Answer:
the transition state will resemble the products more than the reactants
Explanation:
Since the free-radical bromination of methane involves the following reactions
Br₂ → 2 Br• , ΔH⁰ (per mole) = +192 kJ , Ea (per mole) = + 192 kJ
CH₄ +Br• → CH₃• + HBr , ΔH⁰ (per mole) = +67 kJ , Ea (per mole) = + 75 kJ
CH₃• +Br₂ → CH₃Br + Br• , ΔH⁰ (per mole) = -101 kJ , Ea (per mole) = -4 kJ
without considering the Br dissociation (initiation reaction) the rate-determining step is the second equation ( highest Ea) .
Then since the reaction is endothermic (ΔH⁰ (per mole) = +67 kJ) , the transition state will resemble the products more than the reactants ( Hammond postulate)