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.
1 answer:
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
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