The Arrhenius Equation

How temperature changes rate constants

kineticsArrheniustemperature
⏱️ About 19 min

A modest temperature increase can double a reaction rate—not magic, just exponential temperature dependence.

💡
The big idea: The Arrhenius equation models how the rate constant varies with temperature through an activation energy barrier.
🎯 By the end, you'll be able to
  • State the Arrhenius equation and define A and Ea
  • Use the two-point Arrhenius relation to find k at a new temperature
  • Interpret an Arrhenius plot of ln(k) vs 1/T
  • Apply a numerically stable reciprocal-temperature-difference technique
📎 Helpful to know first

Arrhenius temperature dependence

Many rate constants increase strongly with temperature. A widely used model is the Arrhenius equation:

k = A exp(−Ea/(R T))

A is the pre-exponential factor, Ea is the activation energy, R = 8.314 J/(mol·K), and T is absolute temperature (K).

\[ k = A\exp\!\left(-\frac{E_a}{RT}\right) \]
Arrhenius equation.
🔑 Linearized Arrhenius form

Taking the natural logarithm gives a straight-line form:

ln(k) = ln(A) − Ea/(R T)

Plotting ln(k) versus 1/T yields a line with slope −Ea/R and intercept ln(A).

\[ \ln(k) = \ln(A) - \frac{E_a}{R}\left(\frac{1}{T}\right) \]
Linearized Arrhenius relationship (ln(k) vs 1/T).
1/T ln(k) y-intercept: ln(A) slope = −Ea/R Straight line on ln(k) vs 1/T
Arrhenius plot (linearized): ln(k) vs 1/T gives slope −Ea/R and intercept ln(A).
⚠️ Use a stable reciprocal difference

When using the two-point Arrhenius form, compute (1/T2 − 1/T1) as (T1 − T2)/(T1·T2) to reduce round-off from subtracting two close decimals.

\[ \ln\!\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) \]
Two-point Arrhenius equation (avoids needing A).
📝 Worked example: A reaction has k1 = 0.020 s^-1 at T1 = 300 K. The activation energy is Ea = 60 kJ/mol. Estimate k2 at T2 = 320 K using R = 8.314 J/(mol·K).
  1. Use ln(k2/k1) = -(Ea/R)*(1/T2 − 1/T1). Convert Ea: 60 kJ/mol = 60000 J/mol.
  2. Compute reciprocal difference with the exact-fraction form: (1/T2 − 1/T1) = (T1 − T2)/(T1*T2) = (300−320)/(300*320) = -20/96000 = -1/4800 ≈ -0.0002083333 K^-1.
  3. Compute Ea/R = 60000/8.314 ≈ 7216.74 K.
  4. Then ln(k2/k1) = -(7216.74)*(-0.0002083333) ≈ 1.50349.
  5. Compute k2/k1 = exp(1.50349) ≈ 4.497.
  6. Compute k2 = k1*(k2/k1) = 0.020*4.497 ≈ 0.0899 s^-1.
✓ 0.0899 s^-1
✏️ Practice: A reaction has k1 = 1.50e-3 s^-1 at T1 = 310 K. The activation energy is Ea = 50 kJ/mol. Estimate k2 at T2 = 330 K using R = 8.314 J/(mol·K).
s^-1
Solution
  1. Convert Ea: 50 kJ/mol = 50000 J/mol.
  2. Compute (1/T2 − 1/T1) using (T1 − T2)/(T1*T2): (310−330)/(310*330) = -20/102300 ≈ -0.0001955034 K^-1.
  3. Compute Ea/R = 50000/8.314 ≈ 6013.95 K.
  4. ln(k2/k1) = -(6013.95)*(-0.0001955034) ≈ 1.17575.
  5. k2/k1 = exp(1.17575) ≈ 3.241.
  6. k2 = 1.50e-3 * 3.241 ≈ 4.861e-3 s^-1.

Check your understanding

1. In an Arrhenius plot of ln(k) vs 1/T, what is the slope?
From ln(k) = ln(A) − (Ea/R)(1/T), slope with respect to 1/T is −Ea/R.
2. If temperature increases, what typically happens to k for an Arrhenius reaction (Ea > 0)?
Increasing T makes −Ea/(RT) less negative, so exp(−Ea/RT) increases.
3. Why is (1/T2 − 1/T1) often computed as (T1 − T2)/(T1*T2)?
The fraction form is algebraically identical but numerically more stable when T1 and T2 are close.
🔎 Related lessons
Next: Elementary vs Non-Elementary Reactions · soon
✅ Key takeaways
  • Arrhenius: k = A exp(−Ea/RT) captures strong temperature dependence.
  • Linear form: ln(k) vs 1/T is a straight line with slope −Ea/R.
  • Two-point form computes k2 from k1 without needing A.
  • Compute (1/T2 − 1/T1) as (T1 − T2)/(T1T2) for better numerical stability.
➡️ Next we’ll connect rate laws to reaction mechanisms and see why some reactions do not follow stoichiometry-based rate predictions.
Want to test yourself on this? Try the Chemical Aptitude test →