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 k1 and k2 a

Question

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 k1 and k2 are the rate constants of first and second reaction respectively at 300 K, then In 1 k k will be …….. . (nearest integer) [R=8.3 J K –1 mol –1]

Prashant Kotian · Accepted Answer

Correct answer: .. n RX (1) A product E  n RX (2) B product E  Assuming ‘A’ same for both reaction. ln k1 = lnA – E 300R ln k1 = lnA – E 300R 1 1 k E E 20 1000 ln k 300R 300R           = 8.032

💡 Solution & Explanation

Practice 22,000+ questions like this