Chemical Kinetics · TOPIC LESSON
Infer concentration dependence from evidence, then test whether a proposed mechanism can explain it.
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BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
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These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
A reaction rate describes how rapidly composition changes. For a disappearing reactant A at fixed volume, an average disappearance rate over an interval is −Δ[A]/Δt. The minus sign makes the reported disappearance rate positive when [A] decreases. A product formation rate is +Δ[P]/Δt. The unit is commonly mol L⁻¹ s⁻¹, abbreviated M s⁻¹.
An instantaneous rate is the slope of the concentration–time curve at a particular moment, written −d[A]/dt for reactant disappearance. An average over a long interval need not equal the rate at its beginning, middle or end. If the reaction slows appreciably, a secant slope cannot be substituted for a tangent slope without an approximation.
For aA+bB→pP, a stoichiometrically normalized rate is r=−(1/a)d[A]/dt=−(1/b)d[B]/dt=(1/p)d[P]/dt under the fixed-volume convention. The normalized value is shared even though different species disappear or appear at different numerical speeds. A formula for k must identify which rate definition it uses.
If volume changes, concentration can change because of dilution or compression as well as reaction. A rigorous amount balance then separates chemical production from volume effects. The simple concentration derivative used here assumes fixed volume or a measurement model that has already corrected for those effects.
r=−(1/a)d[A]/dt=−(1/b)d[B]/dt=(1/p)d[P]/dt
Fixed-volume normalized rate for aA+bB→pP.
ORIGINAL WORKED EXAMPLE
For 2A+3B→C, A disappears at 0.0400 M s⁻¹ at one instant. Find normalized r, B disappearance and C formation rates.
Answer r=0.0200; B disappearance=0.0600; C formation=0.0200 M s⁻¹.
02 / UNDERSTAND & APPLY
A common empirical form is r=k[A]^m[B]^n. The exponents m and n are partial reaction orders with respect to A and B, and m+n is the overall order for that rate law. They describe how the measured rate responds to concentration changes under specified conditions. They are not generally read from the overall balanced equation.
The rate constant k collects the temperature- and medium-dependent proportionality for the chosen law and rate definition. At fixed temperature and unchanged conditions, changing concentrations changes r according to the exponents rather than automatically changing k. A different catalyst, solvent or temperature can change k or even the applicable rate-law form.
An exponent zero means the rate is independent of that concentration over the studied regime, not that the species is absent or unnecessary. A negative exponent describes inhibition in that empirical range: increasing the species lowers the rate. A fractional exponent can arise from a multistep mechanism or other kinetic effects. None implies a fractional molecule participates in one elementary event.
Rate laws can be more complicated than a single product of powers. A saturation expression may show one apparent order at low concentration and another at high concentration. State the range in which the power-law approximation is being used. Extrapolating a zero-order law to zero substrate or a negative-order law to zero inhibitor can produce unphysical predictions.
03 / UNDERSTAND & APPLY
The initial-rate method compares rates near the start of separate experiments with controlled starting concentrations. Early measurements reduce complications from changing reactant amounts, accumulated products and reverse reactions, although they do not automatically eliminate every experimental artifact. All compared runs should have the same temperature, medium and rate definition.
If only A changes, the rate ratio is r₂/r₁=([A]₂/[A]₁)^m. Taking logarithms gives m=ln(r₂/r₁)/ln([A]₂/[A]₁). Doubling A and doubling rate suggests first order in A; doubling A and quadrupling rate suggests second order; no rate change suggests zero order within the measurement precision.
If more than one concentration changes between two runs, write the full ratio rather than attributing the whole rate change to one reactant. Once one exponent is known, divide out its effect to find another. With enough independent experiments, logarithmic equations can be solved for several exponents.
Uncertainty matters. A measured rate ratio of 3.9 when concentration doubles may be consistent with second order within experimental error; it does not necessarily establish an exact exponent of 1.963. Conversely, forcing every dataset to an integer can conceal a genuine fractional order. Use the precision and model specified by the problem.
m=ln(r₂/r₁)/ln([A]₂/[A]₁)
Only A is varied; all other relevant conditions are held fixed.
ORIGINAL WORKED EXAMPLE
At fixed conditions, run 1 has [A]=0.100 M,[B]=0.100 M,r=1.00×10⁻⁴ M s⁻¹. Run2 doubles A only and gives 2.00×10⁻⁴. Run3 doubles B only from run 1 and gives 4.00×10⁻⁴. Find the law and k.
Answer r=k[A][B]²; k=0.100 M⁻² s⁻¹.
04 / UNDERSTAND & APPLY
Because rate has concentration-per-time units, a total-order N power law requires [k]=concentration^(1−N)time⁻¹. Zero-order k has M s⁻¹, first-order k has s⁻¹, second-order k has M⁻¹ s⁻¹ and third-order k has M⁻² s⁻¹. These units follow from dimensional balance, not a separate memorized table.
If concentration is expressed in mol m⁻³ instead of mol L⁻¹, the numerical value of k changes for orders other than one. A comparison of rate constants requires matching concentration and time units. A change from seconds to minutes also changes the numerical k even though the physical kinetics are the same.
The normalized reaction-rate convention can introduce a constant factor relative to a reactant-disappearance law. If 2A→products has r=k_r[A] and r=−½d[A]/dt, then the disappearance law is −d[A]/dt=2k_r[A]. Both have first-order units, but their numerical constants differ by two. Units alone cannot resolve a definition mismatch.
Overall order from k units is valid when the rate is expressed as the simple concentration-power form. An observed constant in a pseudo-first-order or saturation model may include fixed reactant concentrations or other parameters. Identify the equation before inferring a mechanistic order from units.
ORIGINAL WORKED EXAMPLE
A reaction follows r=k[A]^(1/2)[B]². Find overall order, k units in M and s, and the rate factor if A quadruples while B halves.
Answer Order2.5; k units M⁻³ᐟ² s⁻¹; rate halves.
05 / UNDERSTAND & APPLY
An elementary step represents a single molecular event in a mechanism. Its molecularity counts the reacting entities in that event: unimolecular, bimolecular or, less commonly, termolecular. It is a positive integer. A sequence of several elementary steps has no single molecularity merely because its overall balanced equation has coefficients.
For an elementary step under the standard mass-action model, its rate dependence follows its reacting molecular participants. An elementary A+B→products step gives a rate proportional to [A][B]; an elementary2A→products step gives a dependence on [A]², with convention-dependent numerical factors absorbed into the rate constant. This inference is justified by elementary-step status, not by every balanced equation.
An observed overall first-order reaction need not be unimolecular. It may be a bimolecular process with one reactant held effectively constant, or a multistep mechanism with a rate-limiting unimolecular stage. An overall second-order law likewise does not prove that the net equation occurs in one collision.
A mechanism must satisfy two necessary tests: its steps must sum to the correct overall reaction, and the predicted kinetic law must agree with experiment within the stated assumptions. Passing these tests supports a mechanism but does not uniquely prove it; different mechanisms can yield the same rate law.
ORIGINAL WORKED EXAMPLE
A measured rate law is r=k[A]^(1/2). A student calls the reaction “half-molecular.” What is wrong?
Answer Half-order is meaningful; half-molecularity is not.
06 / UNDERSTAND & APPLY
If an irreversible slow elementary step follows fast steps, its rate can often provide an approximate overall rate, but any intermediate concentration in that expression must be related to measurable starting species. Simply reporting r=k[I][B] is incomplete if I is an unmeasured intermediate and the requested law is in terms of reactants.
Consider a hypothetical fast pre-equilibrium2A⇌I with equilibrium constant K=[I]/[A]², followed by slow I+B→P. The slow-step rate is r=k₂[I][B]. If the first step genuinely maintains near-equilibrium while the second consumes I sufficiently slowly, substitute [I]=K[A]² to obtain r=k₂K[A]²[B].
The overall reaction is 2A+B→P after I cancels. In this particular model, the resulting overall orders happen to match those coefficients, but that agreement is a derived outcome, not permission to infer orders from stoichiometry generally. Altering the mechanism can produce a different law for the same net atom balance.
A pre-equilibrium approximation requires the forward and reverse equilibration to be fast relative to the perturbing consumption under the relevant conditions. If I is rapidly drained, its concentration may not track K[A]². A steady-state approximation or a fuller kinetic solution may then be more appropriate.
ORIGINAL WORKED EXAMPLE
For fast 2A⇌I, K=5.00 M⁻¹; slow I+B→P has k₂=0.200 M⁻¹ s⁻¹. Find the effective law and rate at [A]=0.100 M,[B]=0.300 M under the pre-equilibrium approximation.
Answer r=1.00[A]²[B] in M-based units; numerical rate 0.00300 M s⁻¹.
07 / UNDERSTAND & APPLY
Consider the hypothetical steps A⇌I with forward rate k₁[A] and reverse rate k₋₁[I], followed by I+B→P with rate k₂[I][B]. The intermediate balance is d[I]/dt=k₁[A]−k₋₁[I]−k₂[I][B]. If I reaches a low, slowly changing concentration after an initial transient, the steady-state approximation sets that derivative approximately to zero.
Solving gives [I]≈k₁[A]/(k₋₁+k₂[B]) and product rate r≈k₁k₂[A][B]/(k₋₁+k₂[B]). This is not a single power law over all B. At low B, where k₂[B]≪k₋₁, the rate is approximately (k₁k₂/k₋₁)[A][B]. At high B, where k₂[B]≫k₋₁, it approaches k₁[A] and becomes approximately zero order in B.
The high-B regime does not mean B is absent from the mechanism. It means capture of I by B is so effective that production of I from A becomes the limiting supply in this model. The observed order changes because the controlling kinetic balance changes, not because the net reaction suddenly has a different atom count.
Steady state and equilibrium are different approximations. A near-constant intermediate concentration can result from balanced formation and consumption fluxes without the forward and reverse first step being at thermodynamic equilibrium. State the approximation and its expected time or concentration regime rather than treating the two phrases as interchangeable.
r≈k₁k₂[A][B]/(k₋₁+k₂[B])
Steady-state intermediate model for A⇌I, I+B→P after the initial transient.
ORIGINAL WORKED EXAMPLE
A model gives r=k₁[A][B]/(K+[B]), where K has concentration units. What are the limiting orders in B at [B]≪K and [B]≫K?
Answer Low B: first order; high B: zero order in B.
08 / UNDERSTAND & APPLY
A proposed rate law should be checked across more than one concentration comparison. If r/[A]^m[B]^n gives different k values across supposedly identical conditions, the law, data or controlled conditions need re-examination. A changing temperature or an unnoticed catalyst concentration can mimic a different order.
Integrated concentration–time tests provide complementary evidence. First-order disappearance gives a linear ln[A] versus t relation at constant k, while zero order gives a linear [A] versus t relation over its valid range. These tests use the same disappearance-rate convention as their integrated formulas. A normalized rate constant may require conversion before comparing.
Pseudo-first-order experiments deliberately hold one reactant effectively constant, often by large excess. The observed k_obs incorporates its concentration. Repeating at several excess concentrations can recover its underlying order. A single exponential trace by itself cannot reveal every reactant dependence.
Product inhibition, reversibility, diffusion limitation and surface saturation can complicate interpretation. Initial-rate measurements help but do not guarantee one simple mechanism. Use experimental controls and the stated model rather than forcing an elegant integer-power law onto every dataset.
Finally, rate order is not a stability ranking or an equilibrium exponent. A species with a large equilibrium abundance can still react slowly; a low-concentration intermediate can control a pathway. Kinetics measures flux through a mechanism, while equilibrium describes the thermodynamic composition at balance.
ORIGINAL WORKED EXAMPLE
A measured law in a limited regime is r=k[A][I]⁻¹, where I is an inhibitor. If A doubles and I quadruples, what happens to rate?
Answer Rate halves within the stated regime.
09 / UNDERSTAND & APPLY
These original exercises use fixed-volume rates and the definitions stated in each question. Read whether “rate” means a species disappearance rate or normalized reaction rate. Treat supplied exact ratios as ideal instructional data unless uncertainty is explicitly discussed.
Check the result by substituting into every available run and by dimensional balance. If a mechanism is proposed, cancel intermediates from the summed steps and from the final experimentally testable rate law. A plausible story without these checks is not a kinetic derivation.
ORIGINAL PRACTICE · 1
0.0600 M s⁻¹.
Two moles B form per mole A consumed. The normalized rate equals A disappearance because its coefficient is one.
ORIGINAL PRACTICE · 2
Three.
2^m=8 gives m=3. This is an experimental order for the stated range, not automatically elementary molecularity.
ORIGINAL PRACTICE · 3
Zero order in B over the tested regime.
3^n=1 gives n=0. B can still participate in the mechanism or become limiting outside that regime.
ORIGINAL PRACTICE · 4
0.0500 M⁻¹ s⁻¹.
k=r/[A]²=0.00200/0.0400=0.0500. Multiplying k by M² returns M s⁻¹.
ORIGINAL PRACTICE · 5
No.
Only an established elementary-step interpretation justifies direct mass-action exponents from its participants. Overall reaction order must be measured or derived from a supported mechanism.
ORIGINAL PRACTICE · 6
One; the rate is unchanged.
(1/2)²×4=1. The two concentration effects cancel in the stated law.
ORIGINAL PRACTICE · 7
r=k₂K[A][B].
The slow elementary step gives r=k₂[I]. Substitute the fast-equilibrium expression to eliminate the intermediate.
ORIGINAL PRACTICE · 8
No.
A steady intermediate level can be maintained by equal formation and consumption fluxes in a driven kinetic sequence. Equilibrium requires the appropriate forward and reverse balance, not merely a small net derivative.
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