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Write $n=n_1$ and $m=n_2$. The three stoichiometric vectors are $(1,0)$, $(0,1)$ and $(-1,-1)$. The law of mass action gives the respective reaction propensity functionsIn particular, producing the second chemical species does not consume the first chemical species. Since the two reactants in the third chemical reaction are different chemical species, the number of possible pairs is $nm$.

For each Markov chain state, the chemical master equation adds incoming probability flux and subtracts outgoing probability flux:Here $p(n,m,t)=0$ if either index is negative, and the initial condition is $p(n,m,0)=1_{\{n=m=0\}}$. These conventions also give the correct chemical master equation on the coordinate axes: for example, the third chemical reaction has zero outgoing reaction propensity function when either chemical species is absent.

At the empty Markov chain state, only the constant-rate production chemical reaction is possible. Thus the first-passage time ${\tau }_1$ has an exponential distribution of rate $2$:Its expected value can be obtained by integrating this survival function, and its second moment by integrating $2t$ times the survival function:Consequently the requested expected value and variance are

Conditioning on the observed time. The native PDF defines ${\tau }_2$ as the first production time and then conditions on that production occurring at ${\tau }_2$. If the intended information is its numerical value ${\tau }_2=t$, the answer depends on $t$. We first compute this conditional probability using a conditional probability density; conditioning on an exact value of a continuous random variable means a density ratio, since $\mathbb{P}({\tau }_2=t)=0$.

Before ${\tau }_2$, no molecules of the second chemical species are present, so the joint-removal chemical reaction cannot occur. Molecules of the first chemical species arrive as a rate-$2$ Poisson process, and each independently starts a clock with a rate-$1/2$ exponential distribution for producing the first molecule of the second chemical species. Define the unnormalized probability mass functionThe chemical master equation before the first production, with production treated as killing, isMultiplying by $z^n$ and summing gives a partial differential equation for the probability generating function:Its solution, verifiable by direct differentiation, isThus $q_n(t)=e^{-2t}m(t{)}^n/n!$, the survival function is $S(t)=F(1,t)=e^{-2t+m(t)}$, and the conditional distribution of $A_1(t)$ given survival to $t$ is a Poisson distribution of mean $m(t)$.

Production in $(t,t+dt)$ from Markov chain state $n$ has joint probability $(n/2)q_n(t)dt+o(dt)$. Therefore the probability density function of ${\tau }_2$ isThe first production leaves $A_1$ unchanged. Taking the density ratio gives a size-biased distribution of the surviving count:In particular, under this reading the requested conditional probability isThe shift by one matters: an observation made exactly when production occurs preferentially selects larger counts, since their production reaction propensity function is larger.

Without observing the numerical time. If the conditioning merely repeats the definition of ${\tau }_2$, the requested probability is instead the unconditional count at that stopping time. Before production, $A_1$ increases by one at rate $2$, while production occurs at rate $n/2$ in Markov chain state $n$. The competing exponential clocks implyThe first factor requires a second arrival before production from count one; the second requires production before a third arrival. The first arrival itself is certain. Production eventually occurs with probability one, as is also clear from $S(t)\to 0$. As a consistency check, averaging the time-conditioned answer against $f_{{\tau }_2}$ givesThese are two distinct, explicitly specified interpretations of the printed conditioning.

A product-Poisson stationary distribution is suggested by balancing the three reaction fluxes around the directed cycle of reaction complexes. If their mean counts are $a,b$, this complex balance condition iswhich gives $a=4$ and $b=1/2$. This suggests the joint probability mass functionIt is normalized because it is the product of two Poisson distributions. We verify it directly in the chemical master equation, rather than assuming a product-Poisson stationary distribution theorem. The three incoming terms satisfyWith negative-index values set to zero, these identities also hold on the axes. Their sum equals $(2+n/2+nm)\pi (n,m)$, precisely the outgoing probability flux. The stationary average total exit rate is $2+{\mathbb{E}}_{\pi }n/2+{\mathbb{E}}_{\pi }(nm)=6<\infty$, so these stationary balance equations do define an invariant distribution of a continuous-time Markov chain for the nonexplosive model verified below. This is complex balance; it does not assert pairwise detailed balance for the irreversible chemical reactions.

To identify this invariant distribution with the limiting probability distribution in the question, first check that the model is a nonexplosive continuous-time Markov chain. On any finite time interval, $A_1$ is bounded by its initial count plus the finite number of arrivals in a rate-$2$ Poisson process. Conditional on that bound, the rate of producing $A_2$ is bounded, so only finitely many such productions occur. The number of joint removals cannot exceed the available molecules. Thus only finitely many chemical reactions occur in finite time.

The continuous-time Markov chain is also an irreducible Markov chain. From a Markov chain state with both counts positive, repeated joint removals reduce both. On the $A_2=0$ axis, producing one $A_2$ and then removing a pair reduces $A_1$ by one. On the $A_1=0$ axis, adding one $A_1$ and then removing a pair reduces $A_2$ by one. Each prescribed finite sequence has positive probability, so every Markov chain state reaches $(0,0)$. Conversely, arrivals followed by productions reach any $(n,m)$ with $n>0$; to reach $(0,m)$, reach $(1,m+1)$ and remove one pair. The continuous-time Markov chain is therefore positive recurrent, and the continuous-time Markov chain convergence theorem gives $\varphi =\pi$ from the empty initial condition.

The two Poisson factors happen to have cancelling numerical coefficients at $(2,2)$:

The periodic table arranges chemical elements in increasing atomic number, across successive periodic table periods. Its columns are periodic groups, numbered 1–18. The layout places recurring outer electron configurations above one another, making similarities in chemical bonds and chemical reactions easier to recognize. Ordering is by proton count, not by mass; the mass number of a particular isotope does not determine its box. See the official IUPAC periodic table and numbering conventions.

For the first eighteen neutral chemical atoms in their ground states, a compact description counts electrons in successive electron shells. Thus carbon has $(2,4)$, meaning two electrons in the first electron shell and four in the second; the sum $2+4=6$ agrees with its atomic number. Sodium has $(2,8,1)$, whose sum is $11$. These are occupancy counts, not circular trajectories of electrons.

Reading each row from left to right gives:

The counting outline is short: the occupied electron subshells in this range are $1s$, $2s$, $2p$, $3s$, and $3p$, filled in that order for these neutral ground states. Each $s$ electron subshell holds up to two electrons and each $p$ electron subshell up to six, giving $2$, $2+6=8$, and $2+6=8$ along these three rows. This $2,8,8$ summary applies only through argon; it is not a general rule that each electron shell fills completely before the next begins, nor a claim that the third electron shell can hold only eight electrons. See Flowers et al., Chemistry 2e, electronic structure and electron configurations.

The recurring outer counts make the arrangement useful. Lithium and sodium each have one valence electron, whereas fluorine and chlorine each have seven valence electrons. Such similarities help anticipate chemical reactions, without making all properties identical throughout a periodic group. Helium is an instructive exception to an eight-electron shortcut: its two electrons complete the first electron shell, so it belongs above neon and argon in periodic group 18. Hydrogen shares column 1 with lithium and sodium but has its own distinct chemistry. See Flowers et al., Chemistry 2e, periodic-table organization.

The phase rule counts independent intensive controls compatible with coexistence of $P$ distinct bulk thermodynamic phases made from $C$ independent components:Here $P$ counts thermodynamic phases, not pressure, and $F$ does not count phase fractions. For the nonreacting one-component case, $C=1$ and $F=3-P$: a single thermodynamic phase occupies a region with $F=2$, two-thermodynamic phase coexistence a curve with $F=1$, and a triple point an isolated point with $F=0$. Fixing pressure externally removes one adjustable control wherever that constraint is independent.^[ref]

For an elementary count without chemical reactions, take $C$ chemical species. Each thermodynamic phase has $C-1$ independent composition fractions, since they sum to one. Shared temperature and pressure add two controls, giving $2+P(C-1)$. Equality of each species' chemical potential between one reference thermodynamic phase and the other $P-1$ thermodynamic phases supplies $C(P-1)$ constraints. Subtraction gives $2+P(C-1)-C(P-1)=C-P+2$. This argument assumes the constraints are independent and treats interfaces as negligible; chemical reactions, external fields or critical mergers require appropriate qualifications.

In photography, photographic processing uses chemical reactions to develop and stabilize an exposed photographic emulsion.

Photosynthesis captures light energy and transforms it into chemical forms that a biological cell can use. In plants, algae and cyanobacteria, this supports carbon fixation: incorporation of carbon from carbon dioxide into chemical compounds used to build living material. Capturing light and incorporating carbon are distinct processes coupled through metabolism. See OpenStax Biology 2e, 8.1: overview of photosynthesis.

For oxygenic photosynthesis, a simplified net equation writes a six-carbon sugar, glucose, as the representative product:This is net stoichiometry, not a one-step mechanism. A worked bookkeeping check counts six carbon chemical atoms, twelve hydrogen chemical atoms, and eighteen oxygen chemical atoms on each side: the product side has six oxygen chemical atoms in the sugar and twelve in molecular oxygen. The light label supplies energy, not additional chemical atoms.

Crucially, the released molecular oxygen comes from water, not carbon dioxide. Cancelling water made and consumed in different steps hides this distinction in the net equation. The Calvin cycle supplies a net three-carbon triose phosphate, such as glyceraldehyde 3-phosphate, for subsequent sugar synthesis; it does not directly export glucose. See OpenStax Biology 2e, 8.2: light-dependent reactions and OpenStax Biology 2e, 8.3: using light energy to make organic molecules.

The useful achievement is connecting incoming light to both usable energy and new biological material. ATP is an immediate energy currency; NADPH supplies electrons for synthesis; carbon dioxide supplies the carbon incorporated into carbohydrates. These roles are complementary. Conservation of energy still holds: photosynthesis transforms energy, never creates it, and some absorbed energy is dissipated as heat. A plant also carries out cellular respiration, including in biological cells that perform photosynthesis. Gas exchange reflects both processes. Oxygenic photosynthesis remains central to interpreting terrestrial molecular oxygen as an atmospheric biosignature gas. See OpenStax Biology 2e, 8.3: photosynthesis and respiration.

Polymerization converts a monomer, or a mixture of monomers, into a polymer: repeated chemical reactions establish a connected molecular chain or network. The central question is how connections form, not merely whether a material becomes solid. Joining small molecules can produce a thread that can be handled, while the individual molecules remain invisible. See IUPAC, Basic Definitions of Terms Relating to Polymers, definition 2.2.

In step-growth polymerization, complementary reactive groups on monomers and on shorter or longer chains can react with one another. A two-unit chain can join a three-unit chain to make a five-unit chain; growth does not require a specially active end to add only free monomers. An example is the Nylon 6,10 reaction of ${\mathrm{H}}_2\mathrm{N}(\mathrm{C}{\mathrm{H}}_2{)}_6\mathrm{N}{\mathrm{H}}_2$ with $\mathrm{ClC}(=\mathrm{O})(\mathrm{C}{\mathrm{H}}_2{)}_8\mathrm{C}(=\mathrm{O})\mathrm{Cl}$: each amine end can form an amide connection with an acid-chloride end, with formal elimination of $\mathrm{HCl}$. Both reactants have two suitable ends, allowing repeated joining. See Paula Hammond, MIT 10.569 lecture 2: step-growth joining and Hammond, lecture 5, p. 4: acid chloride–amine reaction.

In chain-growth polymerization, a monomer reacts with an active site on a growing chain, and the growth step regenerates an active site. Schematically, $P_j^{\ast }+M⟶P_{j+1}^{\ast }$, where $j$ counts incorporated monomeric units and the star denotes reactivity rather than a particular electric charge. Ethene forming polyethylene is an example: the double chemical bond is converted into single chemical bonds as the carbon backbone grows. Initiation creates the active chains; subsequent reactions can deactivate them or transfer reactivity to another molecule. See IUPAC: chain polymerization and John McMurry, Organic Chemistry, 31.1: chain growth. Whether growth is stepwise or carried by an active site is a different question from whether small byproducts are released: chain-growth polymerization can also release them.

A motivating experiment: drawing nylon from a liquid interface. Feldwinn's UCSB demonstration places $25\mathrm{mL}$ of an aqueous chemical solution of hexamethylenediamine, ${\mathrm{H}}_2\mathrm{N}(\mathrm{C}{\mathrm{H}}_2{)}_6\mathrm{N}{\mathrm{H}}_2$, in a $100\mathrm{mL}$ beaker, then gently adds $25\mathrm{mL}$ of sebacoyl chloride, $\mathrm{ClC}(=\mathrm{O})(\mathrm{C}{\mathrm{H}}_2{)}_8\mathrm{C}(=\mathrm{O})\mathrm{Cl}$, in an organic solvent. The procedure specifies hexane; other parts of that handout list cyclohexane. The layers remain separate. The instructor catches material at their interface with tweezers and winds it slowly onto a glass rod; the handout reports continued spooling of Nylon 6,10. It specifies freshly prepared acid-chloride chemical solution and washing the thread before touching it. These are observations and conditions of a supervised laboratory demonstration, not measurements of chain lengths. See Feldwinn, UCSB Demonstration Library, Demo 044, pp. 1–3.

The mechanistic interpretation is that the monomers encounter one another near the interface, where amide connections produce the film; withdrawing it exposes a region where more material can form. An authored research demonstration independently reports a rapidly formed interfacial film collapsing into a thread on withdrawal, using cyclohexane and an aqueous chemical solution containing potassium hydroxide. See Polymer Day: Outreach Experiments for High School Students, Journal of Chemical Education (2017), nylon demonstration. This is the striking observation: a thread much longer than the beaker can emerge because material is continually supplied to the formation region, rather than because a pre-existing thread was hidden in the liquid.

Useful proposed controls, not results reported in those demonstrations, omit one monomer at a time while retaining the same solvents, liquid amounts, and pulling procedure. The prediction is loss of a renewable drawable Nylon 6,10 film; merely finding cloudiness would not establish polymerization, since other chemical reactions, including acid-chloride hydrolysis, can occur. See McMurry, 21.4: competing acid-chloride hydrolysis. Stopping and restarting withdrawal would probe renewal at the interface. Even successful drawing does not measure the degree of polymerization, establish a chain-length probability distribution, or prove the entire growth mechanism. Chemical identification and a molecular-size measurement would be needed for those stronger conclusions. These controls and limits are inferences from the setup, not invented experimental data.

Why almost complete reaction matters in step growth. For ideal linear step-growth polymerization, assume exactly two reactive groups per starting molecule, equal numbers of complementary groups, only joining of different chains, no rings, no branching, and no loss of chain material. Let $N_0$ count starting monomer molecules, including both reactant species, and let $p$ be the fraction of either type of reactive group consumed. There are initially $N_0$ groups of each type, so $p{N}_0$ connections reduce the number of chain-bearing molecules to $N=N_0(1-p)$; the count excludes small byproducts but includes unreacted monomers. The number-average degree of polymerization, measured in monomeric units, is thereforeThus $p=0.90$, $0.99$, and $0.999$ give $X_n=10$, $100$, and $1000$. Losing the last fraction of unreacted groups matters enormously. See Hammond, lecture 3: balanced step-growth conversion. This counts residues of both feed species; a Nylon 6,10 interior constitutional repeating unit contains two such monomeric units.

For an $AA+BB$ feed with $A_0\le B_0$ initial reactive-group counts, put $r=A_0/B_0$ and now let $p$ denote the fraction of the limiting $A$ groups consumed. Under the same linear, no-ring assumptions, $N_0=(A_0+B_0)/2$ and $N=N_0-p{A}_0$, givingFor $r=0.98$, even complete consumption of $A$ gives $X_n=(1.98)/(0.02)=99$. Stoichiometry controls chain length as well as yield. See Hammond, lecture 2, pp. 2–3: stoichiometric imbalance. The liquid-interface demonstration is not a homogeneous, balanced reactor, so its thread length cannot be inserted into this formula as a molecular chain length.

A contrasting chain count. In ideal living polymerization, assume no chain termination or irreversible chain transfer, all chains initiated before the counted growth, one chain per effective initiator, a fixed chain count $C$, and every consumed monomer molecule incorporated into those chains. If $M_0$ is the initial number of feed monomer molecules and $q$ their incorporated fraction, conservation gives $X_n=q{M}_0/C=qR$ in monomeric units, excluding initiator fragments. For $R=200$, incorporating half the feed gives $X_n=100$. If initiation consumes one monomer molecule per chain, the fixed-chain interval starts at $q=1/R$, rather than $q=0$. See Hammond, lecture 21, p. 1: ideal living-chain count. The mean does not imply identical chain lengths. Living polymerization does not require reversible loss of monomers; the defining absence of chain termination and irreversible chain transfer is explained in Devon Shipp, IUPAC Polymer Education workshop, Radical Polymerizations II, slide 50. Ordinary chain-growth polymerization with new chains, termination, or transfer need not follow the straight line.

The useful connection is between local chemistry and a macroscopic material: selecting reactive ends and their proportions controls connectivity, while the resulting polymer's conformation is a separate problem described by models such as the freely jointed chain or Gaussian chain. Formation of chemical bonds, the count of monomeric units, and the count of statistical segments must be kept distinct.

Consider an original toy AB salt model with $K_{\mathrm{sp},c}=10^{-8}{\mathrm{M}}^2$ stipulated at fixed temperature and fixed activity coefficient product. This is a conditional concentration constant, not measured data for calcium carbonate or any identified material. A stock contains A cations and X anions; another contains Y cations and B anions. All four ions have unit electric charge magnitude. Only AB can form a solid; X and Y remain spectator ions. Assume full dissociation, no other chemical reactions, and a mixed volume equal to the sum of the initial volumes.

Mix two stocks, each of volume $V$,, each at molar concentration $C$ of its indicated salt. Before any precipitation, the mixed volume is $2V$, soMultiplying the two stock concentrations would overestimate the mixed ion product by a factor of four. The saturation threshold is $C=2\sqrt{K_{\mathrm{sp},c}}=2\times 10^{-4}\mathrm{M}$.

If $C=10^{-4}\mathrm{M}$, mixing gives $Q_{c,0}=2.5\times 10^{-9}{\mathrm{M}}^2<K_{\mathrm{sp},c}$: no AB precipitate is favored. If $C=4\times 10^{-4}\mathrm{M}$, mixing gives $Q_{c,0}=4\times 10^{-8}{\mathrm{M}}^2>K_{\mathrm{sp},c}$. Suppose nucleation occurs and bulk AB solid remains at chemical equilibrium. Equal starting inventories and one-to-one stoichiometry require equal remaining concentrations, henceFor two $10\mathrm{mL}$ portions in the second case, each reacting ion starts with $4\mu \mathrm{mol}$; the final $20\mathrm{mL}$ chemical solution retains $2\mu \mathrm{mol}$ of each, and $2\mu \mathrm{mol}$ of AB enters the solid. The respective X and Y inventories stay at $4\mu \mathrm{mol}$. Exactly half of each reacting inventory precipitates in this toy model.

The chemical equilibrium calculation adds a phase constraint to the inventory balance. Write $x=n_{\mathrm{AB},s}/(2V)$ and $c=C/2-x$. In the equilibrated bulk model,When $x>0$, the last condition requires saturation; when $x=0$, the inventory can leave an undersaturated chemical solution. Together these give $x=\max \{0,C/2-\sqrt{K_{\mathrm{sp},c}}\}$. This excludes a persistent supersaturated solution by its explicit chemical equilibrium assumption; it does not model a nucleation delay.

A primary standard supplies a reliably known reacting inventory from which a titrant concentration can be determined.

The defining advantage is known high purity: the weighed portion can be assigned a known amount of substance. A titrant whose concentration is established against that inventory is a secondary standard solution in IUPAC terminology.^[ref]

An accurate balance determines total mass, including any unwanted water or other material. To infer reacting inventory, one also needs known composition, stability during storage and weighing, complete transfer and dissolution, and a sufficiently complete chemical reaction of known stoichiometry. A finely resolved balance cannot establish these chemical facts. Potassium hydrogen phthalate is useful because a suitably prepared portion provides a known acid inventory for sodium hydroxide titrant standardization; the KHP standardization protocol below shows how that claim is put into practice.

Preparation belongs to the particular material and method. For example, the MIT handout specifies drying potassium hydrogen phthalate at $110{}^{\circ }\mathrm{C}$ for $1.5$ hours or overnight, cooling in a desiccator, and weighing by difference; its footnote identifies removal of superficial moisture as the purpose. This is a condition of that potassium hydrogen phthalate procedure, not a universal instruction to heat every primary standard in the same way.^[ref]

A protein is a biological molecule made of one or more polypeptide chains: ordered amino acid residues joined by peptide bonds. Biological translation assembles a chain in the order specified by messenger RNA codons. Protein folding then brings parts of that chain into spatial relationships that support its function. See Cooper, The Cell: protein composition and structure. These processes can overlap in a biological cell; the sequence-to-shape explanation does not require waiting until biological translation finishes before any folding begins. See Cooper, The Cell: protein folding and processing.

The useful connection is from information to physical work. A protein's protein primary structure records an order; its protein tertiary structure can bring distant amino acid residues together at an active site. An enzyme uses such an arrangement to recognize a substrate and accelerate a chemical reaction. Other proteins serve as receptors, membrane transport proteins or structural material. For example, actin assembles into filaments whose mechanical properties can be described by biopolymer mechanics. See Alberts et al., Molecular Biology of the Cell: protein function and binding.

Order matters beyond composition. Let A stand for alanine and M for methionine, and read from the N-terminus to the C-terminus. The illustrative strings AMAMAMAMAMAM and AAAAAAMMMMMM each contain exactly six A and six M amino acid residues. Each has $12-1=11$ peptide bonds when made into one unbranched polypeptide chain. In the first string all eleven neighboring pairs have different letters; in the second only the pair at positions 6 and 7 does. A composition histogram therefore loses information about which amino acid residues are neighbors. Their arrangements can affect protein folding and function; these short invented strings neither specify a known functional protein nor establish what shape either would adopt.

Protein structure is a moving arrangement, not a permanently rigid outline. Substrate binding can change an enzyme's shape through induced fit, and a functional protein can include flexible regions. Sequence, surroundings and interacting molecules must therefore be considered together. See Alberts et al.: binding, conformational changes and protein function.

A protein binding site is a region of a protein structure that interacts with a ligand. Its arrangement of amino acid residues helps determine recognition; connecting loops supported by a beta sheet can contribute to it. An enzyme's active site binds a substrate and accelerates a chemical reaction. See Alberts et al.: protein binding sites and active sites.

A radical chain reaction regenerates a radical during propagation. Initiation supplies free radicals; propagation consumes one and produces another; termination removes chain-carrying free radicals. This explains why one initiating event can lead to many product-forming chemical reactions. The chain-growth polymerization article develops the corresponding growth of polymers.

A radical has one or more unpaired electrons. In a simple atomic orbital or molecular orbital occupancy picture, an unpaired electron occupies a spatial orbital without an opposite-spin partner in that orbital. A dot such as ${\mathrm{H}}^{\bullet }$ marks this electronic property; it does not mean a missing electron or positive electric charge. Free radicals can be neutral or charged. Terminology also has a boundary: metal ions exhibiting paramagnetism are not normally called free radicals just because they have unpaired electrons.^[ref]

Compare three species containing the same single proton. The hydrogen atom ${\mathrm{H}}^{\bullet }$ has one electron in its $1s$ atomic orbital, and is a neutral radical. The hydrogen cation ${\mathrm{H}}^{+}$ has no electrons, so no unpaired electron. The hydrogen anion ${\mathrm{H}}^{-}$ has two electrons paired in its $1s$ atomic orbital. It is charged but is not a radical in its ground state. With $N_p$ protons, $N_e$ electrons, and positive elementary charge $e$, the independent electric charge count isThe electron pair and electric charge questions require different information. An odd $N_e$ ensures that not all electrons can be paired; an even $N_e$ alone does not ensure pairing, as the molecular orbital description of molecular oxygen illustrates.

The distinction matters whenever a chemical reaction mechanism uses dots and charge signs together. Homolysis gives one bonding electron to each fragment, while heterolysis assigns both to one fragment. A radical chain reaction can pass an unpaired electron from one species to another without generating net electric charge. Detection then exploits the unpaired electron's spin and magnetic moment, rather than merely measuring whether the sample carries an electric charge.

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

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

A chemical species consumed in a chemical reaction.

A formal combination of chemical species appearing on one side of a chemical reaction. The coefficients $y_i$ are nonnegative integers. A chemical reaction network can be viewed as a directed graph whose vertices are reaction complexes and whose edges are chemical reactions. The empty complex represents external production or removal.

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.

For a chemical reaction with signed stoichiometric vector entries ${\nu }_i$, positive for products and negative for reactants, the reaction quotient is the product of the species' thermodynamic activities raised to these powers. It equals the equilibrium constant at chemical equilibrium, with all quantities using the same standard state convention.