Arrhenius Equation — Finding Activation Energy from Rate Constants

By Aniket Bhardwaj · 1 September 2026 · Calculator/Formula Guide

Almost every reaction speeds up when you heat it. The Arrhenius equation is the quantitative statement of that everyday fact, and it is the standard route to the activation energy E_a — the energy barrier a colliding pair must clear before they can react. This guide covers the three forms of the equation you actually need, three fully worked problems, and the mistakes that quietly destroy an answer.

The three forms

1. Exponential form: k = A e^−Ea/RT

2. Logarithmic (graph) form: ln k = ln A − E_a/(RT)

3. Two-point form: ln(k₂/k₁) = (E_a/R) × (1/T₁ − 1/T₂)

All three are the same relationship. Use form 1 when you know A and E_a; form 2 when you have several k values and a graph; form 3 when you have exactly two rate constants at two temperatures — which is the most common exam situation.

What each symbol means

Symbol	Meaning	Unit
k	Rate constant at temperature T	depends on order
A	Pre-exponential (frequency) factor — collision rate with correct orientation	same unit as k
E_a	Activation energy	J mol⁻¹ (report in kJ mol⁻¹)
R	Gas constant, 8.314	J mol⁻¹ K⁻¹
T	Absolute temperature	kelvin, K

The exponential factor e^−Ea/RT is the fraction of collisions energetic enough to react. It is a tiny number. For E_a = 50 kJ mol⁻¹ at 300 K: E_a/RT = 50 000 ÷ (8.314 × 300) = 50 000 ÷ 2494.2 = 20.05, so e^−20.05 ≈ 2 × 10⁻⁹. About two collisions in a billion clear the barrier — which is why a modest temperature rise, which lifts that fraction sharply, changes the rate so dramatically.

Worked example 1 — the "rate doubles per 10 K" rule

A reaction's rate constant doubles when the temperature rises from 300 K to 310 K. Find E_a.

k₂/k₁ = 2, so ln(k₂/k₁) = ln 2 = 0.6931.

1/T₁ − 1/T₂ = 1/300 − 1/310 = (310 − 300) ÷ (300 × 310) = 10 ÷ 93 000 = 1.0753 × 10⁻⁴ K⁻¹

Rearranging form 3: E_a = R × ln(k₂/k₁) ÷ (1/T₁ − 1/T₂)

E_a = 8.314 × 0.6931 ÷ (1.0753 × 10⁻⁴) = 5.7626 ÷ 0.00010753

E_a = 53 593 J mol⁻¹ ≈ 53.6 kJ mol⁻¹

So the familiar textbook statement "rate roughly doubles for every 10 °C rise" is a statement about reactions with E_a near 50–55 kJ mol⁻¹ around room temperature. It is a rule of thumb, not a law.

Worked example 2 — predicting k at a new temperature

For a first-order reaction, E_a = 104 kJ mol⁻¹ and k = 2.5 × 10⁻⁴ s⁻¹ at 300 K. Find k at 320 K.

E_a/R = 104 000 ÷ 8.314 = 12 509 K

1/T₁ − 1/T₂ = 1/300 − 1/320 = 0.0033333 − 0.0031250 = 2.0833 × 10⁻⁴ K⁻¹

ln(k₂/k₁) = 12 509 × 2.0833 × 10⁻⁴ = 2.606

k₂/k₁ = e^2.606 = 13.55

k₂ = 2.5 × 10⁻⁴ × 13.55 = 3.39 × 10⁻³ s⁻¹

A 20 K rise made the reaction about 13.5 times faster — because this barrier is twice as high as the one in example 1, the same temperature change has a far bigger effect.

Worked example 3 — reading E_a off a graph

Plot ln k on the y-axis against 1/T on the x-axis. Form 2 says this is a straight line y = mx + c with:

slope m = −E_a/R → E_a = −m × R
intercept c = ln A → A = e^c

A plot of ln k against 1/T gives slope = −6000 K and intercept = 24.0. Find E_a and A.

E_a = −(−6000) × 8.314 = 49 884 J mol⁻¹ ≈ 49.9 kJ mol⁻¹

A = e^24.0 = 2.65 × 10¹⁰ s⁻¹

Note the units of A: for a first-order reaction A carries the same unit as k, here s⁻¹. A quoted "A = 2.65 × 10¹⁰" with no unit is an incomplete answer.

Common mistakes

Celsius instead of kelvin. T must be absolute. Using 27 instead of 300 does not give a slightly wrong answer — it gives a nonsensical one.
Mixing kJ and J. R is 8.314 J mol⁻¹ K⁻¹. If E_a is in kJ mol⁻¹ you must multiply by 1000 first, or use R = 8.314 × 10⁻³ kJ mol⁻¹ K⁻¹ consistently.
Flipping the two-point form. With k₂ on top, the bracket is (1/T₁ − 1/T₂), not (1/T₂ − 1/T₁). If your E_a comes out negative for a reaction that got faster on heating, you have swapped them.
Rounding 1/T too early. The bracket is a difference of two nearly equal numbers, so early rounding destroys it. Keep at least five significant figures in 1/T₁ and 1/T₂.
Thinking a catalyst changes A only. A catalyst gives a new pathway with a lower E_a; it does not change ΔH of the reaction.
Assuming ln k vs T is linear. It is ln k vs 1/T that is linear.

Where it appears in exams

Exam	Typical question
CBSE/ICSE Class 12	E_a from two rate constants; effect of catalyst on E_a
JEE / NEET	Temperature coefficient, ratio k₂/k₁, fraction of effective collisions
IIT-JAM / CUET-PG	Arrhenius plot slope and intercept; units of A by order
GATE / CSIR-NET	Composite E_a of multi-step mechanisms, negative apparent E_a

Stop losing marks on the arithmetic. Give the calculator any two of k₁, k₂, T₁, T₂ and E_a, and it returns the missing one along with the intermediate values — so you can compare each line against your own working.

Open the Arrhenius Equation Calculator →

Chemical kinetics rewards practice more than memory. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India, with home tuition available in Delhi-NCR — details at abcchemistry.in.