. The intermediate starts at zero. Its supply is kf[R] = 0.001 mol L⁻¹ s⁻¹; its two removal coefficients are kr and k₂[S] = 0.1 s⁻¹.
Derive, then challenge, the substitution
Ignoring consumption by step 2 would set kf[R] ≈ kr[I], giving [I]pre = (kf/kr)[R] = 0.00001 mol/L. Substituting into r = k₂[I][S] gives rpre = k₂(kf/kr)[R][S] = 0.000001 mol L⁻¹ s⁻¹. This is an approximation, not an equilibrium guaranteed merely by a reversible arrow.
| Term | Value /mol L⁻¹ s⁻¹ |
|---|---|
| Supply from R | 0.001 |
| Return to R | 0.00063148907 |
| Drain to P | 6.3148907e-7 |
| Accumulation: supply − return − drain | 0.00036787944 |
Relaxation time τ = 1/(kr+k₂[S]) = 0.00999001 s. Selected elapsed time is 1τ = 0.00999001 s. The exact held-input solution gives [I] = 0.0000063148907 mol/L and product rate 6.3148907e-7 mol L⁻¹ s⁻¹. At late time, supply equals return plus drain, not return alone.
Consumption/reversal ratio ε = k₂[S]/kr = 0.001. Even after full relaxation, the approximation overpredicts the exact rate by 0.1% relative to that exact rate. At the selected time, relative error is 58.355868%. Not within the chosen 5% teaching tolerance. No universal validity cutoff is asserted.
Compare early and late times in the reversal-dominant packet, then wait equally many relaxation times with drain equal to reversal. A small ε needs time to relax; waiting longer cannot remove a large ε error. The time axis is scaled by τ, so equal horizontal positions need not mean equal seconds across packets.
Optional derivation and sources
The held-input balance is d[I]/dt = kf[R] − (kr+k₂[S])[I]. With I initially zero its solution is [I]∞(1−exp(−t/τ)), where [I]∞ = kf[R]/(kr+k₂[S]). Dividing by [I]pre gives (1−exp(−t/τ))/(1+ε). The late-time relative overprediction is exactly ε. The system still produces P, so its steady intermediate inventory is not global chemical equilibrium.
OpenStax Chemistry 2e §12.6 explains eliminating an intermediate using a fast reversible step. Constants, reservoirs and tolerance here are original teaching assumptions, not measured kinetics or proof of a real mechanism.
Approximation exceeds the explicitly chosen 5% rate tolerance at this time.
A small consumption/reversal ratio is not enough before relaxation; waiting cannot repair a large persistent drain error. This tolerance is a teaching choice, not a universal chemical criterion.
Captures retain original held-input scenarios and relaxation-time settings. Replay checks approximation error and flux balance; it does not measure chemical equilibrium or validate a real mechanism.
Understand
Apply
Public practice checks a specific claim. It does not establish independent mastery.
Derive the intermediate-free approximate rate law, identify its assumptions, and distinguish a steady intermediate from global equilibrium. Explain why equal relaxation-time multiples do not imply equal seconds.
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