Understanding the Arrhenius Equation in Chemical Kinetics
Quick Summary: What is the Arrhenius Equation?
The Arrhenius equation is a mathematical formula that models the temperature dependency of reaction rates. It provides a quantitative link between standard temperature, pre-exponential frequency factor, and the activation energy required for reactant molecules to successfully collide and form products.
Thermodynamic Foundation and Governing Formulas
According to collision theory, not all molecular collisions lead to a chemical change. Molecules must collide with sufficient kinetic energy to overcome a specific potential energy threshold, known as the activation energy. The Arrhenius expression describes the fraction of molecules possessing energy equal to or greater than this activation threshold at a given temperature.
The classical Arrhenius equation is stated as:
k = A × e−Ea / RT
Logarithmic Form & Temperature Variations
To calculate parameters graphically or compare two distinct kinetic states, the equation is converted to its natural logarithmic form:
ln(k) = ln(A) − [ Ea / RT ]
When comparing the rate constants k1 and k2 at two different temperatures T1 and T2, the pre-exponential factor A cancels out. This yields the highly authoritative two-temperature Arrhenius formula utilized heavily in NCERT, CBSE boards, and JEE exams:
log10(k2 / k1) = [ Ea / 2.303R ] × [ (T2 − T1) / (T1 × T2) ]
Solved Numerical Examples
Example 1: Rate Constant calculation at Room Temperature
Problem: A first-order reaction has a pre-exponential factor A of 4.0 × 1010 s−1 and an activation energy Ea of 100 kJ/mol. Calculate its rate constant k at 300 K.
Solution:
Given: A = 4.0 × 1010 s−1, Ea = 100 kJ/mol = 100,000 J/mol, T = 300 K, R = 8.314 J/mol•K.
Substitute terms: Ea / RT = 100,000 / (8.314 × 300) ≈ 40.09
Formula: k = A × e−40.09 = 4.0 × 1010 × 3.873 × 10−18
Final Answer: k = 1.55 × 10−7 s−1.
Example 2: Determining Activation Energy from Rate Doubling
Problem: The rate of a chemical reaction doubles when the temperature increases from 293 K to 313 K. Find the activation energy (Ea) of the reaction.
Solution:
Given: T1 = 293 K, T2 = 313 K, k2/k1 = 2, R = 8.314 J/mol•K.
Formula: log10(k2/k1) = [ Ea / 2.303R ] × [ (T2 − T1) / (T1 × T2) ]
Substitute: log10(2) = [ Ea / (2.303 × 8.314) ] × [ (313 − 293) / (293 × 313) ]
Solve: 0.3010 = [ Ea / 19.147 ] × [ 20 / 91709 ]
Calculate: Ea = [ 0.3010 × 19.147 × 91709 ] / 20 ≈ 26435.7 × 2 = 52871 J/mol = 52.87 kJ/mol.
Example 3: Calculating Pre-exponential Factor A
Problem: A decomposition reaction at 500 K exhibits a rate constant of 0.05 s−1. If its activation energy is 80 kJ/mol, solve for the frequency factor A.
Solution:
Given: k = 0.05 s−1, Ea = 80 kJ/mol = 80,000 J/mol, T = 500 K.
Formula: A = k × eEa/RT
Evaluate exponent: Ea / RT = 80,000 / (8.314 × 500) = 19.245
Substitute: A = 0.05 × e19.245 = 0.05 × 2.279 × 108
Final Answer: A = 1.14 × 107 s−1.