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

Year 2026 2025 2024 Total
Questions 14 21 8 43

consider the elementary reaction A(g) + B(g) arrow C(g) + D(g) If the volume of reaction mixture is suddenly reduced to (1)/(3) of its initial volume, the reaction rate will become 'x' times of the original reaction rate. The value of x is:

Solution & Explanation

Related Formula

For an elementary reaction, the rate law corresponds directly to its stoichiometry:

R = K[A]¹[B]¹

Concentration (C) is inversely proportional to volume (V):

C = (n)/(V)
Core Logic

Initial rate expression:

R₁ = K[(nA)/(V)]¹[(nB)/(V)]¹

When volume is reduced to (1)/(3)V, the new concentration becomes 3 times the initial concentration:

R₂ = K[(3nA)/(V)]¹[(3nB)/(V)]¹ = 9 · K[(nA)/(V)]¹[(nB)/(V)]¹
Step 1: Calculating the Value of x

Comparing the two rates:

R₂ = 9R₁

Therefore, the value of x is 9.

Pattern Recognition

For a second-order overall elementary reaction (1+1=2), reducing the volume by a factor of n increases the rate by a factor of n². Here n=3, so the rate increases by 3² = 9 times.

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_2025_28_jan_evening

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