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Chemical Kinetics

JEE Main 2025 (Online) 29th January Morning ShiftMCQEasy📈 Topic in 11% papers+4−1
Ques. 1JEE Main 2025, 29 January (Morning Shift)

The reaction A2 + B2 2AB follows the mechanism: A2 k-1 k1 A + A (fast) A + B2 k2 AB + B (slow) A + B AB (fast) The overall order of the reaction is:

  1. 3
  2. 2.5
  3. 1.5
  4. 2

Answer

(3) 1.5

Solution

Rate is decided by the slow step: r = k2[A][B2]

A is an intermediate; from the fast equilibrium: K = k1k-1 = [A]2[A2]

[A] = K1/2[A2]1/2

r = k2K1/2[A2]1/2[B2]

Overall order = 12 + 1 = 1.5

Answer: (3)

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Ques. 2JEE Main 2025, 29 January (Evening Shift)

Drug X becomes ineffective after 50\% decomposition. The original concentration of drug in a bottle was 16 mg/mL which becomes 4 mg/mL in 12 months. The expiry time of the drug in months is . Assume that the decomposition of the drug follows first order kinetics.

  1. 12
  2. 3
  3. 6
  4. 2
Ques. 3JEE Main 2025, 28 January (Morning Shift)

array|c|c| [A]0/ mol L-1 & t1/2/min \ 0.100 & 200 \ 0.025 & 100 \ array For a given reaction R P, t1/2 is related to [A]0 as given in table. Given: 2 = 0.30 Which of the following is true? Choose the correct answer from the options given below:

  1. A and C Only
  2. A, B and D Only
  3. A and B Only
  4. C and D Only
Ques. 4JEE Main 2025, 28 January (Evening Shift)

Consider an elementary reaction A(g) + B(g) C(g) + D(g) If the volume of reaction mixture is suddenly reduced to 13 of its initial volume, the reaction rate will become 'x' times of the original reaction rate. The value of x is :

  1. 3
  2. 9
  3. 13
  4. 19
Ques. 5JEE Main 2025, 28 January (Evening Shift)

For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth? Where N - Number of bacteria at any time, N0 - Initial number of bacteria.

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
Ques. 6JEE Main 2025, 24 January (Morning Shift)

For a reaction, N2O5(g) 2NO2(g) + 12O2(g) in a constant volume container, no products were present initially. The final pressure of the system when 50\% of reaction gets completed is

  1. 5/2 times of initial pressure
  2. 7/2 times of initial pressure
  3. 7/4 times of initial pressure
  4. 5 times of initial pressure
Ques. 7JEE Main 2025, 24 January (Evening Shift)

Given below are two statements : In the light of the above statements, choose the correct answer from the options given below :

  1. Both Statement I and Statement II are false
  2. Statement I is true but Statement II is false
  3. Statement I is false but Statement II is true
  4. Both Statement I and Statement II are true
Ques. 8JEE Main 2025, 24 January (Evening Shift)

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)

Ques. 9JEE Main 2025, 23 January (Morning Shift)

For the thermal decomposition of N2O5(g) at constant volume, the following table can be formed, for the reaction mentioned below. 2N2O5(g) 2N2O4(g)+O2(g) [figure] x= 10-3 atm [nearest integer] Given : Rate constant for the reaction is 4.606 10-2 s-1.

Ques. 10JEE Main 2025, 23 January (Evening Shift)

Which of the following graphs most appropriately represents a zero order reaction?

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
Ques. 11JEE Main 2025, 22 January (Morning Shift)

A B The molecule A changes into its isomeric form B by following a first order kinetics at a temperature of 1000 K. If the energy barrier with respect to reactant energy for such isomeric transformation is 191.48 kJ mol-1 and the frequency factor is 1020, the time required for 50\% molecules of A to become B is picoseconds (nearest integer). [R=8.314 J K-1 mol-1 ]

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