Chemical EquilibriumHard
Question
For the equilibrium, A(g) B(g), ⇋H is −40 kJ/mol. If the ratio of the activation energies of the forward(Ef) and reverse (Eb) reactions is2/3 then :
Options
A.Ef = 80 kJ/mol; Eb = 120 kJ/mol
B.Ef = 60 kJ/mol; Eb = 100 kJ/mol
C.Ef = 30 kJ/mol; Eb = 70 kJ/mol
D.Ef = 70 kJ/mol; Eb = 30 kJ/mol
Solution
ᐃH = Eaf − Eab
⇒−40 = 2x − 3x
⇒Eaf = 80 kJ/mol
Eab = 120 kJ/mol
⇒−40 = 2x − 3x
⇒Eaf = 80 kJ/mol
Eab = 120 kJ/mol
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