The rate of a reaction A doubles on increasing the temperature from 300 to 310 K . By how much, the temperature of reaction B should be increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A.

Step-by-step solution

log   2 = E a R 1 300 - 1 310 .......(i) log   2 =2 E a R   1 300 - 1 T ......(ii) 2 E a R   1 300 - 1 T = E a R   1 300 - 1 310 1 300 + 1 310 = 2 T ⇒         T = 300 × 310 610 × 2 = 304.92 Hence, the temperature of reaction B should be increased from 300   K by 304 . 92 − 300 = 4 . 92   K .