Arrhenius plot

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An Arrhenius plot displays the logarithm of kinetic constants (\ln(k), ordinate axis) plotted against inverse temperature (1/T, abscissa). Arrhenius plots are often used to analyze the effect of temperature on the rates of chemical reactions. For a single rate-limited thermally activated process, an Arrhenius plot gives a straight line, from which the activation energy and the pre-exponential factor can both be determined.

Example:
Nitrogen dioxide decay
2 NO2 → 2 NO + O2
Conventional plot:
k against T
Arrhenius plot: ln(k) against 1/T.

The Arrhenius equation can be given in the form:

k = A e^{-E_a/RT}

or alternatively

k = A e^{-E_a/k_B T}

The only difference is the energy units: the former form uses energy/mole, which is common in chemistry, while the latter form uses energy directly, which is common in physics. The different units are accounted for in using either the gas constant R or the Boltzmann constant k_B.

The former form can be written equivalently as:

\ln(k) = \ln(A) - \frac{E_a}{R}\left(\frac{1}{T}\right)
Where:
k = Rate constant
A = Pre-exponential factor
E_a = Activation energy
R = Gas constant
T = Absolute temperature, K

When plotted in the manner described above, the value of the true y-intercept (at  x = 1/T = 0) will correspond to \ln(A), and the slope of the line will be equal to -E_a/R. The values of y-intercept and slope can be determined from the experimental points using simple linear regression with a spreadsheet.

The pre-exponential factor, A, is an empirical constant of proportionality which has been estimated by various theories which take into account factors such as the frequency of collision between reacting particles, their relative orientation, and the entropy of activation.

The expression e^{-E_a/RT} represents the fraction of the molecules present in a gas which have energies equal to or in excess of activation energy at a particular temperature.

Worked Example[edit]

Based on the red "line of best fit" plotted in the graph given above:

Let y = ln(k[10-4 cm3 mol-1 s-1])
Let x = 1/T[K]

Points read from graph:

y = 4.1 at x = 0.0015
y = 2.2 at x = 0.00165

Slope of red line = (4.1 - 2.2) / (0.0015 - 0.00165) = -12,667

Intercept [y-value at x=0] of red line = 4.1 + (0.0015 x 12667) = 23.1

Inserting these values into the form above:

\ln(k) = \ln(A) - \frac{E_a}{R}\left(\frac{1}{T}\right)

yields:

\ln(k) = 23.1 - 12,667 (1/T)
k = e^{23.1} \cdot e^{-12,667/T}
k = 1.08 \times 10^{10} \cdot e^{-12,667/T}

for:

k in 10-4 cm3 mol-1 s-1
T in K

Substituting for the quotient in the exponent of e:

-Ea / R = -12,667 K
approximate value for R = 8.31446 J K−1  mol−1

The activation energy of this reaction from these data is then:

Ea = R x 12,667 K = 105,300 J mol-1 = 105.3 kJ mol-1.

See also[edit]