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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.

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 succession of radioactive decays in which a daughter can itself be unstable. For a closed unit-branching chain $a\to b\to c$ with stable $c$, the population rate equations are ${\dot{N}}_a=-{\lambda }_a{N}_a$, ${\dot{N}}_b={\lambda }_a{N}_a-{\lambda }_b{N}_b$ and ${\dot{N}}_c={\lambda }_b{N}_b$. They conserve $N_a+N_b+N_c$. Sequential radioactive decay gives the two-member transient solution; coincident nuclear decay constants require the continuous limit containing $t{e}^{-\lambda t}$.

Consider an original reduced rate equation model using the same potentials. Write $s(z)=\cos (2kz)$ and stipulate positive optical pumping rates $R_{+\to -}={\gamma }_p(1-s)$, $R_{-\to +}={\gamma }_p(1+s)$, with ${\gamma }_p>0$. They pump towards $+$ at $z=0$ and towards $-$ at $z=\lambda /4$, consistently with the diagram. These rates retain the relevant symmetry and delay, while omitting excited-state dynamics, quantum coherence and recoil diffusion.

For populations $p_{+}+p_{-}=1$, put $q=p_{+}-p_{-}$. The rate equations conserve population and give $\dot{q}=2{\gamma }_p(s-q)$. The stationary atom has $q=s$; a moving atom cannot follow it instantaneously. On a prescribed trajectory $z=vt$, the periodic solution isSubstitution into $v{q}^{\prime }=2{\gamma }_p(s-q)$ verifies both coefficients. The mean mechanical force is $F=-p_{+}{U}_{+}^{\prime }-p_{-}{U}_{-}^{\prime }=-U_{0}kq\sin (2kz)$. Averaging over a spatial period, with mean squared sine $1/2$ and mean sine times cosine zero, givesThus the force opposes either sign of velocity and vanishes at rest. A small pumping rate makes the small-velocity friction slope large while narrowing its useful velocity range. Holding $v$ fixed while sending ${\gamma }_p$ to zero makes the force vanish; the apparently divergent zero-velocity slope is not unlimited cooling. Because this model has no momentum diffusion, it predicts no equilibrium temperature. It demonstrates the lag mechanism rather than replacing a multilevel quantum calculation.

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.

A first-order reaction has a reaction rate proportional to one reactant concentration in the simple single-species case. For an irreversible $\mathrm{A}⟶\mathrm{B}$ in a closed system at fixed volume, take $[\mathrm{A}](0)=A_0$, $[\mathrm{B}](0)=0$, and constant rate constant $k>0$. Its rate equation and solution areA short check is enough: differentiating $A_0{e}^{-kt}$ gives $-k{A}_0{e}^{-kt}$, and the two concentrations sum to $A_0$, as required by the one-to-one stoichiometry. Thus the same exponential decay describes both remaining reactant and decreasing reaction rate. Fixed $k$ means a fixed fractional loss per unit time, not a fixed absolute consumption rate.

Choose illustrative values $A_0=1\mathrm{mol}{\mathrm{L}}^{-1}$ and $k=0.2{\mathrm{s}}^{-1}$, with $t$ measured in seconds. Then $[\mathrm{A}]=e^{-0.2t}\mathrm{mol}{\mathrm{L}}^{-1}$, $[\mathrm{B}]=(1-e^{-0.2t})\mathrm{mol}{\mathrm{L}}^{-1}$, and $v=0.2e^{-0.2t}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. The half-life isAt each successive half-life, half the remaining reactant is consumed; $t_{1/2}$ is independent of $A_0$ for this first-order reaction. At $t=10\mathrm{s}$, the calculated values are $[\mathrm{A}]\approx 0.135335\mathrm{mol}{\mathrm{L}}^{-1}$, $[\mathrm{B}]\approx 0.864665\mathrm{mol}{\mathrm{L}}^{-1}$, and $v\approx 0.027067\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. These are model predictions for chosen parameters, rather than experimental measurements.

The left panel tracks the two concentrations; the right tracks their instantaneous conversion reaction rate. Their different vertical units matter: $v=k[\mathrm{A}]$, so the numerical curves need not have the same height. This gives a practical way to translate a measured concentration into a forecast of both conversion and waiting time.

Fluorescence quantum yield counts emitted fluorescent photons per absorbed excitation photon. It differs from quantum efficiency, which describes the photodetector.^[ref]

For an original constant-rate model, let $k_{\mathrm{r}}$ be the radiative decay rate and $k_{\mathrm{nr}}$ the sum of competing rates without fluorescent photon emission. An initial population $N_0$ in the emitting excited state obeys the rate equationIntegrating $k_{\mathrm{r}}N(t)$ over all time givesHere every absorbed excitation is assumed to reach this emitting population; additional losses before that stage require another factor. Under steady weak excitation, absorbed photon rate $R_{\mathrm{abs}}$ and total detection fraction $\eta$ give $C=\eta \Phi R_{\mathrm{abs}}$. The count rate therefore mixes excitation, yield, and detection, whereas $\tau$ specifies a decay timescale.

An integrated rate law relates concentration to elapsed time after solving a rate equation with an initial condition. Its validity retains the assumptions of the underlying rate law, including fixed rate constants and appropriate stoichiometry. The second-order reaction reciprocal-concentration relation is one example; a different rate equation generally requires a different integrated rate law.^[ref]

For $\gamma >0$, setting the rate equations to zero gives the unique stationary distribution of this continuous-time Markov chain:The nonzero mean magnetic quantum number represents atomic orientation. Increasing $R/\gamma$ empties the bright state, but leaves two dark populations. For $R=0$, relaxation alone gives $p_i(t)=1/3+[p_i(0)-1/3]e^{-\gamma t}$. When $\gamma =0$ and $R>0$, every distribution with $p_{-}=0$ is stationary; the initial difference between the two dark populations survives. The equal dark populations below follow from the specified unpolarized initial condition.

Starting from $p_i(0)=1/3$, the solution for $R>0$ or $\gamma >0$ isFor pure pumping, take $p_{-}^{\ast }=0$ in this formula. Its redistribution time is $3/(2R)$, longer than the mean scattering interval $1/R$ of a bright atom, because one third of scattering events return it to the same sublevel. For arbitrary initial populations the additional mode satisfies $p_0(t)-p_{+}(t)=[p_0(0)-p_{+}(0)]e^{-\gamma t}$.

For an original numerical example choose $R=6{\mathrm{s}}^{-1}$ and $\gamma =1{\mathrm{s}}^{-1}$, with $\Gamma$ much larger than either. ThenAt $t={\tau }_{\mathrm{pump}}$, $p_{-}\simeq 0.1648$, while $p_0=p_{+}\simeq 0.4176$. If the light is then switched off, relaxation returns all three populations toward $1/3$ with time constant $1/\gamma =1\mathrm{s}$. Faster optical preparation and slower dark recovery measure different combinations of rates.

In an optically thin uniform sample of $N$ atoms, with equal collection efficiency $\eta$ for each decay branch, the detected fluorescence count rate is approximately $S_{\mathrm{fl}}=\eta NR{p}_{-}$. The same scattering removes pump photons, giving ${\Phi }_{\mathrm{out}}\simeq {\Phi }_{\mathrm{in}}-NR{p}_{-}$ when depletion is small. For the chosen example the steady scattering rate is $R/15=0.40{\mathrm{s}}^{-1}$ per atom, compared with the initial $R/3=2{\mathrm{s}}^{-1}$. Pumping lowers this signal fivefold, while increasing transmission. If rapid mixing restores nearly uniform populations, scattering rises again. Reabsorption, strong attenuation, unequal collection efficiencies or coherent RF dynamics require a more detailed model; the reset rate $\gamma$ does not stand for an exact RF Hamiltonian.

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 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.

Rabi oscillations require a coherent drive with a stable phase over several population cycles and an isolated, appreciably coupled pair of quantum states. For large-amplitude oscillations, the Rabi detuning must be comparable to or smaller than the Rabi angular frequency. The population lifetime, optical coherence time, transit time and drive coherence time should all exceed $2\pi /{\Omega }_R$ by enough to resolve the cycles. An ensemble also needs sufficiently uniform Rabi frequencies and Rabi detunings that phase averaging does not wash the oscillations out. The usual sinusoidal formula additionally assumes the rotating-wave approximation.

The Einstein coefficients describe incoherent transition rates, rather than the evolving optical probability amplitudes. Their simplest use is appropriate for broadband, randomly phased radiation, with field correlation time short relative to population evolution, or for rapidly dephased chemical atoms whose optical coherences can be eliminated. The radiation energy density per unit frequency must be smooth over the relevant spectral line. Stimulated upward and downward transitions and spontaneous emission then enter rate equations, without predicting coherent population return. A short observation time can give a small excitation probability even for a coherent field; that condition alone does not justify an Einstein coefficients rate description. Coherent laser irradiation can also admit steady rate equations after optical coherence has relaxed, with the appropriate saturation response.

Define $r=g_2/g_1$ and let $\sigma ={\sigma }_{21}(\omega )$ be the stimulated-emission cross section, averaged over upper magnetic sublevels. The Einstein coefficients relation $g_1{B}_{12}=g_2{B}_{21}$ gives the absorption cross section ${\sigma }_{12}=r\sigma$. Assume equally populated magnetic sublevels within each energy level, a steady pump, and a monochromatic beam. The photon flux is $I/(\hslash \omega )$; define $W=\sigma I/(\hslash \omega )$ and the effective population inversion $D=N_2-r{N}_1$. Then the rate equations areThe $WD$ term includes both stimulated emission and absorption. In steady state, adding the two rate equations gives $N_1=R_2{\tau }_1$. The first equation then yieldsThe directed beam gains energy at rate $\hslash \omega WD$ per volume, giving $dI/dz=\sigma DI$ and the optical gain coefficient $\alpha =\sigma D$. At line centre, write ${\sigma }_0={\sigma }_{21}({\omega }_{21})$. ThereforeHere ${\alpha }_0$ is the small-signal optical gain coefficient and $I_s$ is the saturation intensity, at which the local optical gain is half its small-signal value. Positive optical gain requires $r{A}_{21}{\tau }_1<1$: the lower energy level must empty sufficiently quickly.

The cancellation of ${\tau }_1$ from $I_s$ is specific to this model's steady fixed pump and the fact that every upper spontaneous emission feeds the lower energy level. Adding an extra lower-state accumulation factor to $I_s$ would contradict these rate equations. Alternative conventions using the absorption cross section must replace ${\sigma }_0$ by ${\sigma }_{12}/r$ throughout. Homogeneous spectral broadening permits the whole selected ensemble to share this local saturation response.

For a closed absorbing two-level system, write $N=N_1+N_2$, let $A=1/\tau$, and use ${\sigma }_a={\sigma }_{12}$ for the absorption cross section. This differs from the stimulated-emission cross section convention in part (a). With $r=g_2/g_1$, the rate equations areAt steady state, $N_2=W_{a}N/[A+(1+1/r)W_a]$ for $W_a={\sigma }_{a}I/(\hslash \omega )$. Hence the positive absorption coefficient isFor the given sodium fine-structure transition, the statistical weights are $g_1=2$, $g_2=4$, so $r=2$ and $1+1/r=3/2$. The small-signal transmission gives the optical depthTo account for the finite decrease of intensity through the sample, integrate $dI/dz=-\kappa I$ rather than treating the stronger beam as exactly uniform. If $T=I_{\mathrm{out}}/I_{\mathrm{in}}$, this givesTaking $270\mathrm{W}{\mathrm{m}}^{-2}$ as the incident intensity and $T=0.967$,Using the photon energy $hc/\lambda$ then givesIgnoring the small variation of the stronger intensity would instead give about $1.11\times 10^{-13}{\mathrm{m}}^2$; the integrated value uses the stated transmissions more consistently. The length is not needed to eliminate $N$, but gives the check $N\simeq 9.3\times 10^{14}{\mathrm{m}}^{-3}$.

The resonant absorption cross section is of optical-wavelength-squared size, much larger than a typical atomic geometric collision area. Pressure broadening cross sections involve phase-changing interatomic encounters and need not equal hard-sphere collision areas; their numerical value depends on collision partners and speed. No precise ratio follows from the supplied data. A useful distinction is that resonant absorption involves coherent response on the optical scale, so its cross section need not resemble an chemical atom's geometric area.

For homogeneous spectral broadening, all participating chemical atoms must share the same resonance and single-atom response width. In a dilute sample dominated by natural linewidth, the weak-beam requirement isUsing the sodium atomic mass, equality occurs near $T=0.017\mathrm{K}$, so $T\ll 17\mathrm{mK}$ is a sufficient thermal condition. Spatial Zeeman energy shifts, light shifts and transit effects should also be small, the laser should address the same selected component, and the interaction should last long enough for steady populations. Uniform impact pressure broadening can also be homogeneous, but would change the width and effective cross section; it is not included in the lifetime-only numerical model.

Real sodium also has hyperfine structure and optical pumping among magnetic sublevels. The inferred ${\sigma }_a$ is the effective cross section of the stated $g_J=2J+1$ two-level system model, assuming statistical sublevel populations and no loss into unaddressed states. A measurement on a resolved, polarized closed hyperfine structure component requires its actual transition dipole moment and selection rules; a universal polarization-independent sodium cross section cannot be inferred from this simplified model. The dependence of resonant cross sections and saturation intensities on sublevel preparation is tabulated explicitly for the real sodium D lines.^[ref]

For a closed unit-branching radioactive decay chain, the rate equations areThe first gives $N_a=N_{a0}{e}^{-{\lambda }_{a}t}$. Multiply the second by its integrating factor $e^{{\lambda }_{b}t}$:Integrate from zero to $t$. For ${\lambda }_a\ne {\lambda }_b$,soThe coefficients can individually be negative; the physical population is nonnegative because the production integral is nonnegative. For equal nuclear decay constants ${\lambda }_a={\lambda }_b=\lambda$, the finite limit isA sum of two constant-coefficient copies of the same exponential function cannot represent this limit; the $t$ factor is essential.

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.

For a closed two-level system with total population $N$, excited-state lifetime $\tau$ and statistical sublevel populations, let $W_a={\sigma }_{12}I/(\hslash \omega )$. The steady rate equation $0=W_a[N_1-(g_1/g_2)N_2]-N_2/\tau$ gives absorption coefficient $\kappa =N{\sigma }_{12}/(1+I/I_{s,a})$. For a uniform slab of thickness $L$ and transmission $T=I_{\mathrm{out}}/I_{\mathrm{in}}$, integrating $dI/dz=-\kappa I$ proves $\ln T+I_{\mathrm{in}}(T-1)/I_{s,a}=-N{\sigma }_{12}L$. Resolved sublevel optical pumping requires a more detailed model.

An open pumped two-level system amplifier with fixed upper-state source $R_2$, upper spontaneous emission only into the lower energy level, and lower-state removal rate $N_1/{\tau }_1$. Let $r=g_2/g_1$, $W={\sigma }_{21}I/(\hslash \omega )$ and $D=N_2-r{N}_1$. Steady rate equations give $N_1=R_2{\tau }_1$ and $D=R_2(1-r{A}_{21}{\tau }_1)/(A_{21}+W)$. Consequently the optical gain is $\alpha ={\alpha }_0/(1+I/I_s)$ with ${\alpha }_0={\sigma }_{21}{R}_2(A_{21}^{-1}-r{\tau }_1)$. Lower-state removal controls whether population inversion is possible, while its lifetime cancels from this model's saturation intensity. Different upper-state branching or a nonconstant pump generally changes that conclusion.

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.

Consider an original reduced model with a fixed number $N$ of active emitters. Let $u=N_2/N$ be the upper lasing-state fraction, $1-u=N_1/N$ the ground-state fraction, and $s=n/N$ the number of cavity photons per emitter. Assume the pump state empties rapidly, so its population can be neglected. Measure time by $x=t/\tau$, where $\tau$ is the upper state's spontaneous lifetime. Let $r=W\tau$ be the effective pump rate, $C>0$ the dimensionless stimulated-coupling coefficient, $K>0$ the photon-loss rate in these units, and $0\le \beta \le 1$ the fraction of spontaneous decays feeding this optical normal mode. The rate equations areEach term has a physical role. Pumping transfers emitters out of the ground state; spontaneous emission removes upper-state population. The net stimulated term includes emission from fraction $u$ and absorption by fraction $1-u$, giving their difference $2u-1$. A stimulated transition removes one stored excitation and creates one cavity photon, or does the reverse. The factor $\beta$ accounts for spontaneous emission into the selected mode without pretending that all spontaneous emission enters it.

For negligible cavity light, the upper-state equation is linear. A constant pump applied to $u(0)=0$ givesThus this model first permits population inversion at $r>1$. Positive optical gain is still insufficient for an oscillator: the gain must exceed the complete laser threshold, including photon loss. The small-signal photon growth coefficient changes sign whenIf $C\le K$, even complete inversion cannot exceed the assumed loss in this model. For chosen $C=100$ and $K=20$, the threshold population is $u_{\mathrm{th}}=0.6$ and the pump threshold is $r_{\mathrm{th}}=1.5$. At $r=2$, the unsaturated population first reaches $0.6$ at $x=(\log 10)/3\simeq 0.7675$. These numbers are stipulated examples, not ruby material parameters.