Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by 20 ~kJ ~mol ^ -1 . If k _ 1 and k _ 2 are the rate constants of first and second reaction respectively at 300 K, then k _ 2 k _

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Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by 20 ~kJ ~mol ^ -1 . If k _ 1 and k _ 2 are the rate constants of first and second reaction respectively at 300 K, then k _ 2 k _ 1 will be \_\_\_\_. (nearest integer) [ R =8.3 ~J ~K ^ -1 ~mol ^ -1 ]

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Correct answer: A. A

Given: A_1 = A_2, E_ a1 = E_ a2 + 20 kJ/mol, T = 300 K, R = 8.3 J K⁻¹ mol⁻¹ Using Arrhenius equation: k = Ae^ -E_a/RT k_2 k_1 = e^ -E_ a2 /RT e^ -E_ a1 /RT = e^ (E_ a1 - E_ a2 )/RT E_ a1 - E_ a2 = 20 kJ/mol = 20000 J/mol k_2 k_1 = e^ 20000 8.3 300 = e^ 20000 2490 k_2 k_1 = 20000 2490 = 8.03 8

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