Rate equations (A-level only)

What you'll learn

How a rate equation links reaction rate to reactant concentrations.
How to deduce orders of reaction from initial-rate data.
How to calculate the rate constant, including its units.
How temperature affects $k$ using the Arrhenius equation.

Starting point: rate and concentration

The rate of reaction tells you how quickly reactants are used up or products are formed. In rate-equation questions, rate is usually measured in mol dm⁻³ s⁻¹.

Square brackets mean concentration. For example, $[A][\text{A}]$ means the concentration of A, usually in mol dm⁻³.

Definition

Initial rate

The initial rate is the rate right at the start of the reaction, before the concentrations have changed significantly.

Initial rates are useful because you know the starting concentrations accurately, so you can compare how changing one concentration affects the rate.

The rate equation

For a reaction involving reactants A and B, the rate equation has the form:

rate=k[A]m[B]n\text{rate}=k[\text{A}]^m[\text{B}]^n

This is A-level-only kinetics: unlike simpler rate ideas, the powers in the rate equation come from experimental data, not just from the balanced chemical equation.

Definition

Order of reaction

The order of reaction with respect to a reactant is the power to which that reactant’s concentration is raised in the experimentally determined rate equation.

Definition

Rate constant

The rate constant, $k$ , is the proportionality constant in the rate equation for a particular reaction at a fixed temperature.

On this specification, the individual orders $m$ and $n$ are restricted to 0, 1 and 2.

The three common concentration effects look like this:

Graphs showing zero, first and second order rate relationships with concentration

What the orders mean

Zero order: changing that reactant’s concentration has no effect on rate.
First order: doubling that reactant’s concentration doubles the rate.
Second order: doubling that reactant’s concentration makes the rate four times bigger.

The overall order is the sum of the individual orders. For $rate=k[A]2[B]\text{rate}=k[\text{A}]^2[\text{B}]$ , the overall order is 3.

Key Idea

Orders are experimental

Do not assume the orders from the balanced equation. They are found by comparing rate data from experiments.

Example

Using a known rate equation

A reaction has $rate=k[A]2[B]\text{rate}=k[\text{A}]^2[\text{B}]$ . At a particular temperature, $k = 0.480$ dm⁶ mol⁻² s⁻¹, $[A]=0.100[\text{A}]=0.100$ mol dm⁻³ and $[B]=0.200[\text{B}]=0.200$ mol dm⁻³.

Use the powers in the rate equation: A is squared, while B is first order.
Substitute the values: $rate=0.480×(0.100)2×0.200\text{rate}=0.480 \times (0.100)^2 \times 0.200$ .
Calculate the value: $rate=9.60×10−4\text{rate}=9.60 \times 10^{-4}$ mol dm⁻³ s⁻¹.

Deducing orders from initial-rate data

The usual method is to compare two experiments where only one reactant concentration changes.

If concentration changes by a factor and the rate changes by another factor, match the change to the order:

Concentration changes but rate does not change → zero order.
Concentration doubles and rate doubles → first order.
Concentration doubles and rate quadruples → second order.

Tip

Factor method

When one reactant changes and the others are constant, use: concentration factor raised to the order equals rate factor.

Example

Deducing a rate equation

Initial-rate data are collected at the same temperature:

Experiment 1: $[A]=0.100[\text{A}]=0.100$ mol dm⁻³, $[B]=0.100[\text{B}]=0.100$ mol dm⁻³, rate = $2.00 \times 10^{-4}$ mol dm⁻³ s⁻¹
Experiment 2: $[A]=0.200[\text{A}]=0.200$ mol dm⁻³, $[B]=0.100[\text{B}]=0.100$ mol dm⁻³, rate = $8.00 \times 10^{-4}$ mol dm⁻³ s⁻¹
Experiment 3: $[A]=0.200[\text{A}]=0.200$ mol dm⁻³, $[B]=0.300[\text{B}]=0.300$ mol dm⁻³, rate = $2.40 \times 10^{-3}$ mol dm⁻³ s⁻¹

Compare experiments 1 and 2: $[B][\text{B}]$ is constant, $[A][\text{A}]$ doubles, and the rate becomes four times larger. Since $2^2=4$ , the order with respect to A is 2.
Compare experiments 2 and 3: $[A][\text{A}]$ is constant, $[B][\text{B}]$ triples, and the rate triples. Since $3^1=3$ , the order with respect to B is 1.
Write the rate equation using those powers: $rate=k[A]2[B]\text{rate}=k[\text{A}]^2[\text{B}]$ .

Common Mistake

Using equation coefficients as orders

The balanced equation might suggest possible reactants, but it does not usually tell you the orders. Use the rate data.

Calculating the rate constant and its units

Once you know the rate equation, rearrange it to find $k$ .

For example, if:

rate=k[A]2[B]\text{rate}=k[\text{A}]^2[\text{B}]

then:

k=rate[A]2[B]k=\frac{\text{rate}}{[\text{A}]^2[\text{B}]}

The units of $k$ depend on the overall order:

Overall order 0: units of $k$ are mol dm⁻³ s⁻¹.
Overall order 1: units of $k$ are s⁻¹.
Overall order 2: units of $k$ are dm³ mol⁻¹ s⁻¹.
Overall order 3: units of $k$ are dm⁶ mol⁻² s⁻¹.

Example

Finding k and its units

Using experiment 1 from the previous example, the rate equation is $rate=k[A]2[B]\text{rate}=k[\text{A}]^2[\text{B}]$ .

Rearrange for $k$ : $k=rate[A]2[B]k=\frac{\text{rate}}{[\text{A}]^2[\text{B}]}$ .
Substitute the data: $k=2.00×10−4(0.100)2×0.100=0.200k=\frac{2.00 \times 10^{-4}}{(0.100)^2 \times 0.100}=0.200$ .
The reaction is third order overall, so the units are dm⁶ mol⁻² s⁻¹. Therefore, $k = 0.200$ dm⁶ mol⁻² s⁻¹.

Common Mistake

Forgetting the units of k

The rate constant does not always have units of mol dm⁻³ s⁻¹. Its units change with the overall order of the reaction.

Zero-order reactions from concentration–time graphs

For a zero-order reaction, the rate does not depend on the concentration of that reactant. If you plot concentration against time, the graph is a straight line with a constant negative gradient.

For the disappearance of a reactant in a simple 1:1 situation:

gradientk=\text{magnitude of the gradient}

Example

Finding k from a zero-order graph

A zero-order reactant decreases from 0.160 mol dm⁻³ to 0.040 mol dm⁻³ in 300 s.

Calculate the gradient of the concentration–time graph: $0.040−0.160300−0=−4.00×10−4\frac{0.040-0.160}{300-0}=-4.00 \times 10^{-4}$ mol dm⁻³ s⁻¹.
The gradient is negative because the reactant concentration is decreasing.
For the rate constant, take the magnitude: $\times 10^{-4}$ mol dm⁻³ s⁻¹.

Temperature and the rate constant

Increasing temperature increases the rate constant $k$ . This means the reaction is faster at the same concentrations.

The reason is that a higher temperature gives particles more kinetic energy, so a greater proportion of collisions have energy at least equal to the activation energy, $E_a$ .

Key Idea

Temperature changes k

Changing concentration changes the rate through the concentration terms. Changing temperature changes the value of $k$ .

The Arrhenius equation

The relationship between $k$ and temperature is given by:

k=Ae^{-E_a/RT}

where:

$A$ is the Arrhenius constant.
$E_a$ is the activation energy.
$R$ is the gas constant, usually 8.31 J K⁻¹ mol⁻¹.
$T$ is the temperature in K.

You will normally be given the equation and the value of $R$ when needed.

Common Mistake

Units in Arrhenius calculations

Use temperature in K, not °C. If $R$ is in J K⁻¹ mol⁻¹, then $E_a$ must be in J mol⁻¹, not kJ mol⁻¹.

Example

Calculating k using the Arrhenius equation

For a reaction, $\times 10^{12}$ s⁻¹, $E_a=75.0$ kJ mol⁻¹ and $T = 298$ K.

Convert the activation energy: 75.0 kJ mol⁻¹ = 75000 J mol⁻¹.
Substitute into the exponent: $−EaRT=−750008.31×298=−30.3-\frac{E_a}{RT}=-\frac{75000}{8.31 \times 298}=-30.3$ .
Calculate $k$ : $\times 10^{12} \times e^{-30.3}=8.3 \times 10^{-2}$ s⁻¹.

The Arrhenius plot

The Arrhenius equation can be rearranged into a straight-line form:

ln⁡k=−EaRT+ln⁡A\ln k=-\frac{E_a}{RT}+\ln A

This can also be written as:

ln⁡k=−EaR(1T)+ln⁡A\ln k=-\frac{E_a}{R}\left(\frac{1}{T}\right)+\ln A

So a graph of $ln⁡k\ln k$ against $1T\frac{1}{T}$ is a straight line with gradient $−EaR-\frac{E_a}{R}$ and intercept $ln⁡A\ln A$ .

Arrhenius plot showing ln k against 1 over T with gradient minus Ea over R and intercept ln A

Example

Finding activation energy from a gradient

An Arrhenius plot of $ln⁡k\ln k$ against $1T\frac{1}{T}$ has a gradient of -7250 K.

Use the gradient relationship: $gradient=−EaR\text{gradient}=-\frac{E_a}{R}$ .
Substitute the gradient: $−7250=−Ea8.31-7250=-\frac{E_a}{8.31}$ , so $Ea=7250×8.31=6.02×104E_a=7250 \times 8.31=6.02 \times 10^4$ J mol⁻¹.
Convert to kJ mol⁻¹: $E_a=60.2$ kJ mol⁻¹.

Exam technique

In the exam

Compare experiments where only one concentration changes; use factor changes to find the order.
Never assume orders from the balanced equation unless the question explicitly gives the rate equation.
For $k$ , derive the units from the rate equation; for Arrhenius work, use K for temperature and J mol⁻¹ for activation energy.

Self review

Check yourself

If doubling $[A][\text{A}]$ leaves the rate unchanged, what is the order with respect to A?
For $rate=k[A][B]2\text{rate}=k[\text{A}][\text{B}]^2$ , how would you find the units of $k$ ?
On a graph of $ln⁡k\ln k$ against $1T\frac{1}{T}$ , how is $E_a$ found from the gradient?

What you'll learn

How a rate equation links reaction rate to reactant concentrations.
How to deduce orders of reaction from initial-rate data.
How to calculate the rate constant, including its units.
How temperature affects $k$ using the Arrhenius equation.

Starting point: rate and concentration

The rate of reaction tells you how quickly reactants are used up or products are formed. In rate-equation questions, rate is usually measured in mol dm⁻³ s⁻¹.

Square brackets mean concentration. For example, $[A][\text{A}]$ means the concentration of A, usually in mol dm⁻³.

Definition

Initial rate

The initial rate is the rate right at the start of the reaction, before the concentrations have changed significantly.

Initial rates are useful because you know the starting concentrations accurately, so you can compare how changing one concentration affects the rate.

The rate equation

For a reaction involving reactants A and B, the rate equation has the form:

rate=k[A]m[B]n\text{rate}=k[\text{A}]^m[\text{B}]^n

This is A-level-only kinetics: unlike simpler rate ideas, the powers in the rate equation come from experimental data, not just from the balanced chemical equation.

Definition

Order of reaction

The order of reaction with respect to a reactant is the power to which that reactant’s concentration is raised in the experimentally determined rate equation.

Definition

Rate constant

The rate constant, $k$ , is the proportionality constant in the rate equation for a particular reaction at a fixed temperature.

On this specification, the individual orders $m$ and $n$ are restricted to 0, 1 and 2.

The three common concentration effects look like this:

Graphs showing zero, first and second order rate relationships with concentration

What the orders mean

Zero order: changing that reactant’s concentration has no effect on rate.
First order: doubling that reactant’s concentration doubles the rate.
Second order: doubling that reactant’s concentration makes the rate four times bigger.

The overall order is the sum of the individual orders. For $rate=k[A]2[B]\text{rate}=k[\text{A}]^2[\text{B}]$ , the overall order is 3.

Key Idea

Orders are experimental

Do not assume the orders from the balanced equation. They are found by comparing rate data from experiments.

Example

Using a known rate equation

Use the powers in the rate equation: A is squared, while B is first order.
Substitute the values: $rate=0.480×(0.100)2×0.200\text{rate}=0.480 \times (0.100)^2 \times 0.200$ .
Calculate the value: $rate=9.60×10−4\text{rate}=9.60 \times 10^{-4}$ mol dm⁻³ s⁻¹.

Deducing orders from initial-rate data

The usual method is to compare two experiments where only one reactant concentration changes.

If concentration changes by a factor and the rate changes by another factor, match the change to the order:

Concentration changes but rate does not change → zero order.
Concentration doubles and rate doubles → first order.
Concentration doubles and rate quadruples → second order.

Tip

Factor method

When one reactant changes and the others are constant, use: concentration factor raised to the order equals rate factor.

Example

Deducing a rate equation

Initial-rate data are collected at the same temperature:

Experiment 1: $[A]=0.100[\text{A}]=0.100$ mol dm⁻³, $[B]=0.100[\text{B}]=0.100$ mol dm⁻³, rate = $2.00 \times 10^{-4}$ mol dm⁻³ s⁻¹
Experiment 2: $[A]=0.200[\text{A}]=0.200$ mol dm⁻³, $[B]=0.100[\text{B}]=0.100$ mol dm⁻³, rate = $8.00 \times 10^{-4}$ mol dm⁻³ s⁻¹
Experiment 3: $[A]=0.200[\text{A}]=0.200$ mol dm⁻³, $[B]=0.300[\text{B}]=0.300$ mol dm⁻³, rate = $2.40 \times 10^{-3}$ mol dm⁻³ s⁻¹

Compare experiments 1 and 2: $[B][\text{B}]$ is constant, $[A][\text{A}]$ doubles, and the rate becomes four times larger. Since $2^2=4$ , the order with respect to A is 2.
Compare experiments 2 and 3: $[A][\text{A}]$ is constant, $[B][\text{B}]$ triples, and the rate triples. Since $3^1=3$ , the order with respect to B is 1.
Write the rate equation using those powers: $rate=k[A]2[B]\text{rate}=k[\text{A}]^2[\text{B}]$ .

Common Mistake

Using equation coefficients as orders

The balanced equation might suggest possible reactants, but it does not usually tell you the orders. Use the rate data.

Calculating the rate constant and its units

Once you know the rate equation, rearrange it to find $k$ .

For example, if:

rate=k[A]2[B]\text{rate}=k[\text{A}]^2[\text{B}]

then:

k=rate[A]2[B]k=\frac{\text{rate}}{[\text{A}]^2[\text{B}]}

The units of $k$ depend on the overall order:

Overall order 0: units of $k$ are mol dm⁻³ s⁻¹.
Overall order 1: units of $k$ are s⁻¹.
Overall order 2: units of $k$ are dm³ mol⁻¹ s⁻¹.
Overall order 3: units of $k$ are dm⁶ mol⁻² s⁻¹.

Example

Finding k and its units

Using experiment 1 from the previous example, the rate equation is $rate=k[A]2[B]\text{rate}=k[\text{A}]^2[\text{B}]$ .

Rearrange for $k$ : $k=rate[A]2[B]k=\frac{\text{rate}}{[\text{A}]^2[\text{B}]}$ .
Substitute the data: $k=2.00×10−4(0.100)2×0.100=0.200k=\frac{2.00 \times 10^{-4}}{(0.100)^2 \times 0.100}=0.200$ .
The reaction is third order overall, so the units are dm⁶ mol⁻² s⁻¹. Therefore, $k = 0.200$ dm⁶ mol⁻² s⁻¹.

Common Mistake

Forgetting the units of k

The rate constant does not always have units of mol dm⁻³ s⁻¹. Its units change with the overall order of the reaction.

Zero-order reactions from concentration–time graphs

For a zero-order reaction, the rate does not depend on the concentration of that reactant. If you plot concentration against time, the graph is a straight line with a constant negative gradient.

For the disappearance of a reactant in a simple 1:1 situation:

gradientk=\text{magnitude of the gradient}

Example

Finding k from a zero-order graph

A zero-order reactant decreases from 0.160 mol dm⁻³ to 0.040 mol dm⁻³ in 300 s.

Calculate the gradient of the concentration–time graph: $0.040−0.160300−0=−4.00×10−4\frac{0.040-0.160}{300-0}=-4.00 \times 10^{-4}$ mol dm⁻³ s⁻¹.
The gradient is negative because the reactant concentration is decreasing.
For the rate constant, take the magnitude: $\times 10^{-4}$ mol dm⁻³ s⁻¹.

Temperature and the rate constant

Increasing temperature increases the rate constant $k$ . This means the reaction is faster at the same concentrations.

The reason is that a higher temperature gives particles more kinetic energy, so a greater proportion of collisions have energy at least equal to the activation energy, $E_a$ .

Key Idea

Temperature changes k

Changing concentration changes the rate through the concentration terms. Changing temperature changes the value of $k$ .

The Arrhenius equation

The relationship between $k$ and temperature is given by:

k=Ae^{-E_a/RT}

where:

$A$ is the Arrhenius constant.
$E_a$ is the activation energy.
$R$ is the gas constant, usually 8.31 J K⁻¹ mol⁻¹.
$T$ is the temperature in K.

You will normally be given the equation and the value of $R$ when needed.

Common Mistake

Units in Arrhenius calculations

Use temperature in K, not °C. If $R$ is in J K⁻¹ mol⁻¹, then $E_a$ must be in J mol⁻¹, not kJ mol⁻¹.

Example

Calculating k using the Arrhenius equation

For a reaction, $\times 10^{12}$ s⁻¹, $E_a=75.0$ kJ mol⁻¹ and $T = 298$ K.

Convert the activation energy: 75.0 kJ mol⁻¹ = 75000 J mol⁻¹.
Substitute into the exponent: $−EaRT=−750008.31×298=−30.3-\frac{E_a}{RT}=-\frac{75000}{8.31 \times 298}=-30.3$ .
Calculate $k$ : $\times 10^{12} \times e^{-30.3}=8.3 \times 10^{-2}$ s⁻¹.

The Arrhenius plot

The Arrhenius equation can be rearranged into a straight-line form:

ln⁡k=−EaRT+ln⁡A\ln k=-\frac{E_a}{RT}+\ln A

This can also be written as:

ln⁡k=−EaR(1T)+ln⁡A\ln k=-\frac{E_a}{R}\left(\frac{1}{T}\right)+\ln A

So a graph of $ln⁡k\ln k$ against $1T\frac{1}{T}$ is a straight line with gradient $−EaR-\frac{E_a}{R}$ and intercept $ln⁡A\ln A$ .

Arrhenius plot showing ln k against 1 over T with gradient minus Ea over R and intercept ln A

Example

Finding activation energy from a gradient

An Arrhenius plot of $ln⁡k\ln k$ against $1T\frac{1}{T}$ has a gradient of -7250 K.

Use the gradient relationship: $gradient=−EaR\text{gradient}=-\frac{E_a}{R}$ .
Substitute the gradient: $−7250=−Ea8.31-7250=-\frac{E_a}{8.31}$ , so $Ea=7250×8.31=6.02×104E_a=7250 \times 8.31=6.02 \times 10^4$ J mol⁻¹.
Convert to kJ mol⁻¹: $E_a=60.2$ kJ mol⁻¹.

Exam technique

In the exam

Compare experiments where only one concentration changes; use factor changes to find the order.
Never assume orders from the balanced equation unless the question explicitly gives the rate equation.
For $k$ , derive the units from the rate equation; for Arrhenius work, use K for temperature and J mol⁻¹ for activation energy.

Self review

Check yourself

If doubling $[A][\text{A}]$ leaves the rate unchanged, what is the order with respect to A?
For $rate=k[A][B]2\text{rate}=k[\text{A}][\text{B}]^2$ , how would you find the units of $k$ ?
On a graph of $ln⁡k\ln k$ against $1T\frac{1}{T}$ , how is $E_a$ found from the gradient?

What you'll learn

Starting point: rate and concentration

The rate equation

What the orders mean

Deducing orders from initial-rate data

Calculating the rate constant and its units

Zero-order reactions from concentration–time graphs

Temperature and the rate constant

The Arrhenius equation

The Arrhenius plot

Review Rate equations (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Rate equations (A-level only)

What you'll learn

Starting point: rate and concentration

The rate equation

What the orders mean

Deducing orders from initial-rate data

Calculating the rate constant and its units

Zero-order reactions from concentration–time graphs

Temperature and the rate constant

The Arrhenius equation

The Arrhenius plot

Review Rate equations (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Rate equations (A-level only)

What you'll learn

Starting point: rate and concentration

The rate equation

What the orders mean

Deducing orders from initial-rate data

Calculating the rate constant and its units

Zero-order reactions from concentration–time graphs

Temperature and the rate constant

The Arrhenius equation

The Arrhenius plot

Review Rate equations (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Rate equations (A-level only) Revision Guide

Rate equations (A-level only)

What you'll learn

Starting point: rate and concentration

The rate equation

What the orders mean

Deducing orders from initial-rate data

Calculating the rate constant and its units

Zero-order reactions from concentration–time graphs

Temperature and the rate constant

The Arrhenius equation

The Arrhenius plot

Review Rate equations (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Rate equations (A-level only) Revision Guide