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Follow one glucose through canonical glycolysis, with no diversion of intermediates into other metabolic pathways. The investment stage uses two ATP to prepare the six-carbon substrate. Cleavage and rearrangement provide two glyceraldehyde 3-phosphate molecules, each containing three carbon chemical atoms. Each then yields one three-carbon pyruvate, one NADH, and two ATP in the payoff stage.^[ref]

The material check is $6=2\times 3$: all six carbon chemical atoms remain in the two pyruvates. No carbon dioxide has been released within this boundary. In the ATP ledger, each three-carbon route contains two ATP-forming transfers, soThe first pair of payoff transfers only restores the initial ATP inventory; the second pair creates the net gain. Four newly formed ATP is therefore compatible with only two additional ATP remaining.

For an original scaled example, stipulate complete conversion of $5\mu \mathrm{mol}$ of glucose. The initial $30\mu \mathrm{mol}$ of carbon chemical atoms becomes $10\mu \mathrm{mol}$ of pyruvate, still containing $30\mu \mathrm{mol}$ of carbon chemical atoms. The pathway spends $10\mu \mathrm{mol}$ of ATP, forms $20\mu \mathrm{mol}$ of ATP, and adds $10\mu \mathrm{mol}$ to the ATP inventory before any other ATP use. It also reduces $10\mu \mathrm{mol}$ of NAD+. These are calculated ideal amounts, not observed yields; side routes or incomplete conversion change the outcome.

A common biochemical net chemical equation isHere ${\mathrm{P}}_i$ is inorganic phosphate. The names group biochemical protonation and metal-binding states; the displayed proton and water terms use a common shorthand rather than a unique specification of every chemical species. The robust ledger records two net ADP-to-ATP conversions and two NAD+-to-NADH conversions. Summing the individual chemical equations cancels two invested ATP against four formed ATP and cancels the internal substrates, giving those coefficients.

The Helmholtz free energy iswhere $U$ is internal energy, also denoted $E$ elsewhere here, $T$ is absolute temperature, and $S$ is entropy. For the closed system of fixed composition and only pressure-volume work described above, spontaneous relaxation at fixed temperature and volume cannot increase $F$. Stable thermodynamic equilibrium is a global minimum over the permitted alternatives, including different thermodynamic phases. A rigid vessel fixes the total volume; it does not require each coexisting thermodynamic phase to retain its own volume.

For a constructed comparison at $300\mathrm{K}$, suppose one candidate has $1.5\mathrm{kJ}$ more internal energy but $6\mathrm{J}{\mathrm{K}}^{-1}$ more entropy than another at the same temperature and volume. Its difference isThus the candidate with higher internal energy has lower Helmholtz free energy. This is the practical reason for retaining the entropy term when selecting a state.

For equilibrium states within a smooth branch of fixed composition, the first law of thermodynamics gives $dU=TdS-pdV$, henceThese derivatives extract entropy and pressure; the differential is not an evolution equation for relaxation. If chemical species amounts $n_i$ vary, add ${\sum }_i{\mu }_{i}d{n}_i$, with molar chemical potentials ${\mu }_i$. A closed system can still change chemical species amounts through chemical reactions; the fixed-composition simplification then needs the conserved-inventory constraints of chemical equilibrium. See MIT 5.60, Lecture 14, fundamental equations and chemical potentials.

The minimum principle has a short explanation. At fixed volume, no boundary work occurs, so $Q=\Delta U$. A thermal reservoir at positive temperature $T$ loses entropy $Q/T$. Consequentlyby the Second law of thermodynamics. Here both endpoint states share the imposed temperature. See MIT 5.60, Lecture 12, criteria for spontaneous change.

The microscopic counterpart is $F=-k_{B}T\ln Z$ in the canonical ensemble, with Boltzmann constant $k_B$ and partition function $Z$. The partition function examples compute the balance between internal energy and entropy; the chemical equilibrium example already displays an ideal-mixture Helmholtz free energy minimum.

Keep the stoichiometry attached to the enthalpy. Use the same abstract $A,B,C$ and suppose each of two entries fixes an amount of substance for every species:Here each unit coefficient represents one mole on the common basis. To obtain $A\to C$, halve $R_1$ and reverse $R_2$. Adding those transformed chemical equations givesThe equal amounts of $B$ on opposite sides cancel, leaving $A\to C$. Doubling this final chemical equation would give $2A\to 2C$ and $\Delta H=-14\mathrm{kJ}$; its change per mole of $A$ would remain $-7\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$.

More generally, write each chemical equation as a stoichiometric vector ${\mathbf{\nu }}_j$, with negative entries for reactants and positive entries for reaction products. Finding coefficients $a_j$ with ${\mathbf{\nu }}_{\mathrm{target}}={\sum }_j{a}_j{\mathbf{\nu }}_j$ also gives $\Delta H_{\mathrm{target}}={\sum }_j{a}_j\Delta H_j$, using compatible state and amount conventions. An identical chemical species cancels only with the same thermodynamic phase and composition conditions. A leftover water or dilution step is a real part of the balance.

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.

Dissolved contact is useful because reactants can encounter the catalytic chemical species throughout the mixed liquid, without first reaching a macroscopic solid surface. Recovering a dissolved catalyst from the same chemical solution can instead require a separation. Neither ease of contact nor phase classification determines an activation energy or an equilibrium constant; those depend on the relevant chemical kinetics and endpoint thermodynamics.

Inorganic phosphate, abbreviated ${\mathrm{P}}_i$, denotes the free phosphate pool in biochemical chemical equations, including ATP hydrolysis. Its constituent chemical species depend on pH and protonation; ${\mathrm{P}}_i$ does not specify a single electric charge or proton count. See the IUBMB recommendations on biochemical reactant pools.

Consider an ideal gas in chemical equilibrium with homogeneous, identical adsorption sites at fixed temperature. Each intact molecule occupies one site; only one monolayer is allowed. Occupation of distinct sites is independent: neighboring bound molecules do not change binding or release rates. The simplest Langmuir adsorption model has one adsorbing chemical species, no splitting of molecules, and no competitors. Irving Langmuir's 1918 paper treats simple adsorption alongside more complicated cases; its scope exceeds this particular model.^[ref]

The derivation is a balance between arrivals finding vacancies and departures from occupied adsorption sites. Let $k_{\mathrm{a}}P$ be the arrival-and-binding rate per vacant site and $k_{\mathrm{d}}$ the departure rate per occupied site. Thus $k_{\mathrm{a}}$ has units of inverse pressure per unit time, whereas $k_{\mathrm{d}}$ has units of inverse time. The mean surface coverage obeysAt chemical equilibrium, setting the two terms equal givesThe product $bP$ is dimensionless. Here $P$ is the adsorbing gas's partial pressure, expressed in modern bar, with $1\mathrm{bar}=10^5\mathrm{Pa}$, and $b$ is expressed in ${\mathrm{bar}}^{-1}$. The pascal is the SI unit of pressure. This dimensional $b$ is not itself a dimensionless thermodynamic equilibrium constant. A larger $b$ gives greater surface coverage at the same $P$ and temperature, but does not specify how quickly chemical equilibrium is reached: multiplying both rate coefficients by the same factor leaves $b$ unchanged.

Choose $b=0.5$, $2$, and $8{\mathrm{bar}}^{-1}$ as illustrative parameters at one fixed temperature, without assigning a particular material. At $P=0.5\mathrm{bar}$, substitution gives, respectively,Solving $\theta =1/2$ gives $P_{1/2}=1/b$: the three half-occupation pressures are $2$, $0.5$, and $0.125\mathrm{bar}$. For $b=2{\mathrm{bar}}^{-1}$, reaching $\theta =0.9$ instead requires $bP=9$, or $P=4.5\mathrm{bar}$. If the complete-monolayer capacity is stipulated to be $q_{\mathrm{m}}=10\mathrm{mmol}{\mathrm{g}}^{-1}$, then $q=q_{\mathrm{m}}\theta$ predicts $q=5\mathrm{mmol}{\mathrm{g}}^{-1}$ at $P=0.5\mathrm{bar}$. This last number is an original hypothetical example, not a measured uptake.

The shape follows directly from the finite number of adsorption sites. For $bP\ll 1$, $\theta \approx bP$: nearly every arriving molecule finds a vacancy. For $bP\gg 1$, $1-\theta =1/(1+bP)\approx 1/(bP)$: vacancies become scarce. More formally,Thus surface coverage rises with diminishing slope and approaches one only as $P\to \infty$. This saturation is a site-capacity limit, not a claim that the gas has reached its condensation pressure. A second layer, different site binding strengths, interactions between bound molecules, or a changing surface invalidate parts of the setup. A good fit alone does not establish identical adsorption sites: a restricted pressure range can hide such differences. The Langmuir adsorption model is useful because it makes the capacity and the pressure scale of filling separate, testable quantities.

Light initiates methane chlorination by photodissociation of $\mathrm{C}{\mathrm{l}}_2$ through homolysis. The elementary bookkeeping isHere $h\nu$ denotes the photon energy. The first propagation step replaces a chlorine radical with a methyl radical; the second regenerates a chlorine radical. Adding the two propagation equations cancels both free radicals and gives $\mathrm{C}{\mathrm{H}}_4+\mathrm{C}{\mathrm{l}}_2⟶\mathrm{C}{\mathrm{H}}_3\mathrm{Cl}+\mathrm{HCl}$: chloromethane and hydrogen chloride are made while the chain carrier is regenerated.^[ref]

For example, recombination $2{\mathrm{C}{\mathrm{H}}_3}^{\bullet }⟶{\mathrm{C}}_2{\mathrm{H}}_6$ of methyl radicals makes ethane and a covalent bond from two unpaired Electrons, removing two carriers. Recombination of two ${\mathrm{Cl}}^{\bullet }$ also terminates carriers; in the gas phase, stabilization of the product requires disposal of released energy, for example to a third collision partner. The figure omits that partner. Termination competes with propagation, and further chlorination can consume chloromethane, so this cycle alone does not predict a pure product or a measured reaction rate.^[ref]

The decadic molar absorption coefficient specifies absorbance per geometric length and molar concentration. With $l$ in centimetres and $c$ in $\mathrm{mol}{\mathrm{L}}^{-1}$, $\epsilon$ has units $\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}\mathrm{c}{\mathrm{m}}^{-1}$; its SI unit is ${\mathrm{m}}^2\mathrm{mo}{\mathrm{l}}^{-1}$.^[ref]

The units cancel in $\epsilon lc$. Since $1\mathrm{L}=10^{-3}{\mathrm{m}}^3$ and $1\mathrm{cm}=10^{-2}\mathrm{m}$, one $\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}\mathrm{c}{\mathrm{m}}^{-1}$ equals $0.1{\mathrm{m}}^2\mathrm{mo}{\mathrm{l}}^{-1}$. Specify the wavelength and chemical solution conditions when giving a molar absorption coefficient: changing the absorbing chemical species can change it even at fixed total analytical concentration.

Molar concentration is amount of substance of a named chemical species per final chemical solution volume: $c=n/V$. State whether the count describes dissolved species or a conserved analytical concentration.

The molecular copy number of a chemical species is the nonnegative integer number of its molecules in a specified reactor or compartment.

A monoprotic acid supplies one proton per acid unit in the acid-base chemistry model under consideration. Acetic acid is an example. Several successive proton transfers require several acid dissociation constants and a larger set of chemical species.

A net ionic equation retains the chemical species changed by a chemical reaction and cancels spectator ions present unchanged on both sides. Dissolved strong electrolytes are represented by their separated ions, while a solid product retains its chemical formula. Cancellation preserves both elemental inventories and electric charge conservation.^[ref]

For the calcium carbonate example, the complete ionic account isCanceling the sodium cations and chloride anions yields the net ionic equation above. Neither cancellation nor the conventional sodium chloride formula implies that the remaining dissolved ions have become paired molecules.

For this schematic, let $\xi$ be the amount of substance of AB formed. With no competing chemical reactions or losses,Adding each dissolved inventory to its contribution in the solid recovers its initial total. Each dissolved pair has net electric charge zero, so transferring a pair into neutral AB also preserves electric charge conservation. This is a short stoichiometry argument; choosing the actual $\xi$ additionally needs chemical equilibrium or chemical kinetics information.

NMR monitoring of chain polymerization uses nuclear magnetic resonance signals to follow monomer and other distinguishable chemical species. In a proton experiment, suitable signal integrals can track changes in their relative amounts. Disappearance of a monomer signal is a consumption readout; it is not a direct count of polymerization-active sites. Overlapping signals, baseline errors and competing chemical reactions can obstruct the inference. A pre-reaction reference and independent product characterization make the consumption interpretation stronger.

A nonelectrolyte supplies neutral molecules rather than ions on dissolving in the specified solvent, as illustrated by sucrose in water.

For example, dissolving an idealized $1\mathrm{mmol}$ sucrose inventory gives $1\mathrm{mmol}$ dissolved sucrose molecules, rather than separate cations and anions. Dissolving the same amount of substance of a fully dissociated one-to-one strong electrolyte instead gives $1\mathrm{mmol}$ of each ion. Both can disappear visibly into the solvent; the resulting chemical species distinguish them.^[ref]

A nonelectrolyte is not necessarily insoluble: sucrose provides a soluble counterexample. Nor does the label assert exactly zero measured electrical conductivity. Self-ionization of water, impurities or residual electrolyte from a probe can supply background ions. Comparing with a solvent blank and rinsing between samples, as in the strong electrolyte experiment, tests whether the dissolved substance itself produces appreciable additional mobile ions. This separates visible dissolution from ion production without treating lamp darkness as an absolute absence of electric charge.

An oxidation increases an oxidation state, recording loss from the formal Electron assignment. For $\mathrm{Zn}\to {\mathrm{Zn}}^{2+}+2e^{-}$, the increase is $0\to +2$. Oxidation need not involve oxygen.^[ref]

In the electrochemistry examples here, oxidation is loss of Electrons by a chemical species; it occurs at the anode. That electrode definition holds in both a galvanic cell and an electrolytic cell; it does not fix terminal polarity. The oxidized chemical species serves as the reducing agent in the coupled redox.

Let ${\mu }_i=⟨X_i⟩$. Applying the Markov jump-process generator to either coordinate gives the exact moment equationsIn each equation the joint reaction removes one molecule, at reaction propensity function $C{X}_1{X}_2$. Because the two reactants are different chemical species, the propensity uses their product; no same-species factorial correction is needed. The exact identity $⟨X_1{X}_2⟩={\mu }_1{\mu }_2+\mathrm{Cov}(X_1,X_2)$ shows explicitly where the equations fail to close. Neglecting this covariance relative to the product of means is the small-fluctuation moment closure. It gives the mean-field joint removal of two chemical species:The difference equation $d({\mu }_1-{\mu }_2)/dt=-\beta ({\mu }_1-{\mu }_2)$ actually holds exactly as well as after closure, because both the constant production and the joint removal terms cancel. The nonlinear moment closure will be used for the subsequent phase plane and stationary linear noise approximation calculations, as requested.