Q.21 The rate of a chemical reaction is tripled when the temperature of the reaction is increased from 298 K to 308 K. The activation energy (in kcal mol−1) up to one decimal place for the reaction is __________.
(Given: R = 1.987 cal mol−1 K−1)
Introduction
Numerical problems based on the Arrhenius equation are frequently asked in
chemical kinetics for JEE, NEET, and other competitive examinations. These questions test the
relationship between reaction rate, temperature, and
activation energy.
In this article, we calculate the activation energy of a reaction whose rate increases threefold
when the temperature is raised from 298 K to 308 K.
Given: R = 1.987 cal mol−1 K−1
Key Concept Used
The Arrhenius equation relates the rate constant of a reaction to temperature:
ln(k2/k1) =Ea/R( 1/T1 −1/T2 )
🔢 Step-by-Step Solution
Step 1: Given Values
k2 = 3k1
T1 = 298 K
T2 = 308 K
R = 1.987 cal mol−1 K−1
Step 2: Substitute into Arrhenius Equation
ln(3) =Ea/1.987( 1/298 −1/308 )
Step 3: Evaluate Numerical Terms
ln(3) = 1.0986
1/298 −1/308=10/91784
≈ 1.089 × 10−4
Step 4: Calculate Activation Energy
Ea =1.0986 × 1.987/1.089 × 10−4
Ea ≈ 20041 cal mol−1
Step 5: Convert to kcal mol−1
Ea =20041/1000
= 20.0 kcal mol−1
Final Answer
Activation Energy =20.0 kcal mol−1
Common Mistakes to Avoid
- Using log10 instead of natural log (ln)
- Using temperature difference instead of reciprocal temperatures
- Forgetting to convert cal to kcal
Quick Exam Tip
Remember important logarithmic values:
ln(2) = 0.693 ln(3) = 1.099
Conclusion
This problem illustrates the strong dependence of reaction rate on temperature as described by
the Arrhenius equation. By carefully substituting values and handling logarithms correctly, the
activation energy can be calculated accurately and efficiently in exams.
