Question

The thermal isomerization of cyclopropane to propene has a rate constant of 5.95 × 10−4 s−1 at 500oc. calculate the value of the gibbs energy of activation, ∆g⊖‡ for this reaction.

Answer 1

Gibb  free   energy  of   activation  is  =  -RT  in  K

R=   gas  constant=8.314  j/mol
T=temperature=  500+273=  773  K
K=rate  constant
gibb   free  energy  is therefore=  - 8.314 j/mol /k  x  773k in  5.95 x10^-4=  471730.94j/mol