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First construct the energy cost for heterolysis from the molecular potential minimum. Dissociation to neutral atoms costs $D_e$, ionizing caesium costs $I$, and attachment of the electron to iodine releases EA. ThereforeFor the ground-state reference convention used above, the initial molecule already has vibrational zero-point energy. Thus its ground-state dissociation energy is $D_0=D_e-\frac{1}{2}{k}_B{\theta }_{\mathrm{vib}}$ andAt $3000\mathrm{K}$, $k_{B}T=0.258520\mathrm{eV}$, givingAlternatively, retaining the molecular minimum as reference gives $\Delta {\epsilon }_{\min }=4.36\mathrm{eV}$ and a Boltzmann factor $4.7372\times 10^{-8}$. That convention must use the full vibrational partition function, including $e^{-{\theta }_{\mathrm{vib}}/(2T)}$, in the next part. The ground-state convention instead uses $q_{\mathrm{vib}}^0$. Either consistent convention gives the same final equilibrium constant; using the ground-state reaction energy with the full zero-point factor would count the correction twice. The distinction between $D_e$ and $D_0$ is a distinction between initial states, not a change in the dissociation products.^[ref]

Let ${\rho }_i=N_i/V$ be a molecular number density and define the volume-independent molecular partition function density $f_i=q_i/V$. Then $q_i/N_i=f_i/{\rho }_i$. Also define the reaction reference-energy differenceSubstitution into the equilibrium chemical-potential relation givesAn equilibrium constant expressed using concentration activities $a_i={\rho }_i/c^{\circ }$ is dimensionless:Here every $f_i$ has units ${\mathrm{m}}^{-3}$, and $c^{\circ }$ is the chosen standard concentration, initially one molecule per cubic metre. The factor $c^{\circ }$ is essential: without it the association quotient has units of volume.

For the three elementary transformations just proposed, a qualitative free-energy profile has three transition states and two intermediates. The sugar substituents and indole C3 substituent stay unchanged, so the structural shorthand below suppresses them while retaining the reacting atoms.

The first transition state has a partially broken anomeric C1–O(leaving group) bond and developing oxocarbenium character, $[{\mathrm{O}}_{\mathrm{ring}}-{\mathrm{C}}_1^{\delta +}\cdots {\mathrm{O}}_{\mathrm{LG}}{]}^{‡}$. Its following minimum is the glycosyloxocarbenium ion plus counterion. The second transition state has a forming C1–C2 bond, $[{\mathrm{O}}_{\mathrm{ring}}-{\mathrm{C}}_1\cdots {\mathrm{C}}_2(\mathrm{H}{)}_{\mathrm{ind}}{]}^{‡}$, and developing positive charge in the indole. Its following minimum is the covalently joined arenium ion with an intact C2–H bond. The third transition state has simultaneous C2–H cleavage, B–H formation and restoration of the indole pi bond, $[\mathrm{B}\cdots \mathrm{H}\cdots {\mathrm{C}}_2-{\mathrm{C}}_3{]}^{‡}$. It leads to the aromatic C-linked product and protonated base.

The vertical separations are illustrative because no activation free energies or equilibrium constants are supplied. In particular, the highest barrier cannot be identified experimentally from this drawing alone. protonation of the leaving group is treated as an initial rapid equilibrium in the chosen mechanism; explicitly resolving it would add the corresponding proton-transfer state and intermediate.

The qualitative nucleophilicity order is $p$-OMe $>$ $p$-Me $>$ H $>$ $p$-Cl $\gg$ $p$-NO${}_2$, matching the product-yield order at the stated stopping time. Electron donation raises the availability of the aniline lone pair for attack. A para methoxy group donates by resonance, while methyl is a weaker donor. Chlorine withdraws inductively and gives an aniline less nucleophilic than the unsubstituted compound. A para nitro group strongly withdraws by resonance: electron flow from nitrogen through the ring into the nitro group stabilizes a charge-separated form and greatly reduces nucleophilicity.

The key arrows for the nitroaniline comparison move the nitrogen lone pair into its own aromatic ring, shift the ring $\pi$ bonds toward the para nitro group, and shift a nitro N=O bond toward oxygen. The methoxy donor produces the opposite electronic bias. These are finite-time yields, so they support a kinetic explanation qualitatively; they are not by themselves measured rate constants or equilibrium constants.

Combining ${\Delta }_r{G}^{\circ }={\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }$ with the equilibrium relation givesIf reaction enthalpy and reaction entropy are approximately constant over the measurement interval, a plot of $y=\ln K$ against $x=1/T$ is straight, with slope $-{\Delta }_r{H}^{\circ }/R$ and intercept ${\Delta }_r{S}^{\circ }/R$. An exothermic reaction therefore has a positive slope. In the more general local form, the Gibbs–Helmholtz equation giveswhich is the Van't Hoff equation without assuming a constant enthalpy.

Use a dimensionless equilibrium constant built from thermodynamic activities. Measure sufficiently separated temperatures after verifying true equilibrium, and avoid phase changes or a changing chemical reaction. Temperature-dependent heat capacities cause curvature: the fitted straight-line slope then estimates an interval-dependent enthalpy. Small errors in the equilibrium measurements can produce large slope errors over a narrow $1/T$ range. The entropy intercept is particularly sensitive because $1/T=0$ requires distant extrapolation, and slope/intercept errors are correlated. A numerical line through a few points is not, by itself, evidence that all these assumptions hold.

At coexistence of the two pure solid forms, their chemical potentials are equal and ${\Delta }_r{G}^{\circ }=0$. Treating the tabulated enthalpy and entropy differences as constant givesAbove this temperature the model favours the higher-entropy yellow form; below it the red form is favoured. This is a prediction from the given thermodynamic data, not a guarantee of an immediately visible colour change: kinetic trapping, particle-size effects and temperature-dependent heat capacities are not included. A ratio of masses of two coexisting pure solids is not the equilibrium constant, since their bulk solid activities are each unity.

At chemical equilibrium, the forward and reverse event rates are equal, rather than both reactions stopping:For the dilute-solution concentration model,The ratio is dimensionless because both concentrations and both rate constants have matching units. An exact thermodynamic equilibrium constant uses activities; here the common standard concentration and ideal-solution activity coefficients cancel.

Excess ethanol drives a reversible ester exchange mainly through the actual mixture composition, not through a large new bond-energy advantage. The net reaction exchanges one ester and one alcohol for a closely related ester and alcohol. It preserves the main bond types and has no obvious large exothermicity. Solvation and ion pairing can contribute to its enthalpy, but there is no basis here for asserting a large negative standard reaction enthalpy.

There are two molecules on each side, so an increase in particle count is not the entropy explanation. Instead useWith ethanol as the solvent and little methanol initially, the reaction quotient is small; product formation and mixing lower the Gibbs free energy until the exchange equilibrium is reached. ethanol can be treated approximately as having fixed activity near its pure-solvent value, rather than assigned an arbitrarily enormous activity, while methanol remains dilute. This composition-dependent chemical-potential contribution is the appropriate entropic/mixing argument. It explains forward conversion under these conditions without claiming that ${\Delta }_r{S}^{\circ }$ must be positive, or that complete conversion follows without an equilibrium constant or removal of methanol.

Let $G$ be Gibbs free energy, $H$ enthalpy and $T$ thermodynamic temperature for a fixed amount and composition of material, with the derivative taken at constant pressure. Apply the Gibbs–Helmholtz equation to the standard molar reaction quantities, obtained by summing product minus reactant contributions with their stoichiometric coefficients:Define $K(T)$ as the dimensionless thermodynamic equilibrium constant expressed using activities in specified standard states. Since ${\Delta }_r{G}^{\circ }=-RT\ln K$, where $R$ is the molar gas constant,If the reaction enthalpy is constant over the interval, integration givesThis is the requested Van't Hoff equation. For an exothermic reaction, increasing temperature decreases $K$. Although the name “isochore” is often used in this context, the displayed enthalpy form follows from standard Gibbs energies and the stated constant-pressure Gibbs–Helmholtz identity; the analogous ideal-gas concentration form involves internal energy when the temperature dependence of standard concentration is retained.

The two remaining partial pressures sum to $200-160=40$ bar. Their ratio therefore gives $p_{{\mathrm{N}}_2}=10$ bar and $p_{\mathrm{N}{\mathrm{H}}_3}=30$ bar. With the ideal-gas activity approximation $a_i=p_i/p^{\circ }$ and $p^{\circ }=1$ bar,The equilibrium constant is dimensionless: pressures enter as ratios to the standard pressure. Blindly substituting pressures in pascals into an unnormalized expression would give a different, incorrect thermodynamic $K$.

The standard reaction enthalpy is $2(-45.90)=-91.80\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ per mole of reaction as written, because the elements are in their reference states. ThusThis calculation uses the supplied constant-enthalpy simplification. At 200 bar, an accurate thermodynamic treatment would use chemical fugacities rather than ideal partial pressures; no chemical fugacity coefficients are supplied. The iron catalyst changes the approach to equilibrium, not its thermodynamic value.

Using the equilibrium constant calculated without premature rounding,Then ${\Delta }_r{G}^{\circ }={\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }$ givesThe sign agrees with the reduction in gas stoichiometric amount. The standard reaction entropy is not the absolute entropy of ammonia: its stoichiometric coefficient is two. ThereforeThis is an absolute standard molar entropy inferred within the paper's thermodynamic approximations, not a claim that elemental gases have zero absolute entropy. Only their standard enthalpies of formation are zero in this calculation.

The reverse elementary rate constant is extraordinarily small at 298 K. Its reaction would require hydrogen chloride and a chlorine free radical to produce a hydrogen free radical and chlorine, reversing a highly exothermic propagation step. The value found above makes this channel negligible for the intended reacting mixture compared with forward propagation.

A large equilibrium constant alone does not permit a reverse reaction to be ignored under every conceivable composition: if forward reactants were exhausted, concentration ratios would matter. The simplification applies during the forward chain-reaction regime specified, rather than constituting a universal statement about all late-time or imposed reverse conditions.

For ideal gases and standard pressure $p^{\circ }=1\mathrm{bar}$,Since $p_{\mathrm{ClF}}/p_{\mathrm{Cl}{\mathrm{F}}_3}=9$, this gives $p_{{\mathrm{F}}_2}/p^{\circ }=1/(9K_p)$. ThereforeMore exactly, subtract the tiny fluorine pressure from $1\mathrm{bar}$ and divide the remainder in the ratio $9:1$. The very small fluorine activity reconciles the reactant-rich ClF ratio with the very large equilibrium constant.

With ideal thermodynamic activities $a_i=[i]/c^{\circ }$, the dimensionless acid dissociation constant isThe factor $c^{\circ }$ makes the equilibrium constant compatible with ${\Delta }_r{G}^{\circ }=-RT\ln K_a$. If the concentration quotient without $c^{\circ }$ is used instead, it has concentration units and its numerical value is identical only when the standard concentration is expressed as $1\mathrm{M}$.

The predominant self-sorted distribution is $[\mathrm{Zn}(1){]}^{2+}$, $[\mathrm{Li}(2){]}^{+}$ and free 3, so the reporter remains mainly in its 554 nm state. zinc benefits from the six amine donors of 1 without forcing 2 into its strained zinc-binding conformation. lithium occupies 2 because its lithium affinity exceeds that of 3. Each metal is consumed in the preferred complex, leaving the reporter unoccupied.

This is a dominant-species prediction. A tenfold affinity difference alone would not guarantee quantitative lithium sorting if the absolute affinities were weak or if free metal and all competing complexes had appreciable populations. Concentration and equilibrium constants are not supplied, so an exact species fraction is not justified.

For solar elemental abundances, take a molecular hydrogen- and helium-dominated atmosphere with atmospheric carbon-to-oxygen ratio near $0.55$. The principal oxygen-bearing molecule is generally water; the carbon and nitrogen carriers depend on both temperature and pressure.

At the cool $500\mathrm{K}$ observable upper layers, methane is the expected main carbon reservoir under chemical equilibrium, with abundant water and much less carbon monoxide. At the hotter layers approaching $1400\mathrm{K}$, carbon monoxide becomes the main carbon carrier and water contains much of the oxygen not bound in it. The relevant law of mass action follows fromLower temperature favors the exothermic methane-forming direction, while increasing pressure favors the side with fewer molecules. This explains why the change of dominant carrier cannot be specified by temperature alone.

Nitrogen is distributed between molecular nitrogen and ammonia. Cooling favors ammonia, but low pressure favors molecular nitrogen; it is therefore unsafe to call ammonia dominant throughout the $500\mathrm{K}$ low-pressure region. The exact partition requires the equilibrium constant for ${\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2{\mathrm{NH}}_3$.

Carbon dioxide is usually a minor constituent at solar composition; hydrogen cyanide and acetylene are much less abundant than the principal carbon carriers in this oxygen-rich equilibrium case. Condensation can remove refractory species where a condensation curve is crossed. The robust cool-atmosphere expectation is a molecular hydrogen–helium background with methane and water, changing toward carbon monoxide in hotter layers. Precise mixing ratios require thermodynamic data and an explicitly specified elemental inventory after rainout.

With $n=2$, the standard reaction Gibbs free energy and dimensionless equilibrium constant areFor the reaction as written, $Q=a_{\mathrm{F}{\mathrm{e}}^{2+}}/a_{\mathrm{S}{\mathrm{n}}^{2+}}$, and the Nernst equation gives $E=E^{\circ }-(RT/2F)\ln Q$. Consequently zero cell potential requires $[\mathrm{F}{\mathrm{e}}^{2+}]/[\mathrm{S}{\mathrm{n}}^{2+}]=K\simeq 1.93\times 10^{10}$ under the specified unit-activity-coefficient assumption. Reversing that ratio reverses the conclusion. This mathematical ratio does not by itself guarantee experimentally attainable concentrations without precipitation or other chemistry.

Interpret the supplied free energies as standard reaction free energies and use $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$. The dimensionless equilibrium constants areUnder the ideal-solution approximation, $[C]=K_C[L]$ and $[D]=(K_D/c^{\circ })[L{]}^2$. Set $k_D=K_D/c^{\circ }$, with units ${\mathrm{M}}^{-1}$. Conservation counts two original molecules in each dimer:The physical, positive root can be written without subtractive cancellation asThe resulting concentrations are:

For example, the first row satisfies $[L]+[C]+2[D]=0.100\mathrm{M}$; using $[L]+[C]+[D]$ would incorrectly count dimers. At both concentrations almost all molecular units belong to dimers.

dimer formation is favored by sufficiently high concentration because it joins two molecules. Dilution favors the separated species, whereas $[C]/[L]=K_C$ is concentration-independent in this ideal model. More than half the molecular units are dimerized when $2[D]>[L]+[C]$, equivalently $c_0>(1+K_C{)}^2/k_D\simeq 7.36\times 10^{-7}\mathrm{M}$. The supplied free energies at one temperature do not determine a temperature trend without additional enthalpy information. At $1\mathrm{M}$, neglected nonideal activities could be significant; the table is the requested concentration-based model.

The important change is preorganization of the chloride-binding amide groups. In the hydroxylated host, intramolecular hydrogen bonds from the phenolic OH groups to the amide carbonyl oxygens restrict rotation and favor a geometry in which the amide NH donors converge on the anion-binding cavity. chloride can therefore accept two favorable hydrogen bonds with a smaller conformational penalty than in the unsubstituted host. The hydroxyl groups need not be counted as two additional freely available chloride donors: their role includes organizing the existing donors.

The association constant increases by $5230/195=26.8$. At $298\mathrm{K}$, the corresponding difference in standard Gibbs free energy isThis is a modest molecular free-energy advantage amplified into a substantial change in the equilibrium constant. solvent competition and changes in solvation also contribute to measured binding; the structures chiefly suggest the preorganization explanation.

Use consistent units for the Gibbs free energy:Then $\Delta G^{\circ }=-RT\ln K$ givesThis is a dimensionless thermodynamic equilibrium constant defined through activities. For an ideal dilute $1:1$ association, the numerical concentration association constant is about $4.53\times 10^3{\mathrm{M}}^{-1}$ when concentrations use the $1\mathrm{M}$ standard state. Both favourable enthalpy and positive entropy favour association here; association need not always have a negative entropy change because solvent release can outweigh the loss of translational freedom.