= Consecutive first-order reaction mechanism
{title2=$\mathrm A\xrightarrow{k_1}\mathrm I\xrightarrow{k_2}\mathrm P$}
A <consecutive first-order reaction mechanism> passes material through a <reaction intermediate>. Consider a stipulated <chemical reaction network> $\mathrm A\xrightarrow{k_1}\mathrm I\xrightarrow{k_2}\mathrm P$, with constant positive <rate constants>, fixed <temperature> and <volume>, uniform mixing, and no reverse or competing <chemical reactions>. All three labels carry one conserved unit of material. Set $[\mathrm A](0)=A_0>0$ and $[\mathrm I](0)=[\mathrm P](0)=0$. The <law of mass action> gives
$$
\frac{d[\mathrm A]}{dt}=-k_1[\mathrm A],\qquad
\frac{d[\mathrm I]}{dt}=k_1[\mathrm A]-k_2[\mathrm I],\qquad
\frac{d[\mathrm P]}{dt}=k_2[\mathrm I].
$$
Adding the three <rate equations> proves $[\mathrm A]+[\mathrm I]+[\mathrm P]=A_0$. Each step has a <first-order reaction> <rate law>; this does not make all three <concentrations> simple <exponential decays>.
For $k_1\ne k_2$, the exact solution is
$$
[\mathrm A]=A_0e^{-k_1t},\qquad
[\mathrm I]=A_0\frac{k_1}{k_2-k_1}\left(e^{-k_1t}-e^{-k_2t}\right),\qquad
[\mathrm P]=A_0-[\mathrm A]-[\mathrm I].
$$
To derive the middle expression, substitute the first into the <first-order linear differential equation> for I and multiply by the <integrating factor> $e^{k_2t}$. Then $d(e^{k_2t}[\mathrm I])/dt=k_1A_0e^{(k_2-k_1)t}$, which can be integrated directly with the stated <initial condition>. The material balance supplies P without another integration.
The apparent division by zero at equal <rate constants> is removable. For $k_1=k_2=k$, the same <integrating factor> gives
$$
[\mathrm I]=A_0kte^{-kt},\qquad
[\mathrm P]=A_0\bigl[1-(1+kt)e^{-kt}\bigr].
$$
This is also the <limit> of the unequal-<rate constant> expression. Initially $[\mathrm P]/A_0=(kt)^2/2+O((kt)^3)$, so <reaction product> formation begins with zero slope despite immediate <reactant> loss. By contrast, a direct $\mathrm A\longrightarrow\mathrm P$ <first-order reaction> with the same $k$ has $[\mathrm P]/A_0=1-e^{-kt}$ and nonzero initial slope. Their A traces agree exactly; their P traces do not.
\Image[/consecutive-reaction-intermediate-time-profiles.png]
{title=An accumulating <reaction intermediate> delays <reaction product> formation in a <consecutive first-order reaction mechanism>}
{id=image-consecutive-reaction-intermediate-time-profiles}
{description=Original chosen model with $k_1=k_2=0.20\,\mathrm{s^{-1}}$ and $A_0=1.00\,\mathrm{mmol\,L^{-1}}$. Solid curves follow the exact three-species <rate equations>; the dashed P curve is a separate direct A-to-P model with the same A trace. These are model predictions, not measurements or an identified molecular <reaction mechanism>.}
For the chosen equal-<rate constant> example, I peaks at $t=1/k=5.00\,\mathrm s$, because $d[\mathrm I]/dt=A_0ke^{-kt}(1-kt)$. At that time $[\mathrm A]=[\mathrm I]=A_0/e\approx0.367879\,\mathrm{mmol\,L^{-1}}$ and $[\mathrm P]\approx0.264241\,\mathrm{mmol\,L^{-1}}$; the direct model would already have $[\mathrm P]\approx0.632121\,\mathrm{mmol\,L^{-1}}$. The peak means formation and consumption of I are momentarily equal, not that the whole system has reached <chemical equilibrium>.
More generally the peak is at $t_*=\ln(k_2/k_1)/(k_2-k_1)$, which is positive for either ordering of the two positive <rate constants> and tends to $1/k$ as they coincide. When $k_2\gg k_1$, a brief initial adjustment can be followed by $[\mathrm I]\approx(k_1/k_2)[\mathrm A]$ and approximately equal formation and consumption <reaction rates>. The exact solution states when this small-intermediate approximation can be reasonable. It is inappropriate for the illustrated equal-<rate constant> case: at the peak, over one third of the conserved material is I. Such timing matters if the goal is to collect a useful <reaction intermediate> before it becomes the final <reaction product>.
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