The observation of reaction rates depends on the energy and frequency of collisions between reacting molecule, on the temperature and on whether the collisions have correct geometry is summarized by the Arrhenius equation.
The Arrhenius equation is based on a simple collision model, which states when two molecules colloid they have to overcome a certain degree of energy for the reaction to proceed. If the energy of the system does not match or exceed the activation energy the molecules cannot overcome the energy hump and will not result in a product. Combining the temperature affects with the theory of an activation energy Arrhenius devised his well known equation. $k = k_{0}$$exp^{\frac{-E_{a}}{RT}}$Where E _{a} is the activation energy, R is the gas constant, T is temperature, and k_{0} is the frequency factor, which is a measure of the frequency of collisions between the reactant molecules. |

A large portion of the field of chemical kinetics can be described by the Arrhenius equation.

**k = A$e^{\frac{-E_{a}}{RT}}$**

The Arrhenius equation relates the rate constant k of an elementary reaction to the absolute temperature T; R is the gas constant. The Parameter Ea is the activation energy with dimensions of energy per mole, and A is the pre-exponential factor, which has the units of k. If k is a first order rate constant, A has the units seconds-1, so it is sometimes called the frequency factor.

The Arrhenius equation describes the variation of the reaction rate constant with temperature whenever the activation energy is temperature dependent. There are instances for which the plot of k(T) against 1/T is a straight line, implying that the kinetics are more complicated that predicted by simple Arrhenius theory.

The relations derived from transition state theory and to some extent the Arrhenius equation as well are only valid for elementary reactions. In cases where temperature dependence needs to be described for non elementary reactions, one could use empirical relationships. There are also other Arrhenius like equations proposed in the literature that could be used. The following equation would do equally well as the Arrhenius as the Arrhenius equation.

**k = A exp ($-\frac{B}{T}$)**

with A and B as fit parameters without a physical meaning. Although this seems undesirable, one has to realize that sometimes parameters do not really have a physical meaning.

For a given reaction, the relationship between the rate constant k and temperature T is expressed by the Arrhenius equation.

**k = A$e^{\frac{-E_{a}}{RT}}$**

where

A = Arrhenius constant

E

_{a}= activation energyR = gas constant (8.31 JK

^{-1}mol^{-1})T = absolute temperature (K)

From the integral form of the Arrhenius equation.

**ln k = ln A - $\frac{E_{a}}{RT}$**

(or)

**log k = log A - $\frac{E_{a}}{2.303RT}$**

The activation energy of a reaction can be found by plotting "log k against $\frac{1}{T}$". A straight line graph is obtained, with gradient = $\frac{E_{a}}{2.303RT}$ and y-intercept = log A.

Hence the activation energy E

_{a}and the Arrhenius constant A may be determined.The Arrhenius equation has been used extensively to describe the effect of temperature on a chemical reaction kinetics and can also be used for thermal death. The relationship between k and T is related to the activation energy Ea as determined by Arrhenius equation.

**Example:**The Arrhenius equation is used to determine how reaction rates and diffusion change with temperature.

An underlying assumption of the Arrhenius equation is that the reaction mechanism does not change as a function of temperature. Since accelerated stability testing of pharmaceutical products normally employs a narrow range of temperature it is often difficult to detect non linearity in the Arrhenius plot.

Arrhenius equations problems are given below.

### Solved Examples

**Question 1:**If the coordinates for two points on a line are (1/T = 3.25 $\times$ 10

^{-3}K

^{-1}, ln(k

_{R}/dm

^{3}mole

^{-1}s

^{-1}) = -5.70 and (1/T = 3.50 $\times$ 10

^{-3}K

^{-1}, ln(k

_{R}/dm

^{3}mole

^{-1}s

^{-1}) = -8.15), What is the value of the slope?

**Solution:**

The slope is calculated as follows

slope = $\frac{-8.15-(-5.70)}{3.50\times10^{-3} K^{-1} - 3.25 \times 10^{-3} K^{-1}}$

= $\frac{-2.45}{0.25 \times 10^{-3} K^{-1}}$

= -9.80 $\times$ 10

^{3}K**Question 2:**Find the activation energy for the above problem by using the same slope value ?

**Solution:**

Slope = -$\frac{E_{a}}{R}$

So that it follows

E

_{a}= - slope $\times$ RE

_{a}= - (-9.80$\times$ 10^{3}K) $\times$ (8.314 JK^{-1}mol^{-1})E

_{a}= 81.5 $\times$ 10^{3}J mol^{-1}E

_{a}= 81.5 KJ mol^{-1}