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Arrhenius Equation & Activation Energy

Physical Chemistry #6

§1 Interactive Simulation

Reaction Coordinate
Collision Dynamics
k vs Temperature
Catalyst Effect
Eyring / Two-Step
ln k vs 1/T
k vs T
Boltzmann Distribution
Ea Sensitivity
Half-life vs T
k (s⁻¹)
Eₐ (kJ/mol)
T (K)
A (s⁻¹)
ln k
f = e^(−Ea/RT)



Show energy labels
Show transition state
Show catalyst path
Show Boltzmann tail

§2 The Idea, Step by Step

Start with the kitchen

Milk keeps for a week in the fridge but sours in a day on the counter. Bread browns faster in a hotter oven, and a glow stick glows longer when you put it in the freezer. It is the same chemistry every time — what changes is only the temperature. Heating something does not change what can react; it changes how often the molecules slam together hard enough to actually react.

Build the idea: name the pieces

Every reaction has an energy "hill" the molecules must climb before reactants can turn into products. The height of that hill is the activation energy $E_a$. Only molecules carrying at least that much energy make it over. Warm the mixture and a much larger share of molecules carry that much energy, so the reaction speeds up. We measure that speed with the rate constant $k$, and it obeys one compact rule:

$$k = A\,e^{-E_a/RT}$$

Here $A$ counts how often molecules collide while lined up correctly, $R = 8.314$ J·mol⁻¹·K⁻¹, and $T$ is the absolute temperature. The whole story lives in the exponent. As a worked number, take a reaction with $E_a = 50$ kJ/mol and warm it from 300 K to 310 K: the fraction of molecules above the barrier, $e^{-E_a/RT}$, roughly doubles — so the rate roughly doubles for just a 10° rise.

Deepen it: the precise picture

That factor $e^{-E_a/RT}$ is the Boltzmann fraction — the slice of the energy distribution sitting above the barrier. Because it is exponential, "doubling per 10°" is only a rule of thumb that holds near $E_a \approx 50$ kJ/mol; a stiffer barrier such as $E_a = 103$ kJ/mol makes the rate roughly quadruple per 10°. Taking the natural log straightens the curve into a line, $\ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T}$, so plotting $\ln k$ against $1/T$ gives a slope of $-E_a/R$ — exactly the "ln k vs 1/T" graph below. The sliders map straight onto the symbols: $E_a$ sets the hill's height (and that slope), $T$ slides you along the line, $\log_{10}A$ shifts the whole line up or down, and the catalyst toggle opens a lower pass without moving $\Delta H$.

Try this in the sim above

Set $E_a$ low (about 30) and then high (about 200) at a fixed temperature, and watch the $k$ readout swing by many orders of magnitude. Next hold $E_a$ fixed and drag $T$ from 300 to 600 K while the "Boltzmann Distribution" graph is open — see the energetic tail past $E_a$ swell as you heat. Finally turn on the catalyst path: the barrier drops and $k$ jumps, yet the products sit at the same height, proof that a catalyst speeds the trip without changing the destination.

§3 Equation Derivation

The Arrhenius Equation

$$k = A\,e^{-E_a/RT}$$

Named after Svante Arrhenius (1889), this equation relates the rate constant $k$ to temperature $T$, activation energy $E_a$, and a pre-exponential factor $A$.

Symbol Definitions

SymbolMeaningUnit
$k$Rate constants⁻¹ (1st order) or L·mol⁻¹·s⁻¹ (2nd order)
$A$Pre-exponential (frequency) factorsame as $k$
$E_a$Activation energy — minimum kinetic energy for reactionJ·mol⁻¹ (or kJ·mol⁻¹)
$R$Universal gas constant8.314 J·mol⁻¹·K⁻¹
$T$Absolute temperatureK
$f$Boltzmann fraction: $e^{-E_a/RT}$dimensionless

Step-by-Step Derivation from Collision Theory

Step 1 — Collision frequency
From kinetic theory, the bimolecular collision frequency $Z_{AB}$ per unit volume is: $$Z_{AB} = n_A\,n_B\,\sigma_{AB}\sqrt{\frac{8k_BT}{\pi\mu}}$$ where $\sigma_{AB}$ is the collision cross-section and $\mu$ the reduced mass.
Step 2 — Boltzmann energy distribution
Not all collisions lead to reaction. From the Maxwell-Boltzmann distribution, the fraction of molecules with translational kinetic energy $\geq E_a$ along the reaction coordinate is: $$f = e^{-E_a/RT}$$ This is the Boltzmann factor — exponentially sensitive to $E_a$ and $T$.
Step 3 — Steric factor
Molecules must also collide with the correct orientation. A steric factor $p$ ($0 < p \leq 1$) accounts for this: $$k = p\,Z_{AB}\,e^{-E_a/RT}$$
Step 4 — Define pre-exponential factor A
Collecting all temperature-weakly-dependent terms: $A = p\,\sigma_{AB}\sqrt{8k_BT/\pi\mu}$. Since $A$ varies as $T^{1/2}$ much more slowly than the exponential, treating $A$ as a constant gives the Arrhenius equation: $$\boxed{k = A\,e^{-E_a/RT}}$$
Step 5 — Linearized (Arrhenius plot)
Taking the natural log: $$\ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T}$$ A plot of $\ln k$ vs $1/T$ gives a straight line with slope $-E_a/R$ and intercept $\ln A$. This is used experimentally to determine $E_a$.
Step 6 — Two-temperature form
At temperatures $T_1$ and $T_2$: $$\ln\frac{k_2}{k_1} = -\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$ This allows $E_a$ determination from just two rate constants.

Simulation Variable Mapping

Slider / ControlEquation SymbolEffect
Activation Energy slider$E_a$ (kJ/mol)Changes slope of Arrhenius plot, exponential change in $k$
Temperature slider$T$ (K)Moves position on Arrhenius plot, shifts Boltzmann distribution
Pre-exp factor slider$\log_{10} A$Shifts entire Arrhenius line up/down, changes $k$ at all $T$
ΔH reaction slider$\Delta H_{rxn}$Sets reverse activation energy: $E_{a,\text{rev}} = E_a - \Delta H_{rxn}$
Catalyst toggle$E_a \to E_a'$Lowers $E_a$ by ~40 kJ/mol, dramatically increases $k$

Worked Example

Problem: The decomposition of N₂O₅ has $E_a = 103$ kJ/mol, $A = 4.94 \times 10^{13}$ s⁻¹. Find $k$ at 25°C and 50°C, and the ratio $k_{50}/k_{25}$.

$T_1 = 298$ K: $k_1 = 4.94\times10^{13}\cdot e^{-103000/(8.314\times298)} = 4.94\times10^{13}\cdot e^{-41.57} = 4.4\times10^{-5}$ s⁻¹
$T_2 = 323$ K: $k_2 = 4.94\times10^{13}\cdot e^{-103000/(8.314\times323)} = 4.94\times10^{13}\cdot e^{-38.36} = 1.1\times10^{-3}$ s⁻¹

Ratio: $k_2/k_1 \approx 25$. With a barrier this large the rate roughly quadruples for each 10°C; the familiar "rate doubles every 10°C" rule applies only near $E_a \approx 50$ kJ/mol.
📚 Reference: Atkins & de Paula — Physical Chemistry, 11th Ed., Chapter 17.5: "The Arrhenius Parameters". Also: Levine — Physical Chemistry, 6th Ed., §17.8.

§4 Frequently Asked Questions

📚 FAQ Reference: LibreTexts Chemistry — "The Arrhenius Law" (chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules/Kinetics/Arrhenius_Equation); Chemguide.co.uk — "The Effect of Temperature on Rate of Reaction"

§5 Common Misconceptions

📚 Misconceptions Reference: Nakhleh — J. Chem. Educ. 69, 191 (1992); Bain & Towns — Chemistry Education Research and Practice 17, 297 (2016) "A review of research on the teaching and learning of chemical kinetics"; Taber — Chemical Misconceptions: Prevention, Diagnosis and Cure, RSC, 2002.