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Chemistry · Ch 8 — Chemical Kinetics

Calculating Activation Energy from the Arrhenius Equation

8.11

Calculating Activation Energy from the Arrhenius Equation

When rate-constant data are available at only two temperatures, rather than a

full series suitable for a graph, the Arrhenius equation can still be used to extract the activation

energy directly, without needing to plot anything.

Deriving the two-point form. Writing the logarithmic form of the Arrhenius equation once for each

of two temperatures, T1T_1 (where the rate constant is k1k_1) and T2T_2 (where it is k2k_2):

log⁡k1=log⁡A−Ea2.303 R T1log⁡k2=log⁡A−Ea2.303 R T2\log k_1 = \log A - \frac{E_a}{2.303\,R\,T_1} \qquad\qquad \log k_2 = \log A - \frac{E_a}{2.303\,R\,T_2}

Subtracting the first equation from the second eliminates log⁡A\log A (since the pre-exponential factor is

assumed unchanged between the two temperatures for the same reaction):

log⁡k2−log⁡k1=Ea2.303 R T1−Ea2.303 R T2\log k_2 - \log k_1 = \frac{E_a}{2.303\,R\,T_1} - \frac{E_a}{2.303\,R\,T_2}

log⁡k2k1=Ea2.303 R(1T1−1T2)\boxed{\log\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)}

Using the formula. Given the rate constant (or any quantity proportional to it, such as a half-life

or an initial rate) at two different temperatures, this single equation can be solved directly for

EaE_a: compute log⁡(k2/k1)\log(k_2/k_1) on the left, compute (1T1−1T2)\left(\frac{1}{T_1} - \frac{1}{T_2}\right) on the

right using the two absolute temperatures, and rearrange to isolate EaE_a. This two-point method is

especially convenient when a reaction's rate is reported at its "before" and "after" condition -- for

instance, when it is known that a reaction's rate doubles, or that its half-life falls by a certain …