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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.

How can an invisible change in concentration produce a readable signal? The saponification of ethyl acetate by hydroxide in water gives ethanol and an acetate ion:The rate law used for this assay is $r=k_2[\mathrm{ethyl}\mathrm{acetate}][\mathrm{O}{\mathrm{H}}^{-}]$. Each reactant has coefficient one, so $r$ equals either disappearance rate. If the two initial concentrations are equal, they remain equal under this stoichiometry, giving $-dc/dt=k_2{c}^2$ and $k_{\mathrm{obs}}=k_2$, without a factor of two.

Manual for Advanced Physical Chemistry Laboratory-II by Nasima Arshad (Unit 7, Experiment 2, printed pp. 66–67) specifies a conductivity cell holding about $60\mathrm{mL}$ in constant-temperature baths, with $0.040\mathrm{mol}{\mathrm{L}}^{-1}$ ethyl acetate and sodium hydroxide stocks equilibrated separately. Mixing $30\mathrm{mL}$ of each initiates the run. The readout is electrical resistance: every minute for ten minutes, every two minutes for ten more, then every five minutes for twenty-five. The cell is rinsed with distilled water. A matched sodium hydroxide–water dilution estimates the initial resistance; a potassium chloride standard determines the cell constant. The manual asks for comparison with the extrapolated initial signal.^[ref]

The post-mixing initial concentration is $c_0=0.040(30/60)=0.020\mathrm{mol}{\mathrm{L}}^{-1}$ for each reactant, assuming additive volumes. This matches our illustrative starting concentration, but the prescribed $k_{\mathrm{obs}}=0.50$ is not a measured value from the manual. As hydroxide is replaced by the less mobile acetate ion, electrical conductance decreases. The preceding experiment describes two platinum electrodes and an alternating-voltage Wheatstone bridge to limit polarization at the electrodes. For measured electrical resistance $R(t)$, electrical conductance is $G(t)=1/R(t)$; electrical conductivity is $\kappa (t)=K_{\mathrm{cell}}G(t)$. Keep the probe geometry fixed and use one signal convention throughout.^[ref]

For the equal-starting-concentration model, assume an affine calibration between electrical conductance and remaining hydroxide concentration, with endpoints $G_0$ and $G_{\infty }$ at the same temperature. This givesThe last plot has slope $k_2{c}_0$, with units ${\mathrm{s}}^{-1}$, whereas plotting $1/c$ against time has slope $k_2$, with units $\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$. These are different slopes. This endpoint transformation also shows why raw resistance must not be substituted for conductance.

Useful additional controls are repeat runs, fixed immersion depth, an independently checked final signal, and comparisons of several starting concentrations. Mixing delay obscures early conversion; temperature drift changes both the rate constant and the signal; contamination adds mobile ions. The affine calibration can fail when ionic mobilities change with composition. Near completion, $G-G_{\infty }$ becomes small, so endpoint error is strongly amplified. Those are methodological limits and proposed checks, not new experimental observations. A conductivity trace supports a calibrated rate law over its tested range; it does not independently reveal the full reaction mechanism.

For unequal starting concentrations $a=[\mathrm{A}{]}_0$ and $b=[\mathrm{B}{]}_0$ in a one-to-one chemical reaction, let $x$ be the consumed concentration. Then $[\mathrm{A}]=a-x$, $[\mathrm{B}]=b-x$, andThe natural logarithm acts on a dimensionless quantity; differentiating the left side with respect to $x$ gives $1/((a-x)(b-x))$, verifying the relation. Mixing $20\mathrm{mL}$ of the stated ethyl acetate stock with $40\mathrm{mL}$ of the sodium hydroxide stock would give $a=0.01333\dots$ and $b=0.02666\dots \mathrm{mol}{\mathrm{L}}^{-1}$. The difference $[\mathrm{B}]-[\mathrm{A}]=b-a$ stays constant, rather than their ratio. Neither the equal-concentration reciprocal formula nor a pseudo-first-order reaction approximation is justified solely by that twofold excess.

The disappearance rate of a reactant A is its positive net loss of concentration per unit time, $v_{\mathrm{A}}=-d[\mathrm{A}]/dt$. For the fixed-volume, single-chemical reaction setting of reaction rate, consuming $a$ units of A per conversion gives $v_{\mathrm{A}}=av$. The factor belongs to the stoichiometry; stating it is essential when quoting a rate constant. A negative net disappearance rate means that A is accumulating.

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.

For $x_0>0$, separation of variables gives $1/x=kt+1/x_0$, hence $x(t)=x_0/(1+k{x}_{0}t)$. Setting $x(t)=x_0/2$ gives the half-lifeThis is the same reciprocal decay law as a second-order reaction with the specified disappearance rate coefficient. A positive initial value is needed for a positive half-time; $x_0=0$ instead gives the identically zero solution.

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.