Why Is 66 Added Instead of Subtracted for Reverse Reaction Ea?

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future_vet
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Hello everyone,

Here's my question.
We have a reaction with an overal enthalpy change of -66KJ. The activation energy is 7KJ.
We want the Ea for the reverse reaction.

Formula:
Ea (rev) = ∆E + Ea (fwd)
So I wrote:
Ea (rev) = -66 + 7 = 59 KJ.

But the corrected exercise says: 66+7 = 73KJ.
Why did we add 66, if it's a negative value?

Thank you,

J.
 
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Now try to see all of this practically...the reaction itself is exothermic, the activation energy for the reverse reaction thus is greater then the forward assuming a single "transition state." Draw a free energy diagram, it should help greatly in understanding what's going on here.
 
So since the reverse reaction is ENDOthermic, then the Ea should be greater than in the reaction that was exothermic... Would this be the correct way to think?

~J.
 
vet: Do you know how to draw the energy diagram? It takes just a couple of minutes to learn it, and once you do, problems like this become a piece of cake.

Exothermic reaction, http://www.bbc.co.uk/scotland/education/bitesize/higher/img/chemistry/calculations_1/pe_diags/fig10.gif - the positive x-direction is the direction of the forward reaction.
 
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I do, and I understand why the answer is 73, but I was just puzzled by the equation, since it seemed to lead to the wrong answer...

Joanna.
 
Ea (rev) = ∆Erev + Ea (fwd)

Remember to change the sign --- the reaction is proceeding in the reverse direction, and the sign of the energy change for the reaction is opposite that for the forward.