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Chemical Kinetics appeared 43 times across 3 years — 5% of Chemistry. This question is from Temperature Dependence of Reaction Rate.

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Questions 14 21 8 43
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Consider the following statements related to temperature dependence of rate constants. Identify the correct statements, A. The Arrhenius equation holds true only for an elementary homogenous reaction. B. The unit of A is same as that of k in Arrhenius equation. C. At a given temperature, a low activation energy means a fast reaction. D. A and Ea as used in Arrhenius equation depend on temperature. E. When Ea >> RT, A and Ea become interdependent. Choose the correct answer from the options given below:

Solution & Explanation

Related Formula

Arrhenius equation is given by:

k = A e-(Eₐ)/(R T)

where:

  • k is the rate constant
  • A is the pre-exponential factor (frequency factor)
  • Eₐ is the activation energy
Core Logic

Evaluate each statement:

  • A: Arrhenius equation is an empirical relation that works well for both elementary and complex homogeneous reactions arrow Incorrect.
  • B: Since the exponential term e-Eₐ/RT is dimensionless, the pre-exponential factor A has the exact same unit as the rate constant k arrow Correct.
  • C: For low Eₐ, the term e-Eₐ/RT is large, giving a high rate constant k and a fast reaction arrow Correct.
  • D: A and Eₐ are assumed to be independent of temperature over a narrow range arrow Incorrect.
  • E: A and Eₐ remain independent parameters of the system, not interdependent arrow Incorrect.
Step 1: Select correct statements

Statements B and C are correct, matching Option (3).

Pattern Recognition

The exponential factor e-Eₐ/RT represents the fraction of collisions with energy greater than the activation barrier. As Eₐ decreases, this fraction grows exponentially, explaining why low-activation pathways (like catalyzed reactions) run much faster.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Reference Study Guides

Chemical Kinetics

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_2025_03_april_evening

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