Rate Equations and Orders of Reaction — AQA A-level Chemistry

SPEC 3.1.9.1–3.1.9.2 • RATE EQUATIONS

Rate Equations & Determining Orders

The mathematical relationship between reaction rate and concentration

Why Rate Equations?

The rate of reaction refers to the change in concentration of a reactant or product per unit time. It can be measured by tracking the decrease in concentration of a reactant or the increase in concentration of a product over time. The units are always mol dm⁻³ s⁻¹.

At AS Level, you learned that rate depends on concentration, temperature, and catalysts. Now we go deeper: rate equations tell us the precise mathematical relationship between rate and concentration.

This is powerful because rate equations give us information about reaction mechanisms — the step-by-step molecular pathway of how reactions actually occur. You cannot deduce a rate equation from a balanced equation; it must be determined experimentally.

What You Will Learn

How to interpret and use the rate equation:

Rate = k [A]^{m} [B]^{n}

What orders of reaction (0, 1, 2) mean and how to identify them

How to determine orders experimentally using initial rates and graphs

How orders connect to the rate-determining step of a mechanism

Exam Tip

Rate equations questions appear on Paper 2 every year. Expect to calculate k, determine orders from data, or deduce mechanism information from a rate equation.

The Rate Equation

Definition

A rate equation shows how the rate of reaction depends on the concentration of reactants.

Rate = k [A]^{m} [B]^{n}

Each symbol has a specific meaning. The rate equation is an experimentally determined relationship — you cannot write it just by looking at the balanced equation.

SymbolMeaningUnits / Notes

Rate

Rate of reactionmol dm⁻³ s⁻¹

k

Rate constantVaries (depends on overall order)

[A], [B]

Concentration of reactantsmol dm⁻³

m, n

Orders of reaction0, 1, or 2 (at A-Level)

The Rate Constant k

The rate constant k is a proportionality constant. A larger value of k indicates a faster rate of reaction. The value of k is constant for a given reaction at a fixed temperature, but increasing the temperature exponentially increases the value of k (collisions become more frequent and more energetic).

The units of k depend on the overall order. Find them by rearranging k = Rate / [A]ᵐ[B]ⁿ and substituting units:

Overall OrderRate Equation ExampleUnits of k

0Rate = kmol dm⁻³ s⁻¹

1Rate = k[A]s⁻¹

2Rate = k[A]² or k[A][B]dm³ mol⁻¹ s⁻¹

3Rate = k[A]²[B] or k[A][B][C]dm⁶ mol⁻² s⁻¹

Important Rule

Products are NEVER involved in the rate equation — they have no effect on the rate of reaction. Only the concentrations of reactants (and sometimes catalysts) appear.

Critical Understanding

The orders

m

and

n

are NOT the stoichiometric coefficients from the balanced equation. They must be found by experiment. A reaction 2A + B → products does NOT automatically mean order 2 with respect to A.

Orders of Reaction

Definition

The order of reaction with respect to a reactant is the power to which its concentration is raised in the rate equation.

At A-Level, orders are restricted to 0, 1, or 2. Each order tells you how changing that reactant's concentration affects the rate:

OrderEffect on RateMathematical Relationship

Zero orderChanging [A] has NO effect on rateRate ∝ [A]⁰ = 1

First orderDouble [A] → ×2 rate; Triple [A] → ×3 rateRate ∝ [A]¹

Second orderDouble [A] → ×4 rate; Triple [A] → ×9 rateRate ∝ [A]²

The overall order is the sum of individual orders: if Rate = k[A]²[B]¹, the overall order is 2 + 1 = 3. The overall order determines the units of the rate constant k. Note that since [A]⁰ = 1, zero order terms are often omitted from the rate equation entirely.

How to Think About Orders

Zero order means the reactant doesn't appear in the rate-determining step. First order means one molecule of that reactant is involved in the RDS. Second order means either two molecules collide, or the reactant appears in two steps up to and including the RDS.

Catalysts in Rate Equations

A homogeneous catalyst can appear in the rate equation if it has a measurable effect on the rate. If a chemical appears in the rate equation but is NOT one of the reactants in the balanced equation, it must be a catalyst. This is how you can identify catalysts from rate data.

01Worked Example

For the reaction A + 2B → C, the rate equation is found experimentally to be: Rate = k[A][B]²

(a) State the order with respect to A.

(b) State the order with respect to B.

(c) State the overall order.

(d) If [A] is doubled and [B] is tripled, by what factor does the rate change?

(a)

Order with respect to A = 1 (the power of [A] in the rate equation)

(b)

Order with respect to B = 2 (the power of [B] in the rate equation)

(c)

Overall order = 1 + 2 = 3

(d)

Rate factor = (factor for A)¹ × (factor for B)² = 2¹ × 3² = 2 × 9 = 18

AnswerThe rate increases by a factor of 18.

02Worked Example

The rate equation for a reaction is: Rate = k[NO]²[O₂]

Calculate the rate of reaction when [NO] = 0.020 mol dm⁻³, [O₂] = 0.010 mol dm⁻³, and k = 7.1 × 10³ dm⁶ mol⁻² s⁻¹

1

Write out the rate equation: Rate = k[NO]²[O₂]

2

Substitute values: Rate = 7.1 × 10³ × (0.020)² × (0.010)

3

Calculate: Rate = 7.1 × 10³ × 4.0 × 10⁻⁴ × 1.0 × 10⁻²

4

Rate = 7.1 × 10³ × 4.0 × 10⁻⁶ = 2.84 × 10⁻²

AnswerRate = 2.84 × 10⁻² mol dm⁻³ s⁻¹ (or 0.028 mol dm⁻³ s⁻¹)

Identifying Orders from Graphs

There are two key types of graphs used to identify orders: concentration–time graphs and rate–concentration graphs. Each order produces a distinctive shape that you must recognise.

The concentration–time graph shows how [A] changes as the reaction proceeds. The shape depends on the order:

Visual Recognition

Zero order: straight line. First order: exponential decay that approaches but never reaches zero. Second order: similar curve but with a steeper initial drop and longer tail.

Rate–Concentration Graphs

Plotting rate against concentration gives an even clearer way to identify orders. The shape of the graph directly shows the mathematical relationship:

OrderRate vs [A] GraphMathematical Form

ZeroHorizontal line (rate = constant)Rate = k

FirstStraight line through originRate = k[A]

SecondCurve through origin (parabola)Rate = k[A]²

Distinguishing First and Second Order

If the graph of Rate vs [A] is a straight line through the origin, it's first order. If it curves upward, try plotting Rate vs [A]² — if THAT is straight through the origin, it's second order.

Half-Life and Order

Definition

The half-life (

t_{1/2}

) of a reaction is the time taken for the concentration of a reactant to fall to half its initial value.

The behaviour of half-life provides another diagnostic tool for identifying order:

OrderHalf-Life BehaviourHow to Check

Zero orderHalf-life decreases as reaction proceedsEach successive t₁/₂ is shorter

First orderHalf-life is CONSTANTSame time for [A]₀→½[A]₀ as for ½[A]₀→¼[A]₀

Second orderHalf-life increases as reaction proceedsEach successive t₁/₂ is longer

First-Order Half-Life Formula

For a first-order reaction only, the half-life can be calculated from the rate constant:

t_{1/2} = \frac{l n 2}{k} = \frac{0.693}{k}

This formula only works for first-order reactions because they are the only ones with a constant half-life independent of concentration.

Comparing half-lives across all three orders:

Exam Technique

To test for first order: read off two successive half-lives from a concentration–time graph. If they are equal, the reaction is first order with respect to that reactant.

03Worked Example

The graph shows the concentration of reactant X against time. Use the graph to determine the order of reaction with respect to X.

1

Find the first half-life: [X] drops from 0.80 to 0.40 mol dm⁻³

2

Read from graph: this takes approximately 20 seconds

3

Find the second half-life: [X] drops from 0.40 to 0.20 mol dm⁻³

4

Read from graph: this also takes approximately 20 seconds (from t = 20s to t = 40s)

5

Since successive half-lives are equal, the reaction is first order with respect to X.

AnswerFirst order with respect to X (constant half-life of ~20s)

The Initial Rates Method

There are two main experimental approaches for determining orders: the initial rates method (multiple experiments with different starting concentrations) and the continuous monitoring method (following one reaction over time using graphs). The initial rates method is most common in exam questions.

Method

The initial rate is the rate at time t = 0. It equals the gradient of the tangent to the concentration–time curve at the start of the reaction.

Why use initial rates? At t = 0, we know the exact concentrations of all reactants (the starting concentrations we measured out). As the reaction proceeds, concentrations change and calculations become more complex.

Key Technique

Compare pairs of experiments where only ONE concentration changes. If doubling [A] doubles the rate, order = 1. If doubling [A] quadruples the rate, order = 2. If doubling [A] has no effect, order = 0.

Analysing Initial Rate Data

Exam questions typically present data in a table. The strategy is to compare experiments systematically:

Step-by-Step Strategy

Find two experiments where only one concentration changes

Calculate the factor by which that concentration changes

Calculate the factor by which the rate changes

Determine the order: Rate factor = (Conc factor)^order

For example, consider this data for the reaction A + B → products:

Exp[A] / mol dm⁻³[B] / mol dm⁻³Initial rate / mol dm⁻³ s⁻¹

10.100.102.0 × 10⁻⁴

20.200.108.0 × 10⁻⁴

30.100.204.0 × 10⁻⁴

Comparing Exp 1 and 2: [A] doubles (×2), [B] unchanged. Rate goes from 2.0×10⁻⁴ to 8.0×10⁻⁴ (×4). Since 2² = 4, order w.r.t. A = 2.

Comparing Exp 1 and 3: [A] unchanged, [B] doubles (×2). Rate goes from 2.0×10⁻⁴ to 4.0×10⁻⁴ (×2). Since 2¹ = 2, order w.r.t. B = 1.

Common Error

Always check that ONLY one concentration changes between the experiments you compare. If both change, the analysis becomes much more complex.

04Worked Example

The following data were obtained for the reaction: 2NO + Cl₂ → 2NOCl

Exp[NO] / mol dm⁻³[Cl₂] / mol dm⁻³Initial rate / mol dm⁻³ s⁻¹

10.0500.0501.0 × 10⁻³

20.1000.0504.0 × 10⁻³

30.0500.1002.0 × 10⁻³

(a) Determine the order with respect to NO and Cl₂.

(b) Write the rate equation.

(c) Calculate the rate constant k, including units.

(a)

Order w.r.t. NO: Compare Exp 1 and 2:

[NO] doubles (0.050 → 0.100), [Cl₂] unchanged

Rate: 1.0×10⁻³ → 4.0×10⁻³ (×4)

Since 2² = 4, order w.r.t. NO = 2

•

Order w.r.t. Cl₂: Compare Exp 1 and 3:

[Cl₂] doubles (0.050 → 0.100), [NO] unchanged

Rate: 1.0×10⁻³ → 2.0×10⁻³ (×2)

Since 2¹ = 2, order w.r.t. Cl₂ = 1

(b)

Rate equation: Rate = k[NO]²[Cl₂]

(c)

Using Exp 1: k = Rate / ([NO]²[Cl₂])

k = (1.0 × 10⁻³) / (0.050² × 0.050) = (1.0 × 10⁻³) / (1.25 × 10⁻⁴)

k = 8.0 dm⁶ mol⁻² s⁻¹

Units: mol dm⁻³ s⁻¹ ÷ (mol dm⁻³)³ = dm⁶ mol⁻² s⁻¹

The Rate-Determining Step

Most reactions occur in multiple steps, not all at once. The rate equation gives us clues about the mechanism — the actual molecular pathway.

Definition

The rate-determining step (RDS) is the slowest step in a reaction mechanism. It acts as a "bottleneck" that controls the overall rate.

The orders in the rate equation tell us which species are involved in the RDS (or in steps before it):

Key Principle

Only species that appear in steps up to and including the RDS appear in the rate equation. Species involved only in fast steps AFTER the RDS do not appear.

Example: For the reaction A + 2B → C with rate equation Rate = k[A][B]:

The rate depends on [A] (first order) and [B] (first order)

This suggests 1 molecule of A and 1 molecule of B are involved in the RDS

The second molecule of B must react in a fast step AFTER the RDS

A possible mechanism:

Step 1 (slow, RDS): A + B → intermediate

Step 2 (fast): intermediate + B → C

Identifying the RDS from a Mechanism

You can identify the rate-determining step by matching the species in the rate equation to the mechanism steps. The RDS will be the step where all the species in the rate equation (accounting for their orders) are either reactants in that step, or are produced in fast equilibrium steps before it. Look for which step's reactants match the rate equation.

Important Limitation

The rate equation alone doesn't prove a mechanism — it only provides evidence that is consistent with a proposed mechanism. Other experimental evidence is needed for confirmation.

05Worked Example

The reaction between propanone and iodine in acidic solution is:

CH_{3} COCH_{3} + I_{2} H^{+} CH_{3} COCH_{2} I + HI

The rate equation is found to be: Rate = k[CH₃COCH₃][H⁺]

(a) What is the order with respect to iodine?

(b) What does this tell us about the role of iodine in the mechanism?

(c) Suggest why [H⁺] appears in the rate equation even though H⁺ is a catalyst.

(a)

The order with respect to iodine is zero. Iodine does not appear in the rate equation, meaning changes in [I₂] have no effect on the rate.

(b)

Since iodine is zero order, it must react in a fast step AFTER the rate-determining step. The RDS does not involve iodine molecules.

(c)

Although H⁺ is a catalyst (regenerated at the end), it is involved in the rate-determining step. The RDS involves the protonation of propanone, which requires H⁺. This is why changing [H⁺] affects the rate.

AnswerThe mechanism involves slow protonation of propanone, followed by fast reaction with iodine.

Practice

Question 1:

Exp[A] / mol dm⁻³[B] / mol dm⁻³Rate / mol dm⁻³ s⁻¹

10.100.104.0 × 10⁻⁵

20.100.303.6 × 10⁻⁴

30.200.307.2 × 10⁻⁴

(a) Determine the order with respect to A and B. (b) Write the rate equation. (c) Calculate k with units.

(a)

Order w.r.t. A: Compare Exp 2 and 3: [A] doubles, [B] same. Rate: 3.6×10⁻⁴ → 7.2×10⁻⁴ (×2). Order = 1.

•

Order w.r.t. B: Compare Exp 1 and 2: [B] triples, [A] same. Rate: 4.0×10⁻⁵ → 3.6×10⁻⁴ (×9 = 3²). Order = 2.

(b)

Rate equation: Rate = k[A][B]²

(c)

k = 4.0×10⁻⁵ / (0.10 × 0.10²) = 4.0×10⁻⁵ / 1.0×10⁻³ = 4.0×10⁻² dm⁶ mol⁻² s⁻¹

Practice

Question 2:

(a) If [X] is halved and [Y] is doubled, by what factor does the rate change?

(b) The rate constant k = 2.5 × 10⁴ dm⁶ mol⁻² s⁻¹. Calculate the rate when [X] = 0.040 mol dm⁻³ and [Y] = 0.080 mol dm⁻³.

(a)

Factor for X: (½)² = ¼

•

Factor for Y: (2)¹ = 2

•

Overall factor = ¼ × 2 = ½ (rate halves)

(b)

Rate = k[X]²[Y] = 2.5 × 10⁴ × (0.040)² × (0.080)

•

Rate = 2.5 × 10⁴ × 1.6 × 10⁻³ × 8.0 × 10⁻² = 3.2 mol dm⁻³ s⁻¹

Practice

Question 3:

(a) What is the order with respect to R?

(b) What does this tell us about R's role in the mechanism?

(c) Is R involved before, during, or after the rate-determining step? Explain.

(a)

The order with respect to R is zero. R does not appear in the rate equation.

(b)

Changing the concentration of R has no effect on the rate of reaction.

(c)

R is involved in a fast step AFTER the rate-determining step. The RDS only involves P and Q (one molecule of each, since both are first order). R reacts in a subsequent fast step that does not limit the overall rate.

Note: Being in the balanced equation doesn't mean a species affects the rate!

Summary

Key Formulas

ConceptFormula / Relationship

Rate equation

Rate = k [A]^{m} [B]^{n}

Overall orderSum of individual orders (m + n + ...)

First order testConstant half-life

Rate constant unitsDepends on overall order

Graph Signatures

Order[A] vs TimeRate vs [A]Half-life

0Straight lineHorizontal lineDecreases

1Exponential curveStraight through originConstant

2Steeper curveParabola through originIncreases

Exam Essentials

Orders must be determined experimentally — not from the balanced equation

Use initial rates: compare experiments where only one [conc] changes

Rate-determining step contains species that appear in the rate equation

Zero order species react in fast steps AFTER the RDS