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Chemical Kinetics

JEE Main 2025 (Online) 24th January Morning ShiftMCQEasy📈 Topic in 32% papers+4−1
Ques. 1JEE 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

Answer

(3) 7/4 times of initial pressure

Solution

At constant V,T: p n.

N2O5 left = p02, NO2=2 p02=p0, O2= 12 p02= p04

ptotal= p02+p0+ p04= 74p0

Answer: (3)

Keep practising — Chemical Kinetics

Ques. 2JEE 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. 3JEE 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. 4JEE 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. 5JEE 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. 6JEE 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 ]

Ques. 7JEE Main 2025, 22 January (Morning Shift)

Which of the following statement is not true for radioactive decay?

  1. Half life is 2 times of 1rate constant.
  2. Decay constant increases with increase in temperature.
  3. Decay constant does not depend upon temperature.
  4. Amount of radioactive substance remained after three half lives is 18th of original amount.
Out of syllabus
Ques. 8JEE Main 2025, 22 January (Evening Shift)

Consider the given figure and choose the correct option : [figure]

  1. Activation energy of both forward and backward reaction is E1 + E2 and reactant is more stable than product.
  2. Activation energy of forward reaction is E1 + E2 and product is more stable than reactant.
  3. Activation energy of backward reaction is E1 and product is more stable than reactant.
  4. Activation energy of forward reaction is E1 + E2 and product is less stable than reactant.

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