A certain reaction has an enthalpy of ΔH=49 kJ and an activation energy of Ea=81 kJ. What is the activation energy of the reverse reaction?
Hey guys! I was wondering if anyone could explain the different steps in terms of solving for the regular activation energy and solving for the activation energy of a reverse reaction?
Thank you!
Sapling #17 Wk.9-10
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Re: Sapling #17 Wk.9-10
So for this problem I actually drew a free energy diagram to visualize the situation. Since delta H is positive in your case, you know that the reaction is endothermic. You know that the activation energy is the distance from the reactant to the peak of the curve, and delta H is the difference between the energy of the products and energy of reactants. Therefore all you do is subtract the delta H from the activation energy of the forward reaction (Ea forward - delta H= Ea reverse). Hope this doesn't sound confusing!
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Re: Sapling #17 Wk.9-10
On the graph it is the product to the peak going in the backwards direction from right to left. So this is why the enthalpy change and the activation energy are proportional to the reverse activation energy.
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Re: Sapling #17 Wk.9-10
Use the equation Ea(forward) − Ea(reverse) = ΔH, which in this case would be 81 - 49 = Ea(reverse) .
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Re: Sapling #17 Wk.9-10
Since ΔH=Ea(forwards)-Ea(reverse), you can rearrange this to find the Ea(reverse). Ea(reverse)=Ea(forward)-ΔH.
I honestly think when I look at an activation energy graph, for both an endothermic and exothermic reaction, it helps me conceptualize it a bit better.
I honestly think when I look at an activation energy graph, for both an endothermic and exothermic reaction, it helps me conceptualize it a bit better.
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Re: Sapling #17 Wk.9-10
This graph of the information might help you understand where the values come from and how to get the EA of the reverse reaction
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