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An energy profile plots an explicitly specified energy against a reaction coordinate. Identify whether the vertical axis is potential energy, enthalpy, or Gibbs free energy; these descriptions are related but not interchangeable. A potential-energy profile describes a path through configurations, while a free-energy description also accounts for the relevant thermodynamic conditions.

For the two chosen profiles, the reaction energy barriers from the reactants are $80$ and $50\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. From the reaction products, the same maxima are $100$ and $70\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ higher. Both paths have endpoint difference $-20\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The barrier is measured from the relevant starting endpoint, not from an arbitrary zero of potential energy.

The appeal of the diagram is that it separates two useful questions: which endpoint is lower, and which path is easier to traverse? Catalysis changes the route through transition states while retaining the endpoint thermodynamics of the same overall chemical reaction. A real catalytic route may have several steps and reaction intermediates; the two simple curves do not specify such a reaction mechanism. The plotted potential-energy difference alone is not an equilibrium constant or a calculated rate constant. See Chemistry 2e, 12.7: alternative reaction routes and catalysis.

An enzyme is a biological catalyst: it accelerates a chemical reaction and is regenerated after a successful catalytic cycle. Most enzymes are proteins, but catalytic RNA molecules, called ribozymes, also exist. Many reactions needed by a biological cell would proceed too slowly under its ordinary conditions without enzymes. See Cooper, The Cell: the central role of enzymes as biological catalysts.

The biological beauty is selective speed. A biological cell contains many possible reactants; its enzymes help determine which transformations happen on a useful timescale. In metabolism, one step's reaction product can become another step's substrate, so selective catalysts support an organized flow of material.

The ribosome is a striking biological example: the machinery that builds proteins has RNA at its catalytic centre. The complete ribosome contains both RNA and proteins, but its bond-forming active site is built from RNA. Structural evidence established this distinction in Nissen et al., The structural basis of ribosome activity in peptide bond synthesis, Science 289, 920–930 (2000).

Selection occurs at an active site, the region where substrate binding and catalytic chemistry take place. Shape and chemical interactions favour some substrates over others; binding can also reshape the enzyme, an induced fit. Recognition alone is insufficient: holding a substrate extremely tightly need not make it react quickly. Catalysis requires a favourable route through the transition state, the high-barrier configuration between reactants and reaction products. Positioning reactive groups and stabilizing this configuration relative to the bound substrate can lower that barrier. Specificity is often strong, but need not mean exactly one acceptable substrate. See Cooper, The Cell: mechanisms of enzymatic catalysis.

Read the diagram numerically. Its chosen Gibbs free energies are $0$ for the reactants and $-10\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ for the reaction products. The uncatalysed peak is $60\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, whereas the catalysed peak is $40\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The forward free-energy barriers are therefore $60$ and $40\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$; viewed from the reaction products, the reverse barriers are $70$ and $50\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Both decrease by $20\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, while the endpoint difference stays $-10\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. These are constructed values, not measurements of a particular enzyme.

This separates chemical kinetics from chemical equilibrium. An enzyme changes how rapidly the system approaches chemical equilibrium; it does not change the equilibrium constant for the same overall chemical reaction under the same conditions. Faster catalysis therefore supplies no extra driving energy. In a biological cell, continuing supply of reactants and removal of reaction products can nevertheless sustain a net flow: the surroundings keep that system away from chemical equilibrium. See IUBMB, Symbolism and terminology in enzyme kinetics: basic definitions.

Some enzymes require a cofactor, an additional nonprotein component such as a metal ion or a small organic molecule. Organic helpers are often called coenzymes; they can assist transfers that the protein alone cannot readily perform. The cofactor may remain bound or exchange with the surroundings, depending on the system. A coenzyme can be chemically changed in one reaction and regenerated by another: regeneration of the enzyme does not mean every helper is unchanged in that individual reaction. Some enzymes need no additional cofactor. See Alberts et al., Molecular Biology of the Cell: small molecules associated with proteins and Alberts et al., activated carriers and their regeneration.

How much faster does adding substrate make a reaction? Enzyme kinetics gives a quantitative answer when a model's assumptions fit. For an illustrative reaction obeying Michaelis-Menten kinetics, use the Michaelis-Menten equation with substrate concentration $s$, limiting reaction rate $V_{\max }$ and Michaelis constant $K_M$. At $s/K_M=0.1$, substitution gives $v/V_{\max }=0.1/1.1=1/11\approx 0.0909$. At $s/K_M=1$, it gives $1/2$; at $s/K_M=10$, it gives $10/11\approx 0.9091$. Thus a hundredfold increase in substrate between the first and last examples yields only a tenfold increase in reaction rate.

For a chosen $V_{\max }=110\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$, those fractions mean reaction rates of $10$, $55$ and $100\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$. Doubling substrate from $10K_M$ to $20K_M$ raises the normalized rate only from $10/11$ to $20/21\approx 0.9524$, a relative increase of about $4.76\%$. In this simplest model, the catalytic machinery is already nearly fully occupied. Doubling the amount of active enzyme instead doubles $V_{\max }$ and the predicted reaction rates at fixed free substrate concentration, provided the other conditions are unchanged. This distinction helps decide whether additional substrate or additional catalytic capacity could improve throughput.

The interpretation has limits. The simplest Michaelis-Menten kinetics assumes a single substrate, fixed active enzyme amount, a suitable quasi-steady-state approximation, and initial measurements with negligible reverse reaction from accumulated reaction product. Substantial depletion, interacting binding sites, or inhibition by excess substrate can invalidate this model. Within it, $K_M$ is the substrate concentration giving half the limiting reaction rate; it is not generally the binding dissociation constant $K_d$. See IUBMB, enzyme kinetics: steady-state approximation and substrate dissociation and IUBMB, Michaelis-Menten and non-Michaelis-Menten kinetics. Measuring activity also requires specified temperature, pH and necessary cofactor availability: a weak signal can reflect unsuitable conditions rather than little enzyme. See Brooks et al., Assay Guidance Manual: basics of enzymatic assays for HTS.

Epimerization interconverts diastereomers differing at one stereocenter. It can change a product ratio after the initial bond-forming chemical reaction.

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.

The equivalence point is the composition at which added titrant and the initial sample inventory match the specified chemical reaction's stoichiometry. It is a theoretical inventory condition; the titration endpoint is an observed stopping criterion. For a monoprotic acid consuming one hydroxide per acid unit, initial acid molar concentration $c_A$, sample volume $V_A$, and base titrant molar concentration $c_B$ giveThe proof is amount of substance conservation with one-to-one neutralization; a different stoichiometry requires its coefficients. All volumes must use compatible units. Since $1\mathrm{mL}=10^{-3}\mathrm{L}$, $0.100\mathrm{mol}{\mathrm{L}}^{-1}\times 10.00\mathrm{mL}=0.001000\mathrm{mol}$, not $1\mathrm{mol}$.

An equivalence point does not prescribe pH $7$. A weak acid titrated with a strong base leaves its conjugate base, which reacts with water; the corresponding chemical equilibrium can make the chemical solution alkaline. The neutral strong acid/strong base case additionally depends on temperature, through self-ionization of water.^[ref]

For a closed system with one specified chemical reaction, the extent of reaction $\xi$ records progress from $\xi =0$ through $n_i=n_{i,0}+{\nu }_i\xi$. Here $n_i$ is the amount of substance of species $i$, and the signed stoichiometric vector has negative entries for reactants and positive entries for reaction products. For the water chemical equation in the order ${\mathrm{H}}_2,{\mathrm{O}}_2,{\mathrm{H}}_2\mathrm{O}$, it is $(-2,-1,2)$. The normalization of $\xi$ depends on the chosen coefficients. See IUPAC Gold Book, extent of reaction.

With particle numbers instead of molar amounts, a molecular extent ${\xi }_N$ obeys $d{N}_i={\nu }_{i}d{\xi }_N$, and the chemical potentials are measured per particle. At fixed temperature and volume, the Helmholtz free energy then satisfies $dF=({\sum }_i{\nu }_i{\mu }_i)d{\xi }_N$. An interior chemical equilibrium requires ${\sum }_i{\nu }_i{\mu }_i=0$; the molar convention gives the same relation with chemical potentials per mole.

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.

A functional group is an identifiable group of chemical atoms associated with characteristic chemical reactions. In step-growth polymerization, $A$ and $B$ name compatible reactive groups, not necessarily entire monomers or particular chemical elements.

The functionality in polymerization is the number of reactive functional groups on a molecule for the specified joining chemistry. The linear step-growth molecule count assumes $f=2$ for every feed molecule. A diol supplies two hydroxy groups in the polyurethane example; an impurity supplying only one suitable group can cap a chain. A group count must reflect the selected chemical reaction, not simply every chemically recognizable group in the molecule.

A galvanic cell supplies electrical work from a spontaneous redox chemical reaction under the specified conditions. In discharge through a passive load, oxidation occurs at its negative anode and reduction at its positive cathode. Separating the reacting compartments makes the electrons take an external route, where their transport can deliver useful work. See Flowers et al., Chemistry 2e, §17.2: separated reactions and electrical work.

In the metallic external electrical circuit, electrons move from anode to cathode, while conventional current goes from cathode through the load to anode. Inside the cell, net ionic conventional current closes the loop from the anode compartment toward the cathode compartment. Oppositely moving cations and anions can both contribute to that direction; see electrolyte for the signed current density calculation. A cell can maintain a nonzero terminal electric potential difference near open circuit while delivering almost no power; the Daniell cell voltage and load experiment makes this distinction concrete.

A chemist who studied energy changes during chemical reactions; Hess's law bears his name.

Royal Society of Chemistry biographical notice: edu.rsc.org/resources/on-this-day-aug-08--henri-hess-was-born/10808.article

In the payoff stage of glycolysis, oxidation of glyceraldehyde 3-phosphate is coupled to reduction of NAD+. Each NADH carries a two-electron reducing equivalent. The formal half-chemical equationmakes the electron count explicit: two NADH per glucose carry four electrons. This is redox bookkeeping, not a claim that free electrons travel through the chemical solution or that NADH is itself ATP.

Fermentation can return NADH to NAD+ by reducing an organic substrate; reduction of pyruvate to lactate is one example. In aerobic cellular respiration, reducing equivalents reach an electron transport chain, reducing molecular oxygen. Other forms of cellular respiration use other electron acceptors, so absence of oxygen does not uniquely imply fermentation.^[ref]

The need for NAD+ regeneration follows from a finite-pool argument. In a hypothetical closed preparation with $12\mu \mathrm{mol}$ of initially oxidized NAD+ and no regeneration, at most $6\mu \mathrm{mol}$ of glucose can complete the two NAD+-consuming conversions per glucose. This is an upper bound even with excess glucose, not a predicted stopping time: other limitations can stop the preparation earlier. With an intact fixed nicotinamide adenine dinucleotide pool and no side reactions, converting NAD+ into NADH preserves their sum but depletes the usable oxidized form.

ATP recycling imposes a second condition. Glycolysis needs some initial ATP for investment and consumes two net ADP and two inorganic phosphate per completed glucose. In a closed ideal preparation with only $8\mu \mathrm{mol}$ of initial ADP, no other source of ADP, and enough initial ATP and inorganic phosphate, at most $4\mu \mathrm{mol}$ of glucose can finish. This bound follows from the net ADP consumption, even though investment temporarily produces additional ADP. Initial investment also means that adequate final inventories alone cannot guarantee the route can start.

For continued operation at fixed carrier inventories, other chemical reactions must match the two net ATP formed per glucose with ATP use that restores ADP and inorganic phosphate, or exchange must supply those reactants. ATP hydrolysis coupled to cellular work is one route. Thus glycolysis can sustain useful energy coupling while neither its ATP inventory nor its NADH inventory grows indefinitely; matter cycles while glucose supplies fresh chemical free energy.

Golgi cargo processing includes enzyme-mediated remodeling of attached oligosaccharides. Glycosylation can involve addition of sugars, while processing can also remove them. Different Golgi apparatus regions have different processing enzymes; their substrate requirements constrain the order of possible chemical reactions.^[ref]

An original model makes the importance of ordering explicit. Write a toy carbohydrate chain as $XA$, read with its terminal symbol on the right. Define a trimming function $T$ that removes a terminal $A$, and an extension function $E$ that adds $B$ only when the terminal symbol is $X$. In every other case the corresponding function leaves the chain unchanged. ThenIn particular, $E(T(XA))=XB$ but $T(E(XA))=X$: the two functions do not commute. Each result follows by inspecting the terminal symbol before applying the stipulated rule. These symbols are artificial labels, rather than a natural oligosaccharide sequence or measured enzyme pathway. The example explains why “two processing steps occurred” does not determine the product unless their order and substrate conditions are also specified. In the Golgi apparatus, spatial organization can help arrange successive opportunities for processing.

A periodic group is a vertical column of the periodic table. For example, fluorine and chlorine are in periodic group 17, with seven valence electrons in each neutral ground state chemical atom. A shared outer electron configuration supports similarities in chemical reactions, rather than equality of every property.

A half-reaction is a balanced chemical equation for either the oxidation or the reduction part of a redox chemical reaction, explicitly including exchanged electrons. Match the electron counts before adding the two half-reactions; the net equation then contains no transferred electrons. A half-reaction is bookkeeping, not necessarily an elementary reaction. See Florida State University, lab background: separate electrode reactions and Daniell cell.

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.

A hemiketal is the addition product of an alcohol and a ketone, with hydroxy group and alkoxy substituents on the former carbonyl carbon. Intramolecular formation can temporarily tether an alcohol to a ketone before another chemical reaction changes the carbon skeleton.

Hess's law states that compatible chemical reaction enthalpies add when their equations add. Reversing a chemical reaction reverses its enthalpy change; multiplying the entire chemical reaction equation by a factor multiplies that change by the same factor. The reason is that enthalpy is a state function: intermediate states cancel from the differences. A cycle supplies one independent balance relation; it cannot determine two unknown inputs without further information.

The law is named after Germain Henri Hess.

For specified initial and final thermodynamic states, the change is $\Delta H=H_{\mathrm{final}}-H_{\mathrm{initial}}$. Because enthalpy is an extensive variable, a quoted reaction $\Delta H$ belongs to the amount of substance specified for each species in its chemical equation. Reversing the equation changes the sign; scaling every coefficient scales $\Delta H$; adding compatible equations adds their $\Delta H$ values. A per-mole value must say which chemical equation and normalization it uses. Match temperature, pressure, thermodynamic phases, chemical solution compositions and reference states before cancellation; the same formula in a gas and a liquid represents different states.^[ref]

The value of Hess's law is that an inaccessible reaction enthalpy can be obtained from measurable transformations. The algebra does not require those transformations to be steps of the actual reaction mechanism. Neither an activation energy nor a reaction rate follows from the endpoint difference: a catalyst may change the reaction rate while preserving the compatible initial and final enthalpies.

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.

Hydrolysis is a chemical reaction involving water that breaks a chemical bond. The free energy released by suitably coupled ATP hydrolysis can drive active transport; the available free energy depends on the actual composition and conditions.