The Arrhenius Equation

Learning Objectives (OCR Spec 5.1.1)

By the end of this lesson you should be able to:

• State the Arrhenius equation and define each term
• Explain the meaning of the frequency factor (pre-exponential factor A) and activation energy Ea
• Rearrange the Arrhenius equation into linear (ln k) form
• Determine Ea graphically by plotting ln k against 1/T

The Arrhenius Equation

Svante Arrhenius proposed in 1889 that the rate constant depends on temperature according to:

k = A e^(-Ea / RT)

where:

• k = rate constant (units depend on overall order)
• A = pre-exponential factor (frequency factor), same units as k
• Ea = activation energy (J mol^-1) — minimum energy required for reaction
• R = gas constant = 8.314 J K^-1 mol^-1
• T = temperature in kelvin

The equation expresses two physical ideas:

1. e^(-Ea/RT) is the Boltzmann factor — the fraction of molecules with energy ≥ Ea at temperature T. This is the probability that a collision has enough energy to react.
2. A is the pre-exponential factor, which captures the frequency of collisions and the fraction with correct orientation. It is only weakly temperature-dependent compared with the exponential.

Multiplying the two gives the rate constant k: collisions per unit time x probability that they react.

Why T Affects k So Dramatically

Even though T appears only in the exponent, a small change in T changes -Ea/RT a lot (because Ea tends to be tens or hundreds of kJ mol^-1 while RT is only a few kJ mol^-1). A 10 K rise in temperature near room temperature roughly doubles k for a typical reaction with Ea around 50 kJ mol^-1.

Example calculation:

• At 298 K, with Ea = 50 000 J mol^-1: -Ea/RT = -50000/(8.314 x 298) = -20.18, Boltzmann factor ≈ 1.72 x 10^-9.
• At 308 K: -Ea/RT = -50000/(8.314 x 308) = -19.52, Boltzmann factor ≈ 3.31 x 10^-9.

Ratio: 3.31/1.72 ≈ 1.92, i.e. k roughly doubles. That matches the rule of thumb.

The Linear Form

Taking natural logarithms of both sides of k = A e^(-Ea/RT):

ln k = ln A + (-Ea/RT)
ln k = -Ea/R x (1/T) + ln A

This has the form y = m x + c, where:

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

So a plot of ln k (y) against 1/T (x) should be a straight line with gradient -Ea/R and intercept ln A. This is the standard OCR method for determining Ea experimentally.

Determining Ea Graphically

1. Run the reaction at several different temperatures.
2. At each temperature, determine k using any appropriate method (initial rates, half-life etc.).
3. Tabulate ln k and 1/T (with T in kelvin).
4. Plot ln k (y) against 1/T (x).
5. Draw a line of best fit.
6. Measure the gradient.
7. Ea = -gradient x R (in J mol^-1; convert to kJ mol^-1 by dividing by 1000).

Worked Example — Finding Ea

A reaction is studied at five temperatures, with these k values:

T / K	1/T / K^-1	k / s^-1	ln k
300	3.33 x 10^-3	1.0 x 10^-3	-6.91
310	3.23 x 10^-3	2.0 x 10^-3	-6.21
320	3.13 x 10^-3	3.9 x 10^-3	-5.55
330	3.03 x 10^-3	7.6 x 10^-3	-4.88
340	2.94 x 10^-3	1.4 x 10^-2	-4.27

Plot ln k (y) against 1/T (x). The line is approximately straight with gradient:

Gradient = (ln k_2 - ln k_1)/(1/T_2 - 1/T_1) = (-4.27 - (-6.91))/(2.94 x 10^-3 - 3.33 x 10^-3) = 2.64 / (-3.9 x 10^-4) ≈ -6770 K

Then:

Ea = -gradient x R = -(-6770) x 8.314
   = 56 300 J mol^-1
   = 56.3 kJ mol^-1