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A rate law specifies how a reaction rate depends on concentrations under stated conditions. A common form iswhere $k$ is a rate constant and the exponents are partial orders of reaction. Determine them experimentally, or derive them from a justified reaction mechanism. Do not infer $\alpha =a$ and $\beta =b$ from the net chemical equation: its coefficients describe consumption ratios, whereas the exponents describe sensitivity to concentration. See IUPAC Gold Book: rate law and Flowers et al., Chemistry 2e, §12.3: rate laws.

A rate law becomes a prediction of time evolution when combined with stoichiometry in a rate equation, for example $d[\mathrm{A}]/dt=-av$. Agreement with a measured rate law is useful evidence for a reaction mechanism, but several routes can give the same dependence.

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.

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.

Protein kinase activity adds modification, while protein tyrosine phosphatase activity can remove it. An original one-site model shows why a receptor phosphorylation assay cannot determine either process from a single modified fraction. Let $f(t)\in [0,1]$ be the fraction of a fixed receptor population carrying one selected phosphotyrosine residue. Stipulate constant effective first-order rate constants $k_{+},k_{-}>0$, with ATP availability absorbed into $k_{+}$, and no receptor synthesis, degradation or trafficking:Here $f_{\ast }$ is the steady state fraction and $\tau$ the time constant. The ordinary differential equation has the exponential function solutionThe balance equation and a single relaxation step give this result; it is an illustrative model, rather than a fit or a claim about every RTK.

With stipulated rate constants $(k_{+},k_{-})=(2,1){\min }^{-1}$, $f_{\ast }=2/3$ and $\tau =1/3\min$. Doubling both rate constants leaves $f_{\ast }=2/3$ but changes $\tau$ to $1/6\min$. The balanced phosphate turnover per site,doubles from $2/3$ to $4/3{\min }^{-1}$. Identical steady state modification therefore permits different catalytic fluxes. Removing the addition term instead gives $f(t)=f(0)e^{-k_{-}t}$; this illustrates why a perturbation that stops protein kinase activity can expose removal dynamics, subject to the model's unchanged-removal assumption. A maintained steady state with ongoing ATP consumption is not 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.

An enzyme accelerates particular chemical reactions, with catalytic selectivity arising from recognition and from the relative ease of the subsequent chemistry. Strong binding alone does not guarantee fast conversion. Nor does catalysis change the chemical equilibrium for an unchanged net chemical reaction. These distinctions explain how enzymes can direct the flow of material without creating an extra source of free energy.^[ref]

For an original model of competing routes, let a substrate $S$ become either a desired reaction product $P$ or another reaction product $Q$:The labels represent hypothetical chemical species with the same elemental composition, not molecular structures. Assume fixed volume, irreversible first-order reactions, constant effective rate constants, and initially $100\mu \mathrm{mol}$ of $S$ with neither reaction product present. The model approximates a limited operating regime; real enzyme kinetics can include saturation and reverse reactions.

The reaction-rate equations give $d{n}_S/dt=-(k_P+k_Q)n_S$, $d{n}_P/dt=k_P{n}_S$ and $d{n}_Q/dt=k_Q{n}_S$, where $n$ denotes amount of substance. Consequently,With $k_P=0.08$ and $k_Q=0.02\mathrm{mi}{\mathrm{n}}^{-1}$, the eventual split is $80\mu \mathrm{mol}$ of $P$ and $20\mu \mathrm{mol}$ of $Q$. At ten minutes, the calculated amounts are approximately $36.8$, $50.6$ and $12.6\mu \mathrm{mol}$ for $S$, $P$ and $Q$. Selectively increasing $k_P$ fourfold changes the eventual desired fraction to $0.32/0.34=16/17\approx 94.1\%$. Increasing both rate constants fourfold instead preserves the $80\%$ split and only shortens the conversion time. Speed and selectivity answer different questions.

The bookkeeping matters independently of that model. In ideal alcoholic fermentation, complete conversion of $3\mathrm{mmol}$ of glucose yields $6\mathrm{mmol}$ each of ethanol and carbon dioxide. Of the initial $18\mathrm{mmol}$ of carbon chemical atoms, $12$ remain in ethanol and $6$ enter carbon dioxide. This is a chosen material-balance example using the net stoichiometry explained under Saccharomyces cerevisiae; it gives neither an observed yield nor the mechanism of the individual steps. Equal amounts of the two reaction products also do not mean equal masses.

Transition state theory relates an elementary reaction's rate constant to a Gibbs energy of activation. The Eyring expression used in the catalyst example assumes every passage counted at the transition state succeeds. More generally, a factor accounting for unsuccessful passage multiplies it. Comparing barriers alone gives the quoted speedup only when the remaining prefactors agree. Free-energy profiles can depend on temperature, so the example's values at $300\mathrm{K}$ do not determine a temperature curve. See the IUPAC Gold Book: transition state theory.