Activation energy calculator

How do you find activation energy from two rate constants?

From the two-temperature Arrhenius equation: ln(k₂ ÷ k₁) = (Ea ÷ R)(1/T₁ − 1/T₂), so Ea = R ln(k₂ ÷ k₁) ÷ (1/T₁ − 1/T₂), with R = 8.314 J/(mol·K) and temperatures in kelvins (OpenStax Chemistry 2e).

Example

For 2HI → H₂ + I₂, OpenStax gives k = 3.52 × 10⁻⁷ at 555 K and 3.95 × 10⁻² at 781 K. Solving gives Ea ≈ 185 kJ/mol; the book, working from a graph, reports about 180 kJ/mol.

Can I predict a rate constant at a new temperature?

Yes — choose “Find k₂”, enter Ea, one known rate constant and both temperatures.

What is the Arrhenius equation?

k = A e^(−Ea/RT): the rate constant grows exponentially with temperature. A is the frequency factor and Ea the minimum energy colliding molecules need to react.

Does the unit of k matter?

No, as long as both rate constants use the same unit — only their ratio enters the calculation.

Why do chemists plot ln k against 1/T instead of using two points?

Taking the logarithm of k = A e^(−Ea/RT) gives ln k = ln A − (Ea ÷ R)(1/T), a straight line whose slope is −Ea ÷ R and whose intercept is ln A (OpenStax Chemistry 2e, section 12.5). Fitting a line through five or more measurements averages out experimental scatter, while two points give an Ea that can swing with a small error in either k. The calculator’s two-point answer is best for a quick estimate or for checking a graph.

How much does a 10-degree rise speed up a reaction?

It depends on Ea; there is no single factor. For an assumed activation energy of 50 kJ/mol, warming from 25 °C to 35 °C multiplies the rate constant by 1.9. A reaction with a larger Ea gains more per degree, and one with a smaller Ea gains less. Enter your own Ea and the two temperatures with “Rate constant k₂”.

The equation and its units

The Arrhenius equation, k = A·e^(−Ea/RT), links the rate constant k to the temperature T (OpenStax Chemistry 2e, section 12.5). Dividing the equation at two temperatures removes A and leaves ln(k₂ ÷ k₁) = (Ea ÷ R)(1/T₁ − 1/T₂). Because R is 8.314 J/(mol·K), Ea comes out in J/mol, and the calculator shows kJ/mol and kcal/mol as well. Temperatures must be absolute. The rate constants may be in any unit, such as s⁻¹ or L/(mol·s), provided they match and describe the same reaction; and the equation treats Ea and A as constant between the two temperatures.

Worked example

The practice problem after OpenStax Example 12.13 concerns 2N₂O₅ → 4NO + 3O₂, with k = 1.66 L/(mol·s) at 650 K and 7.39 L/(mol·s) at 700 K. Then ln(7.39 ÷ 1.66) = 1.494 and 1/650 − 1/700 = 1.10 × 10⁻⁴ K⁻¹, so Ea = 8.314 × 1.494 ÷ (1.10 × 10⁻⁴) = 113 kJ/mol, matching the textbook’s 1.1 × 10⁵ J/mol. Going the other way, choose “Rate constant k₂” with Ea = 113 kJ/mol, k₁ = 1.66 and 650 K to 700 K, and you recover k₂ = 7.39. For the hydrogen iodide data in the second answer above, the first and last of the five points give 185 kJ/mol, while the graph in the book gives about 180; averaging many points, as a graph does, is what makes it more reliable.

Common mistakes

The commonest is using Celsius for the Arrhenius temperatures: the N₂O₅ data at 377 °C and 427 °C, typed as 377 and 427 instead of 650 and 700 K, give 40 kJ/mol instead of 113. Unit handling is second. R is in joules, so Ea must be in J/mol in a hand calculation, while this calculator asks for kJ/mol in the “Rate constant k₂” mode; typing J/mol there would inflate Ea a thousandfold and give an absurd k₂. Third, the order of the pairs: k₁ must go with T₁ and k₂ with T₂, and mismatching a pair gives a wrong value. Fourth, relying on only two points when more data are available; the slope of a ln k versus 1/T graph is the more reliable value.

Sources: OpenStax Chemistry 2e, section 12.5 “Collision Theory” (Arrhenius equation, two-point and graphical forms, R = 8.314 J/(mol·K), Example 12.13 and its practice problem). Related: integrated rate law calculator, half-life calculator, Gibbs free energy calculator and equilibrium constant calculator.