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A consecutive first-order reaction mechanism passes material through a reaction intermediate. Consider a stipulated chemical reaction network $\mathrm{A}\stackrel{k_1}{\to }\mathrm{I}\stackrel{k_2}{\to }\mathrm{P}$, 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 givesAdding 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 isTo derive the middle expression, substitute the first into the first-order linear differential equation for I and multiply by the integrating factor $e^{k_{2}t}$. Then $d(e^{k_{2}t}[\mathrm{I}])/dt=k_1{A}_0{e}^{(k_2-k_1)t}$, 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 givesThis 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 $\mathrm{A}⟶\mathrm{P}$ 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.

For the chosen equal-rate constant example, I peaks at $t=1/k=5.00\mathrm{s}$, because $d[\mathrm{I}]/dt=A_{0}k{e}^{-kt}(1-kt)$. At that time $[\mathrm{A}]=[\mathrm{I}]=A_0/e\approx 0.367879\mathrm{mmol}{\mathrm{L}}^{-1}$ and $[\mathrm{P}]\approx 0.264241\mathrm{mmol}{\mathrm{L}}^{-1}$; the direct model would already have $[\mathrm{P}]\approx 0.632121\mathrm{mmol}{\mathrm{L}}^{-1}$. 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 $t_{\ast }=\ln (k_2/k_1)/(k_2-k_1)$, 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.

An elementary reaction is one microscopic step in a reaction mechanism, with no intervening reaction intermediate needed within that step. Its molecularity counts the participating reacting entities. In an ideal dilute, uniformly mixed model, the law of mass action gives the directional reaction rate from the step's reactant concentrations: $\mathrm{A}⟶\mathrm{P}$ has $r=k[\mathrm{A}]$, while $\mathrm{A}+\mathrm{B}⟶\mathrm{P}$ has $r=k[\mathrm{A}][\mathrm{B}]$. Applying those powers to a net chemical equation requires separate evidence that it is elementary.^[ref]

For a reversible step $\mathrm{A}+\mathrm{B}\rightleftharpoons \mathrm{I}$, apply that rule separately: $r_f=k_f[\mathrm{A}][\mathrm{B}]$ and $r_r=k_r[\mathrm{I}]$, giving net production of I as $r_f-r_r$. Their difference can vanish at chemical equilibrium while both directional reaction rates remain nonzero. For $2\mathrm{A}⟶\mathrm{P}$, the elementary normalized reaction rate is $r=k[\mathrm{A}{]}^2$, whereas A's disappearance rate is $2r$; see second-order reaction for the complete factor-of-two example.

Calling a step slow does not remove the need for its actual reactant concentrations. If a reaction intermediate enters its rate law, determine that concentration from the other rate equations or a justified approximation. In the consecutive first-order reaction mechanism, an intermediate maximum is only one instant at which its production and consumption match. Replacing its entire evolution by that one balance would lose the initial buildup and delayed reaction product formation. A successful overall first-order reaction fit likewise does not establish an elementary event: both worked reaction mechanisms above provide ways for microscopic detail to remain hidden in one reactant trace.

A reaction intermediate is formed during a reaction mechanism and consumed in a subsequent step, cancelling from the net chemical equation. It can accumulate, as I does in the consecutive first-order reaction mechanism. A reaction intermediate can also connect transformations in energy coupling. Unlike a catalyst, it need not be present before conversion begins.^[ref]

A reaction intermediate is distinct from a transition state: the former can persist between steps, whereas the latter describes passage through a barrier. The existing potential energy surface and energy profile provide the configuration-space picture; the time-dependent concentration of an intermediate instead measures an ensemble of reacting entities. Low concentration or short lifetime can make detection difficult, so failure to detect a proposed reaction intermediate is not by itself proof of absence. Conversely, detecting a chemical species does not establish that it carries the proposed route to the final reaction products; it could belong to a competing route.

A reaction mechanism describes how reactants become reaction products through specified elementary reactions. Adding the steps must recover the overall chemical equation; intermediate chemical species cancel. That accounting is necessary, but the proposed steps must also predict a rate law compatible with experiment. Several reaction mechanisms can pass the same test.^[ref]

For example, the stipulated steps $\mathrm{A}+\mathrm{B}⟶\mathrm{I}$ and $\mathrm{I}⟶\mathrm{P}$ sum to $\mathrm{A}+\mathrm{B}⟶\mathrm{P}$. I is a reaction intermediate: it is formed and then consumed. The catalyst C in the existing catalytic cycle instead starts available, participates, and is regenerated; its bound form can be a reaction intermediate. Both can disappear from the summed chemical equation, so cancellation alone does not distinguish their roles. Cancellation is algebraic bookkeeping, not an assertion that every step finishes instantaneously.

The net chemical equation supplies consumption ratios, not a clock. For the stipulated two-step example, directional reaction rates $r_1=k_1[\mathrm{A}][\mathrm{B}]$ and $r_2=k_2[\mathrm{I}]$ give $d[\mathrm{I}]/dt=r_1-r_2$ and $d[\mathrm{P}]/dt=r_2$. Initially $[\mathrm{I}]=0$ can mean zero reaction product formation while A already disappears. In particular, $-d[\mathrm{A}]/dt=d[\mathrm{P}]/dt$ requires negligible intermediate accumulation here. A single elementary reaction with the same net chemical equation would instead predict immediate formation at $k[\mathrm{A}][\mathrm{B}]$. Measuring both sides can reveal a distinction hidden by the net equation.

This makes reaction mechanisms useful for designing observations: follow a candidate reaction intermediate, measure individual reaction products, or vary one starting concentration at a time. The consecutive first-order reaction mechanism quantifies an intermediate's rise and fall; the parallel first-order reaction mechanism shows how identical reactant traces can conceal different reaction products. The acetone iodination rate-law experiment connects such reasoning to a real laboratory protocol. The existing reaction coordinate, potential energy surface, and energy profile describe configurations and barriers along a route; a concentration plotted against time describes a different observable.