The Arrhenius equation: temperature and the rate constant

Introductory College⏱ 10 min readChemical Kinetics

Master the Arrhenius equation k = A·e^(−Ea/RT): what each term means, how to find activation energy from rate data, and why rates roughly double every 10 °C.

Quick answer

The Arrhenius equation relates the rate constant k to temperature: k = A·e^(−Ea/RT), where A is the frequency factor, Ea the activation energy, R the gas constant, and T the kelvin temperature.

$k = A\,e^{-E_a/RT}$

Unit: k: depends on order; Ea: J/mol; T: K

Example: A reaction with Ea = 50.0 kJ/mol has k₂/k₁ = e^((50000/8.314)(1/300 − 1/310)) ≈ 1.9 when warmed from 300 K to 310 K — the rate nearly doubles for a 10 K rise.

What you will learn

Write the Arrhenius equation and define k, A, Ea, R, and T.
Use the two-temperature form to calculate Ea or predict k at a new temperature.
Explain the linearized form ln k = −Ea/RT + ln A and how its slope gives Ea.
Connect the exponential form to collision theory and the '10-degree' rule of thumb.

The equation and its parts

The Arrhenius equation is the quantitative heart of chemical kinetics:

$k = A\,e^{-E_a/RT}$

Each symbol carries meaning:

k — the rate constant at temperature T
A — the pre-exponential (frequency) factor: how often molecules collide with the right orientation
Ea — the activation energy, the barrier height
R — the gas constant, 8.314 J/(mol·K)
T — absolute temperature in kelvin

The exponential term $e^{-E_a/RT}$ is the fraction of molecules with energy exceeding Ea — straight from the Boltzmann distribution. A bigger barrier or a lower temperature shrinks this fraction exponentially, which is why rates are so temperature-sensitive.

The linear form: finding Ea from data

Taking the natural logarithm straightens the curve:

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

Plotted as ln k versus 1/T, data fall on a straight line with slope −Ea/R and intercept ln A. This is how chemists actually measure activation energies: run the reaction at several temperatures, plot, and read Ea from the slope.

With only two temperatures, the two-point form skips the graph:

$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

Worked example: the 10-degree doubling rule

Consider a reaction with Ea = 50.0 kJ/mol warmed from 300 K to 310 K:

$\ln\frac{k_2}{k_1} = \frac{50000}{8.314}\left(\frac{1}{300} - \frac{1}{310}\right) = 6014 \times 0.0001075 = 0.647$

$\frac{k_2}{k_1} = e^{0.647} \approx 1.9$

The rate nearly doubles for a 10 K rise — the origin of the familiar rule of thumb. Note how the temperatures had to be in kelvin and Ea converted to joules to match R = 8.314 J/(mol·K).

What the exponential really buys you

The equation also explains catalysts quantitatively: a catalyst lowers Ea, which sits in a negative exponent — so even a modest cut in Ea multiplies k enormously. Dropping Ea by just 10 kJ/mol at 300 K multiplies the rate by $e^{10000/(8.314 \times 300)} \approx e^{4.0} \approx 55$ .

The same math explains food spoilage: refrigeration from 25 °C to 5 °C (298 K → 278 K) with Ea ≈ 60 kJ/mol cuts the rate to roughly one-quarter, which is why the fridge buys days of freshness.

Variables and units

Symbol	Name	Unit	Meaning
$k$	Rate constant	depends on order	Rate constant at temperature T
$A$	Frequency factor	same as k	Pre-exponential (frequency) factor; collision frequency times orientation probability
$Ea$	Activation energy	J/mol	Energy barrier for the reaction
$R$	Gas constant	J/(mol·K)	Gas constant, 8.314 J/(mol·K)
$T$	Temperature	K	Absolute temperature — must be in kelvin

Common mistakes

✗ The Arrhenius equation says rate increases linearly with temperature.

The dependence is exponential: k = A·e^(−Ea/RT). That exponential is why a modest temperature rise can double or triple a rate, and why the '10-degree doubling rule' exists.

✗ A in the Arrhenius equation is just a fudge factor.

A (the pre-exponential or frequency factor) has physical meaning: it combines collision frequency with the fraction of collisions in the correct orientation. For simple gas-phase reactions it is close to the collision rate; for bulky molecules it is far smaller.

✗ You can plug Celsius temperatures into the Arrhenius equation.

T must be in kelvin, always. Because T sits in an exponent as 1/T, using Celsius gives wildly wrong answers — and the error is not a simple offset, since the equation is nonlinear in T.

Real-world applications

Shelf-life prediction: food and drug companies measure k at high temperatures and extrapolate down to storage temperature.
Materials science: predicting how fast polymers degrade or metals corrode at service temperatures.
Industrial optimization: finding the temperature that balances faster rates against energy costs.
Atmospheric chemistry: modeling how reaction rates vary across the temperature range of the atmosphere.

Frequently asked questions

Is the 'doubling every 10 °C' rule exact?

Only approximately, and only for reactions with Ea near 50–60 kJ/mol around room temperature. The exact factor is e^((Ea/R)(1/T₁ − 1/T₂)), which depends on Ea and the temperature range. It is a handy estimate, not a law.

Which value of R should I use?

R must match the energy units of Ea. With Ea in J/mol, use R = 8.314 J/(mol·K); with Ea in kJ/mol, use R = 0.008314 kJ/(mol·K). T must always be in kelvin — using Celsius is the single most common Arrhenius error.

What does the pre-exponential factor A physically represent?

It folds together how often molecules collide and what fraction of collisions have the right orientation. For simple gas reactions it is roughly the collision frequency; for complex molecules it is much smaller because orientation matters. It is treated as approximately temperature-independent.

How do I find Ea from experimental data?

Take the natural log: ln k = −Ea/RT + ln A. Plot ln k against 1/T and the slope is −Ea/R. Two rate constants at two temperatures are enough to solve for Ea without graphing.