Question

The standard free energy of activation of a reaction A is 73.4 kJ mol–1 (17.5 kcal mol–1) at 298 K. Reaction B is one million times faster than reaction A at the same temperature. The products of each reaction are 10.0 kJ mol–1 (2.39 kcal mol–1) more stable than the reactants. (a) What is the standard free energy of activation of reaction B?

a.) What is the standard free energy of activation of reaction B?

B.) What is the standard free energy of activation of the reverse of reaction A?

C.)What is the standard free energy of activation of the reverse of reaction B?

Answer 1

I would do this.
ln (k2/k1) = Ea/RT
You know k2 is 1E6k1 and you know Ea for A, solve for Ea for B in which k1 is the rate constant for reaction A and k2 is the constant for reaction B.

If products are more stable than reactants then reactions A and B are exothermic and Ea reverse is Ea + 10 kJ/mol.
Check out my thinking before you do anything.