Why we can use ΔH to determine reaction rate in free radical bromination reaction steps?

I am trying to understand why we can use ΔH to determine reaction rate in this a free radical halogenation reaction.

Given: (CH3)2CH2 + Br• → (CH3)2CH• + HBr = 397 - 368 = +29 kJ/mol = ΔH

(CH3)2CH• + Br2 → (CH3)2CHBr + Br• = 193 - 285 = -92 kJ/mol = ΔH

Question posed: Based on ΔH values, which step would you guess would be the slowest?

If reaction rate (kinetics) is based on reaching activation energy and concentration of reactants (K=[A]^a[B]^b), and DOESN'T depend on ΔH, i.e whether a reaction is endo or exothermic, why and how we can use enthalpy (thermodynamics) (endo or exothermic) to find the reaction rate?

Teacher writes following explanation: Step 1 is somewhat endothermic, which Step 2 is significantly exothermic, so Step 1 is very likely to be the slower step.

• Draw a generic reaction profile for a reaction that is endothermic in the forward direction. Let the height of the "hump" be the enthalpy difference between H_products and H_activation complex (i.e., the increase in enthalpy needed to reach the activation state when going in the reverse direction). Your teacher is assuming the height of the hump is the same for both reverse reactions, and thus the larger delta_H_rxn is in the forward direction, the larger the activation enthalpy in the forward direction will be. You should ask your teacher the basis for this assumption. Nov 10 '21 at 20:41