Q57.The equation k = (6. 5 × 1012 s−1)e−26000 K/T is followed for the decomposition of compound A. The activation energy for the reaction is ____ KJ mol−1 . [nearest integer] (Given: R = 8. 314 J K−1 mol−1 )
What This Question Tests
This question tests the ability to extract the activation energy from a given rate constant equation, which is in the form of the Arrhenius equation.
Concepts Tested
Formulas Used
k = A * e^(-Ea/RT)
📚 NCERT Sections This Tests
3.26 — The Decomposition Of Hydrocarbon Follows The Equation
Chemistry Class 11 · Chapter 3
3.26 The decomposition of hydrocarbon follows the equation k = (4.5 × 1011s–1) e-28000K/T Calculate Ea. 87 Chemical Kinetics Reprint 2025-26
3.23 — The Rate Constant For The Decomposition Of Hydrocarbons Is 2.418 × 10–5S–1
Chemistry Class 11 · Chapter 3
3.23 The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5s–1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
3.22 — The Rate Constant For The Decomposition Of N2O5 At Various Temperatures
Chemistry Class 11 · Chapter 3
3.22 The rate constant for the decomposition of N2O5 at various temperatures is given below: T/°C 0 20 40 60 80 105 × k/s-1 0.0787 1.70 25.7 178 2140 Draw a graph between ln k and 1/T and calculate the values of A and Ea. Predict the rate constant at 30° and 50°C.
📋 Question Details
- Chapter
- Chemical Kinetics
- Topic
- Arrhenius equation and activation energy
- Year
- 2022
- Shift
- 29 Jun Shift 2
- Q Number
- Q57
- Type
- Numerical
- NCERT Ref
- Class 12 Chemistry Ch 4: Chemical Kinetics
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