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Kinetics
Reaction pathways determine how quickly chemical change occurs.
What you’ll learn
- Determine a rate law from experimental evidence.
- Distinguish reaction order from equation coefficients.
- Evaluate mechanisms and catalysts.
Before you begin
Reaction rate is change in concentration per unit time, adjusted when comparing different species through stoichiometry. A rate law describes an observed dependence on concentration. The balanced overall reaction does not usually reveal the experimental rate-law exponents.
Explain these starting ideas in your own words. Revisit them whenever a later step feels unclear.
Rates are measured changes over time
Reaction rate can be defined from the disappearance of a reactant or appearance of a product, with stoichiometric coefficients used to express a common reaction rate. A steep concentration–time slope indicates a large instantaneous change, while an interval calculation gives an average. State which species and sign convention are being used.
Collision models require encounters with adequate energy and suitable orientation. Increasing temperature changes the energy distribution and often increases the fraction of productive collisions. Concentration, surface area, and catalysts may also affect rate, but their effects depend on the reaction and conditions.
Read figure values as text
Uncatalyzed pathway: 0: 40; 0.025: 39.90012893065802; 0.05: 40.59066322040751; 0.075: 42.042287963878046; 0.1: 44.20694768281421; 0.125: 47.019029611437205; 0.15: 50.39697930049462; 0.175: 54.245308758464724; 0.2: 58.4569476828142; 0.225: 62.9158798861925; 0.25: 67.5; 0.275: 72.08412011380751; 0.3: 76.54305231718578; 0.325: 80.75469124153527; 0.35: 84.60302069950538; 0.375: 87.9809703885628; 0.4: 90.79305231718578; 0.425: 92.95771203612196; 0.45: 94.4093367795925; 0.475: 95.09987106934197; 0.5: 95; 0.525: 94.09987106934197; 0.55: 92.40933677959248; 0.575: 89.95771203612196; 0.6: 86.7930523171858; 0.625: 82.9809703885628; 0.65: 78.60302069950538; 0.675: 73.75469124153528; 0.7: 68.54305231718578; 0.725: 63.084120113807515; 0.75: 57.50000000000001; 0.775: 51.915879886192506; 0.8: 46.45694768281422; 0.825: 41.24530875846473; 0.85: 36.39697930049463; 0.875: 32.01902961143721; 0.9: 28.206947682814214; 0.925: 25.04228796387805; 0.95: 22.59066322040751; 0.975: 20.900128930658024; 1: 20 • Catalyzed pathway: 0: 40; 0.025: 39.696986550477796; 0.05: 39.78309573927754; 0.075: 40.243895612986115; 0.1: 41.05572809000084; 0.125: 42.18629150101524; 0.15: 43.595435963320426; 0.175: 45.23615200416725; 0.2: 47.05572809000084; 0.225: 48.99704855935631; 0.25: 51; 0.275: 53.0029514406437; 0.3: 54.94427190999916; 0.325: 56.76384799583275; 0.35: 58.40456403667957; 0.375: 59.81370849898476; 0.4: 60.94427190999916; 0.425: 61.756104387013885; 0.45: 62.21690426072246; 0.475: 62.303013449522204; 0.5: 62; 0.525: 61.303013449522204; 0.55: 60.21690426072245; 0.575: 58.75610438701389; 0.6: 56.94427190999916; 0.625: 54.81370849898476; 0.65: 52.404564036679574; 0.675: 49.76384799583275; 0.7: 46.94427190999916; 0.725: 44.0029514406437; 0.75: 41; 0.775: 37.99704855935631; 0.8: 35.05572809000085; 0.825: 32.23615200416725; 0.85: 29.595435963320433; 0.875: 27.18629150101524; 0.9: 25.055728090000844; 0.925: 23.243895612986115; 0.95: 21.783095739277545; 0.975: 20.696986550477796; 1: 20
PAUSE & TRY IT
Why does a larger surface area sometimes accelerate a solid’s reaction?
Reveal answer
It exposes more sites for contact with reacting particles, assuming other conditions are unchanged.
Determine rate laws experimentally
A rate law relates rate to reactant concentrations raised to experimentally determined powers. For an overall reaction, those powers cannot generally be copied from the balanced coefficients. Compare experiments in which one concentration changes while other relevant conditions remain fixed.
If doubling one concentration quadruples the initial rate, its order is two in the tested range. If the rate does not change, the measured order is zero in that range. The rate constant’s units depend on total reaction order, so checking units is a useful way to detect errors.
Integrated laws connect concentration and time
For the common single-reactant models, a linear [A] versus time plot supports zero-order behavior, a linear ln[A] versus time plot supports first order, and a linear versus time plot supports second order. These are tests of models over the measured interval, not guarantees about every possible condition.
A first-order half-life is independent of initial concentration. After each equal half-life interval, half of the remaining reactant survives. Do not subtract half of the original amount repeatedly. Other reaction orders have different half-life dependencies.
PAUSE & TRY IT
Which concentration plot is linear for a first-order process?
Reveal answer
ln[A] versus time, with slope −k.
Mechanisms must fit more than the overall equation
An elementary step describes a single molecular event, and its reactant molecularity can determine that step’s rate expression. A proposed mechanism must add to the overall reaction and be consistent with the measured rate law. Intermediates are formed and later consumed, while a catalyst is consumed and regenerated.
A slow-step model can help derive a rate law, but if that expression contains an intermediate, use the supplied preceding equilibrium or other valid relationship to express it in terms of experimentally controlled species. Matching the overall equation alone does not establish that a mechanism is correct.
PAUSE & TRY IT
How can an intermediate be recognized in a mechanism?
Reveal answer
It is formed in one step and consumed in a later step and cancels from the overall equation.
Catalysts change the path, not the destination
A catalyst provides an alternative pathway with a lower effective activation barrier. It accelerates approach to equilibrium in both directions but does not change the equilibrium constant or the reaction’s net thermodynamic energy change. A heterogeneous catalyst may work through adsorption and reaction at a surface.
On a multistep energy diagram, each peak represents a transition state and each intervening minimum represents an intermediate. Activation energy for a step is measured from the preceding minimum to that step’s peak. The highest point relative to the original reactants is not automatically the largest individual step barrier.
Extract a rate law with controlled comparisons
Compare experiments where one reactant concentration changes while the others remain fixed. If doubling that concentration doubles the rate, the data support first order in that reactant; if the rate quadruples, they support second order; if it is unchanged, they support zero order over that range. When several concentrations change, account for one known dependence before solving for another.
After finding the exponents, calculate k from one experiment and verify it with another. Its units depend on overall order, so a unit check can expose an exponent error. A rate law is tied to conditions such as temperature and mechanism. It is not a universal statement that concentration always affects a reaction the same way.
PAUSE & TRY IT
For first-order decay, does each equal time interval remove the same number of moles?
Reveal answer
No. It removes the same fraction of the amount present at the start of that interval.
Use integrated plots to connect a model with time data
A zero-order process gives a linear concentration-versus-time plot, a first-order process gives a linear ln(concentration)-versus-time plot, and a second-order single-reactant model gives a linear reciprocal-concentration-versus-time plot. Compare the stated models and inspect which transformation linearizes the data. The sign and units of the slope connect the line to k.
For first-order decay, the half-life is independent of initial concentration. Equal fractional decreases occur in equal time intervals, not equal absolute decreases. Other orders have different half-life behavior. A single half-life measurement is usually insufficient to identify a mechanism; use the time pattern or additional concentration experiments.
Read figure values as text
[A]=1×2^(−t/2): 0: 1; 0.16666666666666666: 0.9438743126816935; 0.3333333333333333: 0.8908987181403393; 0.5: 0.8408964152537145; 0.6666666666666666: 0.7937005259840998; 0.8333333333333334: 0.7491535384383408; 1: 0.7071067811865476; 1.1666666666666667: 0.6674199270850172; 1.3333333333333333: 0.6299605249474366; 1.5: 0.5946035575013605; 1.6666666666666667: 0.5612310241546865; 1.8333333333333333: 0.5297315471796477; 2: 0.5; 2.1666666666666665: 0.47193715634084676; 2.3333333333333335: 0.44544935907016964; 2.5: 0.42044820762685725; 2.6666666666666665: 0.3968502629920499; 2.8333333333333335: 0.3745767692191704; 3: 0.3535533905932738; 3.1666666666666665: 0.3337099635425086; 3.3333333333333335: 0.3149802624737183; 3.5: 0.29730177875068026; 3.6666666666666665: 0.28061551207734325; 3.8333333333333335: 0.2648657735898238; 4: 0.25; 4.166666666666667: 0.23596857817042335; 4.333333333333333: 0.22272467953508485; 4.5: 0.21022410381342863; 4.666666666666667: 0.19842513149602492; 4.833333333333333: 0.18728838460958522; 5: 0.1767766952966369; 5.166666666666667: 0.16685498177125427; 5.333333333333333: 0.15749013123685915; 5.5: 0.14865088937534013; 5.666666666666667: 0.1403077560386716; 5.833333333333333: 0.1324328867949119; 6: 0.125; 6.166666666666667: 0.11798428908521168; 6.333333333333333: 0.11136233976754242; 6.5: 0.10511205190671431; 6.666666666666667: 0.09921256574801246; 6.833333333333333: 0.09364419230479261; 7: 0.08838834764831845; 7.166666666666667: 0.08342749088562713; 7.333333333333333: 0.07874506561842957; 7.5: 0.07432544468767006; 7.666666666666667: 0.0701538780193358; 7.833333333333333: 0.06621644339745596; 8: 0.0625
PAUSE & TRY IT
Why must intermediates cancel when elementary steps are added?
Reveal answer
They are formed and consumed within the mechanism rather than appearing in the net overall reaction.
A mechanism must satisfy stoichiometry and kinetics
Elementary steps add to the overall reaction after intermediates cancel. A catalyst is consumed in one step and regenerated later; an intermediate is produced and then consumed. An elementary step’s molecularity can inform its rate expression, but the overall equation’s coefficients cannot be used in the same way unless it truly represents an elementary event.
A proposed slow step must lead to a rate law consistent with experiment, sometimes after substituting a relationship from a preceding equilibrium. An energy diagram can show several activation barriers and intermediates. A catalyst provides another pathway with different barriers; it does not change the overall enthalpy difference or equilibrium constant at a fixed temperature.
Distinguish a measured rate from a rate law
A concentration–time graph gives a changing rate through its slope. A secant gives an average over an interval; a tangent gives an instantaneous value. Reactant disappearance slopes are negative, but a reported disappearance rate is often defined as positive. Stoichiometric coefficients relate species rates, so disappearance of one reactant need not numerically equal formation of one product.
A rate law describes how rate depends on concentrations under specified conditions. Determine orders from controlled comparisons. If doubling one concentration with all others and temperature fixed quadruples rate, its order is two in the tested model. Overall reaction coefficients are not generally the exponents of an experimentally determined rate law.
Changing temperature can change k, so trials at different temperatures cannot isolate concentration order using a simple ratio. Include units: rate has concentration per time, and the units of k depend on overall order. A zero order means rate is insensitive to that concentration within the tested regime, not that the substance is chemically irrelevant.
Choose an integrated model from the graph it predicts
For zero-order behavior, concentration decreases linearly with time. For first-order behavior, ln concentration is linear; for second-order behavior in one reactant, reciprocal concentration is linear. Compare the appropriate transformed plots instead of judging one curved concentration graph by eye. The slope determines k with the sign and units appropriate to the model.
A first-order half-life is independent of initial concentration. After successive equal half-lives, the remaining amount is multiplied by one-half repeatedly; equal amounts are not removed each time. A zero-order half-life depends on initial concentration because the same amount disappears per unit time.
An integrated law links concentrations to elapsed time under its assumptions. Before substitution, establish reaction order, consistent time units and whether the amount or concentration supplied is proportional to the relevant variable. A straight line over a narrow interval is evidence for a model in that interval, not proof of universal behavior at all concentrations.
PAUSE & TRY IT
A first-order concentration falls from 0.80 M to 0.10 M in 30 minutes. What is its half-life?
Reveal answer
The concentration halves three times: 0.80→0.40→0.20→0.10. Therefore the half-life is =10 minutes.
Evaluate a mechanism using independent constraints
Elementary steps add to the overall equation after intermediates and regenerated catalysts cancel. An intermediate is formed and later consumed; a catalyst is consumed in one step and regenerated later. A proposed sequence must account for the observed stoichiometry and be consistent with the measured rate law. Matching only one requirement is insufficient.
For an elementary step, molecularity informs its concentration dependence in the model. A multistep overall equation does not have that privilege. If the slow-step expression includes an intermediate, additional relationships may be needed to express the prediction in measured reactant concentrations. Use only assumptions stated or justified for the mechanism.
A catalyst offers an alternative pathway and accelerates approach to equilibrium. It does not change the reaction’s standard free-energy difference or K at the same temperature. An exothermic reaction can still be slow if its barrier is high. On a multistep energy profile, activation energy for a step is measured from that step’s reactant state to its transition state, not always from the first reactants.
Infer reaction order from controlled comparisons
Compare trials where one reactant concentration changes while others remain fixed. If doubling that concentration quadruples initial rate, its order is two in the tested rate law. If several concentrations change together, use ratios and solve for their effects rather than assigning the whole rate change to one species.
The rate constant’s units depend on overall order because the full rate law must have rate units. Rate constants also depend on temperature. A concentration change alters rate through the concentration factors; it does not ordinarily change k at fixed conditions in the same model.
A balanced overall equation does not generally supply reaction orders. Orders come from experiment or a justified elementary-step mechanism. Distinguish an elementary reaction’s molecularity from the stoichiometry of a multistep overall process.
Test a proposed mechanism against both chemistry and rate
Add elementary steps and cancel intermediates to recover the overall reaction. A catalyst is consumed in one step and regenerated later; an intermediate is produced and then consumed. Both can cancel from the net equation, but their roles differ.
A proposed rate-determining-step argument must be consistent with the experimental rate law. If an intermediate appears in that step’s rate expression, an additional relationship may be needed to express it in measured reactant concentrations. Merely matching the overall equation is insufficient.
A catalyst supplies a different pathway with a lower effective activation barrier under the model. It increases forward and reverse rates toward equilibrium without changing the equilibrium constant at fixed temperature. It does not make an unfavorable equilibrium favorable by changing ΔG°.
PAUSE & TRY IT
Can a mechanism be ruled out even if its steps add to the correct overall equation?
Reveal answer
Yes. It must also be consistent with kinetic evidence and the observed rate law.
FROM IDEA TO APPLICATION
Worked examples
Use initial rates
At fixed [B], doubling [A] doubles rate. At fixed [A], tripling [B] multiplies rate by nine. Find the rate law and total order.
Reveal worked solution
- 2m = 2 gives m = 1.
- 3n = 9 gives n = 2.
- Add the orders to obtain 3.
Rate = k[A][B]2, third order overall. If rate is , k has units M-2s-1.
Repeated half-lives
A first-order process has a 12-minute half-life. What fraction remains after 36 minutes?
Reveal worked solution
- = 3 half-lives.
- Remaining fraction = ()3.
, or 12.5%, remains. This relies on first-order behavior throughout the interval.
Determine two reaction orders
Doubling [A] at fixed [B] doubles the rate. Tripling [B] at fixed [A] increases the rate ninefold. Give the supported rate law.
Reveal worked solution
- The dependence on A is first order because 21=2.
- The dependence on B is second order because 32=9.
- Combine the dependencies with a rate constant.
Rate=k[A][B]2, third order overall under the tested conditions.
Determine two reaction orders
Doubling [A] at fixed [B] multiplies rate by four. Tripling [B] at fixed [A] multiplies rate by three. Predict the rate factor when both changes occur.
Reveal worked solution
- The A dependence is second order because 22=4.
- The B dependence is first order because 31=3.
- The model is rate=k[A]2[B], so simultaneous changes multiply: 4×3=12.
The predicted rate is twelve times the original at unchanged temperature and mechanism.
Orders from two comparisons
Doubling A at fixed B doubles rate; tripling B at fixed A leaves rate unchanged.
Reveal worked solution
- A is first order in the measured law.
- B is zero order over the tested range.
Rate=k[A]; zero order does not mean B is absent from the overall reaction.
Temperature versus concentration
Rate rises after heating a reaction mixture without intentionally changing concentrations.
Reveal worked solution
- Temperature changes the rate constant and the fraction of collisions able to cross the activation barrier.
- This is different from a concentration factor in a fixed-temperature rate law.
A higher rate does not by itself imply a higher reaction order.
MAKE THE DISTINCTION
Common mistakes, clearer reasoning
The trapThe balanced overall equation gives every reaction order.
The better explanationOrders require kinetic evidence except when describing a specified elementary step.
The trapA catalyst changes the equilibrium yield.
The better explanationIt changes how quickly equilibrium is approached, not the equilibrium composition at fixed conditions.
RETRIEVE BEFORE YOU REVEAL
Practice checkpoints
Revisit the quick checks from this guide without looking back. Explain why, then reveal the answer.
1. Which concentration plot is linear for a first-order process?
Reveal answer
ln[A] versus time, with slope −k.
2. How can an intermediate be recognized in a mechanism?
Reveal answer
It is formed in one step and consumed in a later step and cancels from the overall equation.
3. Why does a larger surface area sometimes accelerate a solid’s reaction?
Reveal answer
It exposes more sites for contact with reacting particles, assuming other conditions are unchanged.
4. For first-order decay, does each equal time interval remove the same number of moles?
Reveal answer
No. It removes the same fraction of the amount present at the start of that interval.
5. Why must intermediates cancel when elementary steps are added?
Reveal answer
They are formed and consumed within the mechanism rather than appearing in the net overall reaction.
6. A first-order concentration falls from 0.80 M to 0.10 M in 30 minutes. What is its half-life?
Reveal answer
The concentration halves three times: 0.80→0.40→0.20→0.10. Therefore the half-life is =10 minutes.
7. Can a mechanism be ruled out even if its steps add to the correct overall equation?
Reveal answer
Yes. It must also be consistent with kinetic evidence and the observed rate law.
Key language
- Rate law
- An experimentally supported relationship between rate and concentration.
- Reaction order
- An exponent in a rate law.
- Intermediate
- A species produced and consumed within a mechanism.
- Catalyst
- A regenerated participant that provides an alternative reaction pathway.
Kinetics determines timescale; equilibrium and thermodynamics determine the favored composition.