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`ArarrB, DeltaH= -10 KJ mol^(-1), E_(a(f))=50 KJ mol^(-1)`, then `E_(a)` of `BrarrA` will beA. `40 kJ "mol"^(-1)`B. `50 kJ "mol"^(-1)`C. `-50 kJ "mol"^(-1)`D. `60 kJ "mol"^(-1)` |
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Answer» Correct Answer - D `A rarr B, Delta H_(f) = -10kJ mol^(-1)` (Exothermic) `:. B rarr A, Delta H_(b) + 10 kJ mol^(-1)` (endthermic) `:. (E_(a))_(b) = Delta H_(b) + (E_(a))_(f)` `= 10 + 50` `= 60 kJ mol^(-1)` |
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