Topic 11: Kinetics
- Syllabus
- 2017
- Topic
- —
- Level
- A2
\text{rate}=k[\mathrm{A}]^m[\mathrm{B}]^n
| Term | Precise meaning |
|---|---|
| rate of reaction | change in concentration of a reactant or product per unit time |
| order with respect to A | exponent m found experimentally |
| overall order | sum m+n |
| rate constant, k | proportionality constant at a stated temperature; its units depend on overall order |
| half-life, t1/2 | time for a reactant concentration to fall to half its value |
| rate-determining step | slow step controlling the observed rate |
| activation energy, Ea | minimum energy barrier for a successful route |
| homogeneous catalyst | catalyst in the same phase as reactants |
| heterogeneous catalyst | catalyst in a different phase, so reaction occurs at an interface |
Rate commonly has units mol dm−3 s−1. Rearrange the measured rate equation to find k, then derive its units: zero order gives mol dm−3 s−1; first order gives s−1; second order gives dm3 mol−1 s−1.
The powers in a rate equation are experimental orders, not coefficients copied from the balanced equation. They match molecular numbers only when a justified elementary step controls the rate.
Half-life is the time taken for the concentration of a reactant to halve. For a first-order reaction, equal fractional decreases take equal times, so the half-life remains constant as concentration falls.
| Graph move | Reading |
|---|---|
| choose an initial concentration c | read the time when the curve reaches c/2 |
| start again at c/2 | read the later time when it reaches c/4 |
| compare intervals | similar intervals support first-order behaviour |
N=N_0\left(\frac12\right)^{t/t_{1/2}}
A 100 mg dose with a 20 min half-life undergoes 12 half-lives in 4 h. The mass is 100(1/2)12=0.0244 mg, or 24.4 μg.
One halving interval is not enough to establish constant half-life. Measure at least two successive halvings on the same suitable graph.
| Observable change | Suitable technique | Justification/limit |
|---|---|---|
| coloured species changes | colorimetry | absorbance can be calibrated to concentration; other species must not interfere |
| gas formed or consumed | gas syringe/volume measurement | continuous gas data; apparatus must be gas-tight |
| gas escapes | mass loss on a balance | simple continuous data; unsuitable if no volatile material leaves |
| soluble species with a titratable amount | timed aliquots, quench, then titrate | gives concentration at selected times; quench must stop reaction rapidly |
| conductivity, pH or pressure changes | suitable probe | only when the measured property tracks reaction extent |
The measured signal must change monotonically with the chosen reactant or product and be fast to record compared with the reaction. State what is measured, how it relates to concentration, the sampling frequency and the main source of loss or delay.
Do not select a method merely because the apparatus is available. Colorimetry cannot follow a colourless mixture, and mass cannot decrease in a closed system simply because reaction occurs.
| Method | Procedure | Rate evidence |
|---|---|---|
| initial rate | run separate mixtures, changing one initial concentration while controlling all others | initial gradient, or a clock approximation proportional to 1/t |
| continuous monitoring | record concentration, gas volume, mass or absorbance throughout one run | gradient at any time and a full concentration-time/volume-time curve |
A clock method uses the time to reach the same small, fixed amount of reaction. If the clock reagent is small compared with the main reactants, their concentrations change little before the endpoint, so 1/t is proportional to initial rate.
Keep total volume, temperature and all non-tested initial concentrations constant. Start timing at consistent mixing and use the same endpoint; otherwise 1/t compares different extents rather than rates.
A clock reaction approximates an initial rate; it is not continuous monitoring and does not show how rate changes throughout the whole reaction.
| Order in A | Initial-rate comparison | Rate–[A] graph | [A]–time graph |
|---|---|---|---|
| 0 | changing [A] leaves rate unchanged | horizontal line | straight decrease |
| 1 | doubling [A] doubles rate | straight line through origin | curved decrease with constant half-life |
| 2 | doubling [A] quadruples rate | upward curve | curved decrease with increasing half-life |
Compare runs in which only one concentration changes. If concentration changes by factor f and rate by factor g, solve g=fm. When two concentrations change together, first account for the known order of one before isolating the other.
A concentration–time gradient gives rate, so compare gradients at chosen concentrations. A rate–concentration graph displays the dependence directly. State both the observed factor/shape and the resulting order.
A curved concentration–time graph alone does not prove first order; second-order decay is also curved. Constant half-life or the correct rate–concentration relationship supplies the distinction.
\text{rate}=k[\text{propanone}][\mathrm{H^+}]\quad\text{and is zero order in }\mathrm{I_2}
| Experiment | Controlled comparison | Inference |
|---|---|---|
| change propanone only | initial rate changes in direct proportion | first order in propanone |
| change acid only | initial rate changes in direct proportion | first order in H+ |
| change iodine only | initial rate is unchanged | zero order in iodine |
| distinguish H+ from anion | change chloride with neutral chloride, or use another strong acid | unchanged rate shows chloride is not responsible |
The evidence supports a slow acid-catalysed formation of the enol from propanone, involving propanone and H+, followed by fast reaction of the enol with iodine. Iodine is absent from the rate-determining process, and H+ is regenerated.
Follow iodine loss by timed aliquots and titration, or by calibrated colorimetry. Use iodine in the smaller amount so propanone and acid remain nearly constant during a run.
Zero order in iodine does not mean iodine is absent from the overall reaction; it means changing iodine concentration does not change the observed rate under these conditions.
For an elementary rate-determining step, its reacting species determine the rate expression. A proposed slow step is consistent only when its molecular composition can produce the observed orders.
| Evidence | Mechanism consequence |
|---|---|
| species appears in rate equation | it must participate in, or be linked by a prior fast equilibrium to, the slow step |
| species is zero order | it must not be required in the rate-controlling process |
| exponent 2 | two particles/equivalent concentration factors must influence the slow process |
| intermediate appears in slow step | eliminate it using a justified earlier fast equilibrium before comparing with experiment |
If the observed rate is first order in H2O2, a slow elementary step containing one H2O2 molecule is consistent; a slow collision between two H2O2 molecules predicts second order and is inconsistent.
A matching rate equation supports a proposed slow step but does not prove it uniquely. The steps must also sum to the overall equation and preserve atoms and charge.
| Constraint | Required check |
|---|---|
| observed rate equation | slow-step pathway gives the correct concentration dependence |
| stoichiometric equation | adding all steps cancels intermediates and reproduces the overall equation |
| intermediates | formed in one step and consumed later; absent from the overall equation |
| catalyst | consumed early and regenerated later; absent from the overall equation |
| chemistry | arrows, bonds, charges and plausible species remain consistent |
For an overall A+B+CightarrowP reaction with extrate=k[A][B], a possible scheme is slow A+BightarrowX followed by fast X+CightarrowP. Adding the steps cancels X and gives the overall equation.
Work from both ends: use the rate law to constrain the slow step, then add fast steps needed to account for the remaining overall reactants and products. Reject any scheme that leaves an intermediate uncancelled.
The balanced overall equation cannot by itself reveal a mechanism or rate law. Several step sequences may have the same net stoichiometry.
| Hydrolysis route | Rate equation | Rate-determining event | Structural fit |
|---|---|---|---|
| SN1 | extrate=k[RX] | slow C–X heterolysis forms a carbocation; nucleophile attacks later | tertiary halogenoalkane stabilises the carbocation |
| SN2 | extrate=k[RX][OH−] | one concerted attack as C–X breaks | primary halogenoalkane has less steric hindrance |
If doubling [RX] doubles rate and doubling [OH−] also doubles rate, the reaction is first order in each and supports SN2. If changing [OH−] has no effect while rate follows [RX], the evidence supports SN1.
The labels 1 and 2 describe molecularity of the rate-determining substitution route, not the number of experimental steps or the class of the carbon atom.
\ln k=-\frac{E_a}{R}\left(\frac{1}{T}\right)+\ln A
| Quantity | Treatment |
|---|---|
| temperature | convert °C to K, then calculate 1/T in K−1 |
| rate constant | calculate lnk |
| graph | plot lnk on y against 1/T on x |
| gradient | −Ea/R, with unit K |
| activation energy | Ea=−extgradientimesR; convert J mol−1 to kJ mol−1 |
\ln\left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)
Use R=8.31 J K−1 mol−1, so the calculated Ea is initially in J mol−1. The Arrhenius line has a negative gradient but activation energy is reported as a positive barrier.
Do not plot k against 1/T and then use gradient =−Ea/R; the linearised y-variable must be lnk.
| Stage | Particle-level change |
|---|---|
| adsorption | gaseous reactants attach at active sites on the solid |
| activation | interactions with the surface weaken relevant reactant bonds |
| reaction | adsorbed species follow an alternative pathway with lower activation energy |
| desorption | products leave, freeing active sites for another cycle |
A finely divided catalyst exposes more active surface area, so more reactant particles can be adsorbed at once. The catalyst is heterogeneous because solid and gas are different phases, and separation from the surface stops this catalytic route.
A catalyst changes the pathway and rate, not the overall enthalpy change or equilibrium constant. It accelerates forward and reverse reactions without changing equilibrium composition.
| Practical | Measurement sequence | Rate information |
|---|---|---|
| 9a iodine–propanone | mix iodine, propanone and acid at controlled temperature; remove timed aliquots; stop further reaction rapidly; titrate remaining iodine with standard thiosulfate | iodine concentration against time, then rate/order comparisons |
| 9b Harcourt–Esson iodine clock | mix fixed iodine-forming reagents with a small fixed amount of thiosulfate and starch; record time to permanent blue-black | same iodine amount is formed at endpoint, so 1/t approximates initial rate |
\ce{I2 + 2S2O3^{2-} -> 2I^- + S4O6^{2-}}
Thiosulfate removes iodine as it forms; it does not slow the main iodine-producing reaction. Once thiosulfate is exhausted, iodine remains and forms the blue-black starch complex. Keep temperature, total volume and non-tested concentrations constant.
Use consistent mixing/start time, repeat each condition and compare concordant endpoint times. In the aliquot method, sampling time and complete quenching are critical; in the clock, subjective colour judgement and delay in mixing affect precision.
The clock endpoint is a fixed extent, not completion of the main reaction. It supports relative initial rates only while the main reactant concentrations change negligibly before the colour appears.
| Stage | Controlled action |
|---|---|
| prepare | use identical reagent concentrations and volumes for every run |
| equilibrate | bring separate reagents to the chosen water-bath temperature before mixing |
| measure | mix consistently and time to the same visible endpoint |
| repeat | obtain concordant times at several temperatures |
| transform | convert T to K; calculate 1/T and ln(1/t) |
| analyse | plot ln(1/t) against 1/T and use gradient =−Ea/R |
At a fixed endpoint, the same small amount reacts in each run, so relative rate is proportional to 1/t. Any constant proportionality factor changes the intercept of the Arrhenius plot but not its gradient.
Maintain temperature during each run, use the same observer/endpoint and minimise delay between mixing and timing. Include all valid points in a best-fit line and investigate rather than silently discard an anomalous result.
Use kelvin, not degrees Celsius, in 1/T. Treat 1/t as a relative rate under identical endpoint conditions; it is not automatically the rate constant for every reaction.