If the rate constants of a reaction at 500   K and 700   K are 0 . 002   s - 1 and 0 . 06   s - 1 , respectively, the value of activation energy is R = 8 . 314   J   mol - 1   K - 1 ,   log 3 = 0 . 477

Step-by-step solution

We have log K 2 K 1 = E a 2 . 303 R T 2 - T 1 T 1 T 2 Putting the values, we get, log 0 . 06 0 . 002 = E a 2 . 303 × 8 . 314 700 - 500 700 × 500 log 30 = E a 19 . 147 200 350000 1 . 477 = E a 19 . 147 200 3500 E a = 49 . 49   kJ   mol - 1