Question

How would you determine the activation energy of a reaction with the help of Arrhenius equation?

How can activation energy be determined for a reaction with the help of Arrhenius equation?

Answer 1

Graphical method:

\[\ce{log_10 k = - \frac{E_a}{2.303 RT} + log_10 A}\]

This is an equation of the type y = mx + c and represents a straight line. Therefore, if the values of k are plotted against \[\ce{\frac{1}{T}}\], a straight line should be obtained, and the slope of this line should be equal to \[\ce{- \frac{E_a}{2.303 R}}\].

Therefore, in order to determine the energy of ·activation of a reaction, its rate constants at different temperatures are measured. Now, the values of log₁₀ k against \[\ce{\frac{1}{T}}\] are plotted. 

The curve obtained is a straight line, as shown in the figure. The slope of the line is measured. The slope of the line is related to the activation energy of the reaction as follows.

\[\ce{Slope = \frac{CA}{CB} = - \frac{E_a}{2.303 R}}\]

Rate constant method:

If k₁ and k₂ are the rate constants measured at temperatures T₁ and T_2, respectively, then according to \[\ce{log_10 k = - \frac{E_a}{2.303 RT} + log_10 A}\], we have

\[\ce{log_10 k_1 = - \frac{E_a}{2.303 RT_1} + log_10 A}\]    ...(i)

\[\ce{log_10 k_2 = - \frac{E_a}{2.303 RT_2} + log_10 A}\]    ...(ii)

Subtracting eq. (i) from eq. (ii), we get

\[\ce{log_10  k_2  - log_10 k_1 = \frac{E_a}{2.303 R} ({\frac{1}{T_1}} - {\frac{1}{T_2}})}\]

or, \[\ce{log_10 \frac{k_2}{k_1} = \frac{E_a}{2.303 R} (\frac{1}{T_1} - \frac{1}{T_2})}\]    ...(iii)

Equation (iii) can therefore be used to determine the energy of activation (E_a) of a reaction given the values of its rate constants, k₁ and k₂, measured at two distinct temperatures, T₁ and T₂, respectively.