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Chemical Kinetics appeared 43 times across 3 years — 5% of Chemistry. This question is from Arrhenius Equation and Activation Energy.

Year 2026 2025 2024 Total
Questions 14 21 8 43

For a reaction taking place in three steps at same temperature, overall rate constant K = K₁K₂K₃ . If Ea₁ , Ea₂ and Ea₃ are 40, 50 and 60 kJ/mol respectively, the overall Ea is ______ kJ/mol.

Numerical Answer Type:
Enter a numerical value Answer: 30 to 30 +4 marks

Solution & Explanation

Related Formula
K = A e-Eₐ/RT
Core Logic

Given the relationship between the rate constants:

K = (K₁ · K₂)/(K₃)

Substituting the Arrhenius equation for each rate constant:

A · e-Eₐ/RT = A₁ · e^-Eₐ₁/RT · A₂ · e^-Eₐ₂/RTA₃ · e^-Eₐ₃/RT

Combining the exponential terms using rules of exponents:

A · e-Eₐ/RT = ((A₁ · A₂)/(A₃)) · e^ -(Eₐ₁ + Eₐ₂ - Eₐ₃)RT
Step 1: Equating Activation Energies

By comparing the powers of e on both sides, the overall activation energy Eₐ is related to the individual steps as follows:

Eₐ = Eₐ₁ + Eₐ₂ - Eₐ₃

Substitute the given values (Eₐ₁ = 40, Eₐ₂ = 50, Eₐ₃ = 60 kJ/mol):

Eₐ = 40 + 50 - 60

Eₐ = 90 - 60

Eₐ = 30 kJ/mol
Pattern Recognition

When rate constants are multiplied or divided (K = K₁^a K₂^b / K₃^c), the corresponding overall activation energy follows the linear combination of the exponents: Eₐ = a Eₐ₁ + b Eₐ₂ - c Eₐ₃.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Reference Study Guides

More Chemical Kinetics Previous-Year Questions — Page 9

Q81 jee_main_2024_30_jan_morning First Order Reactions
The rate of first order reaction is 0.04 mol L⁻¹s⁻¹ at 10 minutes and 0.03 mol L⁻¹s⁻¹ at 20 minutes after initiation. Half life of the reaction is ________ minutes. (Given 2=0.3010, 3=0.4771)
Numerical Answer. Answer: 24 to 24.1

Solution

Related Formula

Rate = k[A]

[A] = [A]₀ e-kt t1/2 = (ln 2)/(k)
Core Logic

For a first order reaction, rate is directly proportional to concentration.

R₁ = k[A]₁₀ = k[A]₀ e-k(10 × 60) R₂ = k[A]₂₀ = k[A]₀ e-k(20 × 60)
Step 1: Setting up equations
0.04 = k[A]₀ e-600k (1) 0.03 = k[A]₀ e-1200k (2)
Step 2: Solving for k

Dividing equation (1) by (2):

(0.04)/(0.03) = e-600ke-1200k (4)/(3) = e600k

Take natural log on both sides:

ln((4)/(3)) = 600k k = (ln(4/3))/(600) s⁻¹
Step 3: Calculating half life
t1/2 = (ln 2)/(k) = (ln 2)/((ln(4/3))/(600)) = (600 ln 2)/(ln 4 - ln 3) seconds

Convert to minutes by dividing by 60:

t1/2 = (10 ln 2)/(ln 4 - ln 3) minutes

Substitute log values (since ln x = 2.303 x, the 2.303 cancels out):

t1/2 = 10 × ( 2)/( 4 - 3) minutes t1/2 = 10 × (0.3010)/(2(0.3010) - 0.4771) t1/2 = 10 × (0.3010)/(0.6020 - 0.4771) = 10 × (0.3010)/(0.1249) t1/2 = 24.099 ≈ 24 minutes
Chapter Mix

Class 12 Chemistry: Chemical Kinetics

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_2024_29_jan_morning

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