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Consider a reaction that is first order ...

Consider a reaction that is first order in both direction
`A overset(k_f)underset(k_b)hArr B`
Initially only A is present , and its concentration is `A_0`. Assume `A_t` and `A_(eq)` are the concentrations of A at time 't' and at equilibrium, respectively. The time 't at which `A_t = (A_0 + A_(eq))//2 is`,

A

` t=( I n (3/2))/((K_f +k_b) )`

B

`t=(I n)( (3/2))/((k_f - k_b))`

C

`t= (In 2)/( (k_f + K_b))`

D

`t=( I n2)/( (K_f - K_b))`

Text Solution

Verified by Experts

The correct Answer is:
C

Given at time ` t=t A_t = (A_0 +A_(eq))/(2)`
and `x_(eq) = A_0 - A_(eq)`
Now ` t= ( 1)/(k_f +k_b) ln ((x_c)/(x_(e-x)))=((l n 2)/(k_f +k_b ))`
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