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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.

For one chemical reaction in a closed system of fixed volume $V$, with uniform concentrations and no competing reactions or appreciable accumulation of unlisted species, choose the chemical equationBrackets denote amount of substance per volume, for example $[\mathrm{A}]=n_{\mathrm{A}}/V$. The stoichiometry gives one instantaneous reaction ratewhere $\xi$ is the existing extent of reaction. The minus signs make forward consumption positive; dividing by the coefficients makes measurements of different chemical species agree. Units can be $\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. See IUPAC Gold Book: rate of reaction and Flowers et al., Chemistry 2e, §12.1: chemical reaction rates.

For the illustrative chemical equation $2\mathrm{A}+\mathrm{B}⟶3\mathrm{C}$, a measured $d[\mathrm{C}]/dt=0.060\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ implies $v=0.020\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$, $d[\mathrm{A}]/dt=-0.040\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$, and $d[\mathrm{B}]/dt=-0.020\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. These numbers express stoichiometry, without supplying a rate law.

Each derivative is the slope of a concentration curve at one instant. Over a finite interval, $-\Delta [\mathrm{A}]/(a\Delta t)$ instead gives an average reaction rate. For A converting reversibly to B, the net reaction rate is $v=d[\mathrm{B}]/dt=J_f-J_r$; both directional reaction rates can remain positive when $v=0$ at chemical equilibrium.

A second-order reaction has overall order of reaction two: a rate law may depend on $[\mathrm{A}{]}^2$, or on $[\mathrm{A}][\mathrm{B}]$ with partial orders of reaction one and one. Doubling all relevant reactant concentrations multiplies the reaction rate by four at fixed conditions. This experimentally determined dependence does not establish molecularity or a unique reaction mechanism. For an elementary reaction obeying the law of mass action, molecularity and overall order of reaction agree; a net chemical equation alone cannot establish that the conversion is elementary.^[ref]

For the single-concentration model, explicitly choose the disappearance rate conventionHere $k_{\mathrm{obs}}$ is the coefficient for loss of A, held fixed at the specified temperature and other conditions. Its subscript labels the chosen observed disappearance rate; it does not imply a pseudo-first-order reaction. Assume a uniformly mixed closed system, fixed volume, and no appreciable reverse or competing chemical reactions. The integrated rate law isThe reciprocal-concentration plot is a straight line with slope $k_{\mathrm{obs}}$ and intercept $1/[\mathrm{A}{]}_0$. A brief derivation uses the chain rule: $d(1/[\mathrm{A}])/dt=-[\mathrm{A}{]}^{-2}d[\mathrm{A}]/dt=k_{\mathrm{obs}}$, then applies the initial condition. This is a hyperbolic decay, rather than the exponential decay of a first-order reaction.^[ref]

As an original prescribed example, choose $[\mathrm{A}{]}_0=0.020\mathrm{mol}{\mathrm{L}}^{-1}$ and $k_{\mathrm{obs}}=0.50\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$. At $t=100\mathrm{s}$, the reciprocal-concentration calculation is $50+50=100\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}$, giving $[\mathrm{A}]=0.010\mathrm{mol}{\mathrm{L}}^{-1}$. The initial disappearance rate is $0.50(0.020{)}^2=2.0\times 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$; after $100\mathrm{s}$ it is $5.0\times 10^{-5}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. Halving the concentration quarters the disappearance rate.

Dimensional analysis checks every formula: $k_{\mathrm{obs}}[\mathrm{A}{]}^2$ has units $\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$, $k_{\mathrm{obs}}t$ has units $\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}$, and $k_{\mathrm{obs}}[\mathrm{A}{]}_{0}t$ is a dimensionless quantity. Changing concentration units changes the numerical rate constant: the same example is $[\mathrm{A}{]}_0=20\mathrm{mol}{\mathrm{m}}^{-3}$, $k_{\mathrm{obs}}=5.0\times 10^{-4}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$, giving $[\mathrm{A}](100\mathrm{s})=10\mathrm{mol}{\mathrm{m}}^{-3}$.

For an elementary reaction $2\mathrm{A}⟶\mathrm{P}$, instead define the stoichiometrically normalized reaction rate $r=k[\mathrm{A}{]}^2$. Two A entities disappear per P formed, soConsequently the reciprocal-concentration slope is $2k$, not $k$. In the prescribed example, the normalized rate constant would be $k=0.25\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$; initially $r=1.0\times 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ while the disappearance rate is twice that. With $[\mathrm{P}](0)=0$, stoichiometry gives $[\mathrm{P}](100\mathrm{s})=([\mathrm{A}{]}_0-[\mathrm{A}])/2=0.0050\mathrm{mol}{\mathrm{L}}^{-1}$. Both rate constants have the same dimensions, so a units check alone cannot catch their factor-of-two difference. The symbol $r$ denotes the same normalization as $v$ in reaction rate.

The half-life, meaning the time to reach half a specified starting concentration, follows by setting $[\mathrm{A}]=[\mathrm{A}{]}_0/2$:For $r=k[\mathrm{A}{]}^2$ in $2\mathrm{A}⟶\mathrm{P}$, this becomes $t_{1/2}=1/(2k[\mathrm{A}{]}_0)$. Its units are seconds because $k_{\mathrm{obs}}[\mathrm{A}{]}_0$ has units ${\mathrm{s}}^{-1}$. Doubling the starting concentration halves the first half-life; a quoted half-life needs its starting concentration. This differs from a first-order reaction, whose half-life is independent of starting concentration.^[ref]

Within the prescribed run, $[\mathrm{A}]$ reaches $0.010$, $0.0050$, and $0.0025\mathrm{mol}{\mathrm{L}}^{-1}$ at $100$, $300$, and $700\mathrm{s}$. The successive halving intervals are $100$, $200$, and $400\mathrm{s}$: each begins at a smaller concentration. More generally, reaching fraction $f$ of the initial concentration, with $0<f<1$, requires $t=(f^{-1}-1)/(k_{\mathrm{obs}}[\mathrm{A}{]}_0)$, so $90\%$ consumption takes nine initial half-lives, rather than a fixed sequence of equal waiting times.

A straight reciprocal-concentration plot tests this particular single-concentration rate equation. It does not uniquely identify a reaction mechanism, nor does curvature rule out every second-order reaction: two independently changing reactant concentrations require their own integrated rate law. At low concentration, a small measurement error $\delta [\mathrm{A}]$ becomes approximately $-\delta [\mathrm{A}]/[\mathrm{A}{]}^2$ in the reciprocal, magnifying late-time errors. Use the line as a diagnostic; a nonlinear regression of the original concentrations can retain their measurement scale and explicit error model.

A zero-order reaction has a reaction rate independent of the relevant reactant concentration over the range where its rate law applies. For $\mathrm{A}⟶\mathrm{B}$, $[\mathrm{A}]=A_0-kt$ is valid only before depletion at $t=A_0/k$. The constant-consumption model must stop or change there.