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Chemical Kinetics appeared 42 times across 3 years — 4.9% of Chemistry. This question is from Arrhenius Equation and Activation Energy.

Year 2026 2025 2024 Total
Questions 14 20 8 42

Consider a complex reaction taking place in three steps with rate constants k₁ , k₂ and k₃ respectively. The overall rate constant k is given by the expression k = k₁k₃k₂ . If the activation energies of the three steps are 60, 30 and 10 kJ mol ⁻¹ respectively, then the overall energy of activation in kJ mol ⁻¹ is . (Nearest integer)

Numerical Answer Type:
Enter a numerical value Answer: 20 to 20 +4 marks

Solution & Explanation

Related Formula

From the Arrhenius equation, rate constants vary exponentially with temperature:

k = A · e-Eₐ / RT

When rate constants combine multiplicatively or via roots, their corresponding activation energies combine linearly.

Core Logic

Given the overall rate constant expression:

k = ((k₁ · k₃)/(k₂))1/2

Substitute the Arrhenius expression (kᵢ = Aᵢ · e^-Eₐᵢ/RT) for each rate constant:

A · e-Eₐ/RT = [ (A₁ · e^-Eₐ₁/RT) · (A₃ · e^-Eₐ₃/RT)A₂ · e^-Eₐ₂/RT]1/2

Equating the exponential terms yields the linear relationship for the overall activation energy (Eₐ):

(Eₐ)/(RT) = (1)/(2) ( Eₐ₁RT + Eₐ₃RT - Eₐ₂RT) Eₐ = Eₐ₁ + Eₐ₃ - Eₐ₂2

Substitute the given activation energy values (Eₐ₁ = 60, Eₐ₂ = 30, Eₐ₃ = 10 kJ/mol):

Eₐ = (60 + 10 - 30)/(2) = (40)/(2) = 20 kJ mol⁻¹

The overall activation energy is 20 kJ/mol.

Pattern Recognition

Shortcut: Convert the rate constant algebraic expression directly into an activation energy formula by swapping k for Eₐ, turning multiplications into additions, divisions into subtractions, and powers into multipliers. Here, k = (k₁ k₃ / k₂)1/2 translates directly to Eₐ = (1)/(2)(Eₐ₁ + Eₐ₃ - Eₐ₂).

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Reference Study Guides

More Chemical Kinetics Previous-Year Questions — Page 9

Q82 jee_main_2024_31_jan_evening First Order Kinetics
r = k[A] for a reaction, 50% of A is decomposed in 120 minutes. The time taken for 90% decomposition of A is ________ minutes.
Numerical Answer. Answer: 398.5 to 399.5

Solution

Related Formula
k = 0.693t1/2 t = (2.303)/(k) ( (a)/(a - x) )
Core Logic

Since r = k[A], the reaction follows first-order kinetics. The half-life (50% decomposition) is t1/2 = 120 minutes.

Step 1: Calculating for 90% decomposition

For 90% completion of the reaction, [A]₀ = 100 and [A]ₜ = 100 - 90 = 10.

t = (2.303)/(k) (100)/(10) t = 2.303( 0.693t1/2 ) (10) t = (2.303 × 120)/(0.693) × 1 t = 398.78 minutes

Rounding off to the nearest integer, we get 399 minutes.

Pattern Recognition

For a first order reaction, t90% ≈ 3.32 × t50%. 120 × 3.32 = 398.4, so roughly 399.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q73 jee_main_2024_31_jan_morning First Order Reactions
Integrated rate law equation for a first order gas phase reaction is given by (where Pᵢ is initial pressure and Pₜ is total pressure at time t)
  • A. k = (2.303)/(t) × (Pᵢ)/((2Pᵢ - Pₜ))
  • B. k = (2.303)/(t) × (2Pᵢ)/((2Pᵢ - Pₜ))
  • C. k = (2.303)/(t) × ((2Pᵢ - Pₜ))/(Pᵢ)
  • D. k = (2.303)/(t) × (Pᵢ)/((2Pᵢ - Pₜ))

Solution

Core Logic

Consider a general gas phase reaction: A arrow B + C

Initial (t=0): Pᵢ 0 0 At time t: Pᵢ - x x x

Total pressure at time t:

Pₜ = (Pᵢ - x) + x + x = Pᵢ + x

x = Pₜ - Pᵢ

Partial pressure of A at time t (PA): PA = Pᵢ - x

PA = Pᵢ - (Pₜ - Pᵢ) = 2Pᵢ - Pₜ

For a first-order reaction:

k = (2.303)/(t) (P₀)/(Pₜ)

Here, P₀ = Pᵢ and the pressure of the reactant at time t is PA.

k = (2.303)/(t) (Pᵢ)/(2Pᵢ - Pₜ)
Chapter Mix

Class 12 Chemistry: Chemical Kinetics

More Chemical Kinetics Questions — jee_main_2025_24_jan_evening

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