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The acid dissociation constant quantifies the chemical equilibrium for loss of one proton from a chemical acid. In water, write the actual proton transfer as $\mathrm{HA}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{A}}^{-}$. Its full dimensionless equilibrium constant isThe usual aqueous acid dissociation constant uses the solvent convention with water thermodynamic activity approximated by one:Here hydrated hydrogen ions are understood; ${\mathrm{H}}^{+}$ is shorthand, not an isolated bare proton in the chemical solution. All thermodynamic activities and $K_a$ are dimensionless quantities, using consistent standard states. If water thermodynamic activity differs appreciably from one, retain it in the full equilibrium constant instead of silently mixing conventions.

With activity coefficients ${\gamma }_i$ and reference concentration $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$, put $a_i={\gamma }_i[i]/c^{\circ }$ for each solute. ThenThe last expression is the dilute ideal concentration approximation, not the definition in a nonideal chemical solution. Omitting $c^{\circ }$ instead produces a quotient with units of concentration; its numerical value depends on the chosen units. Do not take a logarithm of that dimensional quotient. Conditional concentration quotients can change with ionic strength even at fixed temperature, whereas the thermodynamic acid dissociation constant retains the activity coefficient corrections. The acid dissociation constant entry distinguishes these conventions.

At the same temperature, solvent and standard states, a larger acid dissociation constant, equivalently a smaller pKa, indicates a stronger tendency to donate a proton. It does not specify an analytical concentration, a pH, or a reaction rate. Composition and the weak acid mass and charge balances are needed to predict pH; chemical kinetics addresses how quickly the chemical equilibrium is approached.

ATP is a nucleotide consisting of adenosine, itself adenine joined to ribose, carrying three phosphate groups. From the ribose outward, these groups are labelled $\alpha$, $\beta$, and $\gamma$. ATP connects energy-supplying processes to energy coupling for synthesis and active transport; the same carrier can be regenerated and reused. ^[ref].

The net hydrolysis commonly used in energy coupling isHere ${\mathrm{P}}_i$ denotes inorganic phosphate. At buffered pH $7$, this biochemical shorthand groups the appropriate protonation states into pools and leaves proton exchange with the buffer solution implicit. An explicit, charge-balanced choice is ${\mathrm{ATP}}^{4-}+{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{ADP}}^{3-}+{\mathrm{HP}{\mathrm{O}}_4}^{2-}+{\mathrm{H}}^{+}$, without implying these are the only chemical species present. See the IUBMB recommendations for biochemical thermodynamics.

The quoted textbook biochemical standard state value is approximatelyThis convention uses fixed pH $7$, temperature $298.15\mathrm{K}$ ($25{}^{\circ }\mathrm{C}$), reference pressure $100\mathrm{kPa}$ ($1\mathrm{bar}$), unit reference thermodynamic activities for dissolved reactants and reaction products, and water thermodynamic activity approximated by one. The prime indicates the fixed-pH biochemical convention. See OpenStax Biology 2e, 6.4: standard ATP hydrolysis value and Biology 2e, 6.2: biochemical standard conditions. This is a reference value, rather than a universal value inside a biological cell; precise tables must also specify chemical solution conditions and any magnesium ion binding convention. Different reference conventions and data treatments give different tabulated values, as discussed in Li et al., 2010: a database of thermodynamic quantities for biochemical reactions.

An exergonic reaction has a negative change in Gibbs free energy, so the overall ATP hydrolysis is favourable under those reference conditions. Breaking a chemical bond by itself costs energy; forming new chemical bonds can release energy. The net change compares all reactants with all reaction products, including their interactions with water and the entropy contribution. Calling the phosphate connection a "high-energy bond" describes a large favourable change in Gibbs free energy on hydrolysis, rather than energy released by bond breaking alone. See 7.5 Strengths of Ionic and Covalent Bonds and Cooper, The Cell: ATP and metabolic energy.

Composition changes the driving force. At the same fixed pH, temperature, pressure, and binding convention, write the molar reaction change asHere $R$ is the molar gas constant, $T$ the absolute temperature, and $a$ a dimensionless thermodynamic activity for a biochemical pool. Constant water thermodynamic activity is absorbed into the reference term; fixed pH accounts for buffered protons. The shorthand $\Delta G$ means transformed molar reaction Gibbs free energy change, ${\Delta }_r{G}^{\prime }$, rather than a finite sample's total change. The natural logarithm acts on a dimensionless reaction quotient, never on concentrations with units. See the IUBMB biochemical thermodynamics recommendations and the general equilibrium constant relation.

For a stipulated ideal dilute chemical solution, let $a_i=[i]/c^{\circ }$, with $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$. Choose $[\mathrm{ATP}]=5\mathrm{mmol}{\mathrm{L}}^{-1}$, $[\mathrm{ADP}]=0.5\mathrm{mmol}{\mathrm{L}}^{-1}$, and $[{\mathrm{P}}_i]=1\mathrm{mmol}{\mathrm{L}}^{-1}$, ignoring activity coefficient corrections and magnesium binding. These are illustrative inputs, not a claim about a particular biological cell. ThenandIncreasing $Q$ tenfold raises $\Delta G$ by $RT\ln 10\approx 5.708\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ at this temperature. Thus the ATP to ADP ratio and available inorganic phosphate matter, even with the same standard state value. These are local molar predictions at the stipulated composition, rather than a constant total yield as an isolated sample reacts.

Hexokinase provides a concrete connection between sugar chemistry and energy coupling: it transfers a terminal phosphate group from ATP to glucose, forming a phosphorylated sugar and adenosine diphosphate. The enzyme accelerates the transfer, while consuming ATP changes the net chemical reaction being carried out. The donated group reaches the substrate through the catalytic route.^[ref]

For a symbolic teaching model, write $S$ for an unphosphorylated sugar, $S\text{-}P$ for its phosphorylated form, and $P_i$ for inorganic phosphate. These are labels for chemical states, not complete elemental formulas. Using one consistent reference convention at fixed pH and temperature, stipulate the following molar changes in Gibbs free energy:All three numbers are chosen illustrative values, not measured properties of glucose or a particular enzyme. The prime marks a biochemical reference convention with pH held fixed; explicit proton and metal-binding bookkeeping is suppressed consistently. The first two lines are a thermodynamic decomposition: cancelling water and inorganic phosphate gives the net transfer. They do not assert that the reaction mechanism first releases free inorganic phosphate and then recaptures it.

Because Gibbs free energy is a state function, consistently defined reaction changes add. A working reaction mechanism must also connect the desired conversion to ATP consumption; independently allowing ATP hydrolysis beside the sugar does not establish energy coupling. The general distinction is developed under energy coupling. ATP-dependent phosphorylation of a membrane protein can likewise connect a chemical change to active transport.^[ref]

Actual Gibbs free energy changes depend on thermodynamic activities. Even a favourable reference sum can be overcome by accumulation of reaction products or depletion of ATP; it supplies no prediction of the reaction rate. A living biological cell also regenerates ATP. Sustained operation therefore requires a connected metabolism, not an inexhaustible stock of a special molecule.

Define side $P$ as the higher proton electrochemical potential reservoir and side $N$ as the synthesis-side reservoir. For transfer $P\to N$ at common temperature $T$, define the positive downhill proton-motive force magnitude $\Delta p_{\mathrm{drive}}$ byHere $a$ is thermodynamic activity, $\psi$ is electric potential, $R$ is the molar gas constant and $F$ is the Faraday constant. Thus both a higher proton thermodynamic activity and a more positive electric potential on side $P$ contribute to the driving magnitude. This definition fixes the sign; conventions for a signed proton-motive force differ. The calculation under chemiosmosis separates these two contributions numerically.

Assume fixed reservoirs, tight mechanical energy coupling, one transported proton per $c$ subunit per revolution, and three ATP produced per revolution. Neglect leaks, slipping and membrane transport of the chemical reagents. Let $g_{\mathrm{ATP}}>0$ be the local molar Gibbs free energy cost of ATP formation at the stipulated composition, pH and magnesium-binding convention. It is not necessarily the standard state value. Adding changes in the state function gives, per mole of complete turns,For this model net synthesis is thermodynamically favoured when $\Delta G_{\mathrm{turn}}<0$. Equality is a reversible balance, not a promise of a useful reaction rate. A negative value establishes an available driving margin; it does not specify how fast the enzyme runs.

Choose $n=10$, $g_{\mathrm{ATP}}=50.0\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, $F=96485\mathrm{C}\mathrm{mo}{\mathrm{l}}^{-1}$ and $\Delta p_{\mathrm{drive}}=0.180\mathrm{V}$. These are illustrative inputs, not measurements of the preparation in the experiment below. Thenand $\Delta G_{\mathrm{turn}}=-23.673\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Per ATP, the ideal input is $57.891\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, leaving a driving margin of $7.891\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The balance point isAt $0.140\mathrm{V}$ the same synthesis cycle instead has $\Delta G_{\mathrm{turn}}=+14.921\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$: the chosen reservoirs favour its reverse.

The machine is reversible: sufficient ATP hydrolysis can drive opposite rotation and pump protons uphill. Actual direction also depends on inhibition and available substrates; a weak proton-motive force need not make every intact enzyme immediately run backward. Respiratory uncoupling supplies a different failure mode: a parallel proton leak dissipates the gradient without productive shaft rotation.^[ref]

The motor budget $10/3$ is not the whole cost of ATP delivered outside a mitochondrion. Import of ADP and inorganic phosphate, and export of ATP, also draw on the electrochemical gradient.^[ref] For a deliberately simplified system budget, assign these transports together one additional downhill-proton equivalent per delivered ATP, valued at the same $F\Delta p_{\mathrm{drive}}$. This is a chosen accounting convention, not a claim of one extra literal proton through $F_0$, nor a universal transport stoichiometry. The total becomes $10/3+1=13/3$ equivalents and $75.258\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ of gradient expenditure per delivered ATP at $0.180\mathrm{V}$. The ratio of stored chemical free energy to that expenditure is $50/75.258\approx 66.4\%$, compared with $50/57.891\approx 86.4\%$ for the motor alone. This system budget changes the resource expenditure, not the motor's balance point $\Delta p_{\ast }$. Real transport terms require their own electrochemical potentials, and leaks add further cost.

A pressure unit defined by $1\mathrm{bar}=10^5\mathrm{Pa}$. In a logarithmic gas thermodynamic activity, a pressure expressed in bars must still be divided by the same-unit reference pressure to form a dimensionless ratio.

For the conjugate base ${\mathrm{A}}^{-}$ accepting a proton from water,This base dissociation constant uses the same water thermodynamic activity approximation and standard state conventions as the acid dissociation constant. Multiplying the two relations cancels the partner thermodynamic activities:Here $K_w$ is the aqueous ion product associated with self-ionization of water. At fixed temperature, a larger acid dissociation constant gives a smaller base dissociation constant for its own conjugate base; this reciprocal rule compares matching conjugate acid-base pairs. A weak acid does not necessarily have a conjugate base that reacts almost completely with water. For example, stipulated $K_a=10^{-5}$ and $K_w=10^{-14}$ give $K_b=10^{-9}$.^[ref]

The value of $K_w$ changes with temperature; $10^{-14}$ is the familiar approximate value near $25{}^{\circ }\mathrm{C}$, not a universal constant. If water is retained explicitly in both full equilibrium constants, their product instead equals $a_{{\mathrm{H}}_3{\mathrm{O}}^{+}}{a}_{{\mathrm{OH}}^{-}}/a_{{\mathrm{H}}_2\mathrm{O}}^2$, the full equilibrium constant for $2{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{OH}}^{-}$. The acid dissociation constant, base dissociation constant and self-ionization of water convention must agree.

For a nonideal chemical solution, write its thermodynamic activities on a mole fraction basis as $a_A=x{\gamma }_A$ and $a_B=(1-x){\gamma }_B$. Use composition-independent pure-component reference chemical potentials at the same temperature and pressure:The activity coefficients describe departure from the ideal solution under that reference convention.^[ref]

At fixed temperature and pressure, insert these expressions into the Gibbs-Duhem equation. The ideal solution terms cancel because $xd\ln x+(1-x)d\ln (1-x)=dx-dx=0$. What remains isThus two independent fits to measured activity coefficients can each look smooth while failing to describe one possible Gibbs free energy. This is a necessary consistency condition; satisfying it alone does not establish experimental accuracy or thermodynamic stability.

For an original algebraic example, suppose at one fixed temperature and pressure that $\ln {\gamma }_A=\alpha (1-x{)}^2$, with dimensionless constant $\alpha$. The constraint requires $d\ln {\gamma }_B/dx=2\alpha x$. With the pure-B normalization ${\gamma }_B(0)=1$, integration givesFor $\alpha =1/2$ and $x=1/4$, the derivatives of the two logarithms are $-3/4$ and $1/4$; their weighted sum is $-3/16+3/16=0$. Choosing $\ln {\gamma }_B=\alpha x$ instead would give a weighted sum $3/16$ at that composition, contradicting the Gibbs-Duhem equation. These are invented model functions, with no claim to fit a particular mixture.

The fixed-temperature, fixed-pressure restriction matters. Along a path with changing temperature or pressure, the general equation contains $SdT-Vdp$, and the reference chemical potentials and $RT$ also change. Across phase coexistence, apply the relation separately within each smooth thermodynamic phase; do not differentiate a discontinuous switch between phases as if it were a single composition curve.

Stipulate pKa $5$ and $K_w=10^{-14}$; these are model inputs, not measurements of an identified substance. At pH $5$ the pair contribution is $0.575646\dots C$. For analytical concentrations $C=0.010$ and $0.100\mathrm{mol}{\mathrm{L}}^{-1}$, respectively, it isThe free hydrogen ion and hydroxide contribution is the same at this pH, ${\beta }_{\mathrm{water}}=(\ln 10)(10^{-5}+10^{-9})\mathrm{mol}{\mathrm{L}}^{-1}=0.0000230282\dots \mathrm{mol}{\mathrm{L}}^{-1}$ per pH. The full buffer capacities are therefore approximately $0.00577949$ and $0.0575877\mathrm{mol}{\mathrm{L}}^{-1}$ per pH. Identical pH fixes the free hydrogen ion thermodynamic activity; it does not fix the inventory of weak acid and conjugate base available for neutralization.

At one unit of pH on either side of pKa, $r=10$ or $0.1$, giving ${\beta }_{\mathrm{pair}}/{\beta }_{\mathrm{pair},\max }=40/121\approx 0.331$. The partners still form a buffer solution, but its local resistance is only about one third of the maximum for that analytical concentration. At fixed pH, the pair contribution scales exactly with $C$; the full buffer capacity contains the additional $C$-independent water contribution. Consequently, extreme dilution does not justify dropping self-ionization of water.

A buffer solution contains appreciable quantities of a weak acid and its conjugate base, and resists changes in pH caused by small quantities of added chemical acid or chemical base. The conjugate base accepts added protons; the weak acid supplies protons when a chemical base consumes them. Consuming most of either partner undermines the buffering action. See Flowers et al., Chemistry 2e, buffers.

For a separate toy calculation, stipulate pKa $5$ for a weak acid $\mathrm{HA}$; this is an input, not measured material data. Initially $1\mathrm{L}$ of buffer solution contains $0.10\mathrm{mol}$ of $\mathrm{HA}$ and $0.10\mathrm{mol}$ of its conjugate base ${\mathrm{A}}^{-}$. In the ideal concentration approximation, the Henderson-Hasselbalch equation gives initial pH $5$. Add $0.010\mathrm{mol}$ of hydroxide with negligible volume change. The dominant neutralization has one-to-one stoichiometry:Each mole of consumed hydroxide removes one mole of weak acid and forms one mole of conjugate base and water. Thus the inventories after neutralization areThe shared volume cancels from the concentration ratio. The Henderson-Hasselbalch equation therefore predictsThis small increase illustrates how conversion between abundant partners absorbs added chemical base. The calculation neglects the tiny free hydronium and hydroxide inventories, changes during re-establishment of chemical equilibrium, self-ionization of water, and activity coefficient corrections. The displayed inventories approximate the final chemical species distribution; they are not an exact solution of all chemical equilibrium conditions.

A buffer solution has a finite buffer capacity: its weak acid inventory supplies protection against added chemical base, and its conjugate base inventory supplies protection against added chemical acid. Here an addition approaching $0.10\mathrm{mol}$ of hydroxide uses up the available $\mathrm{HA}$ in the dominant stoichiometry calculation. Close to that limit, the neglected chemical equilibrium contributions matter; once $\mathrm{HA}$ is exhausted, inserting a zero denominator into the Henderson-Hasselbalch equation cannot predict the actual pH. Excess hydroxide and the remaining chemical species require a fresh chemical equilibrium calculation. This is why a buffer solution stabilizes conditions for chemical reactions over a finite range of additions. See Flowers et al., Chemistry 2e, buffer capacity and added base.

For the pair $\mathrm{HA}/{\mathrm{A}}^{-}$, rearranging the acid dissociation constant gives the activity relationWith thermodynamic activity of water approximated by one and ideal dilute thermodynamic activities for both partners, this yields the familiar Henderson-Hasselbalch equation:The concentrations are at chemical equilibrium; using initial values requires negligible equilibration changes. Below, also neglect self-ionization of water and competing chemical reactions.

Stipulate $\mathrm{p}{K}_a=4.76$ for an illustrative weak acid, without identifying it with a measured substance. A conjugate base to weak acid concentration ratio of two givesEqual concentrations give a ratio of one, whose logarithm is zero, hence pH $4.76$. Multiplying the ratio by ten raises pH by one: ratios $0.1$, $1$, and $10$ give pH values $3.76$, $4.76$, and $5.76$. The ratio controls the ideal prediction, while the absolute quantities control how much added chemical acid or chemical base the buffer solution can accommodate.

A catalyst makes a chemical reaction proceed faster by participating in an alternative reaction mechanism and being regenerated. This makes repeated conversion possible: the same catalyst can help many successive reactants become reaction products. An enzyme is a biological catalyst. The overall standard change in Gibbs free energy is unchanged; see the IUPAC Gold Book: catalyst.

Regeneration does not mean that the catalyst sits unchanged throughout every step. For example, a schematic reaction mechanism can temporarily bind A to a catalyst C in a reaction intermediate AC:Adding the steps cancels both the reaction intermediate AC and the catalyst C from the net chemical equation. AC is produced and later consumed; C is used and later recovered. These letters describe possible roles, not a measured reaction mechanism. The alternative route is useful only if its chemical kinetics permits faster conversion. See Flowers et al., Chemistry 2e, §12.7: catalysts and alternative mechanisms.

For a worked barrier example, consider an idealized reversible A-to-B conversion at fixed temperature $T=300\mathrm{K}$. Stipulate standard molar Gibbs free energy levels $G_{\mathrm{A}}^{\circ }=0$ and $G_{\mathrm{B}}^{\circ }=-20\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, with effective transition state levels $80$ without a catalyst and $50\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ with a catalyst. Use the same standard states and the same zero for both routes; the catalytic route includes the regenerated C at both endpoints. Unlike the separate potential-energy profile example, this illustration uses Gibbs free energy at one specified temperature.

The Gibbs energy of activation is measured from the starting endpoint in the chosen direction. Thus the four barriers areBoth directional barriers fall by $30\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The endpoint difference stays ${\Delta }_r{G}^{\circ }=-20\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$.

To turn those barriers into rate constants, use the Eyring form of transition state theory for this effective first-order reaction pair, with a common prefactor and unit probability of successful passage:Here $k_{\mathrm{B}}$ is the Boltzmann constant, $h$ is the Planck constant, and $R$ is the molar gas constant. The common prefactor cancels from the comparisons. These Gibbs energies of activation are not the empirical activation energy $E_a$ of the Arrhenius equation. With the stipulated value $R=8.314462618\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, the speedup isThe exponent is a dimensionless quantity: convert $30\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ to $30000\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ before dividing by $RT$. The exponential function explains why a modest barrier reduction can have a dramatic kinetic effect. This number follows from the chosen free-energy barriers and common prefactor; it is not a universal catalytic multiplier or a prediction across temperatures.

Assume ideal thermodynamic activities with a common concentration standard state, and directional rate laws $J_f=k_f[\mathrm{A}]$ and $J_r=k_r[\mathrm{B}]$. At chemical equilibrium, equal reaction rates give $K=[\mathrm{B}]/[\mathrm{A}]=k_f/k_r$ for this model. The barriers giveThe catalyst speeds up both directions, so it changes how quickly chemical equilibrium is approached without changing this equilibrium constant. For a general chemical reaction, the thermodynamic relation is $K=\exp [-{\Delta }_r{G}^{\circ }/(RT)]$ with consistent standard states; an arbitrary multistep reaction mechanism need not permit $K=k_f/k_r$ for two measured rate constants. At the same temperature, pressure, and thermodynamic activities, a catalyst also leaves the actual overall reaction change in Gibbs free energy unchanged. See 13.3 Shifting Equilibria: Le Châtelier’s Principle (§13.3: a catalyst does not shift equilibrium). To drive an unfavourable overall conversion, a process such as energy coupling must change the combined transformation.

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

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

Consider two internally equilibrated reservoirs A and B at the same imposed temperature and pressure. They may have different compositions or represent different phases. Transfer a small positive amount of substance $\delta n$ of the same chemical species from A to B, holding the other inventories fixed. To first order, their total Gibbs free energy changes byHence exchange toward lower chemical potential lowers Gibbs free energy. If transfer is allowed in both directions and the two reservoirs contain the exchangeable component, thermodynamic equilibrium requires equal chemical potentials for that component. See MIT 5.60, Lecture 14, pages 1–2.

An impermeable partition can maintain unequal chemical potentials; equality is a condition on an allowed exchange. Two pure samples of the same ideal gas already have equal chemical potentials at the same temperature and pressure. Different compositions at a common pressure can instead give different chemical potentials for one component through its thermodynamic activity.

Equality concerns the same exchangeable component in different places, not all different chemical species. For a permitted chemical reaction, the corresponding interior chemical equilibrium criterion is ${\sum }_i{\nu }_i{\mu }_{\mathrm{m},i}=0$, with signed stoichiometric vector entries ${\nu }_i$. The existing chemical equilibrium and constrained Gibbs minimization for chemical equilibrium sections develop those composition constraints.

For a pure ideal gas at temperature $T$ and pressure $p$,where $R$ is the molar gas constant and $p^{\circ }$ specifies the reference pressure of the standard state. The ratio inside the natural logarithm is dimensionless. The chemical potential difference therefore depends on a ratio of pressures, not an absolute difference. See MIT 5.60, Lecture 14, page 4.

For a chemical species in an ideal gas mixture, replace $p$ by its partial pressure. More generally write ${\mu }_{\mathrm{m},i}={\mu }_{\mathrm{m},i}^{\circ }+RT\ln a_i$ using its thermodynamic activity $a_i$ and a consistent standard state. For an ideal dilute chemical solution, $a_i=c_i/c^{\circ }$ with reference concentration $c^{\circ }$; the thermodynamic activity section describes departures from this idealization.

As a computed comparison, take identical chemical species at $T=300\mathrm{K}$, with $p_A=2p^{\circ }$ and $p_B=p^{\circ }$. Stipulate $R=8.314462618\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. ThenHalving the pressure gives the same decrement whether starting from $2p^{\circ }$ or $20p^{\circ }$, provided the ideal gas approximation remains valid. This comparison uses different pressures; it is not the common-pressure reservoir setup of chemical potential equality for particle exchange. The fixed-volume example in chemical potential and Helmholtz free energy supplies an exchange experiment with unequal initial pressures.

Chemiosmosis couples transmembrane ion movement down an electrochemical gradient to a chemical transformation. For mitochondrial ATP synthesis, protons return from the intermembrane space to the mitochondrial matrix through ATP synthase.^[ref]

An original thermodynamic example separates the electrical and chemical contributions. At fixed temperature $T$, use ideal dilute thermodynamic activities $a$ and the electric potential $\psi$ on each side. The molar electrochemical potential difference for inward transfer of a singly charged proton isHere $R$ is the molar gas constant, $F$ the Faraday constant, and $\ln$ the natural logarithm. The sign convention is destination minus origin: a negative difference makes inward transfer favourable. Both lower proton activity and a more negative electric potential inside favour inward movement.

Stipulate $T=300\mathrm{K}$, $a_{\mathrm{matrix}}/a_{\mathrm{outside}}=0.1$, ${\psi }_{\mathrm{matrix}}-{\psi }_{\mathrm{outside}}=-0.12\mathrm{V}$, $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $F=96485\mathrm{C}\mathrm{mo}{\mathrm{l}}^{-1}$. These illustrative inputs give approximately $-5.743$ and $-11.578\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ for the chemical and electrical terms, hence $-17.321\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ per transferred proton. For a hypothetical device coupling three inward protons per ATP to an ATP-formation cost of $45\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$,The sign follows by adding changes in the state function Gibbs free energy. It establishes thermodynamic feasibility for this chosen model, not its reaction rate. Three is a stipulated coupling count, not a universal ATP synthase stoichiometry; transport costs and leaks can also consume the available free energy.

A group of sheet-silicate minerals; an idealized Mg-rich composition used in weathering models is $\mathrm{M}{\mathrm{g}}_5\mathrm{A}{\mathrm{l}}_2\mathrm{S}{\mathrm{i}}_3{\mathrm{O}}_{10}(\mathrm{OH}{)}_8$. Its alteration to gibbsite or kaolinite depends on dissolved silica and the magnesium-to-proton thermodynamic activity ratio.

A conjugate base is what remains after a chemical acid donates one proton. In the conjugate acid-base pair $\mathrm{HA}/{\mathrm{A}}^{-}$, $\mathrm{HA}$ is the conjugate acid of ${\mathrm{A}}^{-}$; accepting a proton reverses the conversion. For example, acetic acid $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$ loses a proton to give its conjugate base $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{\mathrm{O}}^{-}$. Partners differ by one proton and one unit of electric charge; they need not always be neutral and negative. Hydronium/water and water/hydroxide are also conjugate acid-base pairs.

At a prescribed pH, the ideal dilute acid dissociation constant determines the partition of a monoprotic acid inventory between its chemical species:These dimensionless fractions sum to one. They are the weak acid dissociation fraction and its complement when the only acid-bearing chemical species are $\mathrm{HA}$ and ${\mathrm{A}}^{-}$. Equal fractions occur at $\mathrm{pH}=\mathrm{p}{K}_a$ in this ideal approximation. For nonideal chemical solutions, equality of thermodynamic activities need not mean equality of concentrations.

For stipulated pKa $5$, the conjugate base fractions at pH $4$, $5$ and $6$ are $1/11$, $1/2$ and $10/11$. Here an external reservoir sets pH; its ions must supply the additional electric charge balance. The diagram omits those ions and water populations. It does not represent three pure $\mathrm{HA}$ samples at identical analytical concentration with freely chosen pH. In the latter experiment, the weak acid mass and charge balances determine pH as well as the partition.

A Daniell cell uses the zinc/${\mathrm{Zn}}^{2+}$ and ${\mathrm{Cu}}^{2+}$/copper pairs. In the separated-compartment teaching version, a zinc electrode contacts an aqueous chemical solution containing ${\mathrm{Zn}}^{2+}$, and a copper electrode contacts one containing ${\mathrm{Cu}}^{2+}$. A salt bridge supplies an ionic connection; a wire and load supply an electronic connection. During discharge,The zinc electrode loses material and copper is deposited at the cathode. Each half-reaction conserves chemical atom counts and electric charge: the first has net charge zero on each side, the second $+2e-2e=0$ on the left and zero on the right. The net reaction has $+2e$ on both sides, with no electrons left in its chemical equation. The Zn/Cu teaching cell and these half-reactions are described in Florida State University, Redox Reactions in Voltaic Cells: Construction of a Potential Series, lab background.

Creating one mole of ${\mathrm{Zn}}^{2+}$ adds $2F$ of positive electric charge to the anode compartment's ionic inventory; consuming one mole of ${\mathrm{Cu}}^{2+}$ removes $2F$ of positive electric charge from the cathode compartment. Here $F$ is the Faraday constant. In an ideal bookkeeping example using a potassium nitrate salt bridge, two moles of nitrate anions entering the anode compartment compensate the first change, and two moles of potassium cations entering the cathode compartment compensate the second. Other ions in the cell can share the actual transport. The two compensating inventory changes do not mean that $4F$ flowed through the load: the same series electric current transports $2F$ through each cross-section of the circuit per mole of net reaction. See salt bridge.

Under standard-state conditions near $298\mathrm{K}$, tabulated reduction potentials give a cell electromotive force of approximately $1.10\mathrm{V}$. This is a reversible reference value, not a promised loaded reading; actual composition matters through thermodynamic activity. The tabulated copper and zinc values and their subtraction are given in Purdue University, Electrochemical Reactions: standard-state reduction potentials. For the net chemical equation as written, the molar Gibbs free energy change obeysThus $\mathcal{E}=1.10\mathrm{V}$ corresponds to about $-212\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ of net reaction. At fixed temperature and pressure, $-{\Delta }_{r}G$ is the maximum reversible electrical work per mole for this cell reaction. It is not automatically all the heat released by a finite-current cell. See 17.4 Potential, Free Energy, and Equilibrium (§17.4: cell potential and free energy).

Energy coupling links an endergonic reaction to an exergonic reaction through a reaction mechanism, so that the combined transformation has a favourable change in Gibbs free energy. ATP hydrolysis is a common driving chemical reaction. A shared reaction intermediate, produced in one step and consumed in another, can physically connect the transformations; merely placing two independent chemical reactions in the same chemical solution does not establish energy coupling. See Alberts et al., Molecular Biology of the Cell: coupled reactions and activated intermediates.

For a computed example, stipulate an endergonic reaction $\mathrm{A}⟶\mathrm{B}$ with $\Delta G_1^{\circ \prime }=+20\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ under precisely the same biochemical standard state conditions as the quoted ATP hydrolysis value. Assume a reaction mechanism couples one ATP hydrolysis to each A-to-B conversion. The formal sum isThe sum $+20+(-30.5)=-10.5$ gives an exergonic reaction. The $+20$ is a chosen teaching value, not a measured property of an identified substance. Outside these reference conditions, add the actual $\Delta G_1$ and $\Delta G_2$, each evaluated using consistent thermodynamic activities; mixing a standard value for one chemical reaction with an unrelated cellular value for the other gives no justified prediction.

Why do the changes in Gibbs free energy add? Gibbs free energy is a state function. Adding consistent reaction changes cancels any shared reaction intermediate and leaves the difference between the combined reactants and combined reaction products. This explains the thermodynamic accounting; it does not construct the reaction mechanism. In many ATP-dependent routes, phosphorylation transfers a phosphate group to a substrate or an enzyme, and subsequent transformation consumes that modified material. The shared material is what connects the steps. Other reaction mechanisms use ATP differently. See Alberts et al., Molecular Biology of the Cell: ATP-driven biosynthesis.

The useful achievement is that a common regenerated nucleotide can connect many different processes in metabolism. The net negative change in Gibbs free energy permits the coupled conversion, but supplies no reaction rate: enzymes can lower activation energy barriers without changing the endpoint difference. Energy coupling changes which overall chemical reaction is being considered; catalysis accelerates an available route. See OpenStax Biology 2e, 6.2: free-energy changes and reaction barriers.

A historical landmark is Fritz Lipmann's ^[ref]. Its sections on the phosphate cycle, generation and utilization of phosphate-bond energy, and group transfer organized these processes around a reusable chemical carrier. The historical terminology should be read with the modern distinction between net hydrolysis and the cost of breaking a chemical bond.

For a specified chemical reaction and standard state convention, the dimensionless equilibrium constant is the equilibrium value of the reaction quotient $Q={\prod }_i{a}_i^{{\nu }_i}$, where $a_i$ denotes thermodynamic activity and ${\nu }_i$ a signed stoichiometric vector entry. Thermodynamic activities incorporate the reference state and departures from ideal behavior. See IUPAC Gold Book, equilibrium constant.

In the ideal concentration model for A converting to B, use $a_{\mathrm{A}}=[\mathrm{A}]/c^{\circ }$ and $a_{\mathrm{B}}=[\mathrm{B}]/c^{\circ }$ with the same reference $c^{\circ }=1\mathrm{M}$. Their ratio gives $Q=[\mathrm{B}]/[\mathrm{A}]$ and $K=0.75/0.25=3$. Equality $K=k_f/k_r$ applies to this elementary first-order pair; it is not a universal ratio of two measured rate constants for an arbitrary multistep chemical reaction.

At a specified constant temperature, with consistent standard states, the standard molar reaction Gibbs free energy isHere ${\Delta }_{r}G$ is the local slope per mole of extent of reaction at fixed temperature and pressure, rather than the finite total Gibbs free energy change between two mixtures. At equilibrium ${\Delta }_{r}G=0$, while ${\Delta }_r{G}^{\circ }$ generally remains nonzero. For $K=3$ at $298\mathrm{K}$, it is about $-2.72kJ/mol$. For positive activities, $Q<K$ favors forward conversion and $Q>K$ favors reverse conversion; neither sign supplies a reaction rate. See MIT 5.111, Lecture 18, reaction free energy and equilibrium.

A battery's charge reservoir means a finite amount of electric charge can pass through the external electrical circuit before its usable chemical state is depleted. It does not mean a large net electric charge is sitting inside the casing. For one net cell chemical reaction transferring $n$ electrons per reaction event, an ideal reaction inventory obeyswhere $F$ is the Faraday constant and $\xi$ the extent of reaction in moles, assuming all the electric current serves that reaction. At fixed temperature and pressure, its reversible electromotive force obeys ${\Delta }_{r}G=-nF\mathcal{E}$: a decrease in Gibbs free energy provides the maximum reversible electrical work per mole. The electromotive force also depends on composition; the Daniell cell illustrates this with thermodynamic activity.^[ref]

As an original finite-reservoir exercise, stipulate usable $Q_{\max }=360\mathrm{C}$ and hold the preceding $3\mathrm{V}$, $1\Omega$, $4\Omega$ parameters constant until an abrupt cutoff. Then $t=Q_{\max }/I=600\mathrm{s}$, available electrical work is $\mathcal{E}{Q}_{\max }=1080\mathrm{J}$, load work is $V_L{Q}_{\max }=864\mathrm{J}$, and internal dissipation is $216\mathrm{J}$. This explicitly finite idealization predicts ten minutes, not an experimentally measured runtime. Real accessible charge depends on discharge conditions and the chosen terminal-voltage cutoff. Charge alone is also not an energy: converting it to delivered work requires the terminal electric potential difference throughout discharge.