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.

Question

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., 4 . 92 K, 9 . 84 K, 19 . 67 K, 2 . 45 K

MARKS App · Accepted Answer

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 .

Search any question & find its solution

Looking for more such questions to practice?