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Saturday, 31 October

JEE Main 2017 · ChemistryMultiple choiceSingle correctEasyCalculation

JEE Main 2 April 2017 (offline), Chemistry Q39

Question 39 of 90 in this shift, Chemistry question 9 of 30.

Two reactions R1R_1 and R2R_2 have identical pre-exponential factors. Activation energy of R1R_1 exceeds that of R2R_2 by 10 kJ mol−1^{-1}. If k1k_1 and k2k_2 are rate constants for reactions R1R_1 and R2R_2 respectively at 300 K, then ln⁡(k2/k1)\ln(k_2/k_1) is equal to : (R=8.314 J mol−1K−1R=8.314\ \mathrm{J\,mol^{-1}K^{-1}})
  1. (1)6
  2. (2)4Official answer
  3. (3)8
  4. (4)12

Official answer

Option 2

CBSE answer key (25/04/2017, used for result).