A reaction that is first order with respect to substance A and first order with respect to substance B is second order overall. Its rate equation is rate = k[A][B], and if concentrations are measured in mol dm⁻³ and rate in mol dm⁻³ s⁻¹, the rate constant k has units dm³ mol⁻¹ s⁻¹. You can deduce all of this from a small initial-rates table without knowing a detailed reaction mechanism.
The difficulty is that a student may correctly see a doubled reaction rate and still draw the wrong conclusion. If doubling one reactant concentration doubles the rate, that reactant is first order; if it quadruples the rate, second order; if rate stays unchanged under the specified conditions, zero order. But the order with respect to each reactant is not the same as the total order, and the unit of k follows the total order, not the chemical equation’s coefficient unless an elementary rate law is explicitly justified.
This H1 Chemistry Tutor Hougang lesson follows the 2027 Singapore–Cambridge H1 Chemistry syllabus 8873 and focuses on the part of reaction kinetics that often causes avoidable marks to be lost: finding individual orders, calculating k, checking units, and predicting a new initial rate. The quick information is:
- General rate equation: rate = k[A]ᵐ[B]ⁿ, where m and n are experimentally supported orders.
- Overall order: m + n, not automatically the sum of coefficients in the balanced reaction.
- Zero-order k units: mol dm⁻³ s⁻¹, when rate uses concentration per second.
- First-order k units: s⁻¹.
- Second-order k units: dm³ mol⁻¹ s⁻¹.
- Doubling test: Rate unchanged → order zero; doubled → order one; quadrupled → order two, for one concentration changed and all relevant other conditions held fixed.
- Important scope: H1 8873 permits simple rate equations with individual exponents 0, 1 or 2; integrated rate-equation forms are explicitly not required.
What is the rate of reaction?
A reaction rate expresses how quickly a reactant concentration decreases or a product concentration increases, after accounting for the appropriate stoichiometric coefficient if a common reaction rate is defined.
Concentration in H1 Chemistry is often measured in mol dm⁻³. If a concentration changes by 0.0200 mol dm⁻³ over 10.0 s in a simple interval, its average rate of change in magnitude is 0.00200 mol dm⁻³ s⁻¹.
An initial rate is the rate close to the start of the reaction, where starting concentrations are specified. It is not the average rate over the entire time the reaction runs.
Rate is not the same as rate constant
The reaction rate depends on reactant concentrations through the rate equation. For example, if rate = k[A][B], increasing [A] at fixed [B], temperature and catalyst can increase the rate.
The rate constant k is a proportionality factor for the stated rate law at specified conditions. It is related to temperature and the reaction pathway, and can change when temperature or the catalyst changes.
A student who says “doubling [A] doubles k” in a fixed-temperature rate-law question has confused the measured reaction rate with the constant used to calculate it.
Why a balanced chemical equation is insufficient
Suppose a reaction is represented by A + 2B → products. You cannot automatically conclude rate = k[A][B]² from the overall balanced equation.
An overall reaction can involve multiple steps, and its rate law is generally an experimental kinetic relationship. Stoichiometric coefficients and kinetic orders are different things.
The revised H1 syllabus restricts examinable rate-law constructions to simple single-step cases with individual orders zero, one or two, but students should still respect the data supplied rather than guessing an exponent from a familiar-looking equation.
The initial-rates method
In a series of experiments, change the initial concentration of one reactant while keeping the others and the relevant conditions constant. Record the initial rate.
Compare two rows where only one concentration changes. The ratio of rates tells how strongly the rate depends on that reactant’s concentration.
Repeat the comparison for another reactant. Once individual orders are known, write the full rate equation and calculate k from any reliable experimental row.
A four-row practice dataset
Consider a hypothetical reaction involving reactants A and B at constant temperature. The supplied initial rates are measured in mol dm⁻³ s⁻¹.
Experiment 1: [A] = 0.100 mol dm⁻³; [B] = 0.200 mol dm⁻³; initial rate = 0.00600 mol dm⁻³ s⁻¹.
Experiment 2: [A] = 0.200 mol dm⁻³; [B] = 0.200 mol dm⁻³; initial rate = 0.0120 mol dm⁻³ s⁻¹.
Experiment 3: [A] = 0.100 mol dm⁻³; [B] = 0.400 mol dm⁻³; initial rate = 0.0120 mol dm⁻³ s⁻¹.
Experiment 4: [A] = 0.200 mol dm⁻³; [B] = 0.400 mol dm⁻³; initial rate = 0.0240 mol dm⁻³ s⁻¹.
The data are invented for teaching; they are not measurements from a named laboratory experiment.
Step one: compare Experiments 1 and 2
Between Experiments 1 and 2, [B] stays at 0.200 mol dm⁻³. Only [A] changes, from 0.100 to 0.200 mol dm⁻³: a factor of two.
The initial rate changes from 0.00600 to 0.0120 mol dm⁻³ s⁻¹: also a factor of two.
For rate = k[A]ᵐ[B]ⁿ, doubling [A] gives 2ᵐ times the rate when other factors are fixed. Here 2ᵐ = 2, so m = 1. The reaction is first order with respect to A.
Step two: compare Experiments 1 and 3
Between Experiments 1 and 3, [A] stays constant at 0.100 mol dm⁻³. [B] doubles from 0.200 to 0.400 mol dm⁻³.
The rate doubles from 0.00600 to 0.0120 mol dm⁻³ s⁻¹. Therefore 2ⁿ = 2 and n = 1.
The reaction is also first order with respect to B. The full rate law is rate = k[A][B].
Step three: find the overall order
The individual orders are one with respect to A and one with respect to B. Their sum is two.
The reaction is second order overall. This does not mean it is second order with respect to A alone or B alone.
The distinction matters in answer wording. If a question asks “order with respect to B”, answer one; if it asks “overall order”, answer two.
Step four: calculate the rate constant
From Experiment 1, rate = k[A][B] means k = rate/([A][B]).
Substitute the data: k = 0.00600/((0.100)(0.200)) = 0.00600/0.0200 = 0.300.
Before reporting a final answer, derive the units. The bare number 0.300 is incomplete when the question requests a rate constant with units.
Step five: derive the units of k
The rate has units mol dm⁻³ s⁻¹. The product [A][B] has units (mol dm⁻³)², or mol² dm⁻⁶.
Divide: (mol dm⁻³ s⁻¹)/(mol² dm⁻⁶) = dm³ mol⁻¹ s⁻¹.
Therefore k = 0.300 dm³ mol⁻¹ s⁻¹ in this practice dataset. The derived unit agrees with second order overall.
Verify k using another row
Use Experiment 4, where [A] = 0.200, [B] = 0.400 and rate = 0.0240.
k = 0.0240/((0.200)(0.400)) = 0.0240/0.0800 = 0.300 dm³ mol⁻¹ s⁻¹.
The same value is obtained because the hypothetical data were constructed to follow one consistent rate law at fixed conditions. In a real dataset, small deviations may occur because of experimental uncertainty.
Checking k in a second row is a powerful audit of both the inferred orders and the arithmetic.
Predict a new initial rate
Suppose a new experiment starts with [A] = 0.150 and [B] = 0.300 mol dm⁻³, at the same temperature and conditions.
Rate = k[A][B] = 0.300 × 0.150 × 0.300 = 0.0135 mol dm⁻³ s⁻¹.
Notice that the rate uses the concentration-per-time units, not the units of the rate constant. Keeping rate and k units separate prevents an error in the final line.
The shortcut using ratios
Instead of recalculating k, you could compare the new conditions with Experiment 1. [A] increases by a factor of 1.5 and [B] increases by a factor of 1.5.
Since the reaction is first order in each, the new rate is 1.5 × 1.5 = 2.25 times the original rate. Thus 0.00600 × 2.25 = 0.0135 mol dm⁻³ s⁻¹.
Both methods should agree. If they do not, check concentration ratios or a misplaced exponent.
What zero order means
For a reactant A that is zero order, the rate-law factor [A]⁰ equals one within the range where the rate law applies and A is present.
Changing [A] alone may therefore have no effect on the initial rate under those conditions.
It does not mean A is chemically irrelevant or that a reaction continues at the same rate after all A is gone. Zero order describes measured dependence over a specified regime, not an inexhaustible reaction.
A zero-order practice dataset
Consider a different hypothetical reaction at fixed temperature with [B] constant.
When [A] = 0.100, the measured initial rate is 0.00400 mol dm⁻³ s⁻¹. When [A] doubles to 0.200, the initial rate remains 0.00400.
If the conditions are otherwise comparable, the rate is zero order with respect to A in the tested range.
The resulting rate law may still depend on another reactant, such as rate = k[B], so zero order in A is not necessarily zero order overall.
What first order means
If a reactant is first order, doubling its concentration at fixed other conditions doubles the rate. Tripling its concentration triples the rate.
A first-order rate law in one reactant may have the form rate = k[A]. The rate constant k then has units s⁻¹ when rate is mol dm⁻³ s⁻¹.
For simple first-order concentration–time data, a constant fraction is consumed over equal time intervals and the half-life is independent of the initial concentration.
What second order with respect to one reactant means
A different rate law might be rate = k[A]². In this case, doubling [A] multiplies the rate by four.
Tripling [A] multiplies the rate by nine. This is not the same as the earlier rate = k[A][B] example, even though both laws are second order overall.
A student must distinguish “second order in A” from “second order overall”. Either can be correct depending on the experimental relationship.
Another practice table: A squared, B zero
Take an original dataset where Experiment 1 has [A] = 0.100, [B] = 0.200 and rate = 0.00500 mol dm⁻³ s⁻¹.
Experiment 2 doubles [A] to 0.200 with [B] unchanged, and the rate becomes 0.0200, four times as large.
Experiment 3 doubles [B] to 0.400 with [A] unchanged at 0.100, but the rate stays 0.00500. Therefore the rate law is rate = k[A]²[B]⁰, usually simplified to rate = k[A]².
Calculate k for this second example
k = rate/[A]² = 0.00500/(0.100)² = 0.00500/0.0100 = 0.500.
The total order is two, so the units are dm³ mol⁻¹ s⁻¹. The complete result is k = 0.500 dm³ mol⁻¹ s⁻¹.
This is a useful comparison: rate = k[A][B] and rate = k[A]² have different concentration dependencies but the same total order and the same k units.
Why zero-order k has rate units
For an overall zero-order rate law, rate = k. No concentration factor remains in the denominator when k is isolated.
Therefore k has the same units as rate: mol dm⁻³ s⁻¹, provided rate is measured in concentration per second.
This is a good unit checkpoint. A student who reports s⁻¹ for zero order has probably memorised first-order units without deriving them.
Why first-order k has reciprocal-time units
For rate = k[A], k = rate/[A].
Divide mol dm⁻³ s⁻¹ by mol dm⁻³, leaving s⁻¹.
The concentration units cancel completely, which is why the familiar first-order rate constant has inverse-time units.
Why second-order k carries volume per mole
For total order two, rate = k[A]² or k[A][B]. Either way the concentration product contributes units mol² dm⁻⁶.
Dividing rate by that product leaves dm³ mol⁻¹ s⁻¹. The rate constant’s units reflect how many powers of concentration are needed in the rate law.
You do not need to memorise a separate unexplained list if you can derive k = rate divided by the concentration factors.
A general units expression
For a total order r when concentration is in mol dm⁻³ and time is in seconds, the rate constant’s units can be written as (mol dm⁻³)^(1−r) s⁻¹.
At r = 0, that gives mol dm⁻³ s⁻¹. At r = 1, it gives s⁻¹. At r = 2, it gives dm³ mol⁻¹ s⁻¹.
This is a compact unit-checking aid, not a substitute for understanding the rate law. If the rate is expressed with different units, adjust the derived units accordingly.
Why time units also matter
A rate recorded in mol dm⁻³ min⁻¹ is not the same numerical rate as one in mol dm⁻³ s⁻¹ for the same concentration change.
If a rate is given per minute and concentrations in mol dm⁻³, a second-order k derived without converting time will have dm³ mol⁻¹ min⁻¹ units.
Convert minutes to seconds only when the final question requests a per-second constant. A student can be chemically correct about order and still lose marks by silently changing the time unit.
The factor-of-four trap
Suppose Experiment 2 doubles both A and B simultaneously, and the rate becomes four times larger.
You cannot determine from that one comparison alone whether the reaction is first order in each, second order in A and zero in B, or another combination producing the same factor.
Use rows in which one concentration changes at a time, or enough independent comparisons to solve the separate exponents.
The total rate multiplier is evidence, but it does not uniquely identify individual orders without suitable experimental controls.
Comparing rows where both variables change
Sometimes an unfamiliar table does not provide a convenient pair differing in only one concentration.
You can still form rate ratios and use known orders or solve simultaneous simple equations, provided enough independent information exists.
For example, if A and B both double while the rate increases fourfold, and a separate comparison establishes first order with respect to A, the remaining factor of two must be due to B, making B first order.
Do not claim more than the data identify. Missing independent comparisons may leave the individual orders underdetermined.
Initial rate versus average rate
The average rate over a time interval comes from total concentration change divided by elapsed time. The initial rate is the instantaneous gradient near the beginning of the concentration–time graph.
A concentration–time graph may curve because concentrations and therefore rates change as the reaction proceeds.
A tangent at the start estimates the initial gradient. A straight line from the first point to a much later point gives an average gradient, not necessarily the initial one.
Why gradients often have negative signs
A reactant concentration usually decreases over time, so its concentration–time graph has a negative gradient. A rate defined as disappearance of reactant is reported as a positive magnitude: rate = −d[A]/dt for a simple one-to-one disappearance definition.
A product concentration may increase, giving a positive gradient. The stoichiometric coefficients determine how the common reaction rate relates to individual species changes.
In H1 exam responses, read whether the question asks for the signed gradient or the positive reaction rate. They are not automatically the same number.
The role of stoichiometric coefficients
Consider 2A → B. For each mole of B formed, two moles of A disappear.
The common reaction rate can be defined as −(1/2)d[A]/dt = d[B]/dt. A student who uses the raw disappearance gradient of A as the formation rate of B without adjusting by two obtains the wrong comparison.
This distinction helps connect kinetics to the mole concept. The slope of a particular concentration curve and the rate of the whole reaction depend on how the reaction is defined.
Initial-rates data should be controlled
A valid concentration comparison holds temperature, catalyst, measurement method and other relevant conditions constant.
If Experiment 2 doubles [A] but also heats the mixture, a rate increase cannot be assigned solely to the change in [A].
That is why the wording “all other conditions unchanged” matters in a well-designed question. Rate-law deduction is an experimental inference, not simply arithmetic from an arbitrary table.
Temperature changes k
Raising temperature commonly increases the rate constant for an ordinary activated reaction because a larger fraction of molecular encounters can follow the reaction pathway, as described qualitatively through the Boltzmann distribution.
For rate = k[A][B], you could have identical concentrations at two temperatures but different rates because k differs.
When comparing initial-rate experiments to determine orders, do not mix temperatures unless the problem gives sufficient information to account for k’s change.
Catalysts change the reaction pathway
A suitable catalyst can provide a lower-activation-energy route, raising the rate constant for the catalysed pathway under the specified conditions.
It does not change the overall reaction enthalpy or the equilibrium constant at a fixed temperature simply by being present.
The relationship between k, reaction mechanism and temperature is separate from the effect of changing a concentration within one fixed-rate-law dataset.
An important limit: order is empirical
Orders such as zero, one and two in the rate equation are supported by rate measurements over the stated conditions. They are not labels that automatically describe every conceivable concentration range.
A rate law may be valid only within the investigated regime, especially for systems involving catalysts, saturation or changes of mechanism.
For H1 problems, follow the simplified model and data given. For scientific literacy, recognise the limitations of extending a fitted law beyond the range where it was tested.
Why a constant half-life points to first order
For a simple first-order reaction under fixed conditions, the half-life is independent of initial reactant concentration.
A concentration–time curve can show the amount halving from 0.80 to 0.40, then from 0.40 to 0.20, in equal time intervals.
A zero-order reaction instead gives a straight-line concentration decrease under the simple model, while a second-order single-reactant process generally has a different half-life dependence. Integrated rate equations are explicitly outside the named H1 requirement.
The common table-reading mistake
Suppose a student looks at Experiment 1 and Experiment 4 in the first dataset, where both reactants double and rate quadruples, and concludes that A is second order while B is zero order.
That conclusion happens to fit this one comparison but contradicts the two independent comparisons in Experiments 1–2 and 1–3.
The repair is to isolate changes one by one. If both concentrations change, the rate ratio alone does not reveal each individual order.
Another mistake: adding concentration values
For rate = k[A][B], doubling both concentrations does not simply add the increases to produce an arbitrary plus-two effect.
The factors multiply: 2 × 2 = 4. If the reaction were second order in A and first in B, doubling both would multiply the rate by 2² × 2 = 8.
Write the factors using exponents before substituting. That keeps the rate-law algebra under control.
Another mistake: mixing the units of k with rate
Our main practice dataset gave k = 0.300 dm³ mol⁻¹ s⁻¹. The predicted rate for new concentrations was 0.0135 mol dm⁻³ s⁻¹.
If the learner reports rate = 0.0135 dm³ mol⁻¹ s⁻¹, they have copied the constant’s units onto a physically different quantity.
A final unit audit is often as valuable as a numerical calculator check.
A five-mark model solution
Given four initial-rate experiments, first use a pair with constant [B] to find order in A. Then use a pair with constant [A] to find order in B.
Write rate = k[A]ᵐ[B]ⁿ. Find the overall order m+n, calculate k using one experimental row, and derive its units from rate divided by the concentration terms.
If asked to predict another rate, substitute the new concentrations or use the correct rate multipliers. Include significant figures appropriate to the source data.
A new challenge: changing units of concentration
Suppose the rate law is second order overall but concentrations are expressed in mmol dm⁻³ rather than mol dm⁻³. The numerical value and formal units of the constant must be consistent with that concentration convention.
Do not copy k = 0.300 dm³ mol⁻¹ s⁻¹ directly into a calculation using values written as 100 and 200 mmol dm⁻³ without converting them or adjusting k.
The simplest safe method in A-Level work is to convert concentrations to mol dm⁻³ first, then use the conventional units derived for k.
A second challenge: tripling A
For a rate law rate = k[A]², what happens if [A] triples while all other relevant conditions stay fixed?
The rate becomes 3² = 9 times as large. It does not merely triple.
If rate = k[A][B] with B unchanged, tripling A would instead triple the rate. The exponent—not the reactant name—determines the effect.
A third challenge: concentration B becomes half
Suppose rate = k[A][B]. If B is halved at fixed A, rate becomes half its original value.
If the law were rate = k[A]² with zero dependence on B under the specified regime, halving B would not change the predicted rate as long as that regime remains applicable.
Again, the same concentration change can have very different effects depending on the experimentally determined order.
How to connect the topic to a real reaction
Reaction kinetics is used in industrial processing, medicines, atmospheric chemistry and food systems. Chemists need to know how strongly reaction speed responds to reactant amounts and conditions.
The simple H1 laws are teaching models. A real reactor may involve mixing, transport, changing temperature and multiple reaction steps that complicate an elementary rate equation.
That does not weaken the classroom method. It teaches students to identify variables and use measurable relationships before moving to more complex systems.
A seven-day revision plan
Day one: define rate, rate constant and individual order. Day two: practise the doubling, quadrupling and unchanged-rate tests.
Day three: deduce m and n from a four-row initial-rate table. Day four: calculate k and derive zero-, first- and second-order units independently.
Day five: predict unseen rates when two reactants change. Day six: interpret concentration–time gradients and the first-order half-life without integrated formulas. Day seven: solve a new table with changed values and identify the first incorrect inference.
A learner who can calculate k but cannot state its units needs a unit-repair exercise, not another long list of identical rate tables.
A focused tutorial session
I would begin by giving each student two experimental rows and asking what was deliberately held constant. Next, the learner identifies the exponent for one reactant and explains the rate multiplier.
Then students combine the orders, calculate k and audit units. The last stage is an unfamiliar table where the rows are deliberately scrambled, preventing reliance on the first pair of lines.
The tutor should mark the exact failed step: experimental comparison, exponent algebra, concentration units, calculation or interpretation of what k means.
What parents can ask a student
Ask why a reaction can be first order with respect to A and first order with respect to B but second order overall. The answer should say the exponents add.
Then ask why the rate constant has dm³ mol⁻¹ s⁻¹ units instead of s⁻¹. The student should divide rate units by two powers of concentration and show what remains.
If the learner can reconstruct those ideas without notes, they are more ready for an unfamiliar data-based question than someone who only remembers that k is “some constant”.
H1 Chemistry 8873 curriculum scope
The 2027 H1 Chemistry syllabus includes rate = k[A]ᵐ[B]ⁿ for simple cases where the individual exponents are zero, one or two. It requires the initial-rates method, zero- and first-order concentration–time graphs, calculation of initial rates, and the first-order half-life concept.
The document explicitly states that integrated forms of rate equations are not required. Students should therefore prioritise experiments, graphs, units and conceptual connections rather than spending H1 revision time on unnecessary advanced integrated expressions.
See the official SEAB 2027 H1 Chemistry syllabus 8873.
Frequently asked questions
How do I calculate the overall reaction order?
Add the exponents in the experimentally determined rate equation. For rate = k[A][B], overall order is two.
If doubling A doubles the rate, what order is A?
First order with respect to A, when the other relevant conditions and concentrations are unchanged.
If doubling A quadruples the rate, what order is A?
Second order with respect to A.
If doubling A does not change the rate, what order is A?
Zero order with respect to A in the concentration range and conditions where that law applies.
What are the units of k for first order?
s⁻¹ if rate uses mol dm⁻³ s⁻¹ and concentration mol dm⁻³.
What are the units of k for second order?
dm³ mol⁻¹ s⁻¹ under the same rate and concentration unit conventions.
Why does k change with temperature?
Temperature affects the molecular energy distribution and the rate of crossing the activation-energy barrier; the rate constant usually changes as a result.
Does doubling a reactant concentration double k?
No. Within a fixed rate law at the same temperature and catalytic pathway, k stays constant while the rate changes.
Is total order always equal to the reactants’ balanced coefficients?
No. The orders are kinetic relationships; do not infer them automatically from an overall stoichiometric equation.
Are integrated rate laws examined in H1 8873?
The 2027 syllabus explicitly says their integrated forms are not required.
Continue through the Hougang Chemistry library
For prior rate-graph ideas, read H1 first-order half-life, zero-order graphs and initial rates. To connect k with the role of a catalyst, use H1 enzymes, catalytic converters and activation energy. For real-world heterogeneous catalysis, see H1 NOx catalytic converters and heterogeneous catalysis.
Families can explore JC1 H1 Chemistry small-group learning and JC2 H1 Chemistry tutorials, or consult eduKate Singapore’s parent–student enquiry information for current arrangements.
The final self-check is straightforward. Look at a table where two concentrations and the rate change, find two independent comparisons, write the rate law, calculate k and derive its units from scratch. When all four steps agree, the rate equation has become a tool you can use rather than a formula you hope to remember.