Standard Free Energy of Activation of a Reaction

[8008jsmith]

Homework Statement

The standard free energy of activation of a reaction A is 83.7 kJ/mol at 298K. Reaction B is 10 million times faster at the same temperature. The products of each reaction are 10 kJ/mol more stable than the reactants.

(a) What is the standard free energy of activation of reaction B
(c) What is the standard free energy of activation of the reverse of reaction B.

2. The attempt at a solution

I used the equation attached in the image to solve for rate B and I got 60.9 and then I added 10 for the reverse. Is that the correct way to go about this question?

 

[Bystander]

If it works, and you're comfortable with it, understand what you've done, go with it.

 

[8008jsmith]

I had the right formula, I was just using it wrong. For anyone's future reference: Take the log of the rate ratio and then solve for ΔG of each activation energy you need. To get the reverse just add the energy of the products to each reaction energy you're trying to find. Therefore:

log (1/10,000,000) = ΔG_B - 83.7 kJ/mol / 2.3(0.008314 kJ/K mol)(298K)

-7 = ΔG_B - 83.7 kJ/mol /5.69 kJ/mol

Now just solve for ΔG_B