Did you know? The rate constant k can have units mol dm⁻³ s⁻¹, s⁻¹ or dm³ mol⁻¹ s⁻¹ depending on the reaction’s kinetic order. These are dictated by the concentration powers in its rate equation, not arbitrary symbols to memorise.
This original Hougang JC1 H2 Chemistry guide follows the 2027 SEAB 9476 Reaction Kinetics syllabus. It derives the rate-constant units for zero-, first- and second-order laws, tests mixed-order equations and works entirely with clearly fictional initial-rate records. Integrated rate equations are not required for the H2 learning outcome.
EDUKATESG · HOUGANG · JC1 H2 CHEMISTRY
14 chapters · Rate units table · Transfer practice · FAQs · Syllabus
Complete chapter index
- 1. Why the same symbol k can carry different units
- 2. Start with the correct concentration-per-time rate
- 3. Use kinetic order, not chemical equation coefficients
- 4. Derive the general unit formula for k
- 5. Zero order: the constant itself has rate units
- 6. First order: concentration cancels from the constant
- 7. Second order: k carries inverse-concentration units
- 8. A mixed second-order law has the same total k units
- 9. Zero order in one reactant does not alter total units
- 10. Third overall order as an optional extension
- 11. Use initial-rate ratios before computing k
- 12. The units of k are an independent plausibility test
- 13. Keep rate changes distinct from changes in the rate constant
- 14. Connect first-order k to half-life with clear syllabus scope
1. Why the same symbol k can carry different units
A first-order reaction and a second-order reaction can both be written using a rate constant named k, but k does not have one universal unit. In a zero-order rate law, k has units of concentration per time; in a first-order law it has units of inverse time; in a second-order law it has units of inverse concentration per time. These differences are forced by dimensional balance.
Consider rate=k[A], with concentration measured in mol dm⁻³ and time in seconds. The concentration term already provides mol dm⁻³, so k must provide s⁻¹. For rate=k[A]², the concentration factor supplies mol² dm⁻⁶, leaving k to supply dm³ mol⁻¹ s⁻¹ if the result is still to have the correct rate units.
The 2027 H2 Chemistry syllabus 9476 Section 8 includes rate equations, orders and calculating k from supplied initial-rate data. Units are therefore a meaningful independent audit of a proposed equation, not an arbitrary decoration after a calculation.
Student question: Can k have different units in different reactions?
Answer standard: Yes. The overall order fixes k’s dimensions once the rate and concentration units are chosen.
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2. Start with the correct concentration-per-time rate
In a familiar kinetic experiment, reaction rate is expressed as a change in concentration divided by elapsed time. If concentration has unit mol dm⁻³ and time has unit s, then rate is measured in mol dm⁻³ s⁻¹. This should be established before substituting any numbers.
An original paper record might show [A] decreasing by 0.020 mol dm⁻³ over ten seconds. The average magnitude of its disappearance rate in that interval is 0.0020 mol dm⁻³ s⁻¹. A student’s numerical answer of 0.0020 without units is incomplete, and writing the concentration unit without the seconds incorrectly drops the time dependence.
A full reaction-rate definition can include stoichiometric factors when comparing the formation and consumption of different species. That detail changes the normalised numerical rate, not the fundamental concentration/time dimensions. The dimensional rule for k is based on the units used for the rate as actually defined in the question.
Student question: What is the unit of the rate from a molarity-versus-seconds graph?
Answer standard: mol dm⁻³ s⁻¹.
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3. Use kinetic order, not chemical equation coefficients
The order in a rate equation is an experimentally determined concentration exponent or an exponent justified by a given elementary-step mechanism. In rate=k[A]^m[B]^n, m is the partial order in A and n is the partial order in B. The overall order is m+n.
An overall balanced reaction may have a coefficient two before A, but that does not prove its observed rate is second order in A. A multistep reaction can have a slow elementary step and faster steps; the observed kinetics are linked to its mechanism. The H2 syllabus permits simple cases with partial orders zero, one or two, rather than universal assignment by chemical stoichiometry.
The concentration response helps find the exponents. Doubling a zero-order species changes nothing in the ideal rate model. Doubling a first-order species doubles rate; doubling a second-order species multiplies rate by four, with all other conditions fixed. A tutor should ask for those causal meanings before teaching the unit formula.
Student question: Where does the overall order of a rate law come from?
Answer standard: The sum of the justified concentration powers, not automatically the balanced equation.
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4. Derive the general unit formula for k
If rate=k[A]^m[B]^n and all species concentrations are in mol dm⁻³, the units of k are the rate units divided by a total concentration factor to the power m+n. This can be written [k]=(mol dm⁻³ s⁻¹)/(mol dm⁻³)^(m+n).
Let r=m+n. Cancelling the concentration powers gives [k]=(mol dm⁻³)^(1−r) s⁻¹. The exponent 1−r is the entire dimensional reason zero-, first- and second-order k values have different units.
It is important to label the assumptions. If a question uses concentration in mol L⁻¹ and time in minutes, a different but equivalent unit expression may appear. Always use the specific units supplied in the data and make conversions consistently rather than writing s⁻¹ automatically when the time values are in minutes.
Student question: What general units does k have for overall order r?
Answer standard: (mol dm⁻³)^(1−r) s⁻¹ under the stated convention.
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5. Zero order: the constant itself has rate units
For rate=k[A]^0, the A factor equals one. Thus rate=k and the units are mol dm⁻³ s⁻¹. A fictional rate of 2.0×10⁻³ mol dm⁻³ s⁻¹ gives k=2.0×10⁻³ mol dm⁻³ s⁻¹ under this particular zero-order law.
An illustrative table might list [A]=0.10, 0.20 and 0.40 mol dm⁻³, with all three initial rates equal at 0.0020 mol dm⁻³ s⁻¹. That supports zero order in A over the tested region. It does not prove A has no chemical role, nor does it imply the same rate law applies at concentrations far outside those measured.
A zero-order behaviour can arise, for example, when a catalyst surface or another part of a chemical system is effectively saturated. A syllabus answer should discuss the rate-law result and the provided mechanism evidence, rather than invent a specific catalyst or claim that zero order is impossible because a reactant must be consumed.
Student question: What are zero-order k units?
Answer standard: mol dm⁻³ s⁻¹.
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6. First order: concentration cancels from the constant
For rate=k[A], k=rate/[A]. Dividing mol dm⁻³ s⁻¹ by mol dm⁻³ cancels the concentration dimension and gives s⁻¹. This is why a first-order rate constant is an inverse-time quantity under the familiar convention.
In an invented problem, the initial rate is 0.0050 mol dm⁻³ s⁻¹ when [A]=0.10 mol dm⁻³. Then k=0.0050/0.10=0.050 s⁻¹. If [A] doubles to 0.20 mol dm⁻³ at the same temperature and mechanism, the expected rate is 0.010 mol dm⁻³ s⁻¹, while k remains 0.050 s⁻¹.
Do not copy the original rate units onto k after dividing. A numerically plausible ‘0.050 mol dm⁻³ s⁻¹’ answer would have the wrong dimensions for this first-order model. Asking the learner to cross out the common mol dm⁻³ units explicitly can prevent the mistake.
Student question: For rate=k[A], rate=0.0050 M s⁻¹ and [A]=0.10 M, what is k?
Answer standard: 0.050 s⁻¹.
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7. Second order: k carries inverse-concentration units
For rate=k[A]², the concentration term has units mol² dm⁻⁶. Dividing the rate units mol dm⁻³ s⁻¹ by it yields dm³ mol⁻¹ s⁻¹. The reciprocal concentration factor is necessary because two concentration powers have already appeared.
Suppose the initial rate is 0.0020 mol dm⁻³ s⁻¹ at [A]=0.10 mol dm⁻³. Then k=0.0020/(0.10)²=0.20 dm³ mol⁻¹ s⁻¹. At [A]=0.20 mol dm⁻³, rate becomes 0.20×0.040=0.0080 mol dm⁻³ s⁻¹, four times the original value as second order predicts.
Do not confuse an inverse-concentration dimension with a negative concentration. The unit dm³ mol⁻¹ is merely the reciprocal of mol dm⁻³. It describes how k must be expressed so that multiplication by the concentration squared returns the correct reaction rate dimension.
Student question: What are second-order k units?
Answer standard: dm³ mol⁻¹ s⁻¹.
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8. A mixed second-order law has the same total k units
Compare rate=k[A]² with rate=k[A][B]. The first is second order in A alone, while the second is first order in each of two reactants. Both are overall order two and therefore both have dm³ mol⁻¹ s⁻¹ as the correct k unit under the same conventions.
The rate responses differ. Doubling B leaves rate=k[A]² unchanged if B is not part of that law, whereas doubling B doubles rate=k[A][B] when A and k are fixed. Equal k units are therefore not enough to identify a unique mechanism or to say which reactant controls the rate.
An original mixed-law problem gives rate=0.0060 M s⁻¹ at [A]=0.20 M and [B]=0.10 M. Then k=0.0060/(0.20×0.10)=0.30 dm³ mol⁻¹ s⁻¹. Both the numerical division and concentration-power cancellation confirm the same result.
Student question: What is k for rate=k[A][B] using the supplied 0.0060, 0.20 and 0.10 values?
Answer standard: 0.30 dm³ mol⁻¹ s⁻¹.
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9. Zero order in one reactant does not alter total units
Consider rate=k[A]²[B]^0. The power of B is zero and [B]^0=1, so B contributes no concentration dimension and no rate dependence over the range where this model applies. The total order is two and k again has dm³ mol⁻¹ s⁻¹ units.
A common mistake is to count two chemical species and conclude that the overall order must be at least two. That is not how kinetic order works. A term raised to zero is dimensionless, and a reactant may participate through an earlier equilibrium or other step while not appearing in the observed rate expression.
For dimensional analysis, remove zero-order concentration factors first. Then sum only the actual exponents and derive k. The resulting units reflect the mathematical rate law, not the number of letters used to name chemical species in the original balanced equation.
Student question: Does [B]^0 contribute any dimensions to the units of k?
Answer standard: No, since it equals one.
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10. Third overall order as an optional extension
A rate law rate=k[A]²[B] has overall order three, even though neither individual exponent exceeds two. The concentration factor has total dimensions (mol dm⁻³)³. Therefore [k]=(mol dm⁻³)⁻² s⁻¹, conventionally dm⁶ mol⁻² s⁻¹.
For example, with [A]=0.20 M, [B]=0.10 M and k=0.50 dm⁶ mol⁻² s⁻¹, rate=0.50×(0.20)²×0.10=0.0020 mol dm⁻³ s⁻¹. The final unit cancels correctly. A student should see that the same systematic procedure works even beyond total order two.
This is a dimensional extension, not a requirement to infer that every three-reactant collision is likely or that a complicated mechanism has a particular order. A full rate law requires kinetic evidence or a justified rate-determining mechanism, with the syllabus’s stated simple-case limits respected.
Student question: What units correspond to an overall third-order rate constant?
Answer standard: dm⁶ mol⁻² s⁻¹.
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11. Use initial-rate ratios before computing k
An initial-rates experiment changes one concentration at a time while keeping temperature and other factors fixed. If doubling A quadruples rate, that supports m=2 in a power-law model because 2²=4. If doubling B leaves rate unchanged, that supports n=0. The law becomes rate=k[A]².
Only after identifying the orders should the learner substitute measured values to obtain k. If they assume first order without testing, they can generate a numerical k from one row but it will generally disagree with the value from a second row. This inconsistency is useful evidence that their proposed rate law is wrong.
A tutor can ask for two independent k calculations from different experiments. Under the same kinetics and precision assumptions, the results should agree appropriately. If they do not, check the data, exponents and any experimental limitations before concluding that the reaction constant itself spontaneously changed.
Student question: What orders are suggested when doubling A quadruples rate and doubling B leaves it unchanged?
Answer standard: Second in A and zero in B.
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12. The units of k are an independent plausibility test
Suppose a student proposes rate=k[A] but reports k in dm³ mol⁻¹ s⁻¹. Multiplying that by [A] in mol dm⁻³ gives s⁻¹, not mol dm⁻³ s⁻¹. The rate law and unit cannot both be correct under the stated conventions.
Likewise a proposed second-order law with k in s⁻¹ leaves an extra concentration factor on the calculated rate. This type of dimensional mismatch can reveal a mistake even if the numeric answer looks sensible to a calculator. Units describe the physical quantity being evaluated, not optional annotation.
However, correct dimensions do not prove a mechanism. Different rate laws with the same overall order share k units, and several physical models might have consistent dimensions. The student still needs experimental or mechanism evidence to support the partial orders and the proposed chemical reaction path.
Student question: Can dimensionally correct k units alone prove a proposed chemical mechanism?
Answer standard: No. They check consistency, while kinetics and mechanism require other evidence.
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13. Keep rate changes distinct from changes in the rate constant
When concentration is altered but temperature and mechanism remain unchanged, k is a constant for the stated rate law. Rate can double or quadruple because the concentration factors change. It is misleading to say k must have doubled simply because the measured rate doubled.
Temperature changes, catalysts and mechanism changes can affect the numerical value of k. A catalyst may provide a different pathway with a different activation energy, while a temperature change alters molecular thermal energy populations. Under a single rate-law expression, the units of k still follow its overall order and the chosen unit conventions.
A tutor should have learners write what was changed in a paper scenario. If only [A] doubled, use the existing k and rate exponent. If temperature changed, the question may require fresh kinetic data or a supplied model. The source of a rate change matters as much as its numerical magnitude.
Student question: Does doubling [A] automatically double k?
Answer standard: No, not at fixed temperature with unchanged mechanism.
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14. Connect first-order k to half-life with clear syllabus scope
The 2027 H2 syllabus includes recognising that a first-order reaction’s half-life does not depend on initial reactant concentration. A common mathematical relation in the ideal first-order model is t½=ln2/k, with a dimensionless ln2 and k measured in inverse time. The resulting half-life therefore has units of time.
If k=0.050 s⁻¹, the model gives t½≈0.693/0.050≈13.9 s. The same result applies at different initial [A] for the stated ideal first-order reaction, which illustrates its half-life property. A second-order reaction does not generally share the same concentration-independent half-life.
The official syllabus states that integrated forms of rate equations are not required. Therefore the tutor can use first-order half-life knowledge without demanding a lengthy integrated-rate-law derivation. The goal of this article remains unit analysis and correct interpretation of simple rate constants.
Student question: Why does t½=ln2/k have time units when k is first order?
Answer standard: ln2 is dimensionless, so dividing by s⁻¹ gives seconds.
Worked evidence: rate-constant dimensions
The table uses fictional data, concentrations in mol dm⁻³, rates in mol dm⁻³ s⁻¹ and time in seconds. It is not a record of a physical experiment.
| Order | Rate law | Total order | Units of k | Fictional calculation |
|---|---|---|---|---|
| Zero-order | rate=k | 0 | mol dm⁻³ s⁻¹ | Example k=0.0020 M s⁻¹ |
| First-order | rate=k[A] | 1 | s⁻¹ | 0.0050/0.10 = 0.050 s⁻¹ |
| Second-order | rate=k[A]² | 2 | dm³ mol⁻¹ s⁻¹ | 0.0020/(0.10²)=0.20 |
| Mixed second | rate=k[A][B] | 2 | dm³ mol⁻¹ s⁻¹ | 0.0060/(0.20×0.10)=0.30 |
| Second + zero | rate=k[A]²[B]^0 | 2 | dm³ mol⁻¹ s⁻¹ | Zero B term contributes no power |
| Optional third | rate=k[A]²[B] | 3 | dm⁶ mol⁻² s⁻¹ | 0.0020/(0.20²×0.10)=0.50 |
| Unit conversion | k=0.050 s⁻¹ | 1 | 3.0 min⁻¹ | Multiply inverse-seconds value by 60 for inverse minutes |
For overall order r, [k]=(mol dm⁻³)^(1−r) s⁻¹. Use this after deducing the partial orders from rate evidence. A correct unit checks algebra but does not uniquely determine a mechanism.
Four independent transfer checks
- Given rate=k[A], rate 0.0050 M s⁻¹ and [A]=0.10 M, compute k=0.050 s⁻¹.
- Given rate=k[A]², rate 0.0020 M s⁻¹ and [A]=0.10 M, compute k=0.20 dm³ mol⁻¹ s⁻¹.
- Given rate=k[A][B], [A]=0.20 M, [B]=0.10 M and rate 0.0060 M s⁻¹, compute k=0.30 dm³ mol⁻¹ s⁻¹.
- For the optional overall third-order rate=k[A]²[B], find k units dm⁶ mol⁻² s⁻¹ and justify why multiplying by [A]²[B] restores the rate unit.
Tutor checklist
- Confirm 2027 H2 Chemistry 9476 Section 8 and the student’s current kinetics chapter.
- State the rate units before solving for k.
- Deduced kinetic powers must come from rate evidence or justified mechanism, not overall equation coefficients.
- Sum partial orders before deriving k dimensions.
- Use separate zero, first, second and mixed-order worked examples.
- Check at least two initial-rate rows with the same proposed k.
- Distinguish rate changes from changes in k at fixed temperature and mechanism.
- Mark overall third-order dimensions and integrated-law derivations as optional depth rather than unqualified H2 requirements.
- Finish with an unfamiliar rate law and check the learner’s units as well as numbers.
Frequently asked questions
What is the zero-order k unit?
mol dm⁻³ s⁻¹ with molar concentration and seconds.
What is the first-order k unit?
s⁻¹.
What is the second-order k unit?
dm³ mol⁻¹ s⁻¹.
Do rate=k[A]² and rate=k[A][B] have the same k units?
Yes. Both have overall order two, although their partial orders differ.
Why is k not always expressed in s⁻¹?
Concentration powers in the rate law change the factor needed to return concentration/time dimensions.
Can concentration change rate without changing k?
Yes. At fixed temperature and mechanism, rate can change through the concentration factors while k stays constant.
Are integrated rate equations required for H2 Chemistry 9476?
No. The syllabus explicitly excludes integrated forms, while requiring first-order half-life knowledge.
How should a mixed-order k be calculated?
Divide measured rate by the product of all relevant concentration powers, then derive the remaining units.
Related Chemistry guides
- JC2 H2 Chemistry | Initial Rate Law and Reaction Order
- JC1 H2 Chemistry | Temperature, Catalysts and Rate
- Hougang Chemistry Tuition and Tutor Library
Bring the student’s examination year, current kinetics notes and a recent incorrect rate-law attempt when discussing tutorial support. Confirm teacher, class availability, prices and location directly. No result is guaranteed.
Official syllabus and model boundaries
- SEAB 2027 H2 Chemistry 9476 Section 8 Reaction Kinetics
- SEAB 2027 school-candidate syllabus directory
- Immutable eduKateSG Clementi small-group teaching reference
- Hougang Chemistry Tuition and Tutor Library
The official syllabus includes rate equations, kinetic orders, initial-rate calculations, rate constants and first-order half-lives; integrated forms are not required. The numerical data are original hypothetical practice. Every unit derivation assumes stated rate and concentration dimensions and must be converted when a different unit convention is used.