[ThatGuy]

This is not a homework question, rather I am trying to clarify a statement made in my textbook.

Diagrams (a) and (b) are attached.

They give an example of a reaction the has a low enthalpy change of ΔH = 25 KJ. Now, they say that the reverse reaction for this is more likely to be successful than that of a reaction that has a ΔH = 100 KJ. No explanation for why this is is given. They simply tell me to look at which one has a lower ΔH.

If I look at a ΔH diagram (a) or (b); I can clearly tell that the activation energy for the reverse reaction of diagram (b) will be much lower compared to that of (reverse (a))... wouldn't it then be easier for (reverse reaction of b) to react... so I am kind of confused as to why the activation energy of the reverse reaction is simply not considered.

[Borek]

You are right about activation energies. Chances are they are the way they are just because the plot is lousy, and it was not intended this way.

Perhaps what they mean is the fact that the reverse reaction b is strongly (especially compared to a) exothermic, which is often the driving force behind spontaneity. But that's just a guess, hard to tell without seeing whole chapter.

[ThatGuy]

You are right about activation energies. Chances are they are the way they are just because the plot is lousy, and it was not intended this way.

Perhaps what they mean is the fact that the reverse reaction b is strongly (especially compared to a) exothermic, which is often the driving force behind spontaneity. But that's just a guess, hard to tell without seeing whole chapter.

Perhaps, however, these plots are just ones I drew on paint - but they are the same ones from my textbook. It's kind of weird because no explanation is provided.