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A specified limiting composition used to describe a solid solution in a mineral. Its thermodynamic activity depends on the chosen standard state and activity model. For example, pyrope is a magnesian garnet endmember, while phlogopite is a magnesian biotite endmember.

For a cell reaction transferring $n$ electrons, $E=E^{\circ }-(RT/nF)\ln Q$, with dimensionless reaction quotient $Q$ and consistently chosen standard states.

Let $r=a_{\mathrm{ADP}}/a_{\mathrm{ATP}}$ and use a one-molar standard state, so the inorganic phosphate activity is approximately $a_P=0.006$. With the source's stated standard hydrolysis value,The downhill proton-transfer Gibbs free energy is $F\Delta p$ for the signed driving force $\Delta p=-0.150$ V. Ideal reversible balance of one complete turn requiresSolving yieldsAt 298.15 K the estimate is $3.39\times 10^2$. Temperature is not specified in this paragraph; the first value explicitly adopts physiological temperature rather than pretending it is supplied. The unexpectedly large ADP/ATP ratio follows from using the printed standard value $-50$ kJ/mol with the stated dilute inorganic phosphate. One must not silently replace that value by another convention or treat it as the already concentration-corrected cellular hydrolysis Gibbs free energy.

This estimate assumes ideal thermodynamic activities, consistent pH and magnesium conventions, tight coupling of ten protons to three ATP, no leak or transport overhead, and operation close to reversible balance. A working cell is not in global equilibrium. Net synthesis at this fixed driving force requires $\Delta G_{\mathrm{turn}}<0$, hence $r$ above this reversible-balance threshold; finite chemical kinetics and dissipation prevent the stoichiometry alone from determining an exact physiological ratio.

Use the supplied $-50\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ as a standard transformed Gibbs free energy of ATP hydrolysis, rather than replacing it with a different commonly quoted standard value. Approximate thermodynamic activities by concentrations divided by $c^{\circ }=1\mathrm{M}$, at an explicitly assumed $T=300\mathrm{K}$. The dimensionless reaction quotient isConsequentlyand the reverse synthesis cost is $g_{\mathrm{ATP}}=63.22\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Substituting bare millimolar numbers in this quotient without the standard state would give the wrong logarithm.

For outside-to-inside proton movement, the membrane potential is $\Delta \psi =-0.125\mathrm{V}$ and ${\mathrm{pH}}_{\mathrm{in}}-{\mathrm{pH}}_{\mathrm{out}}=0.5$, so $[H^{+}{]}_{\mathrm{in}}/[H^{+}{]}_{\mathrm{out}}=10^{-0.5}$. The molar electrochemical potential change isThus the available proton-motive force magnitude is $0.15476\mathrm{V}$, supplying $14.93\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ per transported proton. Assuming one proton per c-subunit per revolution, three ATP per revolution and no leakage, slippage or transport overhead,The smallest integer permitting synthesis is therefore $\boxed{n_c=13}$ in this ideal budget. A ring of thirteen gives only a small driving margin; inefficiency or sustained finite output could require a larger ring. The result is an energetic lower bound and near-reversible estimate, not a unique structural determination from these data. Reasonable temperatures near $300$–$310\mathrm{K}$ leave the minimum integer thirteen. Additional membrane transport costs have been excluded explicitly.

In water, separated benzene molecules require separate hydration environments around nonpolar surface. Bringing them together buries part of that surface and reorganizes the surrounding water. The resulting solvent contribution favours association relative to cyclohexane, where benzene was already well solvated by dispersion. This solvent-mediated contribution is the hydrophobic effect, not formation of a new covalent bond or a claim that benzene has become a conventional hydrogen-bond acceptor like oxygen.

The supplied constants imply a thirtyfold enhancement. With matched standard states,at an illustrative 298 K. The water value is still slightly below $1{\mathrm{M}}^{-1}$, so enhancement relative to cyclohexane does not mean a negative standard association free energy for a 1 M standard state, nor does it imply that dilute benzene is all dimer. No universal “entropy only” explanation is needed: solvent enthalpy and entropy both contribute and depend on temperature.

Association in solution includes replacement of solute–solvent contacts, not just formation of the isolated dimer. cyclohexane solvates the benzene surface well through dispersion. Much of the London dispersion gained on bringing two benzene molecules together replaces comparable benzene–cyclohexane contacts, leaving only a modest net gain before the translational and orientational entropy penalty. cyclohexane cannot provide the complementary donor/acceptor interactions available when two waters hydrogen-bond, so water association gains a more distinctive interaction relative to its separated solvated monomers.

Using the same concentration standard state, $K^{\circ }=K_c{c}^{\circ }$ and $\Delta G_{\mathrm{assoc}}^{\circ }=-RT\ln K^{\circ }$. For illustration at 298 K and $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$, the ratio of the two supplied association constants corresponds toThus the water-molecule pair can associate more favourably in this solvent even though the stipulated isolated-pair binding energies are similar. Gas-phase potential energy alone omits both differential solvation and the entropy of association.

Transfer is favourable when new water–solute interactions compensate the cavity and reorganization costs of accommodating the solute. The Gibbs energy also includes the loss of solute–alkane interactions; it is not simply a count of hydrogen bonds. isobutane cannot compensate those costs with strong directional interactions, explaining its positive transfer Gibbs energy. acetone and both amines can accept hydrogen bonds from water and have favourable electrostatic solvation, making their transfer moderately favourable. isopropylamine can additionally donate to water, consistent with a somewhat greater preference for water than trimethylamine.

acetamide combines a strong carbonyl acceptor with N–H donors and has the most favourable hydration in the group, accounting for its much more negative transfer Gibbs energy. Its pure solid lattice is not an extra sublimation penalty to insert here: the stated process starts from a dilute alkane solution and ends in dilute water, rather than dissolving the pure crystal. The nonmonotonic relationship between hydrogen-bond acceptor parameter and transfer Gibbs energy is expected because donor strength, nonpolar surface, solvent cavity formation, and all other solvation contributions also enter.

For matched standard states, $\Delta G_{\mathrm{tr}}^{\circ }=-RT\ln K_{\mathrm{tr}}$, with $K_{\mathrm{tr}}$ the water-to-alkane activity ratio. As an illustrative 298 K calculation, the hydrocarbon value implies $K_{\mathrm{tr}}\simeq 7.0\times 10^{-4}$, whereas the amide value implies approximately $3.6\times 10^4$. These are standard-state partition preferences, not absolute aqueous solubilities. amine protonation could introduce an additional pH-dependent contribution to an experimentally measured distribution coefficient, but no such speciation information is supplied and it is not needed for the molecular comparison.

The standard enthalpy of formation is the enthalpy change on forming exactly one mole of the specified compound in its stated standard state from its constituent chemical elements in their reference states at the same specified temperature. The reference states are normally the stable elemental forms at the standard pressure. A standard state does not itself mean 298 K; temperature must be stated separately.

For a reversible temperature change at constant pressure and fixed composition, $\mathrm{d}{S}_m=\delta q_{\mathrm{rev},\mathrm{m}}/T$ and $\delta q_{\mathrm{rev},\mathrm{m}}=C_{p,m}\mathrm{d}T$. HenceWith $T_0=298\mathrm{K}$, termwise integration yieldsThe last sign follows from integrating $e{T}^{-3}$, not $e{T}^{-2}$. Every term has units $\mathrm{J}{\mathrm{K}}^{-1}\mathrm{mo}{\mathrm{l}}^{-1}$; the logarithm has a dimensionless argument. Inserting 800 K gives an entropy increase of $41.733\mathrm{J}{\mathrm{K}}^{-1}\mathrm{mo}{\mathrm{l}}^{-1}$, and thereforeThe standard pressure remains fixed during this integration; the 200-bar equilibrium mixture is not the standard state. Differentiating the integrated expression recovers $C_p/T$, providing a check on all five terms. Using the ammonia heat-capacity polynomial here does not retroactively replace the explicitly requested constant-reaction-enthalpy approximation in the preceding equilibrium calculation.

At the one-bar standard state, pure phases have unit thermodynamic activities. The Gibbs energy of reaction balance is ${\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }=0$, givingThis uses the stipulated temperature-independent reaction enthalpy and entropy differences, taken per mole of the written reaction.

At the one-bar standard state, pure mineral endmembers and water vapour treated as an ideal gas all have unit thermodynamic activity. The Gibbs energy of reaction is ${\Delta }_r{G}^{\circ }={\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }$. Setting it to zero givesThe quantities are taken per mole of the reaction as written. The simplified result uses the stipulated temperature-independent enthalpy and entropy differences.

The pKa is the negative base-ten logarithm of the dimensionless acid dissociation constant. A lower pKa means a larger acid dissociation constant and a stronger tendency to donate a proton under the same standard state, solvent, and temperature conventions. The Henderson-Hasselbalch equation places equal weak acid and conjugate base thermodynamic activities at pH equal to pKa. See Flowers et al., Chemistry 2e, the Henderson-Hasselbalch equation.

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.

Here is an explicitly stipulated elementary kinetic model, not experimental data. A well-mixed ideal closed system has constant temperature and volume. Alongside an uncatalysed elementary reaction, allow an illustrative catalytic encounter that returns C immediately:Both channels obey the law of mass action. In the second, the forward reaction rate is ${\alpha }_f[\mathrm{A}][\mathrm{C}]$ and the reverse reaction rate is ${\alpha }_r[\mathrm{B}][\mathrm{C}]$. C cancels in each encounter, so its free concentration $c_C$ stays constant without an external supply. This toy elementary reaction encodes regeneration without binding; it is not a claim that real catalytic cycles universally consist of one encounter. A binding cycle can accumulate a reaction intermediate, in which case constant total C does not imply constant free $c_C$.

Adding the parallel reaction rates gives effective first-order reaction rate constantsBecause $k_f^u/k_r^u={\alpha }_f/{\alpha }_r=3$, each channel balances at the same $[\mathrm{B}]/[\mathrm{A}]=3$, and so does their sum. At $c_C=0$ the pair is $(0.030,0.010){\mathrm{s}}^{-1}$; at $c_C=0.001\mathrm{mol}{\mathrm{L}}^{-1}$ it is $(0.300,0.100){\mathrm{s}}^{-1}$. Both give $K=3$ with ideal thermodynamic activities and a common concentration standard state. At $T=300\mathrm{K}$ their compatible standard reaction change in Gibbs free energy is ${\Delta }_r{G}^{\circ }=-RT\ln 3\approx -2.7403\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Choosing an unrelated reverse rate constant would break this consistency.

Let $S=[\mathrm{A}]+[\mathrm{B}]$ be the conserved concentration, and define the dimensionless quantity $x=[\mathrm{B}]/S$. ThenSolving this first-order linear differential equation givesThe chemical relaxation time is $25\mathrm{s}$ without C and $2.5\mathrm{s}$ with the chosen C. From either $x_0=0$ or $x_0=1$, the exponential decay of the deviation leads to the same chemical equilibrium $x=3/4$. Reducing that deviation to one percent takes $\ln (100)\tau$: about $115.13\mathrm{s}$ versus $11.51\mathrm{s}$.

At this chemical equilibrium, conversion continues in both directions. For $S=1\mathrm{mol}{\mathrm{L}}^{-1}$, the uncatalysed channel has equal forward and reverse reaction rates $0.0075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. Adding C gives total directional reaction rates $0.075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ each, yet zero net reaction rate. Faster molecular exchange and unchanged composition coexist. This exact calculation explains how catalysis can matter enormously within an observation window while leaving the final equilibrium constant intact.

Equivalent chemical bonds in an intact molecule need not have equal successive bond-dissociation enthalpies. Once one chemical bond is removed, the next reactant has a different electronic quantum state and may relax to a different molecular geometry. The successive steps compare different species. This distinction also explains why a tabulated average chemical bond strength is not necessarily the cost of the first cleavage.^[ref]

For an invented species $A{B}_2$, stipulate the following gas-phase steps, with all enthalpies at the same temperature and the same standard states:The symbols stand for hypothetical species with consistently chosen electronic states. They are not values for water or another measured molecule. By Hess's law, the total enthalpy of atomization for $A{B}_2(g)\to A(g)+2B(g)$ is $700\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$; the arithmetic mean per broken chemical bond is $350\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Neither individual step equals that mean. Nor must the second step be cheaper: here it is deliberately more expensive. An exact sequence sums exactly; substituting averages taken from other molecules generally gives only an estimate.

The solubility product is the equilibrium constant for dissolution of a specified solid into its dissolved ions. For a pure bulk AB solid at the chosen temperature, pressure and standard states,The pure bulk solid's thermodynamic activity is one in this convention. Coexisting solid and chemical solution at chemical equilibrium give this equality. If all available solid dissolves, the final chemical solution may instead be undersaturated; do not impose equality merely because a salt was initially added.^[ref]

Using the thermodynamic activity definition $a_i={\gamma }_i{c}_i/c^{\circ }$, with activity coefficients ${\gamma }_i$ and reference molar concentration $c^{\circ }$, gives the independent conversionThus $K_{\mathrm{sp}}$ is dimensionless, while the conditional concentration form $K_{\mathrm{sp},c}$ has units ${\mathrm{M}}^2$ for this one-to-one model. Treating $K_{\mathrm{sp},c}$ as constant requires a justified fixed or approximately fixed activity coefficient product. Changing ionic strength can change that product even at fixed temperature; changing temperature can change the equilibrium constant itself. Specify the solid thermodynamic phase too: different crystal polymorphs need not have the same solubility product.

For transfer of the same species $X$ from a gaseous reference state into a solvent $S$, define the molar solvation free energy byHere ${\mu }^{\circ }$ is a standard chemical potential. Negative solvation free energy means that transfer into the stated solvent lowers the Gibbs free energy relative to the stated gaseous reference. Include solute-solvent interactions, making room in the solvent, and its reorganization; counting only a favorable pair interaction omits part of the transfer.^[ref]

State the temperature, gaseous reference pressure, dissolved reference concentration or other standard state convention, and species identity. Reference-state changes alter the quoted solvation free energy. For example, with an ideal dilute thermodynamic activity $a=c/c^{\circ }$, replacing $c^{\circ }$ by $\alpha c^{\circ }$ while preserving the physical chemical potential requires ${\mu }^{\circ }$ to increase by $RT\ln \alpha$. Thus the corresponding solvation free energy increases by $RT\ln \alpha$ if the gaseous reference stays fixed. This follows directly from $\mu ={\mu }^{\circ }+RT\ln a$, with $R$ the molar gas constant.

At fixed temperature and pressure, compare fully specified initial and final states usingThe entropy change is for the whole selected system, including both solute and solvent. Dispersing a solute can increase accessible arrangements while orienting nearby solvent molecules restricts others. One picture of an ordered hydration shell cannot determine the sign of the total entropy change. A negative Gibbs free energy change favors the specified process; it does not determine its reaction rate.^[ref]

An illustrative enthalpy accounting path separates solute constituents, reorganizes the solvent, and establishes their interactions:This is a chosen decomposition between compatible states, not a measured sequence. If a gaseous-ion-to-chemical solution solvation term already includes solvent reorganization, do not add that reorganization a second time. Lattice enthalpy from solution calorimetry gives the alternative Hess's law route that combines lattice separation with the complete hydration of ions term.^[ref]

The invented inputs are $+100$ for separation and $+20$ for solvent reorganization, in $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Interaction terms of $-150$ in solvent A and $-90$ in solvent B give $\Delta H_A=-30$ and $\Delta H_B=+30$. Stipulated total entropy changes of $-50$ and $+150\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ give $-T\Delta S_A=+15$ and $-T\Delta S_B=-45\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ at $300\mathrm{K}$. Both selected processes therefore have $\Delta G=-15\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. This deliberately constructed comparison shows why the signs of the enthalpy and Gibbs free energy changes need not agree; it does not predict either solvent's solubility.

For incremental dissolution at a particular composition, the relevant chemical potential difference also depends on thermodynamic activities. At chemical equilibrium that driving difference vanishes. A quoted standard Gibbs free energy change and the finite change between arbitrary mixtures answer different questions; use compatible standard states and the reaction quotient when discussing solubility.

The motivating calorimetry experiment is especially instructive. A weighed potassium chloride sample dissolves in a known amount of water in a platinum-lined, shielded calorimeter; a quartz-oscillator thermometer and measured electrical heating provide the readouts. Electrical calibration before and after the reaction, comparison with a resistance thermometer, and checks on residual water and particle size constrain errors. The inspected authored article reports positive enthalpy of solution at $298.15\mathrm{K}$ for the specified dilution, despite complete dissolution. This is a combined enthalpy balance, not an isolated solvation free energy or direct single-ion hydration of ions measurement. Shield tracking and final concentration matter; the existing lattice enthalpy from solution calorimetry discussion gives the numerical result and the conversion limits.^[ref]

Solvent-specific stabilization means that the same species need not have the same solvation free energy in different solvents. Chemical polarity, hydrogen bonds, and the cost of reorganizing each solvent all matter. Water can stabilize ions by ion-dipole attractions; a favorable isolated attraction alone still cannot decide the total dissolution balance.^[ref]

With a common gaseous reference and compatible dissolved standard states, subtraction removes that reference:A negative result favors solvent B under those reference conventions. This comparison assumes the same species $X$: if protonation, ionization, or formation of ion pairs changes its identity, include a chemical equilibrium for each corresponding chemical reaction.

The reversible cell voltage with reactants and products in specified standard states. It is the cathode reduction potential minus the anode reduction potential.