MCAT content library / Chemistry & physics / Rate laws: infer reaction order and test a proposed mechanism
Use controlled concentration comparisons to build a rate law, then ask which elementary-step models can reproduce it.
About 14 minutes, including the worked examples and checks.
Useful foundations
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Concept: define the measured rate before finding its law
A reaction equation tells you which amounts can be converted. A rate law describes how quickly conversion occurs under specified conditions. Those are different questions. An overall coefficient is not automatically an exponent in the rate law. This lesson uses concentration comparisons to test a kinetic model, within the chemical-kinetics scope of AAMC Category 5E.1
Consider the original illustrative reaction 2A + B → P in a well-mixed solution of constant volume. Let r mean the rate of P formation in mol L−1 s−1, also written M/s. The corresponding relationships are r = d[P]/dt = −½ d[A]/dt = −d[B]/dt. A disappears twice as quickly as P appears. A negative concentration slope is therefore compatible with a positive reported reaction rate.2
An instrument may measure absorbance or another signal rather than concentration directly. A calibration connects that signal to concentration; its slope over time can then estimate a rate. A secant across a finite interval gives an average rate. A tangent gives the instantaneous rate. An initial rate refers to the beginning of a run, before substantial reactant depletion or product accumulation changes the conditions.
Pattern: isolate one concentration change
The four idealized runs below use the same temperature, solvent, final volume, mixing method and measurement procedure. Each begins with negligible P. Concentrations and rates are supplied as exact teaching values; they are not measurements from a published experiment.
| Run | Initial [A] (M) | Initial [B] (M) | Initial r (M/s) |
|---|---|---|---|
| 1 | 0.100 | 0.200 | 0.00200 |
| 2 | 0.200 | 0.200 | 0.00800 |
| 3 | 0.100 | 0.400 | 0.00200 |
| 4 | 0.200 | 0.400 | 0.00800 |
Start with r = k[A]m[B]n. Runs 1 and 2 hold B constant. Doubling A multiplies the rate by four, so 2m = 4 and m = 2. Runs 1 and 3 hold A constant. Doubling B leaves the rate unchanged, so 2n = 1 and n = 0. The resulting law is r = k[A]2: second order in A, zero order in B, second order overall.3
Run 4 checks the combined prediction. If only runs 1 and 4 were available, their fourfold rate change after doubling both reactants would establish only m + n = 2. It would not distinguish second order in A from first order in each reactant. Controlled comparisons supply that distinction.
Zero order in B does not mean B is absent, unnecessary or never consumed. It means the measured rate does not depend on B concentration over the tested conditions. If B becomes scarce enough to change which step limits conversion, the same empirical law may cease to apply.
Rule: make the dimensions and the predictions agree
Using run 1, k = 0.00200/(0.100)2 = 0.200 M−1 s−1. Multiplying that unit by M2 restores M/s. All four runs give the same k. At [A] = 0.150 M and [B] = 0.300 M, the law predicts r = 0.200 × (0.150)2 = 0.00450 M/s.
For overall order q, a concentration-based rate constant has units M1−q/s when time is in seconds. Changing the concentration unit changes the numerical value assigned to k, so compare constants only with their units. At fixed conditions, changing reactant concentrations changes the rate through the concentration factors; it does not require a new k. Temperature or a changed catalytic pathway can alter k. Fractional and negative empirical orders are also possible; the integer examples here are not a universal restriction.3
Application: test the steps, not only the net equation
Two hypothetical mechanisms both sum to 2A + B → P. Each is proposed with its first step much slower than the second, with negligible reverse reaction under these conditions.
| Model | Elementary steps | Predicted dependence |
|---|---|---|
| M1 | 2A → I (slow); I + B → P (fast) | r ∝ [A]2 |
| M2 | A + B → J (slow); J + A → P (fast) | r ∝ [A][B] |
I and J are intermediates: each is produced and then consumed, so it cancels from the net equation. For an elementary step, reactant molecularity supplies its concentration dependence. Under the stated slow-first-step approximation, M1 fits the observed powers, whereas M2 predicts a B dependence that the table lacks. M2 also predicts only a twofold response when A doubles. Its correct net equation does not rescue those conflicting predictions.4
Agreement leaves M1 as a candidate, not a uniquely demonstrated molecular sequence. Other mechanisms can produce the same rate law. If a slow step follows a rapid reversible step, its rate may contain an intermediate concentration that must first be related to measured reactants. Simply replacing that intermediate with whichever reactant looks familiar is not a valid derivation.
Application: check a concentration-time trace
A calculated continuation of run 1 provides a second consistency check. Volume and temperature remain fixed, B remains sufficient for the assumed regime, and the reaction law continues to apply. The trace shows A concentration, while the table also reports r at each instant.5
| Time (s) | [A] (M) | Instantaneous r (M/s) |
|---|---|---|
| 0 | 0.10000 | 0.002000 |
| 10 | 0.07143 | 0.001020 |
| 25 | 0.05000 | 0.000500 |
| 50 | 0.03333 | 0.000222 |
At 25 s, A has halved, so the instantaneous rate is one quarter of its initial value. Over the entire first 25 s, however, the average reaction rate is ½ × (0.100 − 0.050)/25 = 0.00100 M/s. An interval average need not equal either endpoint’s instantaneous rate. The factor ½ is essential because the plotted species is A rather than P.
A curve that becomes less steep is consistent with slowing consumption, but its appearance alone does not establish a particular order. The concentration comparisons and a quantitative time model do that work. In an actual experiment, repeated runs, suitable time resolution and uncertainty estimates would determine whether apparently equal rates support a zero-order approximation. Temperature drift, altered catalyst concentration or a changed calibration would offer competing explanations for a rate difference.
ACTIVE PRACTICE · CHEMICAL KINETICS
Infer the Rate Law, Then Challenge It
The lesson’s reaction is 2A + B → P. Runs 1–4 report rates 0.00200, 0.00800, 0.00200 and 0.00800 M/s at [A] = 0.100 or 0.200 M and [B] = 0.200 or 0.400 M. Temperature and measurement conditions are held constant.
Predicted initial product-formation rate: M/s, or × Run 1.
Model: r = k[A]² with k = 0.200 M⁻¹s⁻¹. The zero-order dependence on B applies only in the stated measured regime; B must still be available for the reaction to proceed.
Check your reasoning
Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.
Check 1
A separate reaction is measured at one temperature. At [X] = 0.050 M and [Y] = 0.100 M, its initial rate is 0.0010 M/s. Tripling X alone raises the rate to 0.0030 M/s; doubling Y alone raises it to 0.0040 M/s. Which law matches all three results?
A. r = k[X][Y]
Review this choice. This gives the observed threefold response to X but predicts only a twofold response to doubling Y.
B. r = k[X][Y]²
Correct. Tripling X multiplies the rate by three, establishing first order in X. Doubling Y multiplies it by four, establishing second order in Y.
C. r = k[X]²[Y]
Review this choice. This predicts a ninefold response to tripling X and only a twofold response to doubling Y; both disagree.
D. r = k[X]²[Y]²
Review this choice. The Y dependence fits, but the squared X term predicts a ninefold response instead of the observed threefold response.
The principle to carry forward
Infer each exponent from a comparison that holds the other concentration fixed.
Check 2
For another reaction, r = k[M]²[N]. At [M] = 0.200 mol/L and [N] = 0.100 mol/L, r = 8.00 × 10⁻⁴ mol L⁻¹ s⁻¹. Which value of k has both the correct number and dimensions?
A. 0.200 mol L⁻¹ s⁻¹
Review this choice. This is the dimension of the rate itself. Multiplying it by a concentration cubed would not give the stated rate unit.
B. 0.200 s⁻¹
Review this choice. An inverse-time unit fits a first-order concentration law, but the supplied law is third order overall.
C. 0.200 L mol⁻¹ s⁻¹
Review this choice. This is the usual unit for a second-order law. Here [M]²[N] has units mol³/L³.
D. 0.200 L² mol⁻² s⁻¹
Correct. The concentration product is (0.200)²(0.100) = 0.00400 M³. Dividing 0.000800 M/s by it gives 0.200 M⁻² s⁻¹, equivalent to L² mol⁻² s⁻¹.
The principle to carry forward
Obtain the unit by dividing the measured rate unit by the complete concentration factor.
Check 3
For X + Y → P, doubling X alone doubles the initial rate and tripling Y alone triples it. Consider two proposed elementary-step models with negligible reverse reactions. Model 1: X → I is slow, then I + Y → P is fast. Model 2: X + Y → J is slow, then J → P is fast. Each first step is proposed to control the overall rate. Which assessment follows?
A. Both models are consistent with the concentration effects.
Review this choice. Both sum to the net equation, but Model 1 predicts no Y dependence under its stated slow-first-step approximation.
B. Model 1 is consistent; Model 2 is inconsistent.
Review this choice. Model 1 predicts r proportional to [X], whereas Model 2 predicts r proportional to [X][Y]. The measured Y response favors Model 2.
C. Model 2 is consistent; Model 1 is inconsistent.
Correct. The observed first-order responses to both reactants match Model 2. Model 1 would leave the rate unchanged when only Y changes. Consistency does not prove Model 2 is unique.
D. Neither model is consistent with the concentration effects.
Review this choice. Model 2 predicts exactly the supplied separate concentration responses. The presence of its fast second step does not invalidate that prediction.
The principle to carry forward
A proposed mechanism must reproduce both the net stoichiometry and the observed kinetic dependence under its stated assumptions.
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Sources
1. AAMC: C/P Content Category 5E
2. OpenStax Chemistry 2e: Chemical Reaction Rates
3. OpenStax Chemistry 2e: Rate Laws