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

[future_vet]

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.

 

[GCT]

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.

 

[future_vet]

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.

 

[Gokul43201]

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.

 

[future_vet]

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.

 

[Bystander]

Ea (rev) = ∆E_rev + 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.