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JEE Main Chemistry Chemical Kinetics 2026 JEE Main 2026 (28 January Shift 2)

JEE Main Chemistry Question (2026) — Solution

Question

A B (first reaction) C D (second reaction) Consider the above two first-order reactions. The rate constant for first reaction at 500 K is double of the same at 300 K. At 500 ~K , 50 \% of the reaction becomes complete in 2 hour. The activation energy of the second reaction is half of that of first reaction. If the rate constant at 500 K of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at 300 K is \_\_\_\_ 10^ -1 hour ^ -1 (nearest integer).

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

For A K_1 B (2) = E_ a_1 R [ 1 300 - 1 500 ] E_ a_1 = 2 R 1500 2 E_ a_2 = E_ a_1 2 = 2 R 1500 4 (K_1)_ at 500 K = 2 2 (K_2)_ at 500 K = 2 Now for C K_2 D [ (K_2)_ at 500K (K_2)_ at 300K ] = ( 2 R 1500 4 ) 1 R [ 1 300 - 1 500 ] (K_2)_ at 300 K = 2 2 = 0.49 (K_2)_ at 300 K = 4.9 10^ -1 . So, answer is 5.

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Related: Chemistry — Chemical Kinetics · All PYQ Banks