The rate of a reaction doubles when its temperature changes from $300 K to 310 K$ . Activation energy of such a reaction will be:

$(R = {8.314 JK}^{-1} {mol}^{-1} and log 2 = 0.301)$

Question

The rate of a reaction doubles when its temperature changes from 300 K to 310 K . Activation energy of such a reaction will be: ​ R = 8.314 JK -1 mol -1 and log 2 = 0.301

TestHub · Accepted Answer

log k 2 k 1 = − E a 2 .303 R ( 1 T 2 − 1 T 1 ) log k 2 k 1 = E a 2 . 3 0 3 R T 2 - T 1 T 1 T 2 log 2 = E a 2 . 3 0 3 × 8 . 3 1 4 × 1 0 3 0 0 × 3 1 0 E a = 0 . 301 × 2 . 303 × 8 . 314 × 300 × 310 10 = 53603 . 9 J / mol E a = 53.6 kJ mol - 1 .

Chemistry - Chemical Kinetics Question with Solution | TestHub

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