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 &#160; to &#160; 310 K . Activation energy of such a reaction will be: ​ R = 8.314 JK -1 mol -1 &#160; and &#160; log 2 = 0.301

Aicharya · Accepted Answer

log k 2 k 1 = &#8722; E a 2 .303 R ( 1 T 2 &#8722; 1 T 1 ) log k 2 k 1 = E a 2 . 3 0 3 R T 2 - T 1 T 1 T 2 log&#160; 2 = E a 2 . 3 0 3 &#215; 8 . 3 1 4 &#215; 1 0 3 0 0 &#215; 3 1 0 E a = 0 . 301 &#215; 2 . 303 &#215; 8 . 314 &#215; 300 &#215; 310 10 = 53603 . 9 &#160; J / mol E a = 53.6 kJ mol - 1 .

The rate of a reaction doubles when its temperature changes from 300 K   to   310 K . Activation…

Solution

The rate of a reaction doubles when its temperature changes from 300 K &#160; to &#160; 310 K . Activation…

Solution

The rate of a reaction doubles when its temperature changes from 300 K to 310 K . Activation…