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At any instant total no. of moles is constant (since whatever moles of A is lost is converted to same moles of B).

so, [A] + [B] = [Ao]............(1)

rate of formation of [B] = K1.[A]

rate of disappearance of [B] = -K2.[B]

so d[B]/dt = K1.[A] - K2.[B].......... (2)

Now put [A] = [Ao] - [B] from (1) in (2) :

d[B]/dt = K1.{[Ao]-[B]} - K2.[B]

d[B]/dt + (K1+K2).[B] = K1.[Ao]

solving this differential equation :

e^(K1+K2)t.[B] = K1.[Ao].e^(K1+K2)t/(K1+K2) + c

where c is constant of integration.

We know initial conditions : at t=0, [B] = 0

Putting this we get : c = -K1.{Ao]/(K1+K2)

so [B] = K1.[Ao].(1 - e^-(K1+K2)t) / (K1+K2)

When [B] = [Beq]/2

[Beq]/2 = K1.[Ao].(1 - e^-(K1+K2)t) / (K1+K2)

solving this for t we get :

t = {ln(1/(1-a))}/(K1+K2) where a = (K1+K2).[Beq]/2.K1.Ao

The answer can be expressed in various forms.

One more condition which you can use is at equilibrium net rate of reaction is 0 (forward rate = backward rate).

so, d[Beq]/dt = K1.[Aeq] - K2.[Beq] = 0

so K1.[Aeq] = K2.[Beq]

We can use this result to manipulate our answer in different form.

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