Rate Constant and the Arrhenius Equation — AQA A-level Chemistry

SPEC 3.1.9.1 • RATE EQUATIONS

Rate Constants & the Arrhenius Equation

Understanding how temperature affects reaction rates mathematically

Why Does k Change with Temperature?

In Lesson 9.01, we learned that the rate constant

k

appears in every rate equation. But

k

is not actually constant — it changes with temperature. Understanding this relationship is crucial for predicting and controlling reaction rates.

Key Point

The rate constant

k

increases as temperature increases. This is why reactions go faster at higher temperatures, even when concentrations stay the same.

Recall from earlier kinetics work that increasing temperature has two effects:

1

Molecules move faster — more frequent collisions

2

More molecules exceed activation energy — more successful collisions

The second effect is far more important. Even a small temperature increase dramatically increases the proportion of molecules with energy

\geq E_{a}

. This is described mathematically by the Arrhenius equation.

Rule of Thumb

For many reactions near room temperature with

E_{a} \approx 50

kJ mol⁻¹, a 10 K rise in temperature approximately doubles the rate. This is because the fraction of molecules exceeding

E_{a}

roughly doubles with each 10 K increase.

The Maxwell-Boltzmann Connection

The Maxwell-Boltzmann distribution shows how molecular energies are distributed at a given temperature. When temperature increases, the curve shifts right and flattens — crucially, the area under the curve beyond

E_{a}

increases significantly.

Key Insight

The shaded areas show molecules with energy

\geq E_{a}

. At higher temperature (red), this area is much larger — meaning a greater fraction of collisions are successful.

The Arrhenius Equation

The Arrhenius equation quantifies the relationship between the rate constant

k

, temperature

T

, and activation energy

E_{a}

:

The Arrhenius Equation

k = A e^{- \frac{E _{a}}{RT}}

SymbolNameUnitsMeaning

k

Rate constantDepends on orderHow fast the reaction proceeds

A

Arrhenius constantSame as

k

Pre-exponential factor (frequency factor)

E_{a}

Activation energyJ mol⁻¹Minimum energy for reaction

R

Gas constant8.31 J K⁻¹ mol⁻¹Universal constant

T

TemperatureK (Kelvin)Absolute temperature

Understanding Each Term

The exponential term

e^{- E_{a} / RT}

is called the Boltzmann factor. It represents the fraction of molecules with energy greater than or equal to the activation energy. As temperature increases, this fraction increases exponentially.

Why the Negative Sign?

The negative exponent means that as

E_{a}

increases,

k

decreases. High activation energy = slower reaction. As

T

increases, the negative exponent becomes less negative, so

k

increases.

The Arrhenius Constant $A$ is also called the "pre-exponential factor" or "frequency factor". It represents:

The frequency of collisions between reactant molecules

The fraction of collisions with correct orientation

The theoretical maximum rate constant if all collisions were successful

Exam Note

A

has the same units as

k

because the exponential term

e^{- E_{a} / RT}

is dimensionless (pure number with no units).

Worked Example 1

1

Calculate k from A, Ea, and T

A reaction has an Arrhenius constant

A = 4.5 \times 1 0^{11}

s⁻¹ and an activation energy

E_{a} = 85

kJ mol⁻¹. Calculate the rate constant

k

at 350 K. (R = 8.31 J K⁻¹ mol⁻¹)

1

Convert $E_{a}$ to J mol⁻¹:

E_{a} = 85 \times 1000 = 85000

J mol⁻¹

2

Calculate the exponent:

- \frac{E _{a}}{RT} = - \frac{85000}{8.31 \times 350} = - \frac{85000}{2908.5} = - 29.22

3

Calculate $k$ :

k = 4.5 \times 1 0^{11} \times e^{- 29.22} = 4.5 \times 1 0^{11} \times 2.03 \times 1 0^{- 13}

4

Answer:

k = 0.091

s⁻¹ (or

9.1 \times 1 0^{- 2}

s⁻¹)

Worked Example 1b

1b

Calculate Ea from k, A, and T

Calculate the activation energy of a reaction which takes place at 400 K, where the rate constant is

6.25 \times 1 0^{- 4}

s⁻¹.

A = 4.6 \times 1 0^{13}

s⁻¹ and

R = 8.31

J K⁻¹ mol⁻¹.

1

Use the logarithmic form:

ln k = ln A - \frac{E _{a}}{RT}

2

Rearrange for $E_{a}$ :

E_{a} = (ln A - ln k) \times RT

3

Calculate the ln values:

ln (4.6 \times 1 0^{13}) = 31.46

and

ln (6.25 \times 1 0^{- 4}) = - 7.38

4

Substitute:

E_{a} = (31.46 - (- 7.38)) \times 8.31 \times 400

5

E_{a} = 38.84 \times 3324 = 129096

J mol⁻¹ = 129 kJ mol⁻¹

The Logarithmic Form

The exponential form of the Arrhenius equation is difficult to use for graphical analysis. By taking natural logarithms of both sides, we can convert it to a straight-line form.

Derivation

Starting with:

k = A e^{- E_{a} / RT}

Take

ln

of both sides:

ln k = ln A + ln (e^{- E_{a} / RT})

Since

ln (e^{x}) = x

:

ln k = ln A - \frac{E _{a}}{RT}

Rearranging into the form

y = m x + c

:

Straight-Line Form

ln k = - \frac{E _{a}}{R} \times \frac{1}{T} + ln A

Straight Line TermArrhenius Equivalent

y

ln k

x

\frac{1}{T}

Gradient (

m

)

- \frac{E _{a}}{R}

y-intercept (

c

)

ln A

Exam Tip

It is usually easier to use the logarithmic version of the Arrhenius equation when solving problems, especially when calculating

E_{a}

from given values of

k

,

A

, and

T

.

Arrhenius Plots

An Arrhenius plot is a graph of

ln k

against

\frac{1}{T}

. Because the equation is in the form

y = m x + c

, this gives a straight line with negative gradient.

Important

The gradient is negative because of the minus sign in

- E_{a} / R

. To find

E_{a}

, multiply the gradient by

- R

.

Using Arrhenius Plots

Arrhenius plots allow us to determine activation energy and the Arrhenius constant from experimental data.

Finding Ea

1

Calculate the gradient of the line

2

Gradient =

- \frac{E _{a}}{R}

3

Rearrange:

E_{a} = - gradient \times R

4

Remember:

R = 8.31

J K⁻¹ mol⁻¹

Finding A

1

Read the y-intercept from the graph

2

y-intercept =

ln A

3

Calculate:

A = e^{intercept}

Common Mistakes

Forgetting the negative sign when calculating

E_{a}

Using the wrong units for

R

— always use 8.31 J K⁻¹ mol⁻¹

Forgetting to convert

E_{a}

from J mol⁻¹ to kJ mol⁻¹ in your final answer

Using

T

in °C instead of K when calculating

\frac{1}{T}

Worked Example 2

2

Find Ea and A from Arrhenius Plot

An Arrhenius plot has a gradient of −12,500 K and a y-intercept of 28.5. Calculate (a) the activation energy in kJ mol⁻¹ and (b) the Arrhenius constant.

Part (a): Finding $E_{a}$

1

Gradient =

- \frac{E _{a}}{R}

, so

- 12500 = - \frac{E _{a}}{8.31}

2

E_{a} = 12500 \times 8.31 = 103875

J mol⁻¹ = 104 kJ mol⁻¹

Part (b): Finding $A$

3

ln A = 28.5

, so

A = e^{28.5} = 2.39 \times 1 0^{12}

Calculating Ea from Two Data Points

Often in exams, you're given rate constants at two different temperatures and asked to calculate

E_{a}

. You don't need to draw a graph — use the two-point form of the Arrhenius equation.

Two-Point Arrhenius Equation

For two temperatures

T_{1}

and

T_{2}

with rate constants

k_{1}

and

k_{2}

:

ln (\frac{k _{2}}{k _{1}}) = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

Derivation

ln k_{2} = - \frac{E _{a}}{R T _{2}} + ln A

and

ln k_{1} = - \frac{E _{a}}{R T _{1}} + ln A

Subtracting:

ln k_{2} - ln k_{1} = - \frac{E _{a}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})

Which rearranges to:

ln (\frac{k _{2}}{k _{1}}) = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

Worked Example 3

3

Calculate Ea from Two k Values

A reaction has a rate constant of

3.2 \times 1 0^{- 3}

s⁻¹ at 300 K and

4.8 \times 1 0^{- 2}

s⁻¹ at 350 K. Calculate the activation energy.

1

Identify:

k_{1} = 3.2 \times 1 0^{- 3}

at

T_{1} = 300

K;

k_{2} = 4.8 \times 1 0^{- 2}

at

T_{2} = 350

K

2

ln (k_{2} / k_{1}) = ln (\frac{4.8 \times 1 0 ^{- 2}}{3.2 \times 1 0 ^{- 3}}) = ln (15) = 2.708

3

\frac{1}{T _{1}} - \frac{1}{T _{2}} = \frac{1}{300} - \frac{1}{350} = 0.003333 - 0.002857 = 4.76 \times 1 0^{- 4}

K⁻¹

4

E_{a} = \frac{R \times l n ( k _{2} / k _{1} )}{\frac{1}{T _{1}} - \frac{1}{T _{2}}} = \frac{8.31 \times 2.708}{4.76 \times 1 0 ^{- 4}} = 47270

J mol⁻¹ = 47.3 kJ mol⁻¹

Effect of Activation Energy on Rate

The Arrhenius equation shows us how activation energy affects the rate constant:

Key Relationship

Higher $E_{a}$ → Lower

k

→ Slower reaction

Lower $E_{a}$ → Higher

k

→ Faster reaction

This is because a higher

E_{a}

means fewer molecules have sufficient energy to react. This is why catalysts speed up reactions — they provide an alternative pathway with lower

E_{a}

.

Practice

Question 1: A reaction has

A = 2.0 \times 1 0^{10}

s⁻¹ and

E_{a} = 60

kJ mol⁻¹. Calculate the rate constant at 400 K.

1

Convert

E_{a}

:

60 \times 1000 = 60000

J mol⁻¹

2

Calculate exponent:

- \frac{60000}{8.31 \times 400} = - \frac{60000}{3324} = - 18.05

3

k = 2.0 \times 1 0^{10} \times e^{- 18.05} = 2.0 \times 1 0^{10} \times 1.45 \times 1 0^{- 8}

4

Answer:

k = 290

s⁻¹ (or

2.9 \times 1 0^{2}

s⁻¹)

Practice

Question 2: An Arrhenius plot for a reaction has a gradient of −8500 K. Calculate the activation energy in kJ mol⁻¹.

1

Use gradient =

- \frac{E _{a}}{R}

2

E_{a} = - gradient \times R = - (- 8500) \times 8.31

3

E_{a} = 8500 \times 8.31 = 70635

J mol⁻¹

4

Answer:

E_{a} = 70.6

kJ mol⁻¹

Practice

Question 3: The rate constant for a reaction is

1.5 \times 1 0^{- 4}

s⁻¹ at 280 K and

8.9 \times 1 0^{- 3}

s⁻¹ at 320 K. Calculate the activation energy.

1

ln (k_{2} / k_{1}) = ln (\frac{8.9 \times 1 0 ^{- 3}}{1.5 \times 1 0 ^{- 4}}) = ln (59.33) = 4.083

2

\frac{1}{T _{1}} - \frac{1}{T _{2}} = \frac{1}{280} - \frac{1}{320} = 0.003571 - 0.003125 = 4.46 \times 1 0^{- 4}

K⁻¹

3

E_{a} = \frac{8.31 \times 4.083}{4.46 \times 1 0 ^{- 4}} = \frac{33.93}{4.46 \times 1 0 ^{- 4}} = 76100

J mol⁻¹

4

Answer:

E_{a} = 76.1

kJ mol⁻¹

Summary

Key Equations

EquationUse

k = A e^{- E_{a} / RT}

Calculate

k

from

A

,

E_{a}

, and

T

ln k = - \frac{E _{a}}{R} \times \frac{1}{T} + ln A

Straight-line form for Arrhenius plots

ln (k_{2} / k_{1}) = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

Calculate

E_{a}

from two data points

Exam Tips

Always convert

E_{a}

to J mol⁻¹ before calculations (×1000)

Always use temperature in Kelvin

Remember: gradient of Arrhenius plot =

- E_{a} / R

(negative!)

To find

E_{a}

: multiply gradient by

- R

R = 8.31

J K⁻¹ mol⁻¹ — memorise this!

Typical

E_{a}

values are between 40 and 400 kJ mol⁻¹ — use this to sense-check your answers

Examiner Tip

You are not required to memorise the Arrhenius equations — they will be given in the exam. However, you must be able to rearrange them confidently to solve for any variable. Practice rearranging to find

k

,

E_{a}

,

A

, and

T

.