04 // Interactive Formula Compendium // CHEMISTRY-THERMOk = A · exp(−E_a / (R · T))
Arrhenius Chemical Reaction Rate Formula
The Arrhenius equation k = A·e^(−E_a/RT) gives the rate constant k from pre-exponential collision frequency A and the Boltzmann fraction e^(−E_a/RT) of collisions exceeding activation energy E_a.
Formula Reference
Main Formula
k = A · exp(−E_a / (R · T))
Rearranged Forms
ln(k₂ / k₁) = (E_a / R) · (1/T₁ − 1/T₂)
ln(k) = ln(A) − (E_a/R)(1/T)
Variables & Units
| Symbol | Name | Unit |
|---|---|---|
| k | Reaction Rate Constant | s⁻¹ |
| A | Pre-Exponential Factor | s⁻¹ |
| Ea | Activation Energy | J/mol |
| T | Absolute Temperature | K |
Inputs & Parameters
[s⁻¹]
[J/mol]
[K]
Calculated Result (Reaction Rate Constant)
k =0.008742s⁻¹
Formula Notes & Background
- Plotting ln(k) vs. reciprocal temperature 1/T yields a straight Arrhenius line with slope −E_a/R and y-intercept ln(A).
- Catalysts accelerate k by providing an alternate transition-state pathway with lower E_a without altering ΔG° of reaction.
Step-by-Step Examples
#1: First-order isomerization with A = 1.0 × 10¹¹ s⁻¹ and E_a = 75,000 J/mol (75 kJ/mol) at T = 300 K.
A = 10¹¹ s⁻¹E_a = 75,000 J/molT = 300 K
E_a/(RT) = 75,000 / 2,494.34 = 30.068; k = 10¹¹ × e^(−30.068)
→ k = 8.74 × 10⁻³ s⁻¹ (half-life t₁/₂ = ln2 / k ≈ 79.3 s)
#2: Warming the same reaction by 10 K from 300 K to 310 K.
T₁ = 300 K, T₂ = 310 KE_a = 75 kJ/mol
k₂/k₁ = exp((75000/8.31446)(1/300 − 1/310)) = exp(0.970)
→ k₂ / k₁ = 2.64× faster (+164% rate increase)
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Frequently Asked Questions
Why must E_a be in J/mol rather than kJ/mol inside the exponent?+
Because R = 8.31446 J/(mol·K) is in Joules; mixing kJ/mol with J/(mol·K) underestimates the exponent by 1,000×.
Is the 'rate doubles every 10 °C' rule always true?+
Only when E_a ≈ 53 kJ/mol near 300 K; higher activation energies triple or quadruple per 10 K rise.
How does Eyring transition-state theory refine Arrhenius?+
Eyring replaces empirical A with (k_B T / h)e^(ΔS‡/R) and relates E_a = ΔH‡ + RT.