Enthalpy Diagram

[s3a]

I understand the majority of the answer to this question except for #3. I do not understand why the activation energy is 1118kJ. I'd like to know how to apply this universally given different questions which is why I am asking precise what made the ΔH equal 1118kJ.

Question:
http://img91.imageshack.us/img91/852/questionnmx.jpg

Answer:
http://img91.imageshack.us/img91/426/answer.jpg

Any help would be greatly appreciated!
Thanks in advance!

[Xorlev]

Perhaps I don't understand your question, you don't understand what activation energy is?

ΔH is the change in enthalpy, the negative change means that overall, the reaction is exothermic. Activation energy is the energy required to get it over the hump persay, to get it going to release the energy.

[s3a]

Yes I know that "Activation energy is the energy required to get it over the hump persay" but I'd like to know how it was determined numerically that 1118kJ is the activation energy.

Thanks!

[Xorlev]

It's determined with partially experimentally derived data in conjunction with the Arrhenius equation, k=Aexp(-∆E/RT). With the reaction rate at two temperatures, we can cancel A (frequency factor) with some simple algebra.

[s3a]

I don't think I've learnt what you just said. Are you referring to grade 12 material because I am in grade 11 and do not live in the United States so grade 12 isn't part of high school for my location.

[Xorlev]

I wouldn't expect to see this (perhaps briefly mentioned in a General Chem class) until college-level Chemical Kinetics. The activation energy will certainly be given to you if needed.