#### Arrhenius Equation and EA Energy of Activation

Graph

Label the graph and identify the Ea for the catalyzed and un-catalyzed reaction.

Ea = Energy of Activation

The rate constant *k* is affected by the temperature and this dependence may be represented by Arrhenius equation:

–*E*_{a}/*RT*

*k* = *A* e

where the pre-exponential factor *A* is assumed to be independent of temperature, *R* is the gas constant, and *T* the temperature in K. Taking the natural logarithm of this equation gives:

ln *k* = ln A – *E*_{a}/(*RT*)

or

ln *k* = –*E*_{a}/(*RT*) + constant

or

ln *k* = -(*E*_{a}/R)(1/T) + constant

These equations indicate that the plot of ln *k* vs. 1/*T* is a straight line, with a slope of –*E*_{a}/*R*. These equations provide the basis for the experimental determination of *E*_{a}.

**Example 1 **

**The reaction constants k determined at 298 K and 350 K are 0.00123 /(M s) and 0.0394 /(M s) respectively.
(a) Calculate E_{a}.
(b) What is the rate constant at 308 K?**

*Solution*

Let *k*_{1} and *k*_{2} be the rate constants determined at *T*_{1} and *T*_{2}, respectively. Then you have two equations:

ln *k*_{1} = lnA – *E*_{a}/(R *T*_{1})

ln *k*_{2} = lnA – *E*_{a}/(R *T*_{2})

From these, you should be able to derive the following relationships,

= 57811 J/mol

= 57.8 kJ/mol

It is a good idea to manipulate the formula with symbols until you have obtained the desirable form before you substitute numerical values into it. The necessary units are included here to show you the derivation of units for *E*_{a}.

(b) To calculate *k* at 308 K,

= -6.70 + 0.758

= -5.94

*k* = e^{-5.94}

*k* = 0.00263

*Discussion*

An increase of 10 *k* doubles the rate constant in this case.

If *E*_{a} is positive, increasing temperature always leads to an increase in the rate constant.