Rate Equations, Orders & Rate Constants

Last updated: 17 August 2026

Diagnose → Learn → Master: Every study-note section includes 15 free MCQs — start with 5 Diagnostic MCQs to identify gaps in your understanding, then complete 10 Mastery MCQs near the end to check what you have learned.

A balanced chemical equation tells us which substances react and the proportions in which they react, but it does not usually tell us how quickly the reaction takes place. Some reactions are complete almost immediately, while others take minutes, hours or even years.

Chemical kinetics gives us a way to describe these differences mathematically. Instead of asking only whether increasing a reactant concentration makes a reaction faster, we can ask exactly how much faster. The answer is expressed through a rate equation.

Once you can read a rate equation, initial-rate tables, concentration–time data and half-life information become connected parts of the same story. The rate equation can also provide clues about which elementary step controls the speed of a reaction mechanism.

Diagnostic Quiz

What do you already know about reaction rates?

Try these five questions before reading the chapter. They check the concentration, proportionality and equation ideas that make rate equations easier to understand.

5 short starting-point questions
Focus on rate, order and proportional reasoning
Every answer option is explained

Question 1 of 5

Which statement best describes the rate of a reaction?

The total amount of product eventually formed

The final amount of product tells us how far the reaction has proceeded, not how quickly it proceeds.

The change in concentration of a reactant or product per unit time

Correct. Reaction rate describes how rapidly concentration changes with time.

The concentration of a reactant at the start of the reaction

Initial concentration may affect the rate, but the concentration itself is not the definition of reaction rate.

The time required for every reactant particle to disappear

A reaction does not have to consume every reactant particle, and reaction rate is defined through change per unit time rather than total completion time.

Rate measures how rapidly the composition of the reacting mixture changes. It is commonly expressed using a concentration change divided by a time interval.

In the rate equation rate = k[A]²[B], what does the exponent 2 tell you?

rate = k[A]²[B]

Two moles of A must appear in the balanced equation

Stoichiometric coefficients in the overall balanced equation do not normally determine reaction orders.

The overall order of the reaction is 2

The overall order is the sum of all the exponents. Here it is 2 + 1 = 3.

The reaction is second order with respect to A

Correct. The power of [A] gives the order with respect to A.

The concentration of A is always twice the concentration of B

The exponent describes how rate depends on concentration; it does not specify a fixed concentration ratio.

In a rate equation, the exponent attached to a concentration term is the order of reaction with respect to that reactant.

The concentration of A is doubled while all other relevant concentrations and conditions remain unchanged. The rate becomes four times larger. What is the order with respect to A?

Zero order

A zero-order dependence would give no change in rate when [A] is doubled.

First order

A first-order dependence would cause the rate to double, not quadruple.

Second order

Correct. Doubling [A] changes [A]² by a factor of 2² = 4.

Fourth order

Fourth order would give a rate factor of 2⁴ = 16 when [A] is doubled.

Compare the concentration factor with the rate factor. A concentration factor of 2 producing a rate factor of 4 means 2ᵐ = 4, so m = 2.

What is the overall order of the reaction with rate equation rate = k[A][B]²?

rate = k[A][B]²

This is only the order with respect to A.

This is only the order with respect to B.

The orders are added rather than multiplied.

Correct. The overall order is 1 + 2 = 3.

Overall order is obtained by adding the powers of all concentration terms in the rate equation.

A reaction is studied several times at the same temperature. Only the starting concentrations are changed. Which statement about the rate constant, k, is correct?

k must decrease whenever a reactant concentration decreases

Concentration changes the concentration terms in the rate equation, not k itself.

k should remain constant apart from experimental uncertainty

Correct. For the same reaction at a fixed temperature and under the same kinetic conditions, changing concentration does not change k.

k must always have units of s⁻¹

Only a first-order rate constant has units of s⁻¹ when concentration is measured in mol dm⁻³.

k is simply another name for the measured reaction rate

The rate depends on k and the relevant concentration terms. Rate and rate constant are different quantities.

At fixed temperature and unchanged conditions, k is the proportionality constant connecting concentration terms to reaction rate. The rate may change even while k remains constant.

What does rate of reaction mean?

The rate of reaction describes how quickly the concentration of a reactant decreases or the concentration of a product increases with time.

Suppose A is a reactant. As the reaction proceeds, [A] decreases. Its average rate of disappearance over a time interval can be written as:

Rate of disappearance of A = −Δ[A] ÷ Δt

The negative sign is included because Δ[A] is negative for a reactant that is being consumed. The negative sign therefore allows the rate of disappearance to be quoted as a positive quantity.

For a product P, its concentration increases, so:

Rate of formation of P = Δ[P] ÷ Δt

If concentration is measured in mol dm⁻³ and time in seconds, a common unit of rate is:

mol dm⁻³ s⁻¹

Rate and rate constant are not the same quantity.

The reaction rate can change during an experiment because reactant concentrations change. The rate constant, k, remains constant for the same reaction at a fixed temperature and under the same kinetic conditions.

Average rate and instantaneous rate

An average rate describes the concentration change over a finite time interval. An instantaneous rate describes the rate at one particular moment.

On a concentration–time graph, the instantaneous rate is obtained from the gradient of the tangent at that point. The initial rate is the instantaneous rate at the start of the reaction.

Initial rates are especially useful because the starting concentrations are known accurately and very little product has yet accumulated.

The rate equation

For a reaction involving reactants A and B, a simple rate equation may be written as:

rate = k[A]ᵐ[B]ⁿ

The rate equation describes how the measured reaction rate depends on the reactant concentrations.

Symbol	Meaning	Important point
rate	Reaction rate	Commonly measured in mol dm⁻³ s⁻¹
k	Rate constant	Constant at fixed temperature and unchanged kinetic conditions
[A]	Concentration of A	Usually measured in mol dm⁻³
[B]	Concentration of B	Usually measured in mol dm⁻³
m	Order with respect to A	Determined experimentally
n	Order with respect to B	Determined experimentally

What does an order of reaction actually tell us?

An order tells us how strongly the rate responds when the concentration of one reactant is changed while the other relevant conditions remain constant.

For example, if:

rate = k[A]²[B]

the reaction is second order with respect to A;
the reaction is first order with respect to B; and
the reaction is third order overall.

Overall order = 2 + 1 = 3

Orders of reaction are experimental quantities.

Do not simply copy the coefficients from the overall balanced equation into the rate equation. The balanced equation gives the overall stoichiometry, whereas the rate equation describes experimentally observed kinetic behaviour.

Zero-, first- and second-order behaviour

The simplest way to understand reaction order is to imagine changing the concentration of one reactant while keeping everything else constant.

Order with respect to A	Rate dependence	If [A] doubles	If [A] triples
Zero order	rate ∝ [A]⁰	Rate × 1	Rate × 1
First order	rate ∝ [A]	Rate × 2	Rate × 3
Second order	rate ∝ [A]²	Rate × 4	Rate × 9

Zero order with respect to a reactant

If the reaction is zero order with respect to A, changing [A] has no effect on the rate over the range of conditions being studied.

rate ∝ [A]⁰ = 1

Because [A]⁰ = 1, the concentration term does not need to be written explicitly. For example:

rate = k[A]⁰[B] = k[B]

Zero order does not mean that A is absent or that A is not consumed. It means that changing its concentration does not affect the measured rate under the conditions being considered.

First order with respect to a reactant

If the reaction is first order with respect to A:

rate ∝ [A]

The rate changes by the same factor as [A].

Double [A] → rate doubles.
Triple [A] → rate triples.
Halve [A] → rate halves.

Second order with respect to a reactant

If the reaction is second order with respect to A:

rate ∝ [A]²

The concentration factor must be squared.

Double [A] → rate increases by 2² = 4.
Triple [A] → rate increases by 3² = 9.
Halve [A] → rate becomes (1/2)² = 1/4 of its original value.

The useful factor rule: if [A] changes by a factor of x and the rate changes by a factor of xᵐ, then m is the order with respect to A.

The question is therefore not simply “Did the rate increase?” but “By what factor did it increase compared with the concentration factor?”

Finding reaction orders from initial-rate data

In an initial-rates experiment, several runs are performed at the same temperature. The initial concentrations are deliberately changed and the corresponding initial rates are measured.

A reliable comparison method

Choose two experiments in which only one reactant concentration changes.
Find the concentration factor for that reactant.
Find the rate factor between the same two experiments.
Match the factors to zero-, first- or second-order behaviour.
Repeat using another suitable pair to find the order of the other reactant.
Write the complete rate equation.

Worked example: deducing both orders

Consider the following results for a reaction between A and B.

Experiment	[A] / mol dm⁻³	[B] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.100	0.100	1.20 × 10⁻³
2	0.200	0.100	4.80 × 10⁻³
3	0.100	0.200	2.40 × 10⁻³

First compare experiments 1 and 2. [B] is unchanged.

[A]: 0.100 → 0.200 mol dm⁻³ = ×2
Rate: 1.20 × 10⁻³ → 4.80 × 10⁻³ mol dm⁻³ s⁻¹ = ×4

Since 2² = 4, the reaction is second order with respect to A.

Now compare experiments 1 and 3. [A] is unchanged.

[B]: 0.100 → 0.200 mol dm⁻³ = ×2
Rate: 1.20 × 10⁻³ → 2.40 × 10⁻³ mol dm⁻³ s⁻¹ = ×2

The reaction is therefore first order with respect to B.

The complete rate equation is:

rate = k[A]²[B]

The overall order is:

2 + 1 = 3

Choose experiment pairs strategically.

If only one concentration changes, the effect of that reactant can be isolated. Comparing two experiments in which several concentrations change at once is possible, but every concentration change must then be accounted for.

Using ratios when the numbers are less obvious

Suppose B has the same concentration in two experiments. For:

rate = k[A]ᵐ[B]ⁿ

dividing one rate by the other cancels k and the unchanged B term:

rate₂ ÷ rate₁ = ([A]₂ ÷ [A]₁)ᵐ

This form is useful when the concentration change is not a simple doubling or tripling.

Using concentration–time behaviour

Reaction order can also be investigated by following the concentration of a reactant as time passes.

Zero-order concentration–time behaviour

If a reaction is zero order with respect to A, the rate does not depend on [A]. A is therefore consumed at a constant rate while the zero-order behaviour applies.

Zero order: [A] decreases linearly with time

A graph of [A] against time therefore gives a straight line with a constant negative gradient.

First-order concentration–time behaviour

For a first-order reaction:

rate ∝ [A]

As A is consumed, [A] becomes smaller and the reaction slows. A graph of [A] against time is therefore curved rather than a straight line.

The particularly useful feature of a first-order reaction is its constant half-life.

The half-life, t₁/₂, is the time taken for the concentration of a reactant to fall to half its current value.

Suppose [A] falls from 0.800 to 0.400 mol dm⁻³ in 30 s.
It then falls from 0.400 to 0.200 mol dm⁻³ in another 30 s.
It then falls from 0.200 to 0.100 mol dm⁻³ in another 30 s.

Each halving takes the same time, so this is characteristic of first-order behaviour.

For a first-order reaction: t₁/₂ = ln 2 ÷ k = 0.693 ÷ k

Notice what is constant: it is the time required for each halving, not the amount of reactant consumed during each half-life.

Do not identify first order simply because the concentration–time graph is curved.

Other reaction orders can also give curved concentration–time plots. A constant half-life is much stronger evidence for first-order behaviour.

Worked example: finding k from a first-order half-life

A first-order reaction has a half-life of 40.0 s. Calculate k.

k = 0.693 ÷ t₁/₂

k = 0.693 ÷ 40.0 = 1.73 × 10⁻² s⁻¹

Answer: k = 1.73 × 10⁻² s⁻¹

The rate constant, k

The rate constant is the proportionality constant connecting the measured reaction rate to the concentration terms in the rate equation.

For:

rate = k[A]²[B]

rearranging gives:

k = rate ÷ ([A]²[B])

Worked example: calculating k

Use experiment 1 from the earlier initial-rate table:

[A] = 0.100 mol dm⁻³
[B] = 0.100 mol dm⁻³
rate = 1.20 × 10⁻³ mol dm⁻³ s⁻¹

k = (1.20 × 10⁻³) ÷ [(0.100)²(0.100)]

k = 1.20

The reaction is third order overall, so k has units of dm⁶ mol⁻² s⁻¹.

Answer: k = 1.20 dm⁶ mol⁻² s⁻¹

Why do the units of k depend on overall order?

Every rate equation must ultimately produce units of rate. The concentration powers on the right-hand side therefore determine what units k must have.

Overall order	Example form	Units of k
0	rate = k	mol dm⁻³ s⁻¹
1	rate = k[A]	s⁻¹
2	rate = k[A]² or k[A][B]	dm³ mol⁻¹ s⁻¹
3	rate = k[A]²[B]	dm⁶ mol⁻² s⁻¹

For example, for a second-order rate equation:

k = rate ÷ concentration²

Therefore:

units of k = (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)² = dm³ mol⁻¹ s⁻¹

Do not memorise the units of k independently of the rate equation.

First determine the overall order, then derive the units by rearranging k = rate ÷ concentration terms. This also works if the rate equation is unfamiliar.

Does changing concentration change k?

No. If the same reaction is investigated at the same temperature and under the same kinetic conditions, changing reactant concentration changes the rate, but does not change k.

For example, if:

rate = k[A]²

doubling [A] makes the rate four times larger because the [A]² term becomes four times larger. There is no need for k itself to change.

What happens when temperature changes?

Temperature does affect the rate constant. Increasing temperature generally increases k, so the reaction proceeds more rapidly even when the concentrations being compared are the same.

Changing concentration: changes the concentration terms in the rate equation while k remains constant at fixed temperature.

Changing temperature: changes the value of k itself.

When more than one concentration changes

A rate equation allows us to predict the combined effect of changing several concentrations at once.

Suppose:

rate = k[A]²[B]

If [A] is tripled and [B] is halved:

tripling [A] contributes a factor of 3² = 9;
halving [B] contributes a factor of 1/2.

The overall rate factor is:

9 × 1/2 = 4.5

The new rate is therefore 4.5 times the original rate.

Treat each concentration term separately, then multiply the effects.

The rate equation already tells you how strongly each reactant affects the rate. Apply the concentration factor to the correct power for each reactant before combining the factors.

Rate equations and the rate-determining step

Many reactions do not occur in one single molecular event. Instead, the overall reaction takes place through a sequence of elementary steps called a reaction mechanism.

One step may be much slower than the others. This slow step limits the speed at which the overall process can proceed and is called the rate-determining step, or RDS.

A simple example

Consider the overall reaction:

2NO₂ → 2NO + O₂

Suppose a proposed mechanism is:

Step 1: NO₂ + NO₂ → NO₃ + NO slow
Step 2: NO₃ → NO + O₂ fast

Adding the two steps cancels the intermediate NO₃ and gives the overall equation.

Since the first elementary step is slow and involves two NO₂ molecules, this simple mechanism predicts:

rate = k[NO₂]²

If experiment also gives rate = k[NO₂]², the proposed mechanism is consistent with the kinetic evidence.

What a matching rate equation tells us: the proposed mechanism is compatible with the observed kinetics.

What it does not tell us: it does not prove that this is the only possible mechanism. More than one mechanism can sometimes be consistent with the same experimental rate law.

Common mistakes with rate equations and reaction orders

Common mistake	Why it is wrong	Correct approach
Using coefficients from the overall balanced equation as reaction orders	The balanced equation gives stoichiometry, not normally the kinetic dependence on concentration	Determine orders from experimental data
Comparing two experiments in which several concentrations changed and attributing the entire rate change to one reactant	More than one concentration term may have affected the rate	Prefer a pair where only one concentration changes, or account for every change
Saying that zero order in A means A is not involved in the reaction	A may still be consumed even though changing [A] does not affect the rate	State that the rate is independent of [A] under the conditions studied
Finding overall order by choosing the largest exponent	Overall order is the sum of the individual orders	Add all concentration exponents
Changing k whenever a reactant concentration changes	At fixed temperature, concentration changes the concentration terms rather than k	Keep k constant for the same reaction under unchanged kinetic conditions
Giving k units of s⁻¹ for every reaction	The units depend on the overall order	Derive the units from k = rate ÷ concentration terms
Calling k the reaction rate	The rate generally depends on both k and concentrations	Use the complete rate equation to distinguish rate from rate constant
Assuming every curved concentration–time graph proves first-order behaviour	Other reaction orders can also produce curved concentration–time plots	Look for a constant half-life when identifying first-order behaviour
Assuming half-life is constant for every reaction	Constant half-life is a characteristic property of first-order kinetics	Check whether successive halvings take equal times
Claiming that a matching rate equation proves a unique reaction mechanism	Different mechanisms may sometimes give the same observed rate law	State that the proposed mechanism is consistent with the kinetic evidence

Rate-equation checklist

Read the equation

Identify k and each concentration term.
Read each exponent as an individual order.
Add the exponents to obtain overall order.
Remember that a zero-order term is normally omitted.

Use initial-rate data

Choose experiments where one concentration changes.
Find the concentration factor.
Find the corresponding rate factor.
Match the factors to the correct order.

Use graphs and half-life

Zero-order [A]–time behaviour is linear.
A first-order [A]–time graph is curved.
Constant half-life identifies first-order behaviour.
For first order, use t₁/₂ = 0.693/k.

Handle k correctly

Rearrange the rate equation to calculate k.
Derive the units from the overall order.
Do not change k when only concentration changes.
Use experimental rate laws to test proposed mechanisms.

Mastery Quiz

Rate Equations and Reaction Orders Mastery Quiz

Now apply the chapter to initial-rate data, changing concentrations, half-life, rate-constant units and simple reaction mechanisms.

10 application-focused questions
Includes calculations and kinetic reasoning
Every answer option is explained

Question 1 of 10

The following initial-rate data are obtained for a reaction between A and B. Which rate equation is consistent with the results?

Experiment	[A] / mol dm⁻³	[B] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.100	0.100	2.00 × 10⁻³
2	0.200	0.100	8.00 × 10⁻³
3	0.100	0.200	4.00 × 10⁻³

rate = k[A][B]

This predicts that doubling A would only double the rate. Experiment 1 to 2 shows that doubling A makes the rate four times larger.

rate = k[A][B]²

The data show the opposite pattern: A is second order and B is first order.

rate = k[A]²[B]

Correct. Doubling A causes a ×4 rate change, so the reaction is second order in A. Doubling B causes a ×2 rate change, so it is first order in B.

rate = k[A]²[B]²

This would predict that doubling B quadruples the rate, but experiment 1 to 3 shows only a doubling.

Compare experiments 1 and 2 to isolate A and experiments 1 and 3 to isolate B. The resulting orders are 2 in A and 1 in B.

A reaction has the rate equation rate = k[A]²[B]. In one experiment, [A] = 0.100 mol dm⁻³, [B] = 0.100 mol dm⁻³ and the initial rate is 2.00 × 10⁻³ mol dm⁻³ s⁻¹. What is the value and unit of k?

rate = k[A]²[B]

2.00 dm⁶ mol⁻² s⁻¹

Correct. Rearranging gives k = rate/[A]²[B]. Therefore, k = (2.00 × 10⁻³)/[(0.100)²(0.100)] = 2.00. Since the reaction is third order overall, the units are dm⁶ mol⁻² s⁻¹.

2.00 s⁻¹

The numerical value is correct, but s⁻¹ is the unit of k for an overall first-order reaction. This reaction is third order overall.

0.00200 dm⁶ mol⁻² s⁻¹

This treats the measured rate as though it were k. You must divide the rate by [A]²[B].

2.00 dm³ mol⁻¹ s⁻¹

These are the units of k for an overall second-order reaction. Here the overall order is 2 + 1 = 3.

First rearrange the rate equation: k = rate/[A]²[B]. Substituting the experimental values gives k = 2.00. The powers in [A]²[B] add to 3, so the reaction is third order overall and k has units dm⁶ mol⁻² s⁻¹.

A reaction obeys rate = k[A]²[B]. [A] is tripled while [B] is halved. Temperature is unchanged. By what factor does the rate change?

rate = k[A]²[B]

The rate becomes 1.5 times larger

Multiplying 3 by 1/2 ignores the fact that A is second order.

The rate becomes 3 times larger

The A contribution must be squared, so its factor is 9 rather than 3.

The rate becomes 9 times larger

This accounts for tripling A but ignores the halving of B.

The rate becomes 4.5 times larger

Correct. A contributes 3² = 9, while B contributes 1/2. Therefore the overall factor is 9 × 1/2 = 4.5.

Apply each concentration change to its own exponent before multiplying the separate effects together.

The following data are obtained while every condition except [X] is kept constant. What is the order with respect to X?

[X] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
0.100	1.50 × 10⁻³
0.200	1.50 × 10⁻³
0.400	1.50 × 10⁻³

First order

First order would cause the rate to increase in direct proportion to [X].

Zero order

Correct. [X] increases substantially while the measured rate remains unchanged.

Second order

Second order would cause doubling [X] to quadruple the rate.

The order cannot be determined because X is a reactant

Being a reactant does not prevent zero-order behaviour. The unchanged rate directly shows that the rate is independent of [X] over this range.

Zero order means that changing the reactant concentration does not change the rate under the conditions studied.

The concentration of A falls from 0.800 to 0.400 mol dm⁻³ in 30.0 s and then from 0.400 to 0.200 mol dm⁻³ in another 30.0 s. Which conclusion is most appropriate?

The reaction is zero order and k = 30.0 s⁻¹

Equal successive half-lives indicate first-order rather than zero-order behaviour.

The reaction is second order and k = 0.693 s⁻¹

Constant half-life is the characteristic feature being observed here, which indicates first-order kinetics.

The reaction is first order and k = 2.31 × 10⁻² s⁻¹

Correct. The half-life is 30.0 s, so k = 0.693/30.0 = 2.31 × 10⁻² s⁻¹.

No order can be deduced because the concentration is decreasing

The pattern of successive halvings contains exactly the information required to identify first-order behaviour.

Successive concentration halvings taking equal times indicate a constant half-life. For a first-order reaction, k = 0.693/t₁/₂.

A graph of reactant concentration against time is curved downwards as the concentration decreases. Which statement is safest?

The curved graph alone is not enough to prove that the reaction is first order

Correct. More than one reaction order can produce a curved concentration–time graph. Constant half-life provides stronger evidence for first-order behaviour.

Every curved concentration–time graph represents a first-order reaction

Curvature alone is not unique to first-order kinetics.

The reaction must be zero order

A zero-order concentration–time graph is linear while the zero-order behaviour applies.

The value of k must be decreasing continuously

At fixed temperature, the rate can decrease because concentration decreases while k remains constant.

Do not confuse a decreasing rate with a decreasing rate constant. A curved concentration–time plot can arise because the concentration-dependent rate becomes smaller as reactant is consumed.

A rate constant has units dm³ mol⁻¹ s⁻¹. What is the overall order of the corresponding reaction?

Zero order

A zero-order rate constant has the same units as rate: mol dm⁻³ s⁻¹.

First order

A first-order rate constant has units s⁻¹.

Third order

A third-order rate constant has units dm⁶ mol⁻² s⁻¹.

Second order

Correct. Dividing mol dm⁻³ s⁻¹ by one additional mol dm⁻³ concentration factor gives dm³ mol⁻¹ s⁻¹.

The units of k contain information about overall order. For an overall second-order reaction, k has units dm³ mol⁻¹ s⁻¹.

For the same reaction, changing the starting concentrations at 298 K gives essentially the same value of k. At 318 K, a substantially larger value of k is obtained. What is the best interpretation?

k is controlled mainly by reactant concentration

The observations show the opposite: changing concentration at 298 K does not significantly change k.

k is independent of concentration at fixed temperature but changes when temperature changes

Correct. This is the normal interpretation of k for a given rate equation under fixed kinetic conditions.

The reaction order must become zero at 318 K

A larger k does not by itself imply zero-order kinetics.

The reaction rate must be identical at 298 K and 318 K

A larger k at the higher temperature generally produces a larger rate for the same concentrations.

Concentration enters through the concentration terms in the rate law. Temperature affects the value of the rate constant itself.

The overall reaction is 2NO₂ → 2NO + O₂ and experiment gives rate = k[NO₂]². Which proposed mechanism is directly consistent with this rate equation?

Experimental rate equation: rate = k[NO₂]²

Slow: NO₂ → NO + O; fast: O + NO₂ → NO + O₂

A slow elementary first step involving one NO₂ would predict first-order dependence on [NO₂], not second-order dependence.

Slow: NO₂ + NO → products; remaining steps fast

This slow step would introduce a dependence on [NO] that is absent from the experimental rate equation.

Slow: NO₂ + NO₂ → NO₃ + NO; fast: NO₃ → NO + O₂

Correct. The slow elementary step contains two NO₂ molecules and therefore predicts rate = k[NO₂]². Adding the steps gives the stated overall reaction.

Slow: NO + O₂ → NO₃; fast: NO₃ + NO → 2NO₂

This mechanism proceeds towards NO₂ rather than giving the stated overall decomposition, and its slow step does not predict the observed rate law.

A proposed mechanism must reproduce both the overall reaction and the observed kinetic dependence. A mechanism that predicts the wrong rate equation is inconsistent with experiment.

A reaction follows rate = k[P][Q]². [P] is halved and [Q] is doubled at constant temperature. What happens to the rate?

rate = k[P][Q]²

The rate doubles

Correct. Halving P contributes ×1/2, while doubling Q contributes ×2² = ×4. The combined factor is (1/2) × 4 = 2.

The rate is unchanged

The two concentration changes do not cancel because Q is second order.

The rate halves

This considers only the change in P and ignores the larger second-order effect of Q.

The rate becomes four times larger

This considers the Q² term correctly but ignores the halving of P.

For simultaneous concentration changes, calculate the rate factor contributed by each term separately and multiply the factors together.

Frequently Asked Questions

Questions students often ask about rate equations and reaction order

These questions develop the distinctions that become especially important when initial-rate data, half-life and reaction mechanisms are combined in examination questions.

Why can I not simply use the coefficients in the balanced equation as the reaction orders?

The overall balanced equation tells you the net stoichiometry of the reaction. It does not usually reveal the sequence of elementary steps through which the reaction occurs.

The powers in an experimentally determined rate equation describe how the measured rate actually responds to concentration changes, so they must normally be obtained from kinetic data rather than copied from the overall equation.

If a reaction is zero order in A, why is A still called a reactant?

Zero order does not mean that A is absent or never consumed. It means that, over the range of conditions being investigated, changing [A] does not change the measured rate.

A can therefore still appear in the overall reaction even though [A] does not appear explicitly in the rate equation.

Does first order mean that only one molecule of the reactant appears in the balanced equation?

No. First order is a kinetic statement: the rate is directly proportional to that reactant concentration. It is not a statement about the coefficient in the overall balanced equation.

Why do the units of k change when the overall order changes?

The right-hand side of every rate equation must have the same overall units as the reaction rate. Different overall orders contain different powers of concentration, so k must supply different units to make the equation dimensionally consistent.

Can I compare the numerical values of k for reactions of different overall orders?

Such a comparison must be treated carefully because rate constants for different overall orders have different units. A numerically larger k does not automatically mean that one unrelated reaction is faster under every set of concentrations.

The actual reaction rate depends on the complete rate equation as well as the concentrations present.

Why can the rate decrease during a reaction even though k remains constant?

As reactants are consumed, their concentrations usually decrease. If those concentrations appear in the rate equation, the concentration terms become smaller and the rate falls.

At constant temperature, this can happen while k itself remains unchanged.

Does a curved concentration–time graph prove that a reaction is first order?

No. A curved concentration–time graph simply shows that the rate of disappearance is changing. More than one reaction order can produce such a curve.

For a first-order reaction, the particularly useful diagnostic is a constant half-life: each successive halving of concentration takes the same time.

Does a rate equation prove a proposed reaction mechanism?

No. A proposed mechanism must be consistent with the observed rate equation, so an incompatible rate law can be used to reject a mechanism.

However, more than one mechanism can sometimes produce the same kinetic behaviour. Agreement with the rate equation therefore supports a mechanism without necessarily proving that it is unique.

What is the quickest way to approach an initial-rates table in an examination?

First look for two experiments in which only one reactant concentration changes. Compare the factor by which that concentration changes with the factor by which the rate changes.

Repeat with another suitable pair for the next reactant. Only after the orders are known should you write the full rate equation and calculate k.

Rate equations turn experimental observations into a precise description of reaction speed. The powers of the concentration terms tell us how strongly each reactant controls the rate, while the rate constant connects those concentration effects to the measured rate at a particular temperature. Once you can move confidently between concentration factors, initial-rate data, half-life and the units of k, kinetics becomes a logical pattern to analyse rather than a set of rules to memorise.