Rate Equations & Arrhenius Analysis

1. Rate Equations, Orders & The Rate Constant

Chemical kinetics examines the rate at which chemical reactions occur and the molecular pathways (mechanisms) they follow. For any general reaction:

A + B → Products

The relationship between reactant concentrations and reaction rate is expressed mathematically by the rate equation:

Rate = k [A]^m [B]ⁿ

Key Kinetic Definitions

Order of reaction with respect to a reactant: The power to which the concentration of that reactant is raised in the rate equation (m for A, n for B).
Overall order of reaction: The sum of individual reaction orders: Overall Order = m + n.
Rate constant (k): The proportionality constant linking rate to reactant concentrations at a specific temperature. The value of k is independent of concentration but increases exponentially with temperature.

Orders of Reaction (0, 1, and 2)

Zero Order (m = 0): Rate is independent of [A]. Doubling [A] has zero effect on rate (2⁰ = 1).
First Order (m = 1): Rate is directly proportional to [A]. Doubling [A] doubles the rate (2¹ = 2).
Second Order (m = 2): Rate is proportional to [A]². Doubling [A] quadruples the rate (2² = 4); tripling [A] increases rate ninefold (3² = 9).

Deducing the Units of the Rate Constant (k)

Units of k vary depending on overall reaction order. Rearrange the rate equation to isolate k and substitute standard units (Rate in mol dm⁻³ s⁻¹; Concentration in mol dm⁻³):

Overall Order	Representative Rate Law	Rearranged Expression for k	Derived Units of k
Zero	Rate = k	k = Rate	mol dm⁻³ s⁻¹
First	Rate = k [A]	k = Rate / [A]	s⁻¹
Second	Rate = k [A]²	k = Rate / [A]²	mol⁻¹ dm³ s⁻¹
Third	Rate = k [A]² [B]	k = Rate / ([A]² [B])	mol⁻² dm⁶ s⁻¹

2. Determining Reaction Orders Experimentally

Reaction orders cannot be deduced from the balanced stoichiometric equation; they must be determined experimentally.

The Initial Rates Method

A series of experiments is carried out at constant temperature where the initial concentration of one reactant is varied while all other reactant concentrations are kept constant. The initial rate is measured for each run (e.g. from the initial gradient of a concentration-time graph or an iodine clock timer).

Continuous Monitoring

Progress of a reaction can be continuously monitored using:

Gas syringe: Measuring volume of gas evolved over time (e.g. Mg + 2HCl → MgCl₂ + H₂).
Balance: Tracking mass loss over time for reactions producing heavy gases like CO₂.
Colorimetry: Measuring light absorbance as a coloured species is formed or consumed.

A graph of concentration against time provides immediate evidence of reaction order:

Zero order: Constant negative gradient (straight line). Rate is constant regardless of concentration.
First order: Exponential decay curve. Beyond the OxfordAQA specification: a first-order reaction has a constant half-life (t_½), so the time for the concentration to halve does not depend on the starting concentration; t_½ = ln 2 / k = 0.693 / k.
Second order: Rapid initial drop followed by an extended shallow curve. Beyond the OxfordAQA specification: successive half-lives double each time the concentration halves.

Finding a Rate from a Concentration-Time Graph

The rate at any moment is the gradient of the curve at that moment.

Draw a tangent to the curve at the time you want.
Pick two points far apart on the tangent.
Gradient = change in concentration ÷ change in time.
The rate is the size of this gradient. For a reactant the gradient is negative, so drop the minus sign.

The initial rate is the gradient of the tangent drawn at t = 0.

Rate-Concentration Graphs

Plot the rate against the concentration of one reactant. The shape of the graph shows the order with respect to that reactant:

Order 0: a horizontal line. The rate does not change when [A] changes.
Order 1: a straight line through the origin. Rate ∝ [A].
Order 2: an upward curve through the origin. Rate ∝ [A]².

Required Practical 8 (RP8)

RP8 uses both methods on this page: an initial rate method (for example an iodine clock) and a continuous monitoring method (for example collecting gas in a syringe). See RP8: Measuring Reaction Rates.

3. The Rate-Determining Step (RDS)

Most chemical reactions proceed through a multi-step sequence of elementary collisions known as the reaction mechanism. One step in this sequence is significantly slower than all others; this is the rate-determining step (RDS).

Rules Connecting Rate Equations to Reaction Mechanisms

Only species that take part in the rate-determining step, or in fast equilibrium steps that precede it, appear in the experimental rate equation.
The order of reaction with respect to each reactant equals the stoichiometric coefficient of that species in (or prior to) the rate-determining step.
A catalyst appears in the rate equation only if it takes part in the rate-determining step or a step before it. It is regenerated in a later step, so it does not appear in the overall chemical equation.

4. The Arrhenius Equation

The rate constant k increases exponentially with temperature because higher temperatures give a much larger fraction of molecules kinetic energy equal to or greater than the activation energy (E_a). This relationship is quantified by the Arrhenius equation:

k = A × e^{-E_a / RT}

Where:

k: Rate constant (units depend on overall reaction order).
A: Pre-exponential factor (frequency factor / Arrhenius constant), representing the frequency of collisions with correct orientation (same units as k).
E_a: Activation energy in J mol⁻¹.
R: Gas constant = 8.314 J mol⁻¹ K⁻¹.
T: Absolute temperature in Kelvin (K): T(K) = θ(°C) + 273.15.

The Logarithmic Form

Taking the natural logarithm (ln) of both sides converts the exponential equation into a linear equation matching y = mx + c:

ln k = (-E_a / R) × (1 / T) + ln A

Comparing this with y = mx + c:

y: ln k
x: 1 / T
m (gradient): -E_a / R
c (y-intercept): ln A

Using k = Ae^-E_a/RT Directly

You can also put numbers straight into the equation. Follow these steps:

Change E_a into J mol⁻¹ and T into K.
Work out E_a/RT first. It has no units.
Find e^-E_a/RT with the e^x key, then multiply by A.

Worked Example 3 below finds k at two temperatures this way.

5. Arrhenius Graph Analysis

By measuring the rate constant k at several different temperatures and plotting ln k against 1/T, the activation energy and pre-exponential factor can be determined graphically:

Calculating E_a from Gradient: Avoiding the Minus Sign Trap

The line on an Arrhenius plot has a negative gradient. When calculating E_a:
E_a = -(Gradient) × R = -(negative value) × 8.314 J mol⁻¹ K⁻¹.
The two negative signs cancel, giving a positive activation energy. Remember to divide by 1000 if the exam question requests E_a in kJ mol⁻¹!

6. Worked Calculations

Worked Example 1: Deducing Rate Law and k from Initial Rates Data

For the reaction: 2A + B + C → Products, the following initial rate data were obtained:
Exp 1: [A]=0.10, [B]=0.10, [C]=0.10 mol dm⁻³; Rate = 2.0 × 10⁻⁴ mol dm⁻³ s⁻¹
Exp 2: [A]=0.20, [B]=0.10, [C]=0.10 mol dm⁻³; Rate = 8.0 × 10⁻⁴ mol dm⁻³ s⁻¹
Exp 3: [A]=0.10, [B]=0.20, [C]=0.10 mol dm⁻³; Rate = 4.0 × 10⁻⁴ mol dm⁻³ s⁻¹
Exp 4: [A]=0.10, [B]=0.10, [C]=0.20 mol dm⁻³; Rate = 2.0 × 10⁻⁴ mol dm⁻³ s⁻¹
Deduce the rate equation, calculate the value of the rate constant k, and state its units.

Step 1: Determine reaction orders:

Compare Exp 1 and Exp 2: [B] and [C] constant. [A] doubles (×2), Rate quadruples (×4). Since 2² = 4, the reaction is second order with respect to A.
Compare Exp 1 and Exp 3: [A] and [C] constant. [B] doubles (×2), Rate doubles (×2). Since 2¹ = 2, the reaction is first order with respect to B.
Compare Exp 1 and Exp 4: [A] and [B] constant. [C] doubles (×2), Rate remains unchanged (×1). Since 2⁰ = 1, the reaction is zero order with respect to C.

Step 2: State the rate equation:

Rate = k [A]² [B]

Step 3: Calculate k using Experiment 1:

k = Rate / ([A]² [B]) = (2.0 × 10⁻⁴) / ((0.10)² × 0.10) = (2.0 × 10⁻⁴) / (1.0 × 10⁻³) = 0.20

Step 4: Deduce units of k:

Units = (mol dm⁻³ s⁻¹) / ((mol dm⁻³)² × (mol dm⁻³)) = (mol dm⁻³ s⁻¹) / (mol³ dm⁻⁹) = mol⁻² dm⁶ s⁻¹

Final Answer: k = 0.20 mol⁻² dm⁶ s⁻¹

Worked Example 2: Determining Activation Energy from Arrhenius Gradient

An Arrhenius plot of ln k against 1/T for the decomposition of dinitrogen pentoxide yields a straight line with a gradient of -1.24 × 10⁴ K.
Calculate the activation energy (E_a) in kJ mol⁻¹.

Step 1: State the relationship between gradient and activation energy:

Gradient = -E_a / R

Step 2: Rearrange for E_a:

E_a = -(Gradient) × R

E_a = -(-1.24 × 10⁴ K) × 8.314 J mol⁻¹ K⁻¹ = +1.031 × 10⁵ J mol⁻¹

Step 3: Convert to kJ mol⁻¹:

E_a = 1.031 × 10⁵ / 1000 = 103.1 kJ mol⁻¹

Worked Example 3: Using k = Ae^-Ea/RT Directly

A first-order reaction has E_a = 50.0 kJ mol⁻¹ and A = 1.00 × 10¹⁰ s⁻¹. R = 8.31 J K⁻¹ mol⁻¹.
Calculate k at 298 K and at 308 K.

Step 1: Change E_a into J mol⁻¹: 50.0 kJ mol⁻¹ = 50 000 J mol⁻¹.

Step 2: At 298 K:

E_a/RT = 50 000 / (8.31 × 298) = 20.19

k = 1.00 × 10¹⁰ × e^-20.19 = 17.0 s⁻¹

Step 3: At 308 K:

E_a/RT = 50 000 / (8.31 × 308) = 19.54

k = 1.00 × 10¹⁰ × e^-19.54 = 32.8 s⁻¹

Keep the unrounded value of E_a/RT in your calculator. Rounding it first changes the last figure of k.

Conclusion: a 10 K rise roughly doubles k for this reaction (32.8 ÷ 17.0 = 1.9).

Exam-Style Practice Questions

Test your understanding of these core syllabus concepts with targeted questions and detailed explanations.

Question 1: A reaction has the rate equation Rate = k [NO]² [O₂]. If the concentration of NO is doubled while the concentration of O₂ is halved, how does the initial rate change?

A. Rate increases by a factor of 2 B. Rate increases by a factor of 4 C. Rate remains unchanged D. Rate decreases by a factor of 2

Show Answer & Explanation

Correct Answer: A

Doubling [NO] quadruples the rate (2² = 4). Halving [O₂] halves the rate (0.5¹ = 0.5). Combined effect: 4 × 0.5 = 2 times faster.

Question 2: What are the units of the rate constant k for a reaction that is first order overall?

A. mol dm⁻³ s⁻¹ B. s⁻¹ C. mol⁻¹ dm³ s⁻¹ D. mol⁻² dm⁶ s⁻¹

Show Answer & Explanation

Correct Answer: B

k = Rate / [A] = (mol dm⁻³ s⁻¹) / (mol dm⁻³) = s⁻¹.

Question 3: A reaction mechanism consists of two steps: Step 1: NO₂ + NO₂ → NO₃ + NO (slow); Step 2: NO₃ + CO → NO₂ + CO₂ (fast). What is the expected rate equation?

A. Rate = k [NO₂] [CO] B. Rate = k [NO₂]² C. Rate = k [NO₃] [CO] D. Rate = k [NO₂]² [CO]

Show Answer & Explanation

Correct Answer: B

The slow step is the rate-determining step. It involves two molecules of NO₂ colliding, so the rate law is Rate = k [NO₂]². Species in subsequent fast steps (CO) do not enter the rate equation.

Question 4: If temperature is increased from 300 K to 310 K, the rate of many reactions approximately doubles. Why does this occur?

A. The activation energy E_a is halved at the higher temperature B. Collision frequency doubles C. The fraction of molecules with energy ≥ E_a increases substantially D. The pre-exponential factor A increases exponentially

Show Answer & Explanation

Correct Answer: C

The Maxwell-Boltzmann distribution shows that a modest 10 K temperature rise greatly increases the population of molecules possessing kinetic energy greater than or equal to E_a.

Question 5 (3 marks): A first-order reaction has E_a = 75.0 kJ mol⁻¹ and A = 2.00 × 10¹¹ s⁻¹. Calculate k at 350 K. R = 8.31 J K⁻¹ mol⁻¹.

Show Answer & Mark Points

E_a/RT = 75 000 / (8.31 × 350) = 25.79 (1)
k = 2.00 × 10¹¹ × e^-25.79 (1)
k = 1.27 s⁻¹ (1)

Rate Equations & Arrhenius Analysis

1. Rate Equations, Orders & The Rate Constant

Orders of Reaction (0, 1, and 2)

Deducing the Units of the Rate Constant (k)

2. Determining Reaction Orders Experimentally

The Initial Rates Method

Continuous Monitoring

Finding a Rate from a Concentration-Time Graph

Rate-Concentration Graphs

Required Practical 8 (RP8)

3. The Rate-Determining Step (RDS)

Rules Connecting Rate Equations to Reaction Mechanisms

4. The Arrhenius Equation

The Logarithmic Form

Using k = Ae^-E_a/RT Directly

5. Arrhenius Graph Analysis

6. Worked Calculations

Go deeper

Exam-Style Practice Questions

Unit Flashcards

Exam Practice & AI Grader

1. Rate Equations, Orders & The Rate Constant

Orders of Reaction (0, 1, and 2)

Deducing the Units of the Rate Constant (k)

2. Determining Reaction Orders Experimentally

The Initial Rates Method

Continuous Monitoring

Finding a Rate from a Concentration-Time Graph

Rate-Concentration Graphs

Required Practical 8 (RP8)

3. The Rate-Determining Step (RDS)

Rules Connecting Rate Equations to Reaction Mechanisms

4. The Arrhenius Equation

The Logarithmic Form

Using k = Ae-Ea/RT Directly

5. Arrhenius Graph Analysis

6. Worked Calculations

Go deeper

Exam-Style Practice Questions

Unit Flashcards

Exam Practice & AI Grader

Using k = Ae^-E_a/RT Directly