Arrhenius Equation

The Arrhenius equation establishes a link between a chemical reaction’s rate constant and absolute temperature, incorporating the A factor, or pre-exponential factor. This factor represents the frequency of reactant particles colliding in the correct orientation. The equation highlights how reaction rates are influenced by changes in temperature, illustrating the temperature dependency of chemical processes. Let us look at the Arrhenius equation in detail in this article.

What is the Arrhenius Equation?

The Arrhenius equation is written as follows:

k = Ae^-E_a ^{/ RT}

In the above equation,
k is rate constant
A is the pre-exponential factor
E_a is the activation energy for the reaction per mole of the reactants
R is the universal gas constant
T denotes the absolute temperature of the reaction in Kelvin (K)

The above equation may also be written as follows when the energy is taken as energy per molecule of the reactants.

k = Ae^-E_a ^{/ k}_b^T

where
k_b represents the Boltzmann constant.

Arrhenius Equation Graph

We shall plot the Graphical Representation of the Arrhenius Equation using the example of the decomposition reaction of Nitrogen dioxide. The X-axis will represent the Absolute temperature T in Kelvin and the Y-axis will represent the rate constant k. Let us first write the decomposition reaction for nitrogen dioxide.

2NO₂  —>  2NO + O₂

The image given below shows the decomposition of the above reaction,

Arrhenius Plot

We can see in the above graph that as the temperature rises the rate of reaction also increases.

Now we shall plot the graph for the Arrhenius equation. The Arrhenius equation is written as:

k = Ae^-E_a ^{/ RT}

Taking logarithm on both sides, we get

ln k = ln(Ae^-E_a ^{/ RT})

ln k = ln (A) + ln (e^-E_a ^{/ RT})

ln k = ln (A) + -E_a / RT ln (e)

ln k = ln (A) + (-E_a / RT )       [ln (e) = 1]

ln k = -E_a / RT  + ln(A)

The equation thus obtained is of the form y=mx+c where m = -E_a / RT. 

The above equation can be easily plotted to obtain the  Arrhenius Graph. The image given below shows the Arrhenius Graph,

Arrhenius Equation Account for Catalysts

A catalyst reduces the activation energy required for the reaction. Thus the lowered activation energy can be used in the Arrhenius equation to get the rate constant for the catalyzed reaction. We know that the rate of a chemical reaction is directly proportional to the rate constant of that reaction, the decrease in activation energy results in an exponential increase in the reaction rate.

As compared to non-catalytic reactions, the effect of temperature is more on the catalytic reactions. This is because the exponential term -Ea/RT contains the activation energy in the numerator and the absolute temperature in the denominator. Since the activation energy of catalytic reactions is comparatively low, the effect of temperature on rate constants is very large in the corresponding uncatalyzed reactions.

Arrhenius Equation and the Pre-Exponential Factor (A)

We already discussed that A is known as the pre-exponential factor in the Arrhenius equation. This factor deals with intermolecular collisions and can be thought of as the frequency of precisely aligned collisions between molecules with sufficient energy to initiate a chemical reaction.

A = ρZ

where
Z represents the frequency factor
ρ represents the steric factor

The value of A has different values ​​for different reactions and must be determined experimentally. It also depends on the temperature at which the reaction takes place. The units of A depend on the order of reactions. For example, the units of ‘A’ for the second order rate constant are Lmol^-1s^-1 whereas the units of the first order rate constant are s^-1.

Arrhenius Equation without Pre-Exponential Factor

Consider a chemical reaction taking place at two different temperatures T₁ and T₂, and the rate constants for these two reactions are k₁ and k₂ respectively.

We know that the Arrhenius equation can be written as follows for the given temperatures:

ln k = ln (A) – E_a / RT

ln k₁ = ln (A) – E_a / RT₁ …(1)

ln k₂ = ln (A) – E_a / RT₂ …(2)

We can obtain the value of ln(A) from the above equation as follows:

ln (A) = ln k₂ + E_a / RT₂

Using this value of ln (A) in equation 1, the following equation can be formed:

ln k₁ = ln k₂ + E_a / RT₂ – E_a / RT₁

Transporting ln k₂ to the left-hand side, the value of ln k₁ – ln k₂ is as:

ln k₁ – ln k₂ = E_a / RT₂ – E_a / RT₁

The LHS of the equation is of the form ln(x) – ln(y), which can be rewritten as ln (x/y) using the property of logarithms. 

Also, the term E_a / R is common in both the terms on the right-hand side. Therefore, the entire equation can be written as follows:

ln (k₁ / k₂) = E_a / R [ 1/T₂ – 1/T₁]

Solved Examples on Arrhenius Equation

Example 1: The activation energy for a chemical reaction is 2000 J per mole and its A factor is 10 per Mole per second. Find the rate constant of this equation at a temperature of 200 K. 

Solution:

Given:

[Tex]E_a = 2000 \space Jmol^{-1}[/Tex]

A = 10 M-1s-1

T = 200 K

R = 8.314 J.mol-1.K-1

k = ?

We know that,

[Tex]ln(k) =  ln(A) – (\frac{E_a}{RT})[/Tex]

        [Tex]= ln(10)-\frac{2000}{8.314*200}[/Tex]

        = 2.303-(10/8.314)

        = 1.1

Thus the value of K = e^1.1.

Example 2: The rate constant of a chemical reaction is 2.5×10^-7 M⁻¹s⁻¹ at 300K. The rate constant for this reaction is 1.5×10^-6 M⁻¹s⁻¹ at 600K. Calculate the activation energy of the reaction.

Solution: 

Given:

T₁ = 300K, k₁ = 2.5×10^-7 M⁻¹s⁻¹

T₂ = 600K, k₂ = 1.5×10^-6 M⁻¹s⁻¹

Using the Arrhenius equation without a pre-exponential factor

ln (k₁ / k₂) = Ea / R [ 1/T₂ – 1/T₁]

ln (2.5×10^-7/1.5×10^-6} = (E_a / 8.314) [1/600-1/300]

-1.79 = (E_a / 8.314) [-1/600]

E_a = 129.17 Jmol^-1

Example 3: The activation energy for a chemical reaction is 10000 J per mole and its A factor is 50 per Mole per second. Find the rate constant of this equation at a temperature of 700 K. Given ln(50) = 3.912

Solution:

Given:

[Tex]E_a = 100000 \space Jmol^{-1}[/Tex]

A = 50 M-1s-1

T = 700 K

R = 8.314 J.mol-1.K-1

k = ?

We know that,

[Tex]ln(k) =  ln(A) – (\frac{E_a}{RT})[/Tex]

        [Tex]= ln(50)-\frac{10000}{8.314*700}[/Tex]

        = 3.912 – ln(1.718)

        = 3.912 – 0.541

        = 3.371

Thus the value of k = e^3.371

Example 4: It is given that the activation energy of the decomposition of Nitrogen Dioxide at 300K is 111 kJ/mol, and the rate coefficient is 1.0×10^-10. Calculate the pre-exponential factor.

Solution: 

Given:

[Tex]E_a = 111 \space kJ/mol,\\ T= 300K,\\ R= 8.314×10^{-3}kJmol^{-1}K^{-1},\\ k= 1×10^{-10}s^{-1}[/Tex]

Using Arrhenius’s equation,

[Tex]k = Ae^{-{\frac{E_a}{RT}}}[/Tex]

[Tex]1×10^{-10}=Ae^{-\frac{111}{8.314×10^{-3}×300}}[/Tex]

[Tex]10^{-10}= Ae^{-44.50}[/Tex]

Thus,

A = 10^{-10}*e^{44.50}

Example 5: Given that the activation energy of a reaction at 100K is 100 kJ/mol, and the rate coefficient is 2.0×10^-8s^-1. Calculate the pre-exponential factor.

Solution: 

Given:

[Tex]E_a = 100 \space kJ/mol, T=100K, R=8.314×10^{-3}kJmol^{-1}K^{-1}, k=2×10^{-8}s^{-1}[/Tex]

Using Arrhenius’s equation,

[Tex]k = Ae^{-{\frac{E_a}{RT}}}[/Tex]

[Tex]2×10^{-8}=Ae^{-\frac{100}{8.314×10^{-3}×100}}[/Tex]

[Tex]2^{-8}= Ae^{-120.28}[/Tex]

Thus,

[Tex]A = 2^{-8}×e^{120.28}[/Tex]

Check:

• First Order Reaction
• Integrated Rate Laws

Arrhenius Equation – FAQs

What is the Arrhenius Equation?

The Arrhenius equation is expressed as,

k = Ae^-E/RT

where,
k is the rate of chemical reaction, 
A is a constant depending on the chemicals involved,
E is the activation energy, 
R is the universal gas constant,
T is the temperature.

How do you determine activation energy from the Arrhenius Equation?

Activation energy can be determined graphically using an Arrhenius plot where the natural logarithm of the rate constant (lnk) is plotted against the inverse of the temperature (1/T). The slope of this plot, multiplied by the negative universal gas constant (−R), gives the activation energy (Ea)​​.

What does the pre-exponential factor (A) represent?

The pre-exponential factor or frequency factor A in the Arrhenius equation represents the frequency of collisions and their proper orientation necessary for a reaction to occur. It reflects how many molecules are correctly aligned to react when they collide​​.

How does temperature affect the rate constant in the Arrhenius Equation?

As the temperature increases, the rate constant k also increases. This is because higher temperatures provide more energy, reducing the effective activation energy barrier, thus allowing more molecules to have enough energy to react​​.

What is the significance of the exponential term in the Arrhenius Equation?

The exponential term e−Ea/RT in the Arrhenius equation represents the fraction of molecules possessing sufficient energy to overcome the activation energy barrier at a given temperature. This fraction increases with temperature, thus increasing the rate of reaction.

Can the Arrhenius Equation predict rates for all chemical reactions?

While widely applicable, the Arrhenius Equation primarily predicts rates for simple chemical reactions under specific conditions and might not accurately describe reactions with complex mechanisms or those that proceed through multiple steps​.

How do catalysts affect the terms in the Arrhenius Equation?

Catalysts lower the activation energy (Ea) without affecting the pre-exponential factor (A) significantly. This reduction in Ea increases the value of k, thereby speeding up the reaction​.

What is the Arrhenius plot and its use?

An Arrhenius plot graphs ln(k) against 1/T (inverse temperature). This plot is linear, and its slope and intercept are used to determine the activation energy and the pre-exponential factor, respectively​.

How does the Arrhenius Equation relate to real-world applications?

The Arrhenius Equation is crucial in fields like pharmacology and materials science, where it helps in predicting the stability and shelf life of products under various temperature conditions​​.

What are the limitations of the Arrhenius Equation?

One limitation is its assumption that reaction kinetics can always be modeled simply by temperature and activation energy, which may not hold for reactions involving complex or multiple steps​.

Which theory is the basis for the Arrhenius equation?

The Arrhenius equation is based on the concept that the number of effective collisions is proportional to the number of molecules above certain threshold energy and the number of molecules with energies exceeding the threshold energy increases as temperature increases.

What is the use of the Arrhenius equation?

Arrhenius Equation is a formula or expression that explains a relationship between k (the rate constant for a reaction), absolute temperature T and the pre-exponential factor A. 

How does a catalyst increase the rate of the reaction?

A catalyst lowers the activation energy needed for the reaction and thus the reactants react at a faster rate due to the lower energy barrier and hence the rate of reaction increases.

What are the values of the Boltzmann constant and universal gas constants used in the Arrhenius equation?

The value of the Boltzmann constant is 1.38 × 10^-23 m² kg s^-2 K^-1 and that of the universal gas constant is 8.314 JK^-1mol^-1. The Boltzmann constant is used when the activation energy is expressed as per molecule of the reactants whereas the universal gas constant is used when the energy is expressed as per mole of the reactants.

What are the units of the pre-exponential factors for first and second-order reactions?

The units for pre-exponential factor A for first and second-order reactions are s^-1 and Lmol^-1s^-1 respectively.

What is the relation between the rates of reaction at different temperatures?

Let the rates of reaction be k₁ and k₂ and the temperatures are T₁ and T₂ respectively. Thus the relationship between them is as follows:

ln (k₁ / k₂) = Ea / R [ 1/T₂ – 1/T₁]

where,
E_a is the activation energy
R is the universal gas constant