How do you calculate the activation energy for a reverse reaction?
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The activation energy for the following reaction is 125 kJ/mol, and ΔE for the reaction is -216 kJ/mol. NO2(g) + CO(g) NO(g) + CO2(g) What is the activation energy for the reverse reaction [NO(g) + CO2(g) NO2(g) + CO(g)]? i'm not sure at all how to do it. would it just be -125 since it's in reverse?
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Answer:
The activation energy of the reverse reaction should be equal to the activation energy of the forward reaction minus the drive in the reaction (delta E). The answer is 125 kJ/mol + 216 kJ/mol = 341 kJ/mol.
Sean P at Yahoo! Answers Visit the source
Other answers
Ea=216 KJ/mol+125 KJ/mol=341 KJ/mol A potential energy diagram would probably make it easier to see. If you don't know what a potential energy diagram is, it looks like this. http://www.bbc.co.uk/scotland/education/bitesize/higher/img/chemistry/calculations_1/pe_diags/fig10.gif On that diagram (a) would be the activation energy which would be 400-100=300 KJ/mol for the forward reaction.ΔE would be (c) which is 300-100=200 KJ/mol. Now for the reverse reaction (b) is the activation energy which is 400-300=100 KJ/mol. So now draw a diagram like that but make your reactants start 0 meaning zero potential energy. It doesn't matter where you start your reactants at but zero is a good number to work with. Now on your diagram your (a) should go from 0 to 125(so Ea=125-0=125) and your (c) should go from 0 to -216(making Ea=-216-0=-216) Now for the reverse reaction you'll see that the activation energy which is the amount of energy it takes to get up to that peak, is 216+125=341 KJ
Johnny D
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