Two chemical reactions consecutively follow on from one another. Consecutive reactions are generally very common in biochemistry and organic chemistry.

** K _{1} k_{2}**

**A → B → C **

A common situation is for k_{1}, to be a first-order rate constant at which A is transformed to B and k_{2} the first-order rate constant at which B transformed to product C. Assuming that we start at time zero with pure A and no B or C present then this mechanism gives the concentration of B, [B], at any later time, t, as

where [A]o is the initial concentration of A. Find the time, in terms of the rate constants k1 and k_{2}, at which the concentration of B is a maximum [B]_{mas}.

**Here,**

**[B] = [A] _{0} [k_{1} /(k_{2} – k_{1}) (e^{-k1t} – e^{-k2t})**

To find the maximum concentration of B we differentiate [B] with respect to t. Notice that on the right hand side the terms before the bracket are constant and differentiating with respect to t, gives

The maximum in [B] occurs when d[B]/dt = 0, which is either when [A]_{0} k_{1}/(k_{2} — k_{1}) is equal to zero (which it cannot be) or alternatively the term is brackets equals zero.

**k _{1} e^{-k1t} = k_{2} e^{-k2t}**

Taking natural logs allows us to find t at which [B] is a maximum.

**ln (k _{1}) – k_{1}t = ln (k_{2})— k_{2}t**

**ln (k _{1}) – ln (k_{2}) = t (k_{1}—k_{2}**)

**In (k _{1}/k_{2}) = t (k_{1}—k_{2})**

**t = [1 / (k _{1}—k_{2})] In (k_{1}/k_{2})**

The time at which the concentration of B is a maximum depends only on the values of the two rate constants and is independent of the initial concentration of A.

Related Study: