The formula
- Arrhenius equation
k = A·e^(-Ea / RT)- Linear form, for a graph of ln k against 1/T
ln k = (-Ea / R)(1 / T) + ln A- Two-point form, from rate constants at two temperatures
ln(k₂ / k₁) = (Ea / R)(1/T₁ - 1/T₂)- Activation energy from the slope
Ea = -R × slope
What the symbols mean
| Symbol | Meaning | Unit |
|---|---|---|
k | Rate constant at temperature T; k₁ and k₂ are the constants at T₁ and T₂ | depends on the overall reaction order |
A | Frequency factor, covering collision frequency and orientation | same units as k |
Ea | Activation energy, the minimum energy a collision needs to react | J/mol (tables usually print kJ/mol) |
R | Molar gas constant | 8.314 J/(mol·K) |
T | Absolute temperature; T₁ and T₂ are the two temperatures in the two-point form | K |
e | Base of the natural logarithm | 2.718 (dimensionless) |
When it applies
- Temperature is what changed. The rate law handles concentration; this equation handles everything the rate constant does when the thermometer moves.
- Temperatures go in kelvin. Celsius values make the exponent meaningless and can even make it change sign.
- R must match the units of Ea. Using 8.314 J/(mol·K) returns Ea in joules per mole, which then divides by 1000 for the kilojoules a table expects.
- The two-point form needs the same reaction with the same mechanism at both temperatures, so that A and Ea are genuinely shared between the two runs.
Worked example
Problem. A reaction has k = 2.0 × 10⁻³ s⁻¹ at 300 K and k = 8.0 × 10⁻³ s⁻¹ at 320 K. Find its activation energy.
- Use the two-point form, which removes the unknown frequency factor: ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂).
- The rate constants give ln(8.0 × 10⁻³ / 2.0 × 10⁻³) = ln(4) = 1.3863.
- The temperature term is 1/300 - 1/320 = 0.0033333 - 0.0031250 = 2.0833 × 10⁻⁴ K⁻¹.
- Rearrange and substitute with R = 8.314 J/(mol·K): Ea = (1.3863 × 8.314) / (2.0833 × 10⁻⁴) = 11.526 / 2.0833 × 10⁻⁴ = 55,323 J/mol.
- Convert to the usual units: Ea = 55.3 kJ/mol. A 20 K rise quadrupling the rate constant is consistent with a barrier of this size.
Answer. About 55.3 kJ/mol of activation energy.
Common mistakes
- Leaving temperatures in Celsius. Kelvin is required, and 300 K is not 300 °C.
- Swapping the subscripts so that the temperature difference and the rate constant ratio disagree, which returns a negative activation energy.
- Reporting Ea in joules per mole when the question expects kilojoules per mole, a factor of a thousand that is easy to miss.
- Treating the frequency factor A as a fitting fudge. It is a real quantity with the same units as k, and the two-point form exists precisely so that it does not have to be known.
Related formulas
- Half-life formula:
t½ = ln 2 / k ≈ 0.693 / k - Rate law:
rate = k[A]^m[B]^n - Gibbs free energy equation:
G = H - TS