The Arrhenius Equation

We know that the rate constant k increases with temperature, but how exactly are they related? The answer is provided by the Arrhenius equation, one of the most powerful relationships in physical chemistry. It links the rate constant to the temperature, the activation energy, and a pre-exponential factor.

k = Ae^(−Ea/RT)

Where:

k = rate constant (units depend on order)
A = pre-exponential factor (Arrhenius constant), same units as k
e = Euler's number (base of natural logarithms, ≈ 2.718)
Ea = activation energy (J mol⁻¹ -- note: joules, not kilojoules)
R = gas constant = 8.314 J K⁻¹ mol⁻¹
T = absolute temperature (K -- must be in kelvin)

Interpreting the Equation

The exponential term e^(−Ea/RT) represents the fraction of particles with energy ≥ Ea at temperature T. This is the Boltzmann factor. As T increases, Ea/RT decreases, so e^(−Ea/RT) increases -- meaning a larger fraction of particles can overcome the activation energy barrier, so k increases.

The pre-exponential factor A accounts for the frequency of collisions and the fraction with correct orientation. It is approximately constant over moderate temperature ranges.

The Logarithmic Form: ln k vs 1/T

Taking natural logarithms of both sides:

ln k = ln A − Ea/(RT)

This can be rearranged to:

ln k = −(Ea/R) × (1/T) + ln A

This is in the form y = mx + c, where:

y = ln k
x = 1/T
gradient (m) = −Ea/R
y-intercept (c) = ln A

So plotting ln k (y-axis) against 1/T (x-axis) gives a straight line with:

Gradient = −Ea/R (from which Ea can be calculated)
y-intercept = ln A (from which A can be calculated)

This is the most common graphical method for determining activation energy experimentally.

Calculating Ea from the Gradient

Worked Example 1: From a Graph

A student plots ln k vs 1/T and measures the gradient as −5200 K.

Gradient = −Ea/R

−5200 = −Ea / 8.314

Ea = 5200 × 8.314 = 43 230 J mol⁻¹ ≈ 43.2 kJ mol⁻¹

Note: always convert Ea to kJ mol⁻¹ for reporting, but use J mol⁻¹ in calculations with R = 8.314 J K⁻¹ mol⁻¹.

Worked Example 2: From Tabulated Data

A student collects:

T / K	k / s⁻¹	1/T / K⁻¹	ln k
300	1.2 × 10⁻⁴	3.333 × 10⁻³	−9.028
320	5.8 × 10⁻⁴	3.125 × 10⁻³	−7.453
340	2.4 × 10⁻³	2.941 × 10⁻³	−6.032
360	8.6 × 10⁻³	2.778 × 10⁻³	−4.756

Gradient using first and last points:

Gradient = (−4.756 − (−9.028)) / (2.778 × 10⁻³ − 3.333 × 10⁻³) = 4.272 / (−5.55 × 10⁻⁴) = −7697 K

Ea = 7697 × 8.314 = 63 980 J mol⁻¹ = 64.0 kJ mol⁻¹

Calculating Ea from Two Temperatures

If you know k at two different temperatures, you can calculate Ea without plotting a graph. Writing the Arrhenius equation for both temperatures and subtracting:

ln k₁ = ln A − Ea/(RT₁)

ln k₂ = ln A − Ea/(RT₂)

Subtracting: ln(k₂/k₁) = −(Ea/R)(1/T₂ − 1/T₁)

Or equivalently: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂)

Worked Example 3: Standard Two-Temperature Calculation

At 300 K, k₁ = 1.4 × 10⁻³ s⁻¹. At 320 K, k₂ = 5.6 × 10⁻³ s⁻¹. Find Ea.

ln(k₂/k₁) = ln(5.6 × 10⁻³ / 1.4 × 10⁻³) = ln(4.0) = 1.386

1/T₁ − 1/T₂ = 1/300 − 1/320 = (320 − 300)/(300 × 320) = 20/96000 = 2.083 × 10⁻⁴ K⁻¹

Ea/R = ln(k₂/k₁) / (1/T₁ − 1/T₂) = 1.386 / 2.083 × 10⁻⁴ = 6655 K

Ea = 6655 × 8.314 = 55 330 J mol⁻¹ ≈ 55.3 kJ mol⁻¹

Worked Example 4: Predicting k at a New Temperature

If you know Ea and k at one temperature, you can predict k at any other temperature.

Given: Ea = 55.3 kJ mol⁻¹, k₁ = 1.4 × 10⁻³ s⁻¹ at 300 K. Find k at 350 K.

ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂) = (55300/8.314)(1/300 − 1/350)

= 6652 × (3.333 × 10⁻³ − 2.857 × 10⁻³)

= 6652 × 4.762 × 10⁻⁴

= 3.168

k₂ = k₁ × e^3.168 = 1.4 × 10⁻³ × 23.76 = 0.0333 s⁻¹

Worked Example 5: Using Half-Lives to Find Ea

A first-order reaction has t½ = 200 s at 280 K and t½ = 50 s at 300 K.

Step 1: Convert half-lives to k values.

k₁ = 0.693 / 200 = 3.465 × 10⁻³ s⁻¹ at 280 K

k₂ = 0.693 / 50 = 0.01386 s⁻¹ at 300 K

Step 2: Apply the two-temperature Arrhenius form.

ln(k₂/k₁) = ln(0.01386 / 3.465 × 10⁻³) = ln(4.0) = 1.386

1/T₁ − 1/T₂ = 1/280 − 1/300 = (300 − 280)/(280 × 300) = 20/84000 = 2.381 × 10⁻⁴ K⁻¹

Ea = R × 1.386 / 2.381 × 10⁻⁴ = 8.314 × 5822 = 48 400 J mol⁻¹ = 48.4 kJ mol⁻¹

Effect of a Catalyst on the Arrhenius Plot

A catalyst lowers Ea. On an ln k vs 1/T plot:

The catalysed reaction gives a less steep line (smaller magnitude of gradient, since gradient = −Ea/R and Ea is smaller).
The line is shifted upward (larger k values at each temperature).
The y-intercept (ln A) may also change if the catalyst alters the collision geometry.

Both lines are straight, but the catalysed line has a less negative gradient, reflecting the lower activation energy.

The Pre-Exponential Factor A

The factor A represents the maximum possible rate constant -- the value k would have if every collision had enough energy (i.e., if Ea = 0, then k = A). It incorporates:

The collision frequency (how often molecules collide per unit time)
The steric factor (the fraction of collisions with correct orientation)

For most reactions, A is a very large number (often 10⁸ to 10¹³ s⁻¹ for first-order reactions). It is approximately constant over moderate temperature ranges, which is why the ln k vs 1/T plot is a straight line.

Common Pitfalls

Temperature must be in kelvin. Always convert °C to K by adding 273. Using 25 instead of 298 produces catastrophically wrong results.
Ea must be in J mol⁻¹ when using R = 8.314 J K⁻¹ mol⁻¹. If Ea is given in kJ mol⁻¹, multiply by 1000. This is the most common calculation error.
The gradient is negative. The ln k vs 1/T line slopes downward from left to right (since as 1/T increases -- i.e., T decreases -- k decreases). Remember gradient = −Ea/R, so Ea = −gradient × R.
ln, not log₁₀. The Arrhenius equation uses natural logarithms. If you have log₁₀ k values, convert using ln k = 2.303 × log₁₀ k.
Getting the subtraction order wrong. In ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂), note that T₂ > T₁ and k₂ > k₁. If you swap the order, you get a negative Ea, which is physically impossible for a simple reaction.

Checklist for Arrhenius Calculations

Step	Check
1	T in kelvin (not °C)?
2	Ea in J mol⁻¹ (not kJ mol⁻¹)?
3	R = 8.314 J K⁻¹ mol⁻¹?
4	Using ln (not log₁₀)?
5	Gradient is negative (Ea must be positive)?
6	Final answer converted to kJ mol⁻¹ for reporting?

Summary

The Arrhenius equation k = Ae^(−Ea/RT) quantifies the relationship between the rate constant and temperature. Its logarithmic form gives a straight-line plot (ln k vs 1/T) from which Ea can be determined from the gradient (−Ea/R). The two-temperature form allows Ea calculation without a full graph. Catalysts reduce Ea, giving a less steep line on the Arrhenius plot. The pre-exponential factor A represents the collision frequency adjusted for the steric factor.

A-Level Deep Dive: The Arrhenius Equation

Spec mapping

Edexcel 9CH0 specification, Topic 16 — Kinetics II introduces the Arrhenius equation $k = A\exp(-E_a/RT)$ and its logarithmic linearised form $\ln k = \ln A - E_a/(RT)$ , used in Arrhenius plots of $\ln k$ against $1/T$ to determine $E_a$ from the gradient (refer to the official specification document for exact wording). Candidates must be able to use the equation to calculate $E_a$ from two ( $k, T$ ) pairs, to interpret an Arrhenius plot graphically, and to discuss the meaning of the pre-exponential (frequency) factor $A$ . Examined principally on Paper 2 (Topics 11–19) and Paper 3 (synoptic with Core Practicals). The Edexcel data booklet does list the Arrhenius equation and $R = 8.31$ J K⁻¹ mol⁻¹ — this is one of the few kinetics formulae provided.

Worked example with full mark scheme

Question (8 marks):

The rate constant for a reaction is measured at two temperatures.

T / K	k / mol⁻¹ dm³ s⁻¹
298	1.20 × 10⁻³
318	5.40 × 10⁻³

(a) Use the Arrhenius equation to calculate $E_a$ in kJ mol⁻¹. (5)

(b) Calculate the pre-exponential factor $A$ . (2)

Solution with mark scheme:

(a) M1 — write the linearised Arrhenius for both temperatures: $\ln k_1 = \ln A - E_a/(R T_1)$ , $\ln k_2 = \ln A - E_a/(R T_2)$ .

M1 — subtract to eliminate $\ln A$ : $\ln(k_2/k_1) = -E_a/R \cdot (1/T_2 - 1/T_1) = (E_a/R)(1/T_1 - 1/T_2)$ .

M1 — substitute values: $\ln(k_2/k_1) = \ln(5.40 × 10^{-3} / 1.20 × 10^{-3}) = \ln(4.50) = 1.504$ .

$1/T_1 - 1/T_2 = 1/298 - 1/318 = 3.356 × 10^{-3} - 3.145 × 10^{-3} = 2.111 × 10^{-4}$ K⁻¹.

M1 — solve: $E_a = R \cdot \ln(k_2/k_1) / (1/T_1 - 1/T_2) = 8.31 \times 1.504 / 2.111 × 10^{-4} = 5.92 × 10^4$ J mol⁻¹.

A1 — $E_a = 59.2$ kJ mol⁻¹ (3 sig fig).

(b) M1 — substitute back: from $k_1 = A\exp(-E_a/RT_1)$ , $A = k_1 \exp(E_a/RT_1) = 1.20 × 10^{-3} \times \exp(5.92 × 10^4 / (8.31 \times 298))$ .

$5.92 × 10^4 / 2476 = 23.91$ , $\exp(23.91) = 2.42 × 10^{10}$ .

A1 — $A = 1.20 × 10^{-3} \times 2.42 × 10^{10} = 2.91 × 10^7$ mol⁻¹ dm³ s⁻¹.

(c) M1 — sketch with x-axis $1/T$ , y-axis $\ln k$ , two points correctly placed (low $1/T$ → high $\ln k$ ; high $1/T$ → low $\ln k$ ). Gradient = $-E_a/R$ < 0 (slope is negative).

Total: 8 marks (M5 A2 + sketch, split as shown).

Specimen question modelled on the Edexcel 9CH0 Paper 2 format

Question (6 marks): A student measures $k$ at five temperatures and plots $\ln k$ against $1/T$ . The plot is linear with gradient $-7500$ K and y-intercept 22.0.

(a) Calculate $E_a$ from the gradient. (2)

(b) Calculate the pre-exponential factor $A$ from the y-intercept. (2)

Mark scheme decomposition by AO:

Part	Mark	AO	Earned by
(a) M1	1	AO1	Gradient = $-E_a/R$ .
(a) A1	1	AO2	$E_a = -\text{gradient} \times R = 7500 \times 8.31 = 6.23 × 10^4$ J mol⁻¹ = 62.3 kJ mol⁻¹.
(b) M1	1	AO1	y-intercept = $\ln A$ .
(b) A1	1	AO2	$A = e^{22.0} = 3.58 × 10^9$ (units same as $k$ ).
(c) M1	1	AO2	$\ln k = \ln A - E_a/(RT) = 22.0 - 6.23 × 10^4 / (8.31 \times 350) = 22.0 - 21.42 = 0.58$ .
(c) A1	1	AO2	$k = e^{0.58} = 1.79$ (units same as $A$ ).

Total: 6 marks split AO1 = 2, AO2 = 4.

Synoptic links

Connects to:

The Arrhenius Equation