# What is the Arrhenius equation?

The **Arrhenius equation** is an approximation that relates the rate constant of a chemical reaction as a function of temperature. It was established in 1899 by the Swedish chemist Svante Arrhenius (1859-1927). It represents one of the most fundamental contributions in the development of chemical kinetics.

This equation owes its theoretical foundations to numerous scientists, including Wihelmy (1850), Berthelot (1862), and JJ Hood (1885). But the greatest influence on the creation of the Arrhenius equation is attributed to the work of Van’t Hoof (1884), who established a dependence of the equilibrium constants of chemical reactions with temperature.

Based on the Arrhenius equation, the idea has spread that an increase of 10 ° C causes a doubling of the reaction rate. Other interpretations that come from it is that the molecules need a certain activation energy to be able to react with each other.

Likewise, it is pointed out that the Arrhenius equation is another form of exponential decay in which the value of the rate constant decays as a function of the exponent -E _{a} / RT, where E _{a} is the activation energy.

__Equation and explanation__The Arrhenius equation has the following two expressions:

K = Ae ^{-Ea / RT}

Form of the equation used in chemistry and that is related to the moles of reactant.

K = Ae ^{-Ea / kBT}

Form of the equation used in physics and that is related to molecules more, than to moles.

Both are derived from the Van’t Hoof equation for the variation of the equilibrium constant K as a function of temperature. The base equation is:

d (ln k) / dT = E _{a} / RT ^{2}

Which is integrated assuming that E _{a} is independent of temperature. So, we have:

ln k = – E _{a} / RT + ln A

Where ln A is the constant of integration. By applying the exponential function to both sides of the equation, we obtain the Arrhenius equation already entered.

**Components**

**k**

It is the rate constant of a chemical reaction. Its value can be obtained, in addition to the use of the Arrhenius equation, by applying the Law of Speed. Represents the number of collisions between particles that a reaction produces per second.

**A**

It is the so-called pre-exponential factor, which represents the frequency of collisions between the molecules of the reactants with an appropriate geometric orientation that may or may not produce a chemical reaction. If the temperature variation is small, A is usually taken as constant. This factor has its own formula:

A = Z ρ

Where Z is known as the frequency or collision factor, and ρ the geometric or steric factor that indicates the relative orientation of the molecules at the collision point. The constant A has the same units as the rate constant. In the case that the activation energy does not exist, the magnitude of A will be equal to that of k.

**E**

_{a}It is the activation energy, which represents the threshold energy before reaching the transition state: a state where the molecules have an intermediate shape between the molecules of the reactants and the molecules of the products.

The activation energy has as a unit kJ / mol. But in the calculations, J / mol is used as the unit. The minus sign (-) that precedes E _{a} , serves to indicate that its increase produces a decrease in the speed of the reaction, as well as its decrease produces an increase in the reaction speed.

**RT**

Represents the average kinetic energy. Meanwhile, R is the universal gas constant, one of its most used values being 8.31 J · K ^{-1} · mol ^{-1} . YT is the absolute temperature expressed in Kelvin (K).

**and**

It is the base of natural or natural logarithms, having a value of 2.71828.

**e **^{-Ea / RT}

^{-Ea / RT}

It is the fraction of reacting molecules with energy equal to or in excess of the activation energy.

__Applications__

__Applications__

Most of the applications of the Arrhenius equation come from its use in determining the rate constant; and by extension, the speed of the reaction, as well as its activation energy.

For example, a chemical model has been developed based on the Arrhenius equation, which can predict the properties of materials as their temperature changes, being applied in the fields of geology, construction, materials engineering, and in food science.

The Arrhenius equation has been applied, albeit with some criticism, to the kinetics of solid state reactions . It has also been used to characterize the responses of plants to water stress.

The Arrenhius equation served as the basis for the creation of a mathematical model, which quantifies the effect of temperature on the useful life of nickel metal hydride cells or batteries.

Likewise, based on the Arrhenius equation, the decomposition rate of pig waste and cooking oil was established under various moisture contents.

__Solved exercises__

__Solved exercises__

**Exercise 1**

What is the activation energy of a reaction if its rate constant is found to triple as the temperature rises from 600 K to 610 K?

Starting from the Arrhenius equation:

k = ^{AE -Ea / RT}

We solve for factor A:

A = k _{1} / (e ^{-Ea / RT1} )

But since we have two temperatures, T _{1} and T _{2} , there will be two speed constants: k _{1} and k _{2} . Factor A does not change, so we can equalize it for the second temperature:

k _{1} / (e ^{-Ea / RT1} ) = k _{2} / (e ^{-Ea / RT2} )

And clearing E _{a} we will have:

E _{a} = R (ln k _{2} / k _{1} ) / (1 / T _{1} – 1 / T _{2} )

Since k _{2} is three times larger than k _{1} ,

k _{2} / k _{1} = 3

ln (3) = 1.099

And on the other hand:

1 / T _{1} = 1/600 K = 1.66 x 10 ^{-3} K ^{-1}

1 / T _{2} = 1/610 K = 1.64 x 10 ^{-3} K ^{-1}

Substituting then:

E _{a} = (8.31 J K ^{-1} mol ^{-1} ) (1.099) / (1.66 x 10 ^{-3} K ^{-1} – 1.64 x 10 ^{-3} K ^{-1} )

= 456.5 kJ mol ^{-1}

**Exercise 2**

In a gas phase reaction the activation energy is equal to 103 kJ / mol, and the rate constant is 0.085 min ^{-1} . Find the rate constant at 323 K.

From the previous expression we solve for ln k _{2} / k _{1} :

ln k _{2} / k _{1} = (E _{a} / R) (1 / T _{1} – 1 / T _{2} )

Developing the right part of the equation:

ln k _{2} / k _{1} = (103,000 J mol ^{-1} / 8.31 J K ^{-1} mol ^{-1} ) (1/273 K – 1/323 K)

ln k _{2} / k _{1} = 6.99

Taking antilogarithms:

k _{2} / k _{1} = 1.086

k _{2} = (k _{1} ) (1.086)

= (0.085 min ^{-1} ) (1.086)

= 0.092 min ^{-1}