For the equilibrium, A(g)⇌B(g), ΔH = -40 kJ/mol. If the ratio of the activation energies of the forward (Ef) and reverse (Eb) reaction is 2/3, then:

(A) Ef=30kJ/mol; Eb=70kJ/mol
Ef/Eb = 30kJ/mol/70kJ/mol = 3/7
(B) Ef=70kJ/mol; Eb=30kJ/mol
Ef/Eb = 70kJ/mol/30kJ/mol = 7/3
(C) Ef=80kJ/mol; Eb=120kJ/mol
Ef/Eb = 80kJ/mol/120kJ/mol = 2/3
(D) Ef=60kJ/mol; Eb=100kJ/mol
Ef/Eb = 60kJ/mol/100kJ/mol = 3/5
But the given ratio is Ef/Eb = 2/3
Hence, the option C is the correct answer.