Tutorials

# Arrhenius Equation

### Arrhenius equation- Core Concept

In this tutorial, you will learn what the Arrhenius equation is, how to use the equation to determine the activation energy or rate constant of a reaction, and how to derive it. Using the Arrhenius Equation is an important concept in chemical kinetics.

#### What is the Arrhenius equation?

The Arrhenius equation is used to determine the activation energy or rate constant of a chemical reaction. as the temperature changes. If you want to see how the rate constant changes when the temperature changes, the Arrhenius equation is your friend. Where is rate constant, is the energy of activation (energy required to get the reaction started), is the frequency factor, is the gas constant, and is temperature.

#### Derivation of the Arrhenius equation

The Arrhenius equation can be derived from experimental data. When linearizing the equation into the form of y=mx+b, we get Now we want to solve for  Raise both sides to the e to get rid of the natural log   When we are given two temperature inputs, we can use a different form of the equation. #### Visualizing the Arrhenius Equation

By plotting the previous equation, we can find attributes of a reaction. For example, graphing the linearized function by plotting vs. , you can determine the slope which will be and the Arrhenius constant , which will be the y-intercept.

### Example Problems of the Arrhenius Equation:

The rate constant k of a chemical reaction is measured at two different temperatures:

k1= 1.2 x 1011 at 425 degrees celsius

k2= 1.4 x 1011 at 538 degrees celsius

Calculate the energy of activation for this chemical reaction.

#### Solution:

Since we are given two temperature inputs, we must use the second form of the equation: First, we convert the Celsius temperatures to Kelvin by adding 273.15:

425 degrees celsius = 698.15 K

538 degrees celsius = 811.15 K

Now let’s plug in all the values. We know that k1= 1.2 x 1011 and k2= 1.4 x 1011 and now we have the T values.

ln((1.2 x 1011)/(1.4 x 1011))=(-Ea/(8.31 J/Kmol))((1/698.15 K)-(1/811.15 K))

-Ea=-6735

Ea=6735 J/mol

### Example problem 2:

The decomposition reaction 2NO2(g)—–> 2NO(g) + O2(g) has an Ea of 1.14 x 105 J/mol and a rate constant of 7.02 x 10-3 L/mol x s at 500K. At what temperature will the rate be twice as

If you are asked to solve for a different variable, there are a few ways to check your work. The larger rate constant k should always be associated with the higher temperature. The activation energy should always have a positive sign. The reaction rate and rate constant are both larger at higher temperatures so long as all other factors are held constant.

#### Solution

If the reaction rate is twice as fast, k2=2k1= (7.02 x 10-3 L/mol x s) x 2 =1.40 x 10-2 L/mol x s ln((1.40 x 10-2 L/mol x s)/(7.02 x 10-3 L/mol x s))= ((1.14 x 105 J/mol)/8.314 J/mol x K) x ((1/500K)-(1/T2)

0.69029= 13712 K x (0.00200 K-1-1/T2)

0.00195 x 10-5 K-1 = 1/T2

T2=513 K