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-Ea / RT
In the above equation,
k is rate constant
A is the pre-exponential factor
Ea 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-Ea / kbT
where
kb 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,
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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-Ea / RT
Taking logarithm on both sides, we get
ln k = ln(Ae-Ea / RT)
ln k = ln (A) + ln (e-Ea / RT)
ln k = ln (A) + -Ea / RT ln (e)
ln k = ln (A) + (-Ea / RT ) [ln (e) = 1]
ln k = -Ea / RT + ln(A)
The equation thus obtained is of the form y=mx+c where m = -Ea / RT.
The above equation can be easily plotted to obtain the Arrhenius Graph. The image given below shows the Arrhenius Graph,
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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) - Ea / RT
ln k1 = ln (A) - Ea / RT1 ...(1)
ln k2 = ln (A) - Ea / RT2 ...(2)
We can obtain the value of ln(A) from the above equation as follows:
ln (A) = ln k2 + Ea / RT2
Using this value of ln (A) in equation 1, the following equation can be formed:
ln k1 = ln k2 + Ea / RT2 - Ea / RT1
Transporting ln k2 to the left-hand side, the value of ln k1 - ln k2 is as:
ln k1 - ln k2 = Ea / RT2 - Ea / 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 Ea / R is common in both the terms on the right-hand side. Therefore, the entire equation can be written as follows:
ln (k1 / k2) = Ea / R [ 1/T2 - 1/T1]
Also, Check
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:
E_a = 2000 \space Jmol^{-1} A = 10 M-1s-1
T = 200 K
R = 8.314 J.mol-1.K-1
k = ?
We know that,
ln(k) = ln(A) – (\frac{E_a}{RT})
= ln(10)-\frac{2000}{8.314*200} = 2.303-(10/8.314)
= 1.1
Thus the value of K = e1.1.
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} = (Ea / 8.314) [1/600-1/300]
-1.79 = (Ea / 8.314) [-1/600]
Ea = 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:
E_a = 100000 \space Jmol^{-1} A = 50 M-1s-1
T = 700 K
R = 8.314 J.mol-1.K-1
k = ?
We know that,
ln(k) = ln(A) – (\frac{E_a}{RT})
= ln(50)-\frac{10000}{8.314*700} = 3.912 - ln(1.718)
= 3.912 - 0.541
= 3.371
Thus the value of k = e3.371
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:
E_a = 111 \space kJ/mol,\\ T= 300K,\\ R= 8.314×10^{-3}kJmol^{-1}K^{-1},\\ k= 1×10^{-10}s^{-1} Using Arrhenius's equation,
k = Ae^{-{\frac{E_a}{RT}}}
1×10^{-10}=Ae^{-\frac{111}{8.314×10^{-3}×300}}
10^{-10}= Ae^{-44.50} 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:
E_a = 100 \space kJ/mol, T=100K, R=8.314×10^{-3}kJmol^{-1}K^{-1}, k=2×10^{-8}s^{-1} Using Arrhenius's equation,
k = Ae^{-{\frac{E_a}{RT}}}
2×10^{-8}=Ae^{-\frac{100}{8.314×10^{-3}×100}}
2^{-8}= Ae^{-120.28} Thus,
A = 2^{-8}×e^{120.28}
Check: