Arrhenius Plot | IB Chemistry

The Arrhenius equation describes how the rate constant k depends on temperature:

k = A · e^−Ea/RT

k — rate constant (units depend on overall reaction order)
A — pre-exponential (frequency) factor — related to collision frequency and orientation
Ea — activation energy (J/mol) — minimum energy for a successful collision
R — gas constant = 8.314 J mol⁻¹ K⁻¹
T — absolute temperature in Kelvin

By taking the natural log: ln(k) = ln(A) − Ea/(R·T)

This is in the form y = mx + c, where plotting ln(k) vs 1/T gives a straight line with slope = −Ea/R and y-intercept = ln(A).

Graph interpretation — A steeper line = higher Ea. If asked to compare two reactions on the same plot, the one with the steeper slope has the higher activation energy
Finding Ea from slope — slope = −Ea/R, so Ea = −slope × R. Convert to kJ by dividing by 1000
Two-point formula — ln(k₂/k₁) = −Ea/R × (1/T₂ − 1/T₁). IB may give just two data points instead of a full graph
Catalyst effect — A catalyst lowers Ea, so the Arrhenius plot has a shallower slope but higher intercept (more collisions succeed at lower T)
Units trap — Always use T in Kelvin and Ea in J/mol (not kJ) when calculating. Convert at the end
R value — Use R = 8.314 J mol⁻¹ K⁻¹ (from the IB Data Booklet)

Starting from: k = A · e^−Ea/RT

Take natural log of both sides:

ln(k) = ln(A) + (−Ea/R) · (1/T)

Compare with y = c + m·x:

Variable	Arrhenius	Linear Form
y	ln(k)	Dependent variable
x	1/T	Independent variable
m (slope)	−Ea/R	Always negative → line slopes downward
c (intercept)	ln(A)	Where line crosses y-axis

The two-temperature form eliminates A:

ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂)