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Question

For a first order reversible reaction AKfKbB, the initial concentration of A and B are [A]0 and zero respectively. If concentrations at equilibrium are [A]eq. and [B]eq., derive an expression for the time taken by B to attain concentration equal to [B]eq/2.

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Solution

The given reaction is
Kf
AB
Kb

Initial concentration of A=[A]0 & B=0
Concentration at equilibrium are [A]eq & [B]eq respectively.
Now,
Rate=d[A]dt=Kf[A]Kb[B] (i)
At, t=0,[B]=0
[B]=[A]0[A] (ii)
Using (ii) in (i):-
d[A]dt=Kf([A])Kb([A]0[A])
d[A]dt=(Kf+Kb)[A]Kb[A]0 (iii)
At equilibrium, d[A]dt=0 & [A]=[A]eq
(Kf+Kb)[A]eqKb[A]0=0
[A]eq=KbKf+Kb[A]0 (iv)
Now, from eqn(i), at equilibrium, we have:-
Kf[A]eqKb[B]eq=0
[B]eq=KfKb[A]eq
[B]eq=Keq[A]eq (v) (Keq=KfKb)
Now, Integrating eqn(iii) after substituting [A]eq from (iv):-
[A]=[A]eq+C1e(Kf+Kb)t (vi)
;C is integration constant
At, t=0,[A]=[A]0
[A]=[A]eq+([A]0[A]eq)e(Kf+Kb)t (vii)
Now, when [B]=[B]eq/2 (From (ii)) - (viii)
Putting (viii) in (vii):-
[A]0[B]eq2=[A]eq+([A]0[A]eq)e(Kf+Kb)t
e(Kb+Kf)t=2[A]0[B]eq2[A]eq2([A]0[A]eq)
Taking natural log on both sides:-
(Kf+Kb)t=ln(2[A]0[B]eq2[A]eq2([A]0[A]eq))
t=1(Kf+Kb)ln(2[A]0[B]eq2[A]eq2([A]0[A]eq))

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