The Arrhenius Equation
One equation ties the rate constant to temperature and the activation barrier — and lets you measure Ea from data.
⏱️ About 16 min
Why does milk keep for weeks in the fridge but sour in a day on the counter? Same chemistry, a 20-degree difference — and a dramatic change in speed. The Arrhenius equation captures that sensitivity in a single, powerful expression.
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The big idea: The rate constant follows k = A·e^(−Ea/RT). The exponential term is the fraction of collisions energetic enough to react, so raising temperature or lowering the activation energy makes k — and the reaction — larger and faster.
Packaging collision theory into one equation
Collision theory said a reaction needs energy ≥ Ea and the right orientation, with the reactive fraction growing fast as temperature rises. Svante Arrhenius wrote that story as a formula for the rate constant:
\[ k = A\,e^{-E_a / RT} \]
A = frequency (pre-exponential) factor, folding in collision rate and orientation; Ea = activation energy; R = 8.314 J·mol⁻¹·K⁻¹; T = absolute temperature (K).
Reading the equation
The exponential term e^(−Ea/RT) is the heart of it: it is the fraction of collisions with enough energy to clear the barrier. Two things make that fraction bigger — and so make k bigger:
- Higher temperature (T) shrinks Ea/RT, so the exponential rises → faster.
- Lower activation energy (Ea) also shrinks Ea/RT → faster. (That is exactly the lever a catalyst pulls.)
Because the dependence is exponential, small changes in T or Ea produce large changes in rate.
🔑 The two-point form (for finding Ea)
Measure k at two temperatures and you can solve for the activation energy. Taking the log of the Arrhenius equation at two temperatures and subtracting gives the workhorse form below — no need to know A.
\[ \ln\!\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
Two rate constants at two temperatures are enough to pin down Ea.
✨ The Arrhenius plot
Rearranged, ln k = ln A − (Ea/R)(1/T). So a plot of ln k against 1/T is a straight line with slope −Ea/R and intercept ln A. A steep (very negative) slope means a large activation energy.
📝 Worked example: A reaction has Ea = 50 kJ/mol. By what factor does its rate constant change when the temperature rises from 300 K to 310 K? (R = 8.314 J·mol⁻¹·K⁻¹.)
- Use ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂).
- 1/T₁ − 1/T₂ = 1/300 − 1/310 = 0.0033333 − 0.0032258 = 1.075×10⁻⁴ K⁻¹.
- Ea/R = 50000 / 8.314 = 6014 K.
- ln(k₂/k₁) = 6014 × 1.075×10⁻⁴ = 0.647.
- k₂/k₁ = e^0.647 = 1.9.
✓ About 1.9× — the rate nearly doubles for a 10 °C rise, the classic rule of thumb.
✏️ Practice: For a reaction, k = 1.0×10⁻³ s⁻¹ at 300 K and k = 2.0×10⁻³ s⁻¹ at 310 K. Find the activation energy Ea in kJ/mol. (R = 8.314 J·mol⁻¹·K⁻¹.)
kJ/mol
Solution
- The rate constant doubles: ln(k₂/k₁) = ln 2 = 0.693.
- 1/T₁ − 1/T₂ = 1/300 − 1/310 = 1.075×10⁻⁴ K⁻¹.
- Rearrange: Ea = R·ln(k₂/k₁) / (1/T₁ − 1/T₂) = 8.314 × 0.693 / 1.075×10⁻⁴.
- = 5.763 / 1.075×10⁻⁴ = 5.36×10⁴ J/mol = 53.6 kJ/mol.
✏️ Practice: A reaction has Ea = 75 kJ/mol. By what factor does the rate constant increase when the temperature goes from 298 K to 308 K? (R = 8.314; give k₂/k₁.)
×
Solution
- ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂).
- 1/298 − 1/308 = 0.0033557 − 0.0032468 = 1.090×10⁻⁴ K⁻¹.
- Ea/R = 75000 / 8.314 = 9021 K; ln(k₂/k₁) = 9021 × 1.090×10⁻⁴ = 0.983.
- k₂/k₁ = e^0.983 = 2.7 — the rate constant nearly triples over 10 °C.
Check your understanding
1. In k = A·e^(−Ea/RT), increasing the activation energy Ea (all else equal) makes k…
A larger Ea makes the exponent −Ea/RT more negative, so e^(−Ea/RT) shrinks and k gets smaller — a higher barrier means a slower reaction.
2. An Arrhenius plot (ln k vs 1/T) is a straight line. Its slope equals…
ln k = ln A − (Ea/R)(1/T), so the slope is −Ea/R and the intercept is ln A. A steeper downward slope means a larger activation energy.
3. Which change will always increase the rate constant k?
k rises when the exponent −Ea/RT gets less negative — that happens when T increases or Ea decreases. Concentration does not appear in k at all.
✅ Key takeaways
- Arrhenius: k = A·e^(−Ea/RT), with A the frequency factor and R = 8.314 J·mol⁻¹·K⁻¹.
- The exponential term is the fraction of collisions with energy ≥ Ea.
- Higher T or lower Ea makes k larger — the reaction goes faster.
- Two-point form: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂) lets you find Ea from data.
- An Arrhenius plot of ln k vs 1/T is linear with slope −Ea/R.
➡️ The equation shows two ways to speed a reaction: turn up the temperature, or lower Ea. Heating is not always an option — but lowering the barrier is, and that is exactly what a catalyst does. Next up: catalysis.
Want to test yourself on this?
Try the Chemistry practice test →