Question

Calculate the activation of a reaction whose rate constant is tripled by 10degree rise in temperature in the vicinity of 27 degree celsius

Answer 1

12 DKDKX CLOSE SNDF SKF WU TANG IN THE SECOND QUARTER IS THE REASON WHY THE BEST

Answer 2

Answer: the value of activation energy will be 84.93

Explanation: Given T2-T1=10°

T1=27°C=300K

T2=(300+10)K=310K

Also given,

k2=3k1

By the equation,

Log k2/k1=Ea/2.303[T2-T1/T2*T1]

Substituting the above values we get

Log 3k1/k1= Ea/2.303[10/300*310]

Or log 3= Ea/2.303[10/93000]

Therefore , Ea=0.477*2.303*8.314*93000/10

Ea=849386.366/10

Ea=84938.6J mol^-1

Or Ea=84.93 kJmol^-1(ans)