One temperature
With k, A and T given, I’d use the exponential equation directly. Put Ea into joules when R is in J K−1 mol−1.
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Calculate with the Arrhenius equation, turn experimental data into a straight-line plot, and recognise the graph and unit mistakes that cost easy marks.
With k, A and T given, I’d use the exponential equation directly. Put Ea into joules when R is in J K−1 mol−1.
Plot ln k against 1/T. The gradient is −Ea/R, so a steeper negative line means a larger activation energy.
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.
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.
T means absolute temperature. A temperature quoted in °C must be converted before substitution.
The straight-line equation uses ln k. Using log10 changes the gradient and gives the wrong Ea.
Higher temperature increases k. Ea belongs to the reaction pathway; a catalyst provides a pathway with a lower Ea.