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Acetone reacts with iodine to form iodoacetone, iodide, and a hydrogen ion:In water, ${\mathrm{H}}^{+}$ denotes a solvated hydrogen ion; added hydrochloric acid provides acidic conditions.

Moorpark College's protocol mixes acetone, hydrochloric acid and water from burets in matched test tubes; pipet in iodine, swirl and time colour loss against white paper. Use water, then reacted solution, as references. Duplicate each mixture; keep total volume $5.00\mathrm{mL}$ and run temperatures within $2{}^{\circ }\mathrm{C}$.^[ref]

The stock dilutions give $[\mathrm{acetone}{]}_0=0.800$, $[{\mathrm{H}}^{+}{]}_0=0.200$, and $[{\mathrm{I}}_2{]}_0=0.00100\mathrm{mol}{\mathrm{L}}^{-1}$. B and C separately double acetone and acid; D halves iodine. The manual supplies no measured times; it states zero partial order of reaction in iodine and asks students to determine the other exponents.^[ref]

The endpoint measurement is a time $\tau$ to an operational colour threshold, not direct observation of a reaction intermediate. If the iodine disappearance rate $r$ is approximately constant until that endpoint, $r\approx [{\mathrm{I}}_2{]}_0/\tau$. Halving $[{\mathrm{I}}_2{]}_0$ at fixed other concentrations then approximately halves $\tau$ without changing $r$. A longer or shorter endpoint time alone is not a comparison of reaction rates unless the starting concentration is included. For an observed power-form rate law, doubling only acetone gives a rate ratio $2^m$ and hence $m=\ln (r_B/r_A)/\ln 2$; doubling only acid similarly estimates its own partial order of reaction. Those are inference formulas, not asserted experimental results.

A useful proposed reaction mechanism would explain how iodine is consumed after an earlier transformation sets the throughput, making the observed reaction rate insensitive to iodine over a tested range. The observed zero exponent would contradict treating this net chemical equation as a single elementary reaction with directional rate law $k[\mathrm{acetone}][{\mathrm{I}}_2]$. It would not uniquely prove which reaction intermediate or earlier step is involved. Independently detecting a candidate reaction intermediate would provide another kind of evidence.

The excess-reactant approximation has an explicit scale: the baseline iodine inventory is only $0.00100/0.800=0.125\%$ of the initial acetone inventory, and its net hydrogen ion production is $0.00100/0.200=0.5\%$ of the initial acid inventory, assuming only the stated one-to-one chemical reaction. These are calculated bounds on composition change, not measured reaction orders. Mixing delay, a subjective colour threshold, and temperature drift limit the inference. Repeats and a fixed viewing geometry help; a calibrated absorbance trace could test linearity during conversion instead of relying on an endpoint alone. That extension would require a reference blank and a justified Beer-Lambert law calibration, rather than treating visible colour as an exact concentration. A zero-order reaction approximation in iodine must fail at depletion; it cannot imply consumption after no iodine remains.

In a power-form rate law $v=k[\mathrm{A}{]}^{\alpha }[\mathrm{B}{]}^{\beta }$, the partial orders of reaction are $\alpha$ and $\beta$, and the overall order of reaction is $n=\alpha +\beta$. For example, doubling $[\mathrm{A}]$ at fixed $[\mathrm{B}]$ multiplies $v$ by $2^{\alpha }$: factors of one, two, and four correspond to partial orders of reaction zero, one, and two. These exponents can also be fractional or negative. A single constant order of reaction need not describe a rate law that is not a power form. See IUPAC Gold Book: order of reaction.

For a simple irreversible conversion $\mathrm{A}⟶\mathrm{B}$ with rate equation $-d[\mathrm{A}]/dt=k[\mathrm{A}{]}^n$, initial concentration $A_0>0$, and fixed $k>0$, compare zero-order reactions, first-order reactions, and second-order reactions using the dimensionless quantitiesTheir integrated curves, starting at $X(0)=1$, areFor the zero-order reaction, the plotted extension $X=\max (1-\tau ,0)$ stops conversion at depletion. It does not continue to obey $dX/d\tau =-1$ once $X=0$. A negative concentration would be unphysical. The first-order reaction and second-order reaction approach zero without reaching it at finite time in these models.

The common $\tau$ axis compares curve shapes after each model's own time rescaling, rather than equal laboratory times for an identical numerical $k$. In the lower panels, $X$, $\ln X$, and $1/X$ respectively give straight lines with slopes $-1$, $-1$, and $+1$ for the matching order of reaction. The natural logarithm uses a dimensionless quantity. These transformed plots help test a proposed rate law; the zero-order reaction line applies only before depletion. See Flowers et al., Chemistry 2e, §12.4: integrated rate laws.

A rate constant is the coefficient in a specified rate law, held fixed under specified conditions. It can describe an empirical overall chemical reaction, as well as an elementary reaction. Fixed temperature matters: changing temperature can change $k$, even for the same chemical reaction. See IUPAC Gold Book: order of reaction and rate constants and IUPAC Gold Book: rate coefficient.

For $v=k[\mathrm{A}{]}^n$, dimensional consistency requiresA zero-order reaction therefore uses $\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$; a first-order reaction uses ${\mathrm{s}}^{-1}$; a second-order reaction uses $\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$. Comparing bare numerical values of $k$ across different orders of reaction is meaningless without their units and concentrations. Even for the same order of reaction, state whether $k$ multiplies the normalized reaction rate $v$ or a particular reactant disappearance rate: a stoichiometric coefficient can change that convention.

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.