Activation energy and the Arrhenius equation

In a nutshell

The Arrhenius equation is used to calculate the activation energy. This equation links the activation energy with the rate constant. The activation energy can be calculated from the gradient of an Arrhenius plot. Catalysts increase the rate of reaction. 

Equations

The equation that links the activation energy with the rate constant is:  

​$$k=Ae^{-\dfrac{E_a}{RT}}$$

When plotting an Arrhenius plot the gradient can be used to calculate the activation energy using the equation:

$$E_a=-R\>\times\>gradient$$​​

The equation to find the rate is: 

​$$rate \propto \dfrac{1}{time}$$​​

Constants 

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Variable units

Arrhenius equation

The Arrhenius equation is used to calculate the activation energy. 

​$$k=Ae^{-\dfrac{E_a}{RT}}$$​​

The activation energy, $$E_a$$, is the particle's minimum amount of kinetic energy required for it to react. The rate constant, $$k$$, links to the activation energy in the Arrhenius equation. The frequency factor, $$A$$, is the frequency at which the molecules collide and it's just another constant you don't need to know about. 

The Arrhenius equation shows as the rate constant increases, the activation energy decreases. It also shows that the rate constant increases as temperature increases. 

The Arrhenius equation can be rearranged: 

​$$lnK=lnA-\dfrac{E_a}{RT}$$​​

This can be further simplified to:

​$$lnK=constant-\dfrac{E_a}{RT}$$

This allows the Arrhenius equation to be plotted on a graph. The $$y$$​-axis is $$lnK$$, the $$x$$​-axis is $$\dfrac{1}{T}$$. This results in the gradient being $$-\dfrac{E_a}{R}$$.

Working out the gradient from this graph will allow you to work out the activation energy. 

$$E_a=-R\>\times\>gradient$$

Calculate the activation energy from an Arrhenius plot.

procedure

Example 

The Arrhenius plot is shown for a reaction. Calculate the activation energy for this reaction. Round your answer to three significant figures. $$R\>=\>8.31\>J\>K^{-1}\>mol^{-1}$$.

Calculate the gradient:

$$gradient=\dfrac{\triangle y}{\triangle x}=\dfrac{-30}{0.0008\>K^{-1}}=-37 \, 500\>K$$

The gradient equals to: 

$$gradient=-\dfrac{E_a}{R}=-37 \, 500\>K$$​​

Rearrange for the activation energy, $$E_a$$:

$$E_a=-R\>\times\>gradient$$​​

Insert the gas constant and gradient values to calculate $$E_a$$:

$$E_a=-8.31\>J\>K^{-1}\>mol^{-1}\times-37\,500\>K=311 \, 625\>J\>mol^{-1}=311.625\>kJ\>mol^{-1}$$

Round to three significant figures:

​$$E_a=312\>kJ\>mol^{-1}$$ 

The activation energy, $$E_a$$, is $$\underline{312\>kJ\>mol^{-1}}$$.

Calculating E_a from data

Sometimes the Arrhenius plot isn't given. This means you need to draw the Arrhenius plot from the data provided. The temperature values will be given. The time values might be given instead of the rate constant. 

The rate is proportional to:

​$$rate \propto \dfrac{1}{time}$$

This means for the $$y$$​-axis, $$ln \dfrac{1}{t}$$ can be plotted instead of $$lnk$$. 

procedure 

Catalysts

Catalysts increase the rate of reaction. Catalysts lower the activation energy by giving a different reaction pathway. It doesn't change chemically after the reaction. There are two types of catalysts, homogeneous and heterogeneous catalysts.

Homogeneous catalysts

Homogeneous catalysts are catalysts that have the same physical state as the species in the reaction.

Example

The esterification for propyl-ethanoate is catalysed by concentrated sulfuric acid. 

​$$CH_3COOH(aq)+CH_3CH_2CH_2OH(aq)→CH_3COOCH_2CH_2CH_3(aq)+H_2O(l)$$​​

The catalyst, concentrated sulfuric acid, is aqueous which is the same as the reactants.

Heterogeneous catalysts

Heterogeneous catalysts are catalysts that don't have the same physical state as all the species in the reaction.

Example

The production of gaseous ammonia is catalysed by a solid iron catalyst. 

$$N_2(g)+3H_2(g)⇌2NH_3(g)$$

The catalyst, iron, is a solid which is not the same as all the gaseous nitrogen, hydrogen and ammonia.