One temperature
With k, A and T given, use the exponential equation directly. Put Ea into joules when R is in J K−1 mol−1.
Arrhenius equation & plots
Calculate with the Arrhenius equation, turn experimental data into a straight-line plot, and practise the graph and unit checks that these questions require.
Two-temperature practice: the derived relationship is supplied in core questions. Deriving or recalling it unaided is Stretch & challenge, not assumed recall.
Choose the Arrhenius form from the data you have
k = Ae−Ea/RT · ln k = −Ea/R(1/T) + ln A
Several temperatures
Plot ln k against 1/T. The gradient is −Ea/R, so a steeper negative line means a larger activation energy.
AQA gives the required Arrhenius equations when needed; OCR A includes them on the Data Sheet.
OCR A intercept
The y-intercept is ln A and lies at 1/T = 0. Extrapolate the straight line to that point when the plotted axis starts above zero.
Before you accept an Arrhenius answer
J and kJ
R is in joules. Convert Ea to J mol−1 before it goes into the exponential, then convert the final Ea back to kJ mol−1 if required.
Kelvin
T means absolute temperature. A temperature quoted in °C must be converted before substitution.
Natural logarithms
The straight-line equation uses ln k. Using log10 changes the gradient and gives the wrong Ea.
What temperature changes
Higher temperature increases k. Ea belongs to the reaction pathway; a catalyst provides a pathway with a lower Ea.