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Two reactions, R 1 and R 2 have identical pre-exponential factors. Activation energy of R 1 exceeds that of R 2 by 10 kJ mol –1 . If k 1 and k 2 are rate constants for reactions R 1 and R 2 respectively at 300 K, then ln(k 2 /k 1 ) is equal to. (R = 8.314 J mol –1 K –1 ).

1. 6
2. 4
3. 8
4. 12 Students may find similar question in CP exercise sheet : [ JEE Main, Chapter : Chemical Kinetics, Exercise-5, Section [B], Q. No.27 ] [ JEE Advance, Chapter : Chemical Kinetics, Exercise-4, Q. No.25 ]

Correct answer: 4

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