The activation energy of one of the reactions in a biochemical process is 532611 J mol - 1 . When the temperature falls from 310 K to 300 K , the change in rate constant observed is k 300 = x × 10 - 3 k 310 ° . The value of x is [Given: ln 10 = 2 . 3 R = 8 . 3 J K - 1 mol - 1 ]

Step-by-step solution

ln K 2   K 1 = E a R 1   T 1 - 1   T 2 ln K 2   K 1 = 532611 8 . 3 × 10 310 × 300 where K 2 is at 310   K   &   K 1 is at 300   K ln K 2   K 1 = 6 . 9 = 3 × ln 10 ln K 2   K 1 = ln 10 3 K 2 = K 1 × 10 3 K 1 = K 2 × 10 3 So