Chemical kinetics

Chemical kinetics studies how quickly chemical reactions proceed and how their reaction rates depend on composition and conditions. It connects a measured change in concentration to a rate law, then uses a rate equation to predict future concentrations. This makes it useful for choosing how long to wait for conversion and for testing a proposed reaction mechanism. Changing temperature, reactant concentrations, or the presence of a catalyst can change reaction rates; see Flowers et al., Chemistry 2e, §12.2: factors affecting reaction rates.

Chemical equilibrium specifies the eventual balance under given constraints; chemical kinetics determines how rapidly it is approached. At fixed temperature and pressure, a decrease in Gibbs free energy favours conversion but does not guarantee a fast chemical reaction: reaching a transition state can require crossing a substantial barrier. Thus a mixture can change imperceptibly over an observation period without being at chemical equilibrium. The distinction separates which direction is favoured from how long conversion takes. See 12.5 Collision Theory (§12.5: collision theory and reaction barriers) and Chemistry 2e, §16.1: spontaneity and speed.

For a one-dimensional channel, $k(T)Q_r=(2\pi \hslash {)}^{-1}{\int }_0^{\infty }{e}^{-\beta E}P(E)dE$. The reactant partition function and any length normalization must be specified.

A time correlation $C_{fs}(t)=⟨\delta (s_0){\dot{s}}_0\Theta (s_t)⟩$ correlating signed initial flux through a dividing surface with product-side occupation at a later time. Dividing by reactant population gives a rate on a plateau after microscopic recrossing and before macroscopic equilibration.

For competing irreversible pseudo-first-order channels with rates $k_j$, the fraction following channel $j$ is $f_j=k_j/{\sum }_l{k}_l$ when those conditions stay fixed during the intermediate's lifetime. A photochemical destruction rate must count only branches that actually remove the chosen species after fast recycling. For excited oxygen, collisional deactivation followed by ozone reformation is different from reaction with water to make hydroxyl radicals.

An approximation setting the time derivative of a transient population or intermediate concentration to zero when production and loss balance on the timescale considered. It differs from thermal equilibrium.

For $dX/dt=P-kX$ with locally fixed coefficients, $X(t)=P/k+(X_0-P/k)e^{-kt}$. Setting $dX/dt\simeq 0$ requires $1/k$ to be short compared with changes in production, loss and transport. A major reservoir can fail this test even when nearby radicals rapidly adjust.

A set of species interconverted rapidly relative to the production or removal of the set as a whole. Fast interconversion can justify a concentration ratio without establishing that the total family concentration is in steady state.

In a simple stratospheric model, hydroxyl and hydroperoxyl radicals interconvert rapidly. Their total abundance is controlled by slower sources and termination reactions. Atomic hydrogen can be included in a broader family definition when its chemistry is treated explicitly.

For $X\rightleftharpoons Y$ with forward and reverse first-order rates $a,b$, the closed-family partition relaxes at rate $a+b$. Adding losses changes the relaxation eigenvalues. If family abundance evolves much more slowly, a quasi-equilibrium ratio can be useful; setting the entire family derivative to zero needs a separate timescale comparison.

Persistence over the observation time because reaction pathways have sufficiently large barriers. It does not require the state to be the thermodynamic minimum.

Product selection governed by equilibrium free energies after reversible interconversion, in contrast to kinetic control before equilibration.

Approximation that formation and consumption of selected reactive intermediates nearly balance after an initial transient. It does not set their concentrations to zero or imply equilibrium of the stable reagents.

A feedback in which heat release raises temperature sufficiently to accelerate further heat release. Its occurrence depends on kinetics, heat capacity, transport and depletion; exothermicity alone does not prove an inevitable explosion.

Association stabilized by energy transfer to another collision partner. Counting removed reactive intermediates requires their stoichiometric factors, not just the event rate.

A rapid temperature perturbation that shifts equilibrium, allowing relaxation toward the new state to reveal kinetic processes.

The first-order rate constant for dissociation of a specified complex. For a single exponentially dissociating state its mean lifetime is the reciprocal of this constant.

The rate constant for formation of an associated complex, with units of concentration inverse times time inverse for a bimolecular elementary step.

A kinetic isotope effect is a change in a reaction rate constant caused by isotopic substitution. Solvent isotope effects can constrain proton-transfer pathways, but coupling and kinetic masking prevent interpreting every rate ratio as one isolated bond-breaking event.

An isotope effect from a retained labeled bond adjacent to the reaction centre, often influenced by hyperconjugation and its geometry.

An isotope effect from a retained labeled bond at the reaction centre, often reflecting changes in hybridization and bending vibrations.

Under kinetic control, the observed product distribution reflects competing formation rates rather than an equilibrated distribution. Heating may speed access to a lower-Gibbs free energy product, but does not itself prove that the equilibrium favors it.

Collision theory estimates a gas-phase reaction rate by combining encounter frequency with energetic and geometrical requirements for reaction. For distinct hard spheres with contact separation $R$, the collision cross section is $\pi R^2$, and the mean relative speed is $\sqrt{8kT/(\pi \mu )}$. A reactive barrier and orientation restrictions reduce the fraction of collisions that react.

A small increase in temperature can produce a large increase in a reaction rate. In the simplest Arrhenius equation model, warming from $300$ to $310\mathrm{K}$ increases a rate constant with $E_a=80\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ by about $2.81$ times, even though the absolute thermodynamic temperature rises by only about $3.33\%$. This sensitivity helps explain why cooling can slow unwanted chemical reactions and why a hot process can become difficult to control. The comparison assumes the same rate law and reactant concentrations; the activation energy characterizes $k(T)$, rather than every change in the measured reaction rate.

The physical motivation is the distribution of molecular energies. Warming changes the Boltzmann distribution, increasing the relative population of energetically costly configurations. A transition state can therefore become much more accessible than the modest change in average kinetic energy might suggest. The exponential function in an Arrhenius equation captures such sensitivity in a simple model; it is not a universal formula for the fraction above every barrier in a Maxwell-Boltzmann distribution.^[ref]

Activation energy quantifies how strongly a rate constant depends on thermodynamic temperature. For a positive rate constant $k(T)$, the empirical definition iswhere $R$ is the molar gas constant and $k_{\mathrm{ref}}>0$ is a fixed reference with the same units as $k$. The ratio is a dimensionless quantity, keeping the natural logarithm meaningful without changing its derivative. Thus $E_a$ is a molar energy, commonly in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ or $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The Arrhenius equation makes $E_a$ constant over its assumed range. This rate-based definition, and the need for care when identifying it with a threshold on a potential energy surface, are stated in the activation energy.

In a simple barrier picture, reaching a transition state costs energy relative to the starting reactants. The height of that barrier is different from the difference between the reactants and reaction products. An energetically favourable endpoint can still be reached slowly. An energy profile makes the distinction visible, while an Arrhenius plot tests temperature sensitivity from rate constants. A measured $E_a$ need not equal a barrier read from an arbitrarily chosen diagram.

For a computed example, assume a first-order reaction has $k(300\mathrm{K})=0.100{\mathrm{s}}^{-1}$, constant $E_a=80\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, and a temperature-independent pre-exponential factor. The Arrhenius equation givesHence $k(310\mathrm{K})\approx 0.281398{\mathrm{s}}^{-1}$. Its half-life falls from $\ln 2/0.100\approx 6.93147\mathrm{s}$ to $\ln 2/0.281398\approx 2.46322\mathrm{s}$. These are chosen model parameters, not measurements of an identified chemical reaction. The exponent uses kelvin, not unconverted Celsius temperatures, and $80000\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ must match $R$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$.

A microscopic reaction energy barrier belongs to a specified path on a potential energy surface. An experimental activation energy belongs to a specified rate constant and its temperature dependence. A multi-step reaction mechanism can change which step controls the observed conversion, and a temperature-dependent pre-exponential factor contributes to the observed activation energy. Even the Gibbs energy of activation of transition state theory is a different quantity: it incorporates entropy as well as enthalpy and depends on the standard states. These distinctions prevent a fitted Arrhenius plot from being mistaken for a direct microscopic energy measurement.^[ref]

Catalysis illustrates the separation between a path and its endpoints. At the same temperature and with the same reactants, reaction products, and standard states, a regenerated catalyst changes the available reaction mechanism and rate constants, while the overall standard change in Gibbs free energy and the equilibrium constant remain the same:For a reversible first-order reaction pair $\mathrm{A}\rightleftharpoons \mathrm{B}$, consistency at chemical equilibrium requires $k_f/k_r=K$. A model that multiplies only $k_f$ by a catalytic speedup while holding $k_r$ fixed would change this ratio and misrepresent catalysis. This simple ratio applies to that specified rate law; a multi-step catalytic reaction mechanism needs its own rate equations. A potential-energy profile alone cannot determine $K$, because its endpoint difference is not generally the standard change in Gibbs free energy.^[ref]

An apparent activation energy reports the temperature sensitivity of an observed rate constant, without assuming that it equals one microscopic reaction energy barrier. The chain rule converts the empirical activation energy definition into the local slope of an Arrhenius plot:The fixed reference $k_{\mathrm{ref}}$ keeps the natural logarithm dimensionless. A fitted straight-line slope estimates this quantity over the chosen temperature interval; curvature means that one number does not describe every local slope.^[ref]

For the explicit model $k(T)=A(T)\exp [-E_0/(RT)]$ with constant exponent parameter $E_0$, differentiation giveswhere $T_{\ast }$ is fixed. The second term demonstrates why a microscopic interpretation needs more than a slope. If $E_0$ were itself allowed to depend on temperature, differentiation would also contribute $-Td{E}_0/dT$; simply inserting a variable $E_0(T)$ and naming it the measured activation energy is generally incorrect.

An instructive transition state theory example for an effective first-order reaction isHere $k_{\mathrm{B}}$ and $h$ are the Boltzmann constant and Planck constant; $\Delta S^{‡\circ }$ is the entropy of activation, and $\Delta H^{‡\circ }$ the enthalpy of activation. With those two quantities constant over the chosen range and successful passage assumed, direct differentiation gives $E_{a,\mathrm{app}}=\Delta H^{‡\circ }+RT$, while the Gibbs energy of activation is $\Delta G^{‡\circ }=\Delta H^{‡\circ }-T\Delta S^{‡\circ }$. These are three distinguishable quantities. The stated first-order reaction and fixed standard states matter when comparing these formulas with measurements.^[ref]

The Arrhenius equation models the temperature dependence of a rate constant byHere $A$ is the pre-exponential factor, $E_a$ the activation energy, $R$ the molar gas constant, and $T$ the absolute thermodynamic temperature. In its original form, $A$ and $E_a$ are held constant over the range being modeled. It is an empirical relationship, not a universal rule for every reaction mechanism. See the Arrhenius equation.

Eliminating $A$ between two positive temperatures gives the useful predictionFor $E_a>0$, warming increases $k$. The logarithmic sensitivity is $d\ln (k/k_{\mathrm{ref}})/dT=E_a/(R{T}^2)$, so there is no universal doubling per ten degrees. The same $10\mathrm{K}$ rise from $300$ to $310\mathrm{K}$ gives a factor $1.67749$ for $E_a=40\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, versus $2.81398$ for $80\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$.

The useful simplification is that a strong change in reaction rate becomes a straight-line relation after changing variables. Measurements at several temperatures can estimate the slope and test whether one pair of constants describes the chosen range. Curvature can indicate a failure of that approximation; fitting a line does not uniquely establish a microscopic reaction mechanism.

Svante Arrhenius's 1889 paper on sucrose conversion through catalysis by a chemical acid compared the strong observed temperature dependence with the much smaller increase in collision frequency. He discussed van't Hoff's earlier temperature relations and proposed a small population of hypothetical active sucrose, replenished from ordinary sucrose, to interpret the dependence. The inspected translated excerpt derives an inverse-temperature relation in its equations (6)–(7) and then applies it to the reaction rate. That historical hypothesis is distinct from a modern transition state description. See the On the Reaction Velocity of the Inversion of Cane Sugar by Acids. The excerpt omits parts of the original paper; it does not provide a complete original apparatus description or dataset.

The pre-exponential factor must have the same units as the rate constant, since the exponential function supplies a dimensionless quantity. For a rate law $v=k[\mathrm{A}{]}^m$, both $A$ and $k$ have units $(\mathrm{mol}{\mathrm{L}}^{-1}{)}^{1-m}{\mathrm{s}}^{-1}$. Thus a first-order reaction uses ${\mathrm{s}}^{-1}$, whereas a second-order reaction uses $\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$. Calling $A$ a frequency factor does not give it frequency units for every order of reaction. Its value includes the physical and geometrical factors allowed by the chosen reaction mechanism; equal activation energies need not mean equal reaction rates.

To read the blue Arrhenius plot, define the dimensionless quantity $X=1000\mathrm{K}/T$ and $y=\ln [k/k(300\mathrm{K})]$. Its numerical slope is $m_X=-60000/(1000R)\approx -7.21634$, giving $E_a=-(1000\mathrm{K})R{m}_X=60\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The slope against $1/T$, by contrast, is $-7216.34\mathrm{K}$. The orange tangent at $300\mathrm{K}$ gives $E_{a,\mathrm{app}}\approx 64.9887\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$; its finite-interval slope between $280$ and $340\mathrm{K}$ gives about $65.1227\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The distinction between a local derivative and a fitted interval matters even with perfectly noise-free synthetic evaluations.

Two measured positive rate constants give a two-point estimateSeveral temperatures let linear regression test the constant-parameter approximation: inspect the regression residuals, keep the same rate law and composition convention, and account for uncertainty in both $k$ and $T$. A straight line over a short range supports a useful local approximation; it does not uniquely identify a reaction mechanism. A fixed change of units or of $k_{\mathrm{ref}}$ shifts the intercept without changing the inferred activation energy.^[ref]

For a constant-parameter Arrhenius equation $R(T)=A{e}^{-\beta /T}$ at positive absolute temperatures, subtracting logarithms eliminates the Arrhenius prefactor. The ratio of the two resulting equations gives the displayed prediction. It applies to a reaction rate only when composition and other factors relating rate to the rate constant are held fixed. Equal increments of temperature do not give equal multiplicative rate changes, since the linear variable is $1/T$.

The pre-exponential factor multiplies the exponential function in the Arrhenius equation. It has the same units as the rate constant, so for a first-order reaction both have units ${\mathrm{s}}^{-1}$. Its physical interpretation depends on the reaction mechanism; an assumed constant $A$ should not be identified with a universal collision frequency. Comparing activation energies alone does not rank absolute rate constants when the pre-exponential factors differ. See the pre-exponential factor.

An Arrhenius plot places $\ln (k/k_{\mathrm{ref}})$ against $1/T$, using a fixed reference rate constant with the same units as $k$. Under the constant-parameter Arrhenius equation,The slope is $-E_a/R$ and the intercept is $\ln (A/k_{\mathrm{ref}})$. Changing the reference units shifts the intercept but leaves the slope unchanged. If the horizontal coordinate is displayed in $10^{-3}{\mathrm{K}}^{-1}$, its numerical slope must be rescaled before multiplying by $-R$. A temperature-dependent pre-exponential factor or activation energy can make the plot curved.

The Modified Arrhenius equation allows a power of thermodynamic temperature in the pre-exponential factor:Here $A_{\ast }$, $E_0$, $n$, and $T_{\ast }$ are constant model parameters. Using the dimensionless quantity $T/T_{\ast }$ leaves $A_{\ast }$ with the same units as $k$; in the equivalent convention $B{T}^n\exp [-E_0/(RT)]$, $B$ instead has units of $k$ multiplied by ${\mathrm{K}}^{-n}$. The power exponent $n$ here is not an order of reaction. The IUPAC definition uses that equivalent $B{T}^n$ convention.^[ref]

Applying the apparent activation energy definition yields the original short calculationFor the orange synthetic Arrhenius plot, $n=2$, $T_{\ast }=300\mathrm{K}$, and $E_0=60\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Its pre-exponential factor changes by $(310/300{)}^2\approx 1.06778$ from $300$ to $310\mathrm{K}$; combined with the exponential function, $k(310\mathrm{K})\approx 0.231991{\mathrm{s}}^{-1}$, compared with $0.217266{\mathrm{s}}^{-1}$ for the constant-pre-exponential factor model. The diagram uses those models only to reveal the difference between an exponent parameter and a slope-derived activation energy.

A concrete laboratory route to an Arrhenius plot is to infer a rate constant at each controlled temperature from a changing electrical conductivity. The Technical University of Munich's Activation Energy of a First-Order Reaction supplies a teaching protocol, rather than a published set of measured rate constants.

The manual follows tertiary amylbromide conversion, which produces conducting ions. A measuring cell connects to a thermostat; an immersed electrical conductivity probe also measures solution temperature, and a magnetic stirrer mixes it. Use $80\mathrm{mL}$ of $80\%$ by volume ethanol in water, equilibrate, then add $0.5\mathrm{mL}$ of tertiary amylbromide. Record the baseline before addition and set time zero at addition. Repeat with fresh mixtures at $25$, $30$, and $35{}^{\circ }\mathrm{C}$, each within $\pm 0.5{}^{\circ }\mathrm{C}$; record for $30$ minutes at the first two temperatures and until completion at the last. Clean the cell between runs and keep the added volume fixed. The completed $35{}^{\circ }\mathrm{C}$ mixture supplies endpoint electrical conductivity measurements at the other temperatures: one endpoint reading cannot be used unchanged at every temperature.^[ref]

To see what is inferred from the readout, let ${\sigma }_0$, $\sigma (t)$, and ${\sigma }_{\infty }$ be the baseline, current, and completed-mixture electrical conductivities at the same temperature. Assume their baseline-subtracted changes are proportional to converted concentration. Combining that measurement model with a first-order reaction givesThus the electrical conductivity increases toward an endpoint, while the remaining reactant fraction decreases. Fit $-\ln f$ against time to obtain $k$; then fit $\ln (k/k_{\mathrm{ref}})$ against $1/T$ in kelvin to estimate an apparent activation energy. This baseline-aware derivation states the measurement assumptions explicitly; it does not infer a rate constant from the steepness of a raw conductivity trace alone.

The controls distinguish a faster chemical reaction from a larger instrumental response. Measure the actual solution temperature, retain the same solvent composition and initial reactant concentration, and compare endpoint and baseline at that temperature. As further diagnostic checks, repeat runs and inspect regression residuals for departure from a first-order reaction. These checks follow from the model: electrical conductivity also depends on how ions move, a baseline drift can mimic conversion, and a mixing delay can distort early points. Near completion, a small endpoint error produces a large relative error in $f$ and its natural logarithm. A successful fit supports an apparent rate model for this solvent and interval; it does not directly image a transition state or identify a unique reaction mechanism. No measured $k$ or $E_a$ is supplied here.

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.

A curved arrow whose tail marks an electron pair and whose head marks its destination in a proposed reaction step. The pair may start as a lone pair or a bond pair; a full arrowhead represents two electrons.

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.

A parallel first-order reaction mechanism consumes the same reactant through competing routes. Extend the product-split idea in catalytic selectivity by following both reaction products over time. Stipulate irreversible first-order reactions $\mathrm{A}\stackrel{k_P}{\to }\mathrm{P}$ and $\mathrm{A}\stackrel{k_Q}{\to }\mathrm{Q}$ in a uniformly mixed closed system of fixed volume and temperature. Initially only A is present, with concentration $A_0$. With $K=k_P+k_Q$,Here $K$ abbreviates a sum of rate constants, not an equilibrium constant. The rate equations are $-d[\mathrm{A}]/dt=K[\mathrm{A}]$, $d[\mathrm{P}]/dt=k_P[\mathrm{A}]$, and $d[\mathrm{Q}]/dt=k_Q[\mathrm{A}]$. Integrating the last two after solving the first proves the expressions and their conserved total $[\mathrm{A}]+[\mathrm{P}]+[\mathrm{Q}]=A_0$.

Choose two alternative models: model U has $(k_P,k_Q)=(0.30,0.10){\mathrm{s}}^{-1}$; model V has $(0.10,0.30){\mathrm{s}}^{-1}$. Both have $K=0.40{\mathrm{s}}^{-1}$, so every measurement of A alone is identical, even with no measurement error. Yet their P-to-Q concentration ratios at any $t>0$ are $3$ and $1/3$, respectively. This is a precise failure of identifiability from that one observable. A calibrated measurement distinguishing P from Q resolves it within the stipulated two-route model.

At $t=5.00\mathrm{s}$, both models give $[\mathrm{A}]=e^{-2}{A}_0\approx 0.135335\mathrm{mmol}{\mathrm{L}}^{-1}$. Model U gives $([\mathrm{P}],[\mathrm{Q}])\approx (0.648499,0.216166)\mathrm{mmol}{\mathrm{L}}^{-1}$; model V exchanges these. Measuring only the summed reaction product concentration would also fail to distinguish them. Product-specific observations therefore answer a question that total conversion cannot. As with any reaction mechanism, fitting these observables still does not rule out a more elaborate untested route.

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.

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 tetrahedral intermediate has a formerly trigonal carbonyl group carbon bonded to four groups after nucleophilic addition. Its subsequent elimination can restore a carbonyl group.

A reaction coordinate parametrizes changes in molecular configuration along a chemical reaction path. It can involve a bond length, a bond angle, or a combination of geometric changes. It is not elapsed time and is not the bulk extent of reaction: one describes configurations, while the other accounts for how much material has converted. The precise choice matters when interpreting an energy profile. See the IUPAC Gold Book: reaction coordinate.

A hypersurface $s(q)=0$ separating selected reactant and product regions. A rate derived from signed flux and a suitable long-time commitment is independent of its location when the dynamics and normalization are treated consistently.

A potential energy surface assigns potential energy to the configurations of the reacting chemical atoms, for a specified electronic description. A reaction coordinate selects a path through this many-dimensional configuration space; a potential-energy profile then records the potential energy along that path. Different paths can share endpoints but cross different barriers. See the IUPAC Gold Book: potential-energy reaction surface.

A correlation between many-electron states classified by total spin and spatial symmetry along a nuclear path. Orbital occupations, spin adaptation and configuration interaction are needed; one-electron energy curves alone are not a state diagram.

A degeneracy between adiabatic electronic states with characteristic splitting in suitable nuclear-coordinate directions. It can connect to Jahn-Teller dynamics in open-shell triangular molecules.

An energy profile plots an explicitly specified energy against a reaction coordinate. Identify whether the vertical axis is potential energy, enthalpy, or Gibbs free energy; these descriptions are related but not interchangeable. A potential-energy profile describes a path through configurations, while a free-energy description also accounts for the relevant thermodynamic conditions.

For the two chosen profiles, the reaction energy barriers from the reactants are $80$ and $50\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. From the reaction products, the same maxima are $100$ and $70\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ higher. Both paths have endpoint difference $-20\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The barrier is measured from the relevant starting endpoint, not from an arbitrary zero of potential energy.

The appeal of the diagram is that it separates two useful questions: which endpoint is lower, and which path is easier to traverse? Catalysis changes the route through transition states while retaining the endpoint thermodynamics of the same overall chemical reaction. A real catalytic route may have several steps and reaction intermediates; the two simple curves do not specify such a reaction mechanism. The plotted potential-energy difference alone is not an equilibrium constant or a calculated rate constant. See Chemistry 2e, 12.7: alternative reaction routes and catalysis.

A plot of Gibbs free energy along a specified reaction coordinate under stated conditions. A barrier must be measured from the appropriate starting minimum.

A potential-energy profile plots the potential energy of the reacting system along a specified path from reactants through a transition state to reaction products. It is a particular kind of energy profile. For an elementary reaction, a suitable reaction coordinate parametrizes the path; a multi-step reaction mechanism needs the coordinates of its successive steps. See the IUPAC Gold Book: potential-energy profile.

A reaction energy barrier is the height of a specified peak relative to the starting state in an energy profile. State which energy is used and which direction is being considered. A schematic barrier is not automatically the empirical activation energy obtained from the temperature dependence of a rate constant.

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.

Molecularity counts the reacting entities in one elementary reaction: one for a unimolecular step, two for a bimolecular step. An entity can be a molecule, ion, or chemical atom. Solvent entities involved only through solvation are excluded. Molecularity describes a microscopic event; order of reaction describes the concentration dependence of a rate law. Assigning molecularity to a net multi-step chemical equation is therefore inappropriate.^[ref]

Catalysis accelerates a chemical reaction by providing an alternative reaction mechanism with a more accessible barrier through its transition states. A catalyst participates in the route and is regenerated. At fixed temperature, catalysis changes how quickly chemical equilibrium is reached without changing the equilibrium constant; it does not make an unfavourable overall change in Gibbs free energy favourable. ^[ref][ref]

In his 1909 lecture, Wilhelm Ostwald recalled the introduction of the term catalysis by Jöns Jacob Berzelius and emphasized measurable acceleration rather than merely naming an unexplained force. He also discussed reaction intermediates and participation of the catalyst in the steps whose sum gives the net chemical reaction. This is a participant's retrospective account, not evidence that every catalytic reaction mechanism is the same.^[ref]

Hydrogen-bond catalysis preferentially stabilizes a reactive configuration through hydrogen bonds. A hydrogen bond to an intermediate or bound substrate is evidence of an interaction, but its catalytic contribution depends on differential stabilization relative to the starting state.

Metal ion catalysis uses a coordinated metal ion to activate a reagent, orient reactants or stabilize charge. Zinc can lower the acidity constant of coordinated water and support attack by the resulting oxygen nucleophile without a metal redox step.

Proximity and orientation catalysis favors reaction by bringing partners together in a productive geometry. A binding interaction is catalytic only insofar as it improves access to the reactive configuration relative to the starting state.

Electrostatic catalysis preferentially stabilizes developing charge along the reaction pathway. The comparison must be with the bound reactant: equally stabilizing ground and transition states does not lower their activation free-energy difference.

Covalent catalysis creates a temporary covalent bond between catalyst and substrate, then breaks that bond while regenerating the catalyst. Merely forming a substrate–water bond does not constitute covalent catalysis by an enzyme residue.

General base catalysis uses a base to accept a proton during the reacting pathway, for example activating water while it forms a bond. It need not involve an equilibrium pool of free hydroxide.

General acid catalysis uses proton donation from an acid participating in the rate-influencing reaction pathway. It can activate an electrophile or assist protonation of a departing or newly formed group.

Transition state stabilization lowers an activation barrier when a catalytic environment stabilizes the reacting configuration more than the starting bound state. Stronger ground-state binding alone does not demonstrate catalysis.

A catalytic group that accepts a proton during a reaction step, increasing the reactivity of another group such as a hydroxyl nucleophile.

A catalytic cycle follows the changing forms of a catalyst through a reaction mechanism until its starting form is regenerated. The schematic binding cycle in catalyst illustrates the bookkeeping: C is consumed and recovered, whereas AC is a reaction intermediate formed and consumed. Cancelling both gives the net A-to-B chemical equation. Regeneration concerns the completed cycle; it does not imply that C avoids forming chemical bonds or exchanging electrons. An alternative reaction mechanism may have more elementary reactions than the uncatalysed route and still be faster.^[ref]

For that reversible binding cycle, ideal thermodynamic activities give step equilibrium constants $K_1=a_{\mathrm{AC}}/(a_{\mathrm{A}}{a}_{\mathrm{C}})$ and $K_2=a_{\mathrm{B}}{a}_{\mathrm{C}}/a_{\mathrm{AC}}$. Their product is $K_1{K}_2=a_{\mathrm{B}}/a_{\mathrm{A}}=K$: the same cancellations occur in the equilibrium bookkeeping. Choose consistent standard states for the species. A proposed cycle with $K_1{K}_2\ne K$ for the same net conversion at the same temperature would contradict its specified thermodynamics. This statement concerns free-species thermodynamic activities. Appreciable binding in AC can change how the conserved total amount is partitioned among free A, free B and bound AC. An assay counting bound and free material together therefore needs that additional balance; unchanged $a_{\mathrm{B}}/a_{\mathrm{A}}$ does not by itself guarantee identical analytical totals when the amount of C changes.

A directional Gibbs energy of activation measures the climb from a starting state to a transition state. The net standard reaction change in Gibbs free energy measures the difference between endpoints. Changing the climb can strongly change a rate constant while leaving that endpoint difference intact. For two illustrative single-barrier routes with common endpoint levels $G_{\mathrm{A}}^{\circ }$ and $G_{\mathrm{B}}^{\circ }$,Lowering the common transition state level lowers both directional Gibbs energies of activation; it does not lower just the forward climb. The usual statement that a catalyst lowers an activation energy describes a kinetic route, not a change in the net Gibbs free energy. An activation energy inferred from an Arrhenius plot is also distinct from a Gibbs energy of activation at one temperature.

With a common prefactor in transition state theory, subtracting the directional barriers yieldswhere $R$ is the molar gas constant and $T$ is the absolute temperature. This particular rate constant ratio requires the simple compatible directional rate laws and consistent standard states; a general multistep reaction mechanism does not supply two arbitrary measured rate constants whose ratio must equal $K$. At fixed temperature a catalyst preserves $K$ for the same net conversion. The actual reaction change in Gibbs free energy, ${\Delta }_{r}G={\Delta }_r{G}^{\circ }+RT\ln Q$, is likewise unchanged at the same reaction quotient $Q$. Composition changes during the faster approach to chemical equilibrium, so comparing mixtures at the same elapsed time need not compare the same $Q$.^[ref]

Here is an explicitly stipulated elementary kinetic model, not experimental data. A well-mixed ideal closed system has constant temperature and volume. Alongside an uncatalysed elementary reaction, allow an illustrative catalytic encounter that returns C immediately:Both channels obey the law of mass action. In the second, the forward reaction rate is ${\alpha }_f[\mathrm{A}][\mathrm{C}]$ and the reverse reaction rate is ${\alpha }_r[\mathrm{B}][\mathrm{C}]$. C cancels in each encounter, so its free concentration $c_C$ stays constant without an external supply. This toy elementary reaction encodes regeneration without binding; it is not a claim that real catalytic cycles universally consist of one encounter. A binding cycle can accumulate a reaction intermediate, in which case constant total C does not imply constant free $c_C$.

Adding the parallel reaction rates gives effective first-order reaction rate constantsBecause $k_f^u/k_r^u={\alpha }_f/{\alpha }_r=3$, each channel balances at the same $[\mathrm{B}]/[\mathrm{A}]=3$, and so does their sum. At $c_C=0$ the pair is $(0.030,0.010){\mathrm{s}}^{-1}$; at $c_C=0.001\mathrm{mol}{\mathrm{L}}^{-1}$ it is $(0.300,0.100){\mathrm{s}}^{-1}$. Both give $K=3$ with ideal thermodynamic activities and a common concentration standard state. At $T=300\mathrm{K}$ their compatible standard reaction change in Gibbs free energy is ${\Delta }_r{G}^{\circ }=-RT\ln 3\approx -2.7403\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Choosing an unrelated reverse rate constant would break this consistency.

Let $S=[\mathrm{A}]+[\mathrm{B}]$ be the conserved concentration, and define the dimensionless quantity $x=[\mathrm{B}]/S$. ThenSolving this first-order linear differential equation givesThe chemical relaxation time is $25\mathrm{s}$ without C and $2.5\mathrm{s}$ with the chosen C. From either $x_0=0$ or $x_0=1$, the exponential decay of the deviation leads to the same chemical equilibrium $x=3/4$. Reducing that deviation to one percent takes $\ln (100)\tau$: about $115.13\mathrm{s}$ versus $11.51\mathrm{s}$.

At this chemical equilibrium, conversion continues in both directions. For $S=1\mathrm{mol}{\mathrm{L}}^{-1}$, the uncatalysed channel has equal forward and reverse reaction rates $0.0075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. Adding C gives total directional reaction rates $0.075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ each, yet zero net reaction rate. Faster molecular exchange and unchanged composition coexist. This exact calculation explains how catalysis can matter enormously within an observation window while leaving the final equilibrium constant intact.

Catalytic selectivity concerns which reaction products form when competing reaction mechanisms are available. Faster conversion is valuable when a process would otherwise be impractically slow; a suitable catalyst can also favour a useful product within the available time. In an illustrative pair of irreversible first-order reactions A to B and A to D, the initial fraction of converted A going to B is $k_B/(k_B+k_D)$. Multiplying both rate constants by the same factor changes speed but leaves this fraction unchanged; changing their relative values changes the kinetic product split. This is a finite-time kinetic illustration, not a new equilibrium constant.

A catalyst need not accelerate every competing chemical reaction equally. Enzymes, for example, can distinguish substrates through their active sites.^[ref] If every relevant reversible channel eventually equilibrates in a closed system, its final chemical equilibrium still obeys the endpoint thermodynamics. Selective acceleration, product removal, or external driving can make an observed product distribution depend on time and operating conditions. A claim of improved yield must specify those conditions.

Catalyst deactivation is loss of catalytic activity during use. Regeneration in an ideal catalytic cycle does not rule out a separate damaging chemical reaction, blocking of a surface, or physical loss of catalyst from the apparatus. A route can also be slowed by diffusion to the surface or by accumulation of a reaction intermediate. Such effects limit the usefulness of a low schematic reaction energy barrier: measured throughput depends on the whole reaction mechanism and on transport. Demonstrating regeneration requires more evidence than observing a faster first run. For a recovered solid catalyst, a useful further comparison is a second run with fresh reactants, matched recovered amount, and the same mixing and temperature, alongside a fresh-solid run and a blank. Loss during recovery or residual reactants can otherwise imitate catalyst deactivation or apparent reuse. This is a proposed control, not an observation reported for the catalysed peroxide comparison.

In homogeneous catalysis, the catalyst and reactants occupy the same thermodynamic phase.

For a liquid example, specify hydrogen peroxide as the reactant, dissolved iodide as the catalyst, and water as the solvent. If both reacting chemical species remain dissolved in that one liquid thermodynamic phase, this is homogeneous catalysis. The University of Wisconsin–Madison's The Rate of a Reaction identifies iodide-catalysed hydrogen peroxide decomposition in its final example; the aqueous setting here makes the phase roles explicit.^[ref] Its net chemical equation isThe molecular oxygen is a reaction product: its bubbles do not make the dissolved catalytic route heterogeneous catalysis. The classification compares the participating catalyst and reactants, rather than demanding that everything in the vessel share a phase.^[ref]

Being in the same thermodynamic phase describes where the chemical species are; it does not specify their reaction mechanism. A uniform-looking liquid can contain unresolved solid particles, so appearance alone is insufficient. Conversely, suspended solid particles do not establish that their surfaces supply the observed catalysis: dissolved material released from them could contribute. Phase identification and catalytic participation need separate evidence.

The bookkeeping of a catalyst in homogeneous catalysis is particularly transparent in the schematic catalytic cycle A plus C forming AC, followed by AC releasing B plus C. This is the generic model in catalyst, not an asserted iodide reaction mechanism. Suppose a fixed volume initially contains $1.00\mathrm{mmol}$ of A and $0.20\mathrm{mmol}$ of C. At a chosen snapshot, let $0.60\mathrm{mmol}$ of B and $0.12\mathrm{mmol}$ of AC be present, with no other chemical reactions. Each AC contains one A unit and one C unit, soThus free A is $0.28\mathrm{mmol}$ and free C is $0.08\mathrm{mmol}$. B formation already exceeds the initial amount of substance of C by a factor of three, while the total C inventory remains $0.20\mathrm{mmol}$. Adding the two steps cancels C and AC, giving the net A-to-B chemical reaction; counting their constituents proves these balances. Conserved total catalyst does not mean constant free catalyst concentration.