Q59.For a reaction taking place in three steps at same temperature, overall rate constant K = K1K2 . If Ea1, Ea2 and K3 Ea3 are 40, 50 and 60 kJ / mol respectively, the overall Ea is _________ kJ / mol .
What This Question Tests
This question requires combining rate constants for a multi-step reaction to find the overall activation energy, demanding an understanding of the Arrhenius equation for complex reactions.
Concepts Tested
Formulas Used
K = A * exp(-Ea/RT)
E_overall = Σ(n_i * Ea_i) (for product K)
📚 NCERT Sections This Tests
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.
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
📋 Question Details
- Chapter
- Chemical Kinetics
- Topic
- Activation energy of multi-step reactions
- Year
- 2024
- Shift
- 29 Jan Shift 1
- Q Number
- Q59
- Type
- Numerical
- NCERT Ref
- Class 12 Chemistry Ch 4: Chemical Kinetics
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