Activation Energy from an Arrhenius Plot

Ea=R×slopeE_a = -R \times \text{slope}

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Take logarithms of the Arrhenius equation and it straightens out: ln k = ln A − (Ea/R)(1/T). Plot ln k on the vertical axis against 1/T on the horizontal and you get a line whose slope is −Ea/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 Ea = 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.

Activation Energy from an Arrhenius Plot
Ea=R×slopeE_a = -R \times \text{slope}
Where
  • EaE_a= Activation energy
  • slope\text{slope}= Slope of ln k vs 1/T, in kelvin
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