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For the equilibrium, A(g)rarrB(g), trian...

For the equilibrium, `A(g)rarrB(g), triangleH` is - 40 kJ `mol^(-1)` . If the ratio^f the activation energies of the forward `(E_(f))` and reverse `(E_(b))` reaction is (2/3) then:

A

`E_(r) = 60 kJ mol^(-1), E_(b) = 100 kJ mol^(-1)`

B

`E_(r) = 30 kJ mol^(-1), E_(b) = 70 kJ mol^(-1)`

C

`E_(r) = 80 kJ mol^(-1), E_(b) = 120 kJ mol^(-1)`

D

`E_(r) = 70 kJ mol^(-1), E_(b) = 30 kJ mol^(-1)`

Text Solution

Verified by Experts

`A(g) hArr B(g) " "Delta H = -40 kJ`
Since, `(E_(r))/(E_(b)) = (2)/(3)`, therefore, `E_(f) = (2x)/(5)` and `E_(b) = (3x)/(5), E_(b) - E_(r) = +40`
`(3x)/(5) - (2x)/(5) = +40 rArr (x)/(5) = 40 rArr x = 200`
Therefore, `E_(B) = (3x)/(5) = (3 xx 200)/(5) = 120 kJ mol^(-1), E_(f) = (2x)/(5) = (2 xx 200)/(5) = 80 kJ mol^(-1)`
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