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 20text kJ mol^-1. If k_1 and k_2 are the rate constants of first and second reaction respectively at 300 K, then ln frack_2k_1 will be ..... (nearest integer) [R=8.3text J K^-1text mol^-1]

Numerical Answer Type:
Enter a numerical value Answer: 8 to 8 +4 marks

Solution & Explanation

### Related Formula ln k = ln A - fracE_aRT ### Core Logic For Reaction 1: ln k_1 = ln A - fracE_1RT For Reaction 2: ln k_2 = ln A - fracE_2RT (Pre-exponential factor A is the same). Subtracting the first from the second: ln k_2 - ln k_1 = -fracE_2RT - left(-fracE_1RTright) ln left(frack_2k_1right) = fracE_1 - E_2RT Given that E_1 exceeds E_2 by 20text kJ mol^-1, E_1 - E_2 = 20000text J mol^-1. T = 300text K, R = 8.3text J K^-1text mol^-1. ln left(frack_2k_1right) = frac200008.3 times 300 = frac2008.3 times 3 = frac20024.9 ln left(frack_2k_1right) = 8.032 Rounding off to nearest integer gives 8. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

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