Activation Energy, Collision Theory and Arrhenius Equation — Part 2: The Arrhenius equation: visual lesson, ISC Class 12 Chemistry

Lesson

The Arrhenius equation

Part 2 — putting numbers to the effect of temperature on rate.

In this lesson you will

State the Arrhenius equation and explain the meaning of A and Ea
Use the log form and the graph of log k against 1/T to find Ea and A
Calculate activation energy from rate constants at two temperatures
Calculate the effect of a catalyst or a temperature change on the rate constant

Concept

The Arrhenius equation

$k = A\,e^{-E_\text{a}/RT}$ . A is the frequency (pre-exponential) factor, related to the frequency of collisions with proper orientation; it has the same unit as $k$ . $e^{-E_\text{a}/RT}$ is the fraction of molecules with energy at least $E_\text{a}$ (the Boltzmann factor). A larger $E_\text{a}$ gives a smaller $k$ and a reaction more sensitive to temperature.

k = A\,e^{-E_\text{a}/RT}

Table of terms in the Arrhenius equation
Term	Meaning
A	Frequency factor (collisions with proper orientation)
Ea	Activation energy, J mol⁻¹
R	8.314 J K⁻¹ mol⁻¹
T	Absolute temperature, K
e^(−Ea/RT)	Fraction of molecules with energy ≥ Ea

Concept

Log form and the Arrhenius plot

Taking logarithms: $\ln k = \ln A - \dfrac{E_\text{a}}{RT}$ , or $\log k = \log A - \dfrac{E_\text{a}}{2.303RT}$ . A plot of $\log k$ against $1/T$ is a straight line with slope $= -E_\text{a}/2.303R$ and intercept $= \log A$ .

\log k = \log A - \frac{E_\text{a}}{2.303R}\cdot\frac{1}{T}

slope = −Ea ÷ 2.303R

Illustrative line: slope −4000 K (per unit of 1/T), so Ea = 2.303 × 8.314 × 4000 ≈ 76.6 kJ mol⁻¹; intercept log A = 11.

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