Activation Energy from an Arrhenius Plot

Take logarithms of the Arrhenius equation and it straightens out: ln k = ln A − $(E_a/R)(1/T)$. Plot ln k on the vertical axis against 1/T on the horizontal and you get a line whose slope is $-E_a/R$ and whose intercept is ln A. Multiplying the slope by −R (8.314 J/(mol·K)) recovers the activation energy from as many data points as you care to collect, which is far more robust than the two-point method because random scatter averages out across the whole set.

A measured slope of −6448 K gives $E_a$ = 8.314 × 6448 = 53.6 kJ/mol, the same value a doubling-per-10-K experiment would yield. The slope has units of kelvin, and it is always negative for a normal reaction: a positive slope means the reaction speeds up on cooling, which happens only in unusual multi-step systems with a negative apparent activation energy. Watch the axis scaling too: 1/T values cluster in a narrow band (1/300 to 1/320 spans only 0.00021), so plotting the reciprocal to too few decimal places wrecks the slope long before the chemistry does.