I have done an ROP experiment for kobs with various alcohols (initiator), and I got different kobs values. My question is, if all alcohols(initiator) follow the exact reaction mechanism, how will the rate change? I mean, only the induction period will change; the propagation rate will not change.
Please correct me if I am wrong. I also expect a convincing response.
Thank you very much in advance.
Dear @Rajiv, oh no, the kobs are coming!
Sorry for kidding. However, if you expect a convincing respose, it is always good to provide some background details. What is a ROP experiment? What is kobs? Which reaction have you studied?
I assume that by kobs you mean the observed reaction rate constant. In this case, I can tell you that the reaction rate does not only depend on the mechanism, but also on the molecules involved. It's like a marathon where different runners run at different rates. Maybe in your case the heavy alcohols react more slowly because of being heavy (no offense to heavy alcohols). Maybe they are faster because of certain reasons. Here again, more details would help to answer your question appropriately.
The reaction rate depends on the activation energy of the reaction. The two are related by the Arrhenius equation (Arrhenius equation): k=Ae−Ea/RT. The activation energy of a reaction can be influenced by the different substituents surrounding the reactive group in different members of a family of related compounds. By measuring the reaction rate as a function of temperature, it is possible to determine the activation energy of the reaction. Catalysts, e.g. enzymes, increase the reaction rate by lowering the activation energy.
- Badges
- Science topic
- Similar topics
- Agricultural Science
- Horticulture