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JEE Main Chemistry Chemical Kinetics 2026 JEE Main 2026 (02 April Shift 1)

JEE Main Chemistry Question (2026) — Solution

Question

For reaction A P, rate constant k = 1.5 10^3 s^ -1 at 27°C. If activation energy for the above reaction is 60 kJ mol^ -1 , then the temperature (in °C) at which rate constant, k = 4.5 10^3 s^ -1 is _______. (Nearest integer) Given : 2 = 0.30, 3 = 0.48, R = 8.3 J K^ -1 mol^ -1 , 10 = 2.3

Options

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

Answer

A. A

Step-by-step solution

Using the Arrhenius equation: ( k_2 k_1 ) = E_a 2.3 R ( 1 T_1 - 1 T_2 ) Given: k_1 = 1.5 10^3 s ^ -1 k_2 = 4.5 10^3 s ^ -1 T_1 = 27^ C = 300 K E_a = 60 kJ mol ^ -1 = 60000 J mol ^ -1 R = 8.3 J K ^ -1 mol ^ -1 Substituting the values into the equation: ( 4.5 10^3 1.5 10^3 ) = 60000 2.3 8.3 ( 1 300 - 1 T_2 ) 3 = 60000 19.09 ( 1 300 - 1 T_2 ) 0.48 = 60000 19.09 ( 1 300 - 1 T_2 ) 1 300 - 1 T_2 = 0.48 19.09 60000 1 300 - 1 T_2 = 9.1632 60000 = 1.5272 10^ -4 1 T_2 = 1 300 - 1.5272 10^ -4 1 T_2 = 3.3333 10^ -3 - 0.1527 10^ -3 = 3.1806 10^ -3 K ^ -1 T_2 = 1 3.1806 10^ -3 314.4 K Converting to Celsius: T_2 = 314.4 - 273 = 41.4^ C The nearest integer is 41. Answer: 41

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