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The energy change accompanying the equil...

The energy change accompanying the equilibrium reaction `A hArr B` is `-33.0 kJ mol^(-1)`.

Assuming that pre-exponential factor is same for forward and backward reaction answer the following :
The energy of activation for forward and backward reaction (`E_(f)` and `E_(b)`) at 300 K. Given that `E_(f)` and `E_(b)` are in the ratio 20:31,

A

69, 93

B

60, 93

C

93, 60

D

90, 60

Text Solution

Verified by Experts

The correct Answer is:
B
`(E_(r))/(E_(b)) = (20)/(31)` (given). Also `E_(f) - E_(b) = -33` (given) ….(i)
Substituting `E_(b) = (31)/(20)E_(f)`, in Eq. (i) we get
`E_(f) - (31)/(20)E_(f) = -33` or `(-11)/(20) E_(f) = -33` or `E_(f) = (33 xx 20)/(11) = 60 kJ mol^(-1)`
`E_(b) = E_(f) + 33 = 60 + 33 = 93 kJ mol^(-1)`
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