Using the Arrhenius equation to find the activation energy for a reaction

For almost all reactions there is an energy barrier to overcome – this is known as the activa tion energy, E_a, and it is the minimum amount of energy that two colliding particles must possess in order to react. We can illustrate this on a Maxwell-Boltzmann distribution curve:

The area under the curve is proportional to the total number of particles involved and the blue shaded area is proportional to the number of particles with energy greater than E_a (activation energy). Hence the fraction of particles with energy greater than E_a is given by the ratio:

In the mathematical expression shown above that Maxwell and Boltzmann derived for this fraction, R is the gas constant (8.31 JK^-1mol^-1) and T is the temperature in K. The implication of the ‘RT’ term is that the higher the temperature, the greater the probability that a given molecule possesses a particular energy.

So, for particles to react their energy must exceed E_a for that reaction. This suggests that at a given temperature the rate of the reaction is proportional to the fraction of particles with energy greater than E_a:

rate ∝ e ^{(-Ea / RT)}

Since the rate changes during the progress of a reaction, it is more useful to use k, the rate constant.

Consider the general reaction

A_(g) + B_(g) → products

rate = k [A]^x [B]^y at a room temperature of T₁

If the experiment is repeated using the same concentration of A and B at a higher temperature, T₂, the rate increases. Since [A]^x and [B]^y have the same initial value at each temperature the rate constant, k, must increase as temperature increases.

The rate constant is a general measure of the reaction rate at a particular temperature.

The equation now becomes:

k ∝ e ^{^{(-Ea / RT)}}

and so we have the Arrhenius equation:

k = A e ^{^{(-Ea / RT)}}

where A is a constant, sometimes called the Arrhenius constant or pre-exponential constant / factor. The fraction of collisions which have the correct orientation, the size of the particles and their speed all contribute to the value of A.

If we plot a graph of the rate constant against temperature we see that the rate constant increases exponentially with temperature:

The Arrhenius equation is telling us that the rate constant increases with temperature because at a higher temperature, more colliding particles have sufficient energy to overcome E_a and a reaction with a high E_a will be considerably more sensitive to temperature changes than a reaction with a low E_a.

As mentioned above, the term e^{(-Ea / RT)} gives us the fraction of collisions with the required energy to react and as the temperature increases the term (E_a/RT) will become smaller and smaller until it approaches zero.

If (E_a/RT) = 0, then e^{(-Ea / RT)} = 1 and the pre-exponential factor, A, equals the rate constant, k. A gives us the rate constant when there is no energy barrier to the reaction because at this exceedingly high temperature all collisions have sufficient energy to react (although in reality the temperature will be so high that in most cases the reactants will have dissociated into atoms!).

We can linearise the Arrhenius equation by taking natural logarithms of both sides:

Plotting ln k against 1/T (with temperature in K) gives us a straight line graph with a gradient of (-E_a/R) and ln A is the intercept of the y-axis. We can then use the Arrhenius equation to find the activation energy for a reaction (in J mol^-1 because the units for R, the gas constant, are JK^-1mol^-1).

Finally, if we know the value of the rate constant at one temperature we can use the Arrhenius equation to calculate its value at any other temperature.

Subtracting the first equation from the second gives us:

E.g. given that the activation energy for the reaction between methane and hydroxyl radicals in the atmosphere is 20.3 kJ mol^-1, calculate the ratio of rate constants for the reaction at 25°C and -55°C (this is the average temperature at the top of the troposphere). Comment on your answer.

The rate constant for the reaction between methane and hydroxyl radicals is 20.3 times greater at room temperature than it is at the top of the troposphere, which means that assuming the concentrations of methane and hydroxyl radicals are the same in each place, the reaction is 20.3 times faster at room temperature.

Practice questions

Hydrogen iodide decomposes to form iodine and hydrogen

2HI_(g) ⇌ H_{2 (g)} + I_{2 (g)}

In a series of experiments, the rate constant for this reaction was determined at several different temperatures, as shown in the table below:

Temperature / K	Rate constant (k) / dm³ mol^-1 s^-1
635	1.8 x 10^-5
665	1.0 x 10^-4
695	5.0 x 10^-4
715	1.0 x 10^-3
780	1.5 x 10^-2

(a) Use the data given to determine the activation energy, E_a, for the reaction.

(b) Calculate the ratio of the rate constants for the reaction at 300K and 310K and hence comment on the effect of a 10K temperature rise on the rate of the reaction.

The oxidation of iodide ions by peroxodisulphate ions can be followed using a clock reaction. The rate of the reaction is ∝ (1/t), where t is the time taken for a fixed amount of iodine to be produced.

S₂O₈^2-_(aq) + 2I^–_(aq) ⇾ 2SO₄^2-_(aq) + I_2(aq)

The experiment was carried out at varying temperatures and the results are shown below:

Temperature / °C	Time / s
30	205
35	140
40	115
45	75
50	55

Use these results to plot a graph of ln (1/t) against (1/T) and hence determine a value for the activation energy for this reaction.

Answers

(a) The first thing we need to do is calculate ln k and 1/T for each experiment (note the units for 1/T):

Temperature / K	(1/T) / K^-1	Rate constant (k) / dm³ mol^-1 s^-1	ln k
635	0.00157	1.8 x 10^-5	-10.9
665	0.00150	1.0 x 10^-4	-9.2
695	0.00144	5.0 x 10^-4	-7.6
715	0.00140	1.0 x 10^-3	-6.9
780	0.00128	1.5 x 10^-2	-4.2

If we plot a graph of ln k against 1/T, then E_a = – gradient x R.

Although we could calculate the gradient, Δy/Δx, directly from the results table since Δy = 0.00157 – 0.00128 and Δx = -10.9 – (-4.2), we would want to be sure that neither of these results were anomalous, and the only way to do that is to plot the graph and check they lie on the line of best fit.

E_a = 192 kJ mol^-1 for the decomposition of hydrogen iodide.

(b) The rate constant for the reaction is 11.9 times greater at 310K than it is at 300K, which means that a 10K rise in temperature leads to a significantly large increase in the rate of reaction.