We all know that most of the chemical reaction rates increase with an increase in temperature. It has been observed that for every 10^{o}C rise in temperature, the rate of reaction gets doubled. Until 1889, there was no fixed way to relate the variation of the rate of a reaction with temperature. In 1889, the Swedish chemist, Svante Arrhenius proposed an empirical equation that related the rate constant of a chemical reaction with temperature.

**Arrhenius Equation: k = \( Ae^{\frac{-E_a}{RT}} \)**

Where,

k=rate constant of the reaction

A= Arrhenius Constant

E_{a}= Activation Energy

R= Universal Gas Constant

T=Temperature in absolute scale

The term E_{a} is called the activation energy. It is the minimum amount of energy that the reacting molecules must possess so that the effective collisions may result in the formation of products. It is also sometimes referred as the Activation Energy Barrier, and the reacting molecules must cross this barrier to collide effectively and form products. Arrhenius equation is used extensively in an industry by process engineers and reactor designers to model reaction kinetics.

**Consider the Arrhenius Equation:**

k = \( Ae^{\frac{-E_a}{RT}} \)

If we take to log on both sides of the equation, the resulting equation will be as follows:

\( ln k = ln A – \frac {E_a}{R} \frac{1}{T} \) ……….. (1)

Now let’s compare this equation with the equation of a straight line that is y=mx+c, where the m= slope of the line and c= y-intercept.

So we have:

\( y = ln k \)

\( x = \frac {1}{T} \)

\( m = \frac {-E_a}{R} \)

\( c = lnA \)

So a plot of ln k on the y-axis and \( \frac {1}{T} \) on x-axis will give a straight line with its slop equal to –E_{a}/R and y-intercept equal to ln A.

The following figure shows the plot of ln k Vs ( \( \frac {1}{T} \) ).

Arrhenius Equation plays a very important role in the study of reaction. It gives the dependence of rate constant on temperature for a chemical reaction and provides us with a method to quantify the variation in reaction rate with temperature.

Let us now consider a reaction has activation energy E_{a} is occurring at a temperature T_{1} and has a rate constant k_{1}. Assume that the temperature of the reaction is now altered, and it is now T_{2}. No, we have to calculate the rate constant k_{2} at temperature T_{2}.

As per Arrhenius Equation, for first condition:

\( ln K_1 = ln A – \frac {E_a}{R} \frac{1}{T_1}\) ……………. (2)

**For second condition:**

\( ln K_2 = ln A – \frac {E_a}{R} \frac{1}{T_2}\) ……………. (3)

Subtracting Equation (2) from equation (3) we will get,

\( ln~ k_2 – ln~ k_1 \)= \(\frac {E_a}{R} \frac{1}{T_1} – \frac {E_a}{R} \frac{1}{T_2} \)

\( \Rightarrow \) \( ln ~\frac {k_2}{k_1}\) = \(\frac {E_a}{R} ( \frac{1}{T_1} – \frac {1}{T_2} ) \)

The above equation can be used to determine the rate constant k_{2} at temperature T_{2}. Equation (1) is referred as Arrhenius Equation in logarithmic form.