This is not an Arrhenius equation problem.
This is tough to explain without a "blackboard". Here is a link that shows a potential energy diagram for an ENDOTHERMIC equation.
https://www.google.com/search?client=firefox-b-1-d&q=find+activation+energy+for+reverse+endothermic+reaction
In this diagram, for example. The system is at 50 to start and the "hill" is at 250. The Ea for the forward reaction is, of course, 250-50 = 200.
dH for the reaction is 100 - 50 = +50
If you want the reverse Ea, it will be 250 (that's the top of the hll) on the way back up the hill) - 100 (that's the starting point for the reverse) and 250-100 = 150.
So for your problem, Ea fwd is 31. dH for the fwd rxn is 12 so the difference is 20 the atoms would need have to crawl back up the hill if started with products going back to reactants. I hope this helps. I spent forever looking for a diagram I could link to you. Sorry that took so long.
For the elementary process below, the activation energy is 31 kJ mol−1 and the enthalpy of reaction is 12 kJ mol−1. What is Ea for the reverse process, in kJ mol−1 ?
A + B ⇌ AB
how can i solve this question if im not given the temperature, and even without using a graph?
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