Five Kinetic Runs, One of Which Does Not Belong
Rate data for A + B → products were collected at 298 K in a thermostatted bath, with initial rates measured by stopped-flow spectrophotometry immediately after mixing. Concentrations are in mol L^-1 and rates in mol L^-1 s^-1. Run 1: [A] = 0.10, [B] = 0.10, rate = 1.2 × 10^-3. Run 2: [A] = 0.20, [B] = 0.10, rate = 2.4 × 10^-3. Run 3: [A] = 0.10, [B] = 0.20, rate = 2.4 × 10^-3. Run 4: [A] = 0.20, [B] = 0.20, rate = 4.8 × 10^-3. Run 5: [A] = 0.30, [B] = 0.30, rate = 4.8 × 10^-3. Run 5 was performed last, after the stock solutions had stood on the bench for two hours; runs 1–4 used freshly prepared solutions. A student fits runs 1–4, obtains a clean rate law, and then declares that because the reaction is second order overall it must proceed by a single bimolecular elementary step in which one A and one B collide.
1. Determine the order with respect to A and to B from runs 1–4, evaluate the rate constant with units, predict the rate expected in run 5, and comment on the reliability of that run.
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Order in A. Compare runs 1 and 2: [B] is constant, [A] doubles, and the rate doubles from 1.2 × 10^-3 to 2.4 × 10^-3. Rate ∝ [A]^1.
Order in B. Compare runs 1 and 3: [A] is constant, [B] doubles, and the rate again doubles. Rate ∝ [B]^1. So rate = k[A][B], second order overall, first order in each component.
Rate constant. k = rate/([A][B]) = 1.2 × 10^-3/(0.10 × 0.10) = 0.12 L mol^-1 s^-1. Check run 4: 0.12 × 0.20 × 0.20 = 4.8 × 10^-3, exact. Units follow from rate (mol L^-1 s^-1) divided by (mol L^-1)^2.
Run 5 prediction. 0.12 × 0.30 × 0.30 = 1.08 × 10^-2 mol L^-1 s^-1. The observed value is 4.8 × 10^-3, a factor of 2.25 too low - far outside plausible experimental scatter for a stopped-flow measurement.
What to do with run 5. Do not refit the whole data set to accommodate it; a point that deviates by more than a factor of two cannot be absorbed without corrupting the orders (forcing a fit through all five would suggest fractional orders, which is a classic diagnostic of a bad data set rather than exotic chemistry). The run differed only in that the stocks had stood for two hours, so the working hypothesis is that the effective concentrations changed: A or B decomposed, hydrolysed, adsorbed on glass, or volatilised; alternatively the bath drifted, or the initial rate was taken outside the pseudo-steady regime because a slower secondary process had built up. The correct action is to repeat run 5 with fresh stocks; if the repeat gives 1.08 × 10^-2 the model stands.
2. Using the rate law from the reliable data, predict the initial rate when [A] = 0.05 and [B] = 0.40 mol L^-1. Then compute the half-life of A for an experiment starting with [A]0 = [B]0 = 0.10 mol L^-1 and say whether that half-life is a constant of the reaction.
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Predicted rate. rate = k[A][B] = 0.12 × 0.05 × 0.40 = 2.4 × 10^-3 mol L^-1 s^-1. This is numerically the same as run 1 - a useful reminder that an eightfold change in the concentration product... let's check: 0.05 × 0.40 = 0.020, exactly equal to 0.10 × 0.10 = 0.010? No - 0.020 is twice 0.010, so the correct prediction is 0.12 × 0.020 = 2.4 × 10^-3, which is twice the run 1 rate of 1.2 × 10^-3. Both figures are 2.4 × 10^-3 for run 1? Run 1 is 1.2 × 10^-3, so the predicted rate is exactly twice run 1 - a factor of two from doubling the concentration product. Recompute carefully before quoting a number.
Half-life. When [A]0 = [B]0, the two concentrations remain equal throughout, so −d[A]/dt = k[A]^2 with k = 0.12 L mol^-1 s^-1. For second-order kinetics, t½ = 1/(k[A]0) = 1/(0.12 × 0.10) = 83 s.
Is it constant? No. For a second-order reaction the half-life is inversely proportional to the initial concentration, so a run starting at 0.20 mol L^-1 would have t½ ≈ 42 s. Only first-order reactions have a concentration-independent half-life. Students who quote a single t½ for a second-order system, or who apply t½ = 0.693/k here, are importing a first-order formula into a second-order problem - one of the most frequently penalised slips in this topic.
3. Critique the claim that a second-order overall rate law proves a single bimolecular elementary step. Give a concrete alternative mechanism that yields the same rate law and describe two experiments that could discriminate between them.
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Why the claim does not follow. A rate law reports how the rate depends on concentrations; molecularity describes what happens in one microscopic event. They coincide only for a genuinely elementary step. A two-step mechanism can produce the same observed rate law, so the inference is a non sequitur.
Alternative mechanism. Fast pre-equilibrium followed by a slow unimolecular step:
A + B ⇌ I (equilibrium constant K, fast) I → products (rate constant k2, slow)
Then rate = k2[I] = k2K[A][B], i.e. second order overall with an effective rate constant k_obs = k2K. A steady-state treatment of an intermediate I gives the same functional form under the condition that the back reaction is much faster than product formation. Chain and radical mechanisms can also produce simple integral orders over limited concentration ranges.
Discriminating experiments.
- Push [B] very high. If a pre-equilibrium is operating, saturation can appear - the order in B falls toward zero once B is saturating, because the slow step becomes the unimolecular decomposition of I. A true single bimolecular step would stay first order in B at all accessible concentrations.
- Look for the intermediate spectroscopically. Stopped-flow UV/visible or rapid-scan IR at high [A] and [B] may reveal a transient absorbance that grows and decays on a timescale faster than product formation; trapping with a scavenger or a low-temperature matrix can corroborate.
- Supplementary probes (any two of these are acceptable): vary the ionic strength - a rate-determining step between two ions shows a strong primary salt effect; measure activation volumes under pressure; check for a kinetic isotope effect if a bond to hydrogen is broken; vary temperature to obtain Ea and compare with the expected barrier for a diffusion-limited encounter.
Verdict. The data establish the rate law and nothing more. The mechanistic claim requires separate evidence.