Q: What does H2 Chemistry Notes: Topic 8 - Reaction Kinetics cover? A: Nail rate laws, experimental determination of orders, Arrhenius analysis, and catalysis narratives for Core Idea 3 (Reaction Kinetics).
Kinetics questions test both experimental planning and mathematical fluency.
This guide summarises rate-law theory, data analysis routines, and catalytic mechanisms tailored to the 2026 examination format.
Status: SEAB's current H2 Chemistry (9476) syllabus PDF is labelled for 2026, and the current Chemistry Data Booklet is labelled 8873/9476/9813 for use from 2026 in non-practical papers. Core Idea 3 Topic 8 is assessed across Papers 1-3.
The core idea is simple: Kinetics asks how fast a reaction happens and what controls that speed.
Use it as a working check: Orders come from experimental data, not from the balanced equation unless the question proves it is an elementary step.
Then go one layer deeper: Example: if doubling [A] doubles the initial rate while [B] is constant, the reaction is first order in A
Check this topic from memory
Attempt the matching topic bank before reopening the notes. Use each missed idea to decide what to review next.
What this topic tests: Rate laws, order determination, Arrhenius relationships, and catalyst reasoning.
Top mistakes to avoid: Treating coefficient as order by default; poor graph interpretation; superficial catalyst explanations.
20-minute sprint plan: 5 min order-method recap; 10 min initial-rate/graph questions; 5 min catalyst and Arrhenius explanation.
1 Rate Laws and Orders
General rate law for reaction aA + bB -> products:
rate=k[A]m[B]n
k : rate constant (temperature-dependent).
m,n : orders with respect to A,B; determined experimentally.
Overall order = m+n.
1.1 Units of Rate Constant
Overall order
Rate expression
Units of k
0
rate = k
mol⋅L−1⋅s−1
1
rate = k[A]
s−1
2
rate = k[A]2
L⋅mol−1⋅s−1
Always deduce units by inspecting the rate equation provided.
Rate-constant unit checkpoint
Do not memorise only the three rows above. Derive the units from the actual rate law, especially when the powers are mixed across reactants.
Step
Move
Common trap
1
Write the rate units as mol⋅L−1⋅s−1.
Using concentration units only and forgetting time.
2
Write the concentration-unit contribution from every reactant term.
Counting only one reactant when the rate law contains two or more.
3
Divide rate units by the concentration contribution to get units of k.
Copying s−1 just because one reactant is first order.
Worked check: if rate=k[A]2[B], then the overall order is 3. The concentration part has units (mol⋅L−1)3, so
[k]=(mol⋅L−1)3mol⋅L−1⋅s−1=L2⋅mol−2⋅s−1.
Misconception check: the units of k depend on the overall order, not on the number of different reactants in the equation.
2 Determining Orders Experimentally
2.1 Initial Rates Method
Conduct experiments varying one reactant concentration while keeping others constant.
Measure initial rates.
Compare rate ratios to concentration ratios:
If doubling [A] doubles rate → first order; quadruples rate → second order.
Worked comparison routine
Use one pair of experiments at a time. Choose the pair where only one reactant concentration changes.
Step
Question to ask
What to write
1
Which reactant concentration changed?
State the concentration ratio.
2
What happened to the initial rate?
State the rate ratio.
3
Which power connects them?
Match rate ratio = concentration ratio raised to the order.
Example data:
Experiment
[A] / mol⋅L−1
[B] / mol⋅L−1
Initial rate / mol⋅L−1⋅s−1
1
0.10
0.20
2.0×10−4
2
0.20
0.20
8.0×10−4
3
0.20
0.40
8.0×10−4
Compare experiments 1 and 2: [A] doubles while [B] is constant, and rate increases by a factor of 4. Therefore the reaction is second order with respect to A.
Compare experiments 2 and 3: [B] doubles while [A] is constant, and rate is unchanged. Therefore the reaction is zero order with respect to B.
Rate law: rate=k[A]2.
Trap check: do not compare experiments where two concentrations change at once unless all but one order are already known.
2.2 Continuous Monitoring
Plot concentration vs time data:
Zero order: linear decrease; gradient = −k.
First order:[A] decays exponentially; ln[A] vs time linear with gradient −k.
Second order:[A]1 vs time linear with gradient k.
Half-life t1/2 for first-order reactions is constant: t1/2=kln2.
Continuous monitoring graph choice checkpoint
When concentration-time data is provided, test the graph that should become a straight line. The straight-line graph tells you the order; the gradient then gives k, with the sign handled by the axis choice.
Straight-line graph
Order shown
What to read from the graph
Common trap
[A] against time
Zero order
Gradient equals −k.
Reporting k as negative instead of using the magnitude.
ln[A] against time
First order
Gradient equals −k.
Judging first order from the curved [A] against time plot alone.
[A]1 against time
Second order
Gradient equals k.
Taking reciprocal concentration after fitting the wrong graph.
Worked check: if ln[A] against time is linear with gradient −0.023s−1, then k=0.023s−1 and the reaction is first order with respect to A.
Misconception check: a curved concentration-time graph does not mean the data is unusable. Transform the y-axis and look for the straight-line plot.
3 Arrhenius Equation
Temperature dependence captured by:
k=Ae−RTEa
Taking natural logs:
lnk=lnA−REa⋅T1
Plot lnk vs 1/T; gradient = −REa, intercept = lnA.
Activation energy Ea in J⋅mol−1; convert to kJ⋅mol−1
Before reading an Arrhenius graph, check the axes and units:
Graph feature
What to read
Common mistake
Horizontal axis
1/T, with T in kelvin.
Using degrees Celsius before taking the reciprocal.
Vertical axis
lnk, not k.
Taking the gradient from a non-linear k against 1/T plot.
Gradient
−Ea/R.
Dropping the negative sign and reporting a negative activation energy.
Final units
Convert J⋅mol−1 to kJ⋅mol−1
This means a downward-sloping straight line on lnk against 1/T is expected. The activation energy is found from the magnitude of the gradient multiplied by R.
Two-temperature Arrhenius checkpoint
For two-temperature questions, set up the ratio before substituting numbers. This keeps the sign of Ea under control when the second temperature is higher.
Step
What to write
Check
1
Convert both temperatures to kelvin.
A Celsius value must never appear inside 1/T.
2
Put the larger-temperature run as T2 if its rate constant is larger.
Usually k2>k1, so ln(k2/k1) should be positive.
3
Use ln(k2/k1)=−REa(T21−T11)
4
Convert J⋅mol−1 to kJ⋅mol−1
Worked check: if temperature rises and k increases, the left side ln(k2/k1) is positive while (1/T2−1/T1) is negative. The minus sign in the equation is what makes the calculated activation energy positive.
Misconception check: a negative bracket in the two-point equation is not an error. A negative activation energy usually means the sign convention, temperature order, or units have been mishandled.
4 Mechanisms and Rate-Determining Step (RDS)
Link overall rate law to mechanism:
Identify slow step (RDS); its molecularity often reflects orders.
For fast pre-equilibria, express intermediate concentrations using equilibrium constants, substitute into rate expression.
Ensure proposed mechanism matches stoichiometry.
Example: Reaction between NOX2 and CO. If rate law is rate = k[NOX2]X2, mechanism might involve dimerisation of NOX2 as slow step followed by fast reaction with CO.
RDS mechanism checkpoint
When a mechanism is proposed, test it against the experimental rate law before accepting it.
Mechanism feature
What to check
Answer move
Common trap
Slow step contains only reactants from the overall equation
Its reacting particles should match the rate-law concentration terms.
Write the rate law from the slow step, then compare with the experimental law.
Assuming every overall reactant appears in the rate law.
Slow step contains an intermediate
The intermediate must be removed using a fast previous equilibrium or another given relationship.
Substitute the intermediate concentration before comparing orders.
Leaving an intermediate in the final rate law.
A species appears in a fast step after the RDS
It should not control the rate unless it affects an earlier equilibrium.
Explain that the slow step has already limited the rate.
Including a later fast-step reactant in the rate law.
The proposed steps add to the overall equation
Cancel intermediates and catalysts from both sides.
Check stoichiometry after deriving the rate law.
Matching the rate law but forgetting that the mechanism must also add up.
Worked check: if the slow step is NOX2+NOX2NX2OX4, then the rate law predicted by that elementary slow step is
rate=k[NOX2]2.
That is consistent with an experimental second order in NOX2. It does not make CO part of the rate law if CO reacts only in a later fast step.
Misconception check: the balanced overall equation is a stoichiometry statement. The rate law is a mechanism-and-data statement.
Discuss adsorption-reaction-desorption sequence for heterogeneous catalysts, referencing surface area and poisoning effects.
In exam responses, pair catalyst explanations with an energy-profile sketch showing uncatalysed and catalysed pathways: label lower Ea, keep ΔH/ΔG unchanged, and describe adsorption → surface reaction → desorption for heterogeneous systems.
Catalyst answer checkpoint
When a question asks how a catalyst affects rate, separate the energy-profile statement from the mechanism statement.
Question cue
What changes
What does not change
Common trap
"Explain why rate increases"
A lower activation-energy pathway is available, so a larger fraction of particles have enough energy to react.
The enthalpy change and overall reaction equation stay the same.
Saying the catalyst gives particles more energy.
"Draw an energy profile"
The catalysed curve has a lower peak than the uncatalysed curve.
Reactant and product energy levels stay fixed.
Moving the product level and changing ΔH.
"Heterogeneous catalyst"
Reactants adsorb on the surface, bonds weaken, reaction occurs, then products desorb.
The catalyst is regenerated at the end.
Writing only "provides a surface" without adsorption and desorption.
"Equilibrium system"
Forward and reverse reactions both become faster.
Kc, ΔG, and equilibrium position do not change at the same temperature.
Saying more product is formed at equilibrium.
Worked check: for a catalysed exothermic reaction, draw the catalysed pathway with a lower maximum than the uncatalysed pathway, but keep the products below the reactants by the same vertical amount. The catalyst lowers Ea; it does not change ΔH.
Hydrogen peroxide: catalase substrate commonly used in kinetics-rate demonstrations.
For catalase-linked rate questions, track how quickly HX2OX2 disappears (or OX2 appears) as temperature or catalyst loading changes.
6 Worked Example
Question:
The decomposition of NX2OX5 in gas phase follows first-order kinetics. At 298K, k=3.46×10−5s−1; at 318K, k=1.16×10−4s−1. Calculate the activation energy.
State activation energy as 48kJ⋅mol−1 to two significant figures.
7 Practical Contexts
Clock reactions: Measure time for colour change; use reciprocal time as rate proxy.
Gas collection: Monitor volume evolved vs time (link to mole concept).
Colorimetry: Track absorbance of coloured species to determine concentration (Beer-Lambert law).
Hydrolysis follow-up task: Aspirin hydrolysis can be treated as a practical rate-comparison system when concentration is tracked by titration or spectroscopy.
When planning experiments (Paper 4), mention control of temperature (water bath), use of data loggers, and replicates for reliability.
Practical rate proxy checkpoint
Before calculating order from practical data, check whether the measurement is a rate, a rate proxy, or only a final yield. This prevents students from comparing the wrong numbers.
Practical setup
Use as rate measure
What must stay fixed
Common trap
Clock reaction with a fixed colour-change endpoint
t1, where t is the time to the same visible endpoint
Same total volume, same endpoint amount, same temperature
Comparing raw time directly, where shorter time means faster rate
Gas collection against time
Initial gradient of gas volume against time
Same gas measured, same pressure or collection setup, same temperature
Comparing final gas volume when the question asks about initial rate
Mass loss against time
Initial gradient of mass against time
Same balance setup, same open vessel arrangement, same temperature
Using the average rate over the whole run when only the initial rate is needed
Colorimetry
Initial gradient of concentration or absorbance against time
Same wavelength, same cuvette path length, same calibration approach
Treating absorbance as concentration without a consistent calibration or proportionality statement
Worked check: if two clock experiments reach the same colour change after 40s and 20s, the second run is faster. Compare 201 with 401, not 20 with 40, so the rate proxy doubles.
Misconception check: a larger final amount of product does not automatically mean a faster initial rate. Kinetics questions usually need the early slope or a fixed-endpoint time comparison.
Aspirin: ester hydrolysis substrate for temperature-dependent kinetics comparisons.
8 Common Missteps
Assuming order equals stoichiometric coefficient without experimental evidence.
Forgetting to convert minutes to seconds when calculating rate constants.
Failing to justify why catalysts do not affect ΔG but lower Ea.
Mixing up rate and rate constant when temperature changes: k changes; order remains constant.
9 Quick Drills
Experimental data: When [A] doubles and rate quadruples while [B] stays constant, deduce orders and write rate law.
Plot ln[A] vs time for sample first-order data; extract k and calculate half-life.
Explain why poisoning of heterogeneous catalysts reduces rate, referencing adsorption theory and industrial examples (e.g. lead poisoning Pt in catalytic converters).
Common exam mistakes
Mistake: Reading off stoichiometric coefficients as reaction orders without experimental evidence - orders must always be determined from data, not from the balanced equation.
Mistake: Carrying rate constant units as mol⋅L−1⋅s−1 regardless of order - units depend on overall order and must be derived each time from the rate equation.
Mistake: Confusing the rate constant k with the rate of reaction - k is temperature-dependent but concentration-independent; rate changes when concentrations change even at fixed temperature.
Mistake: Plotting [A] vs time and concluding first-order from a smooth curve - the correct test is a linear ln[A] vs time graph with constant half-life, not a visual impression.
Mistake: Forgetting to convert temperature to Kelvin in the Arrhenius equation - using Celsius gives a completely wrong activation energy.
Mistake: Claiming that a catalyst changes the enthalpy change ΔH or equilibrium position - catalysts lower Ea only; ΔH and the position of equilibrium are unchanged.
Mistake: Giving only "provides a surface" for heterogeneous catalysis without describing the adsorption-reaction-desorption sequence - full mechanism marks require all three stages.
Frequently asked questions
Is reaction kinetics tested in Paper 1? Yes. Multiple-choice questions can ask you to identify correct rate expressions, read half-lives from graphs, or select the correct units for k. Always apply the experimental evidence rule - never assume order equals coefficient.
What is the difference between reaction rate and rate constant? Rate of reaction is the change in concentration per unit time and varies with concentration. The rate constant k is a proportionality constant that depends only on temperature (via the Arrhenius equation); it does not change when you alter concentration.
How do I know which graph to plot for determining reaction order? Plot [A] vs time (linear -> zero order), ln[A] vs time (linear -> first order), or 1/[A] vs time (linear -> second order). For a first-order reaction, a constant half-life across the concentration-time curve is also diagnostic.
Does a catalyst affect the value of Kc? No. A catalyst speeds up both the forward and reverse reactions equally, so the equilibrium position and Kc are unchanged. Only a temperature change alters Kc.
Struggling with Reaction Kinetics? Our H2 Chemistry tuition programme covers this topic with structured practice, Paper 4 practical drills, and worked exam solutions.
