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