SCIENCE · INVESTIGATE AND EXPLAIN

Design rate comparisons that distinguish competing orders

Compare controlled initial rates, intersect candidate k intervals and diagnose confounding, underdetermination and inconsistent data.

Learning outcome: Infer bounded candidate orders from same-temperature initial rates, use units and a shared k, and explain which additional comparison discriminates remaining models.

Starting recap

A rate has concentration-per-time units. Initial rates refer to the starting slope, not accumulated product. A power law describes how rate changes with concentrations at fixed conditions.

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Try

Doubling both A and B increases a synthetic rate eightfold. Does that identify which reactant has order 2 and which has order 1?

Open the experiment

Predict

Predict which candidate laws survive diagonal-only records. Choose a controlled change that would distinguish them before exposing the additional record.

Your prediction and reason

Experiment

Compare candidate (1,2) and (2,1) on diagonal-only records, then add the A-only change. Inspect complete controls and the separate public changed-concentration check. Compare bounded rates, changed temperature and inconsistent records. Capture a supported law and a case where no unique supported choice exists.

Identify orders using controlled initial-rate evidence

Original synthetic initial-rate records · candidate comparison · Original synthetic initial product-formation rates at stipulated concentrations and temperatures. Candidate power laws have integer exponents 0–2; this bounded family is not every possible kinetics model. Only same-temperature calibration rows constrain one shared k. Supplied rate bounds are absolute allowed ranges, not confidence levels; no actual experiment, mechanism or reaction extent is inferred.

Original synthetic experimental design: Both concentrations varied together packet Candidate concentration exponents: m=1, n=1 candidate

Reset keeps your prediction, captures and practice history.

Separate reaction order from a confounded concentration change

A and B change together. The eightfold rate increase constrains their combined order but does not separate the two exponents.

Candidate: r = k[A]^1[B]^1; overall order 2. Here r is the supplied initial product-formation rate in mol L⁻¹ s⁻¹, concentrations are mol/L and k has units (mol/L)^-1 s⁻¹. Stoichiometric coefficients do not determine these exponents for a general reaction.

Record	[A]	[B]	T /K	Rate ± bound /mol L⁻¹ s⁻¹	Use
Baseline	0.1	0.1	300	0.002 ± 0	Shared 300 K calibration
Both doubled	0.2	0.2	300	0.016 ± 0	Shared 300 K calibration

Require one shared rate constant

Each eligible record permits k between (rate − bound)/([A]^m[B]^n) and (rate + bound)/([A]^m[B]^n). Intersect those intervals; a law cannot use a different k for each same-temperature row.

Eligible record	Permitted k interval /(mol/L)^-1 s⁻¹
Baseline	0.2 to 0.2
Both doubled	0.4 to 0.4

No shared k for this candidate. Displayed candidate bars use the baseline record's k = 0.2 only to expose disagreement. They are not accepted fitted results.

Compatible order pairs in the bounded family: (m=1, n=2); (m=2, n=1). A unique pair here does not rule out other law families or identify a mechanism. Fractional, negative and larger orders are outside this exercise's candidate list, not chemically impossible.

Sources and evidence limits

OpenStax Chemistry 2e §12.3 explains initial-rate comparisons, orders and temperature-dependent k. The records here are original synthetic values, not that textbook's measured examples. Initial slopes are supplied, not estimated from a finite time window. Measurement technique, concentration uncertainty, instrument calibration and mechanisms remain unmodeled.

2 candidate order pairs remain compatible; design another comparison.

The selected candidate has no common k. Its displayed baseline-anchored predictions are a diagnostic, not a fitted accepted law.

Observation to attach to a capture

Captures retain synthetic rate records and selected candidate exponents. Replay recomputes compatible constant intervals, exclusions and public-check comparisons; it is not a measured experiment or independent validation.

Understand

Try: two changing inputs can hide the explanation

The first two synthetic records double both A and B while rate grows from 0.002 to 0.016 mol L⁻¹ s⁻¹. The factor is 8. A candidate r=k[A]²[B] predicts 2²×2=8; r=k[A][B]² predicts 2×2²=8. Both fit. Choose a concentration change that would separate them before opening the controlled packet.

Predict, then change one concentration

The A-only record doubles A at fixed B and temperature; rate grows fourfold. That supports order 2 in A within the candidate family. The B-only record doubles B and doubles rate, supporting order 1. The total order is 3, not the sum of coefficients in a chemical equation. Rate-law exponents need evidence unless an elementary-step model supplies additional justified assumptions.

Understand the shared constant

For r=k[A]²[B], the baseline gives k=0.002/(0.1²×0.1)=2. Its units are (mol/L)⁻²s⁻¹, because multiplying by three concentration factors must leave concentration/time. The same k must explain all matched calibration rows. Allowing a separate arbitrary k for every row would avoid rather than test the proposed law.

With the supplied±5% rate ranges, each matched row permits k from 1.9 to 2.1. Their intersection is the allowed shared range. These are hard input bounds, not a confidence interval or a fitted probability distribution. The separate public check at A=0.15 and B=0.30 predicts 2×0.15²×0.30=0.0135 mol L⁻¹ s⁻¹. It does not influence the calibration interval; its visible reuse is practice, not unseen testing.

Apply the failure cases

The changed-temperature record cannot share a 300 K k without further evidence. Its rate change does not isolate concentration order; the interface retains the record but withholds a common-temperature prediction. In the inconsistent controlled packet, no candidate has one k compatible with every exact supplied record. Investigate data quality, missing conditions or the model family; do not choose the least-wrong curve and label it verified.

Integer orders 0–2 are an interface candidate limit. Actual rate laws can have other orders or different forms. Even a single surviving candidate does not identify a unique elementary mechanism or prove extrapolation outside the tested conditions.

Continue

Complete the unfamiliar zero-order/second-order practice. Use average-rate windows to distinguish a supplied initial slope from a finite-interval rate, and first-order decay to see a different, assigned time-course law. Those models do not substitute for estimating initial slopes from raw concentration/time observations.

Source and boundaries

OpenStax Chemistry 2e §12.3 provides rate-law and controlled initial-rate conventions. The current AP CED 5.2 connects data to rate laws, orders, units and temperature dependence. All records here are original synthetic teaching values; no actual experimental measurement or independent mechanism validation is claimed. Practical measurement, raw slope estimation and independent learner review remain separate work.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Explain why k must be shared only across matched conditions, why a finite candidate set is not all kinetics, and why no candidate should be forced when intervals conflict. Distinguish public prediction checking from independent validation and a rate law from a mechanism.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Separate a molecular solid, a layer and a connected framework Next: Choose a defensible window for a reaction-rate estimate

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