Asked by Anonymous
The activation energy for the isomerization ol cyclopropane to propene is 274 kJ/mol. By what factor does the rate of this reaction increase as the temperature rises from 256 to 297 oC? (The factor is the ratio of the rates.)
A catalyst decreases the activation energy of a particular exothermic reaction by 60 kJ/mol, to 27 kJ/mol. Assuming that the mechanism has only one step, and that the products are 46 kJ lower in energy than the reactants, sketch approximate energy-level diagrams for the catalyzed and uncatalyzed reactions.
What is the activation energy for the uncatalyzed reverse reaction?
So I'm not sure how to do either of the questions. I've looked through my notes and the book, but I still can't get it. Thanks in advance!
A catalyst decreases the activation energy of a particular exothermic reaction by 60 kJ/mol, to 27 kJ/mol. Assuming that the mechanism has only one step, and that the products are 46 kJ lower in energy than the reactants, sketch approximate energy-level diagrams for the catalyzed and uncatalyzed reactions.
What is the activation energy for the uncatalyzed reverse reaction?
So I'm not sure how to do either of the questions. I've looked through my notes and the book, but I still can't get it. Thanks in advance!
Answers
Answered by
Devron
You are going to have to search graphs that depict the energy of activation with and without a catalyst for you to understand how to sketch the graphs.
Answered by
Anonymous
Is it necessary to graph it?
Answered by
Devron
To understand how to sketch it, yes.
Answered by
Devron
To do the first part, I believe ou have to use Arrhenius equation
ln(k2/k1) = Ea/R(1/T1 - 1/T2)
ln(k2/k1) = Ea/R(1/T1 - 1/T2)