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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 Born-Haber cycle uses Hess's law to infer a lattice energy from measurable enthalpy differences and isolated-species data. The motivating difficulty is that separating a mole of an ionic crystal into mutually separated gaseous ions is not a convenient direct calorimetry experiment. Instead, compare two accounting routes between the same elemental reference states and the same solid. Because enthalpy is a state function, the signed differences close exactly when all inputs describe compatible states. These routes do not assert a reaction mechanism or a temporal sequence of reaction intermediates.^[ref]

For one mole of sodium chloride formula units, the overall formation reaction isThe stoichiometry requires half a mole of diatomic chlorine molecules, supplying one mole of chlorine chemical atoms. A half coefficient counts an amount of substance; it does not describe half of an individual molecule. Write each auxiliary chemical reaction with its own sign:Here $S$ is the sublimation enthalpy, $B$ the enthalpy of atomization of chlorine per mole of chemical atoms, and $I$ the molar ionization enthalpy represented by the ionization energy input in elementary treatments. If the bond-dissociation energy table gives the compatible molar enthalpy for ${\mathrm{Cl}}_2(g)\to 2\mathrm{Cl}(g)$, use half that number for $B$. Both ions in the last step are gaseous, and no aqueous hydration of ions belongs in that step.

Adding these equations cancels the gaseous chemical atoms, the gaseous ions and the free electron, leaving precisely the overall formation reaction. ThusThe closed traversal returns from the solid to the elemental reference states, adding $-F$; therefore $S+B+I+A+L_{\mathrm{form}}-F=0$. Notice that the signed attachment term $A$ is negative. Substituting a positive spectroscopic electron affinity for $A$ would reverse the attachment contribution.

The diagram deliberately uses invented, internally consistent inputs in $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$: $S=100$, $B=120$, $I=500$, $A=-350$ and $F=-400$. Preparation of separated gaseous ions raises the enthalpy by $370$, so $L_{\mathrm{form}}=-770$ and $L_{\mathrm{diss}}=+770$. The closure check is $100+120+500-350-770+400=0$. These numbers are not measured sodium chloride data and do not replace the source's $769\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ example in lattice energy. Every level retains one sodium nucleus, one chlorine nucleus and zero total electric charge; the ionization level explicitly retains the free electron.

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.

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.

Enthalpy is the state function obtained by adding pressure times volume to internal energy:Here $U$ denotes internal energy, also written $E$ in some references. Like internal energy, enthalpy is measured in joules and is an extensive variable. Its definition packages the internal energy change and expansion work into one useful quantity for fixed-pressure calculations. It does not define a body's stored heat. ^[ref]

For a fixed amount and composition of a simple compressible system, the relation between neighboring equilibrium states isHere $T$ is absolute temperature and $S$ is entropy. Thus entropy and pressure are the natural variables of this thermodynamic potential, with partial derivatives $T=(\partial H/\partial S{)}_p$ and $V=(\partial H/\partial p{)}_S$. Fixed composition matters: changing composition adds further terms. The formula follows in one line from the equilibrium first law of thermodynamics identity:The cancellation of $pdV$ is the reason enthalpy replaces volume with pressure as a natural variable. This identity concerns equilibrium states; it does not identify $TdS$ with actual irreversible heat transfer. ^[ref]

For a closed system undergoing a quasistatic process at constant pressure, require fixed amount and composition, only pressure-volume work, no friction, and unchanged bulk kinetic energy and potential energy. Then the actual first law of thermodynamics balance givesThe symbol $Q_p$ means transferred heat for this fixed-pressure process. If the piston example in heat additionally meets these conditions, its $20\mathrm{kJ}$ of incoming heat equals $\Delta H$, while $\Delta U=15\mathrm{kJ}$ and $p\Delta V=W=5\mathrm{kJ}$. The example shows exactly what enthalpy adds to internal energy: the expansion work required at the imposed pressure.

Within a single thermodynamic phase, heat capacity at constant pressure giveswith the amount and composition fixed. Constant $C_p$ gives $\Delta H=C_p\Delta T$; dividing by mass gives specific heat capacity at constant pressure. At equilibrium melting, enthalpy can instead change through latent heat while temperature stays fixed. The sensible heat example therefore adds two single-thermodynamic phase enthalpy changes and one latent heat contribution.

The usefulness of enthalpy lies in matching the imposed constraint: a fixed-pressure experiment naturally reports $\Delta H$, while fixed-volume heating with no other work reports $\Delta U$. Neither equality is universal. Additional work or changing pressure alters the relation between supplied heat and $\Delta H$; irreversible entropy production also prevents identifying actual $\delta Q$ with $TdS$. The separate Joule-Thomson coefficient branch uses enthalpy to describe throttling under its own assumptions.

Entropy is an extensive quantity and a state function of an equilibrium system, measured in $\mathrm{J}{\mathrm{K}}^{-1}$. Differences between equilibrium states are calculated along a reversible thermodynamic process:Here heat entering the system is positive and $T$ is its absolute thermodynamic temperature. The reversible path used for calculation need not be the path actually taken. Dividing the actual heat by a single temperature generally fails for an irreversible process.

For example, put two rigid bodies, each with constant heat capacity $C=100\mathrm{J}{\mathrm{K}}^{-1}$, at $600\mathrm{K}$ and $300\mathrm{K}$ in an otherwise isolated system. Assume each body has a uniform internal temperature, no phase transition occurs, and no work is extracted. Their final thermal equilibrium is at $450\mathrm{K}$, because the internal energy lost by one equals that gained by the other. Using entropy from heat capacity,so $\Delta S_{\mathrm{tot}}=C\log (9/8)=11.78\mathrm{J}{\mathrm{K}}^{-1}$. One body's entropy decreases while the total increases. ^[ref]

In statistical mechanics, Boltzmann's entropy formula counts equally likely microstates compatible with a macrostate; Gibbs entropy handles a specified probability distribution over microstates. If the number of equally likely microstates doubles, entropy increases by $k_B\log 2$. The logarithm makes multiplicative counts additive for independent systems. These are precise statements about states and probabilities, not a universal measure of visible messiness. Calling entropy merely "disorder" obscures the thermal calculation above and hides which microstates are being counted. ^[ref]

The usefulness of entropy is that measurable heat capacities determine its changes, and the Second law of thermodynamics turns those changes into constraints on processes. It also connects macroscopic measurements to microscopic descriptions without requiring us to track every particle.

For the insulated gas-only free expansion of an ideal gas, the equilibrium endpoints are $(T,V_1)$ and $(T,V_2)$ with $V_2>V_1$. Because entropy is a state function, calculate their difference along a hypothetical reversible thermodynamic process: an isothermal expansion using a frictionless piston and a thermal reservoir at $T$. This comparison process receives heat and delivers work, even though the actual free expansion does neither. Its reversible heat increment is $\delta Q_{\mathrm{rev}}=pdV=nRTdV/V$, soHere $R$ is the molar gas constant, and the logarithm acts on a dimensionless ratio. The reversible reference gives the correct endpoint entropy difference because entropy is a state function; putting the actual $Q=0$ into $\Delta S=Q/T$ would misuse a relation requiring reversible heat transfer. For the actual isolated gas, all this increase is entropy production. Thus an adiabatic process can increase entropy and need not be an isentropic process. ^[ref]

In the figure's volume doubling, the ideal gas law gives $p_1=99.768\mathrm{kPa}$ and $p_2=49.884\mathrm{kPa}$, while $\Delta S=R\ln 2\simeq 5.763\mathrm{J}{\mathrm{K}}^{-1}$. The reference process receives and delivers $nRT\ln 2\simeq 1728.85\mathrm{J}$ as heat and work; its thermal reservoir loses the same entropy that the gas gains. The actual free expansion has $Q_{\mathrm{in}}=W_{\mathrm{out}}=0$ and produces that entropy internally. Reversibly compressing the gas back at $300\mathrm{K}$ would require $1728.85\mathrm{J}$ of work input and rejection of equal heat. Restoring the gas alone therefore changes the surroundings: the actual release is irreversible.

For a real gas the final temperature may differ, so the ideal-gas logarithm formula cannot simply be copied. Evaluate $S(T_2,V_2)-S(T_1,V_1)$ using that gas's equilibrium properties or a suitable reversible comparison path. If the walls and water exchange heat with it, the Second law of thermodynamics constrains the total entropy of the isolated apparatus; the gas alone is then not an isolated system.

The first law applies conservation of energy to a thermodynamic system. For a closed system with unchanged bulk kinetic energy and potential energy, writeHere $U$ is internal energy, $Q>0$ is heat entering the system and $W>0$ is work done by the system on its surroundings. Compression therefore gives negative $W$. Some references define work positive when done on the system; their plus sign represents the same balance with a different convention. If bulk kinetic energy or potential energy changes, include it in the total energy balance. ^[ref]

Supplying $1000\mathrm{J}$ of heat while a gas does $300\mathrm{J}$ of work increases its internal energy by $700\mathrm{J}$. Compressing it with $200\mathrm{J}$ of input work in an adiabatic process instead gives $Q=0$, $W=-200\mathrm{J}$ and $\Delta U=200\mathrm{J}$.

The distinction between $d$ and $\delta$ records state versus path: internal energy is a state function, whereas heat and work depend on the process. For an ideal gas expanding reversibly at constant temperature, $W=N{k}_{B}T\log (V_2/V_1)$ and $Q=W$, since internal energy of an ideal gas depends only on temperature. Choose $N{k}_{B}T=1000\mathrm{J}$ and $V_2=2V_1$: then $Q=W=693.15\mathrm{J}$. An insulated free expansion into vacuum between the same equilibrium endpoints has $Q=W=0$. Both have $\Delta U=0$, yet their transfers differ. Their common entropy increase is $N{k}_B\log 2$. ^[ref]

For a simple compressible equilibrium system of fixed composition, combining the laws givesHere $p$ is pressure and $V$ is volume. This identity relates neighboring equilibrium states. Identifying $TdS$ with actual heat and $pdV$ with actual work requires a suitable reversible thermodynamic process; an irreversible piston calculation uses the external resisting pressure for boundary work. This distinction prevents applying a reversible ideal-gas adiabat to an irreversible adiabatic piston compression.

The Euler theorem for homogeneous functions applied to an extensive quantity, together with $dE=TdS-pdV+\mu dN$ give

The Gibbs-Duhem equation constrains changes of the intensive quantities describing one uniform thermodynamic phase. For a mixture with $k$ independently specified component amounts, its molar form isHere $S$ is entropy, $V$ is volume, $T$ is temperature, $p$ is pressure, $n_i$ is the amount of substance of component $i$, and ${\mu }_i$ is its molar chemical potential. This assumes a simple compressible system whose bulk internal energy is an extensive quantity; additional work coordinates require additional terms.^[ref]

The proof exposes the source of the constraint. For internal energy $E(S,V,n_1,\dots ,n_k)$, replication of the whole thermodynamic phase by a factor $\lambda$ gives $E(\lambda S,\lambda V,\lambda n_1,\dots ,\lambda n_k)=\lambda E(S,V,n_1,\dots ,n_k)$. The Euler theorem for homogeneous functions and the fundamental differential of internal energy therefore giveTaking the differential of the first equation produces all the terms in the second, plus $SdT-Vdp+{\sum }_i{n}_{i}d{\mu }_i$. Subtraction proves the Gibbs-Duhem equation. Thus the constraint comes from two descriptions of the same state function, rather than an assumed microscopic interaction model.

At fixed temperature and pressure, divide by $n={\sum }_i{n}_i>0$ and introduce the mole fractions $x_i=n_i/n$:The chemical potentials cannot be varied independently. In a two-component mixture, put $x_A=x$ and $x_B=1-x$. Within a smooth single-thermodynamic phase region with $0<x<1$,One chemical potential fixes the other's change, but determining its absolute value still requires a reference value. This does not say ${\mu }_A={\mu }_B$: equality at phase coexistence concerns the same chemical species in different thermodynamic phases. Nor does the constraint say $dG=0$ whenever temperature and pressure are fixed; changing composition still contributes ${\sum }_i{\mu }_{i}d{n}_i$ to the Gibbs free energy.

Heat is energy transferred across a thermodynamic system's boundary because of a temperature difference. It is measured in joules, like work and internal energy. A hot cup transfers heat to cooler surroundings; its temperature describes its state, while the transferred heat describes a process. The cup stores internal energy, rather than a substance called heat. ^[ref]

For a closed system with unchanged bulk kinetic energy and potential energy, choose $Q>0$ for heat entering and $W>0$ for work done by the system. The first law of thermodynamics givesHere $U$ is the stored internal energy. For example, a sealed piston cylinder receiving $20\mathrm{kJ}$ of heat and delivering $5\mathrm{kJ}$ of work gains $15\mathrm{kJ}$ of internal energy. A closed system retains its matter while exchanging energy; it need not be an isolated system. ^[ref]

Unlike internal energy, heat is not a state function: different paths between the same equilibrium states can transfer different $Q$, with compensating differences in work. The symbol $\delta Q$ records this dependence on the path; there is no stored quantity whose difference is always $Q$. The two paths in first law of thermodynamics illustrate this directly. Supplied heat can raise temperature as sensible heat, change a phase fraction through latent heat, or help supply expansion work. Measuring temperature alone therefore does not determine transferred heat.

For a reversible thermodynamic process, $\delta Q_{\mathrm{rev}}=TdS$, where $T$ is absolute temperature and $S$ is entropy. This identification requires reversibility: an irreversible adiabatic process has $Q=0$ and can nevertheless increase entropy. Likewise $Q=\Delta H$, with enthalpy $H$, requires the particular fixed-pressure conditions in enthalpy. These distinctions make heat useful as a precise boundary-transfer quantity in heat engines, refrigerators and heating calculations.

Historically, James Prescott Joule's paddle experiments showed that stirring water with input work could raise its temperature, helping establish conservation of energy. The lesson is still useful: a temperature rise can result from work even without incoming heat. ^[ref]

Heat capacity measures the heat supplied per small temperature increase along a specified process:The subscript $X$ records the constraint, such as fixed volume or fixed pressure. The units are $\mathrm{J}{\mathrm{K}}^{-1}$. The symbol $\delta Q$ emphasizes that heat is a transfer along a path, rather than a state function with a differential $dQ$. Consequently, an unspecified heat capacity does not determine the required heat.

For a closed system of fixed composition undergoing a quasistatic process, assume that only pressure-volume work occurs, with no friction and no change in bulk kinetic energy or potential energy. The first law of thermodynamics and $H=U+pV$, where $U$ is internal energy and $H$ is enthalpy, then giveThese partial derivatives define heat capacity at constant volume and heat capacity at constant pressure, with the amount and composition also held fixed. Integrating along the selected constraint gives $Q={\int }_{T_1}^{T_2}{C}_X(T)dT$ within a single thermodynamic phase; $Q=C_X\Delta T$ requires constant $C_X$. ^[ref]

For a computed example, take a monatomic gas in the classical ideal gas regime, with amount of substance $n=1\mathrm{mol}$ and stipulated molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. The equipartition theorem gives $U=3nRT/2$, and the ideal gas law gives $pV=nRT$. ThusFor the same temperature increase $\Delta T=10\mathrm{K}$, the two processes requireHere work done by the gas is positive. At fixed pressure, the extra heat supplies expansion work; the internal energy increase is the same because internal energy of an ideal gas depends only on temperature. Their final volume values differ. ^[ref]

The practical value of heat capacity is that a measured heat input and temperature response can determine it. Once its constraint and temperature dependence are known, it predicts heating requirements and supplies entropy changes through entropy from heat capacity. ^[ref]

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.

A practical test compares three transformations involving sodium hydroxide: dissolution of the solid, neutralization of its chemical solution by hydrochloric acid, and reaction of the solid directly with hydrochloric acid. In the ideal compatible-state notation, the hydrogen ion and hydroxide giveThe hydroxide cancels, whereas the newly formed water remains. Sodium and chloride spectator ions must still be included in the full material inventory when matching actual chemical solutions.

The teaching protocol Calorimetry: Heat of Neutralization Reaction & Hess's Law, Procedure A–C, uses nested insulated cups, a cardboard lid and a thermometer. It mixes $75\mathrm{mL}$ each of $2.0\mathrm{mol}{\mathrm{L}}^{-1}$ hydrochloric acid and sodium hydroxide chemical solutions; separately it adds $6.00\mathrm{g}$ solid sodium hydroxide to diluted hydrochloric acid, or to $150\mathrm{mL}$ water. It records initial temperatures, stirs, reads temperature every $30\mathrm{s}$, repeats runs, and extrapolates cooling trends to mixing time. Rapid weighing limits moisture uptake. The Introduction neglects vessel heat capacity and room losses; these are approximations, not a measured calibration. The readout tests whether the inferred enthalpies add, rather than directly measuring a state function.^[ref]

Matching the chemical equations alone is insufficient. The direct-acid recipe uses $0.160\mathrm{mol}$ hydrochloric acid, whereas the solution-mixing recipe uses $0.150\mathrm{mol}$. The dissolution run also does not automatically reproduce the water inventory of the mixed-solution run. Exact closure requires matching initial and final chemical solution compositions and adding any necessary mixing or dilution enthalpies. Agreement after approximate corrections is a limited test, rather than proof of exact cancellation for those recipes.

To improve the calorimetry, use calorimeter calibration for the participating hardware, and estimate calorimeter heat-loss correction under comparable conditions. With an approximately constant total receiving heat capacity $C_{\mathrm{tot}}=m{c}_p+C_{\mathrm{cal}}$, common initial temperature, constant external pressure, only expansion work, negligible escaping material and negligible bulk kinetic energy and potential energy changes, the balance isHere $m$ is the chemical solution mass, $c_p$ its specific heat capacity at constant pressure, $C_{\mathrm{cal}}$ the hardware heat capacity, and $Q_{\mathrm{loss},\mathrm{out}}$ the signed outward heat loss. Different initial temperatures require separate warming terms. Calorimeter calibration and matching chemical solution compositions address different errors.

For an original constructed model, choose $C_{\mathrm{tot}}=120\mathrm{J}{\mathrm{K}}^{-1}$ and $\Delta T=3\mathrm{K}$. Neglect outward heat loss and other corrections. The receiving chemical solution and hardware gain $q_{\mathrm{receiver}}=360\mathrm{J}$, so $q_{\mathrm{rxn}}=-360\mathrm{J}$. If the completed extent of reaction is $\xi =0.020\mathrm{mol}$ for a unit-coefficient chemical equation, the inferred reaction enthalpy per mole of that reaction isThese chosen numbers illustrate the signs and normalization; they are not measured sodium hydroxide data and are independent of the abstract $A,B,C$ example.

A loss of heat to the room makes the uncorrected warming underestimate the magnitude of a heat-releasing reaction. Incomplete dissolution, incomplete reaction, uncertain amount of substance, or a slow thermometer response can also spoil closure. A useful experimental residual is $r=\Delta H_{\mathrm{direct}}-(\Delta H_D+\Delta H_N)$ after all changes have been placed on the same amount and state basis. Compare $r$ with the measurement uncertainty, accounting for shared calibration errors; averaging repeated runs reduces scatter without removing a common bias.

Internal energy $U$ is the energy stored in the microscopic degrees of freedom of a thermodynamic system: random molecular kinetic energy, internal molecular excitations, and interaction potential energy. Its value requires a specified system and an energy reference. In thermodynamic equilibrium, internal energy is a state function; it records the present state, without remembering whether a previous change came from stirring, compression or heat transfer. This makes internal energy the common accounting quantity connecting mechanical experiments to thermal ones.^[ref]

Heat and work describe energy crossing the system boundary during a process. They are not separately stored contents of the system. For a closed system whose bulk kinetic energy and external potential energy have unchanged endpoint values, the first law of thermodynamics givesHere $Q_{\mathrm{in}}$ is positive for incoming heat, and $W_{\mathrm{on}}=-W_{\mathrm{out}}$ is positive for net work done on the system. Two routes between the same equilibrium states have the same $\Delta U$, although their heat and work can differ. The infinitesimal notation $dU=\delta Q_{\mathrm{in}}+\delta W_{\mathrm{on}}$ distinguishes a state function change from path-dependent transfers.^[ref]

For a calculated example, take one mole of a classical monatomic gas in the ideal gas regime, at fixed volume, from $300$ to $310\mathrm{K}$. With stipulated molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, its internal energy change is $3nR\Delta T/2=124.71\mathrm{J}$. A purely thermal route supplies $Q_{\mathrm{in}}=124.71\mathrm{J}$ with $W_{\mathrm{on}}=0$. An ideal insulated stirring route supplies $W_{\mathrm{on}}=124.71\mathrm{J}$ with $Q_{\mathrm{in}}=0$, measured between equilibrium endpoints after the motion dies away. The same stored increment has two different transfer histories; neither history defines a separate kind of stored energy.

The familiar internal energy of an ideal gas is a useful special case. Neglect intermolecular interactions, hold the amount of substance $n$ and composition fixed, and suppose that $f$ independent quadratic terms per particle obey the classical equipartition theorem. Choosing the constant reference contribution as zero givesHere $T$ is absolute temperature, $R$ is the molar gas constant, and $C_V$ is the whole sample's heat capacity at constant volume. A monatomic gas has $f=3$ translational terms. A rigid diatomic molecule has $f=5$ when its two rotational terms are active and vibration is frozen; activating one harmonic oscillator vibration adds two terms, one kinetic energy and one potential energy, giving $f=7$. The counting is of quadratic contributions, so a vibrational coordinate is not just one contribution.^[ref]

This constant-$f$ formula applies only over a regime with those same active terms, classical translational statistics, nonrelativistic motion and unchanged composition. Quantum freezing, quantum degeneracy, chemical reactions or appreciable intermolecular interactions invalidate that simple counting. More generally, at fixed amount and composition an ideal gas can still have $U=U(T)$ with varying heat capacity at constant volume, so use $\Delta U={\int }_{T_1}^{T_2}{C}_V(T)dT$. A nonzero constant reference adds $U_{\ast }$ to $fnRT/2$ without altering $\Delta U$.

An isothermal expansion increases the volume of a thermodynamic system at fixed temperature. For a reversible ideal gas path, the gas delivers positive work and absorbs positive heat, with $Q=W_{\mathrm{by},\mathrm{rev}}=nRT\ln (V_f/V_i)$ and $\Delta U=0$. The heat bath supplies the expansion energy; without sufficient heat transfer, doing work instead lowers the temperature. This transfer is the hot step of a Carnot cycle, and it is why an expanding gas need not cool when its surroundings continuously replenish its internal energy.

An insulated free expansion into an evacuated compartment is different. In the idealization with no heat exchange, no external work and no container energy change, $Q=W_{\mathrm{by}}=0$. Since internal energy of an ideal gas depends only on temperature, its final equilibrium temperature equals its initial value. The transient need not have a uniform temperature or pressure, so equal endpoint temperatures do not make it a reversible isotherm. Do not apply the logarithmic isothermal ideal-gas work formula to this irreversible process. Its entropy still increases by $\Delta S=nR\ln (V_f/V_i)$; entropy change in free expansion evaluates this state function difference using a hypothetical reversible path.^[ref]

For a real gas, even equal-temperature equilibrium endpoints need not have equal internal energy. An insulated free expansion can therefore change the temperature, while a controlled isothermal expansion must provide the heat needed for both its internal energy change and its delivered work. The internal energy at constant temperature article gives the Van der Waals equation example.^[ref]

The Joule-Thomson coefficient is the local temperature response to changing pressure at fixed enthalpy and fixed amount and composition:It describes the equilibrium endpoints of a Joule-Thomson effect experiment meeting the steady-flow energy balance assumptions. With $\Delta p=p_2-p_1<0$, a sufficiently small drop gives $\Delta T\simeq {\mu }_{\mathrm{JT}}\Delta p$. Thus ${\mu }_{\mathrm{JT}}>0$ means cooling, ${\mu }_{\mathrm{JT}}<0$ warming, and ${\mu }_{\mathrm{JT}}=0$ no first-order change. At zero coefficient, a finite drop can still change temperature. For a chosen illustrative local value ${\mu }_{\mathrm{JT}}=0.20\mathrm{K}\mathrm{ba}{\mathrm{r}}^{-1}$ and $\Delta p=-0.50\mathrm{bar}$, the linear prediction is $\Delta T=-0.10\mathrm{K}$; these are stipulated numbers, not measured gas properties.

For a fixed-composition simple compressible system, the measurable form isHere $C_p$ is the whole sample's heat capacity at constant pressure and $\alpha$ its volumetric coefficient of thermal expansion; alternatively use molar volume and molar heat capacity consistently. An ideal gas has $\alpha =1/T$, so ${\mu }_{\mathrm{JT}}=0$. Its enthalpy depends only on temperature; positive heat capacity at constant pressure then makes any such finite throttle preserve temperature. A real gas can have either sign. For $C_p>0$, compare $T\alpha$ with one: the thermal expansion response decides the sign, not insulation alone.^[ref]

A short derivation uses $dH=TdS+Vdp$ and the Maxwell relation $(\partial S/\partial p{)}_T=-(\partial V/\partial T{)}_p$:Setting $dH=0$ gives the coefficient. These are equilibrium state function identities, not a claim of a reversible thermodynamic process inside the throttle. For a finite drop, solve the endpoint enthalpy equality or integratealong the constant-enthalpy equilibrium curve. The coefficient can change sign during that integration, as the Van der Waals throttling model illustrates.

An ice bath can absorb heat while its temperature barely changes: the heat converts ice into water. The latent heat of fusion quantifies this melting budget. For a pure substance at its equilibrium melting temperature $T_m(p)$ and specified pressure $p$, define it here per unit mass:Here $h_l$ and $h_s$ are the specific enthalpies of the coexisting bulk liquid and solid, and $H$ is enthalpy. For ordinary melting, $L>0$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$; an amount $m$ has a total fusion enthalpy $mL$ in joules. A molar fusion enthalpy instead has units $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and equals $ML$, where $M=m/n$ is mass divided by amount of substance in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Always check the normalization before inserting a tabulated value into an energy balance.^{[ref] [ref]}

For a closed system converting completely from solid to liquid at fixed pressure, with only pressure-volume work and negligible changes in bulk kinetic energy and potential energy, the first law of thermodynamics gives $Q_p=\Delta H=mL$. Both endpoints are at $T_m(p)$: warming initially colder ice or subsequently warming the resulting water requires additional sensible heat. Since $H=U+pV$, the corresponding specific internal energy jump is $\Delta u=L-p\Delta v$, where $v=V/m$; $L$ includes the work associated with the change in volume. Heat is transferred energy, whereas enthalpy is a state function. ^[ref]

Let $x=m_l/m$ be the liquid phase fraction by mass, rather than by volume. Neglecting interfacial contributions to enthalpy, equilibrium phase coexistence at fixed pressure givesIn the model above, reaching $T_m$ takes $2.10\mathrm{kJ}$ of sensible heat; the plateau then absorbs $33.40\mathrm{kJ}$, ending at a cumulative $35.50\mathrm{kJ}$. At $18.80\mathrm{kJ}$, half the mass is liquid and the temperature is still $T_m$. The final warming takes another $4.18\mathrm{kJ}$. The rounded $L$ and $c_l$ are inspired by the introductory ice-melting calorimetry values in Case Western's demonstration; constant properties and exact equilibrium throughout are stipulated model assumptions. ^[ref]

Within one thermodynamic phase, specific heat capacity at constant pressure is $c_p=(\partial h/\partial T{)}_p$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$; a sample's heat capacity at constant pressure is $C_p=m{c}_p$, in $\mathrm{J}{\mathrm{K}}^{-1}$. The plateau is a finite enthalpy change at zero temperature change through varying $x$, so neither a finite single-phase $c_p$ nor $Q=m{c}_p\Delta T$ represents it. The heating curve needs separate sensible heat intervals and a fusion contribution. ^[ref]

The plateau assumes a pure substance, a stable bulk solid-liquid phase coexistence pair, fixed pressure, and sufficiently slow, uniform heating to remain near thermodynamic equilibrium. The total mass and composition must remain fixed, with negligible evaporation and no other phase transition. A fast heater can produce temperature gradients, so a local thermometer need not read the equilibrium $T_m$. Neither $T_m$ nor $L$ is universal across pressures or compositions: common-pressure melting-point slope describes the pressure dependence of $T_m$, while mixtures can melt over a temperature range as their coexisting compositions change. Their liquidus must be consulted instead of assuming a pure-substance plateau. ^[ref]

This makes ice useful as a temperature buffer: under these conditions it can absorb $m_{s}L$ before the remaining solid mass $m_s$ disappears, after which the water warms. If a stipulated constant net input power $P$ goes entirely into melting, that buffer lasts $m_{s}L/P$. Reversing the equilibrium conversion releases the same fusion enthalpy during freezing. The beautiful thermodynamic connection is $L=T_m(s_l-s_s)$, with $s$ the specific entropy and $T_m$ absolute temperature: equal Gibbs free energies at phase coexistence allow different enthalpies and entropies. The nearly stationary thermometer hides a finite, measurable change of state. ^[ref]

Maxwell relations turn pressure, volume and temperature measurements into information about entropy. For example, the pressure rise when heating a rigid sealed vessel determines the entropy change when expanding the same sample isothermally:The quantities on the two sides describe different experiments, yet both are mixed partial derivatives of one Helmholtz free energy. This is the useful surprise: a mechanical measurement can reveal a thermal property. Similarly, the coefficient of thermal expansion gives $(\partial S/\partial P{)}_{T,N}=-\alpha V$. ^[ref]

For a simple compressible system, hold particle number $N$ and composition fixed, and allow only pressure-volume work. Write $U$ for internal energy, $H=U+PV$ for enthalpy, $F=U-TS$ for Helmholtz free energy, and $G=H-TS$ for Gibbs free energy. Their natural-variable differentials areThus the first partial derivatives, with the subscript recording the variables kept fixed, areThe symmetry of second derivatives produces all four standard Maxwell relations, including their signs:For a brief derivation of the $F$ relation, differentiate $F_T=-S$ with respect to $V$ and $F_V=-P$ with respect to $T$: $-S_V=F_{TV}=F_{VT}=-P_T$. Each other row follows in the same way. Assume continuous second partial derivatives of the relevant thermodynamic potential in the region considered. At a phase transition, the equilibrium potential may lose this smoothness; apply the identities within each smooth thermodynamic phase, and handle latent heat and phase coexistence separately. If particle numbers vary, the potential differentials acquire chemical potential terms, so the fixed-$N$ constraints must remain explicit. ^[ref]

For an ideal gas with fixed $N$, the ideal gas law gives $P=N{k}_{B}T/V$. At fixed volume, heating changes the pressure at rate $N{k}_B/V$. Consequently an isothermal expansion at positive temperature hasFor one mole with $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $V_2=2V_1$, this is $\Delta S=R\ln 2\simeq 5.763\mathrm{J}{\mathrm{K}}^{-1}$. At $300\mathrm{K}$, a reversible isothermal expansion absorbs $Q=T\Delta S\simeq 1728.85\mathrm{J}$ and performs equal work, since its internal energy is unchanged. The entropy difference is a state function difference: the same equilibrium endpoints give the same $\Delta S$ in an insulated free expansion, although there $Q=0$ and the process is irreversible.

The same Maxwell relations let thermal measurements determine the difference between constant-pressure and constant-volume heat capacities. In a smooth single thermodynamic phase,using heat capacity at constant volume, heat capacity at constant pressure, coefficient of thermal expansion and isothermal compressibility. Along fixed pressure, $dV=\alpha VdT$, soThus measured expansion and compression responses determine a heat capacity difference without measuring both heat capacities separately. They do not determine the absolute heat capacities or absolute entropy by themselves; an independent thermal measurement supplies that missing information. For an ideal gas, $\alpha =1/T$ and ${\kappa }_T=1/P$ reduce the result to $C_P-C_V=PV/T=N{k}_B$, which is Mayer's relation.

When the necessary higher partial derivatives exist and commute, differentiating the Maxwell relations once more also givesIntegrating these slopes determines how a heat capacity changes with volume or pressure at fixed temperature. For an ideal gas, both right-hand sides vanish because the ideal gas law is linear in temperature at either fixed volume or fixed pressure. Consequently $C_V$ and $C_P$ depend only on temperature at fixed $N$, although they need not be constant in temperature. ^[ref]

Historically, James Clerk Maxwell's Theory of Heat, first published in 1871, presents these connections using the geometry of isothermal processes and reversible adiabatic processes. His fourth edition, page 167, explicitly relates expansion on heating at fixed pressure to heat emitted during an isothermal compression: the modern $G$ relation gives $\delta Q_{\mathrm{rev}}=TdS=-T\alpha VdP$. This explains why a substance with positive coefficient of thermal expansion emits heat when compressed isothermally, whereas one with negative coefficient absorbs it. ^{[ref] [ref]}

The Second law of thermodynamics restricts which energy-conserving processes can occur. In the Kelvin-Planck statement of the second law, no cyclic engine can have as its sole effect the extraction of heat from one reservoir and its complete conversion into work. In the Clausius statement of the second law, no process can have as its sole effect transferring heat from a colder reservoir to a hotter one. An engine or refrigerator combined with the reverse of a reversible device shows that a violation of either statement would violate the other.

A heat engine takes heat $Q_h>0$ from a reservoir at $T_h$, rejects $Q_c>0$ to one at $T_c<T_h$, and returns its working substance to the initial state. Energy conservation gives $W=Q_h-Q_c$ and efficiency $\eta =W/Q_h=1-Q_c/Q_h$. Carnot's theorem states that no engine between these reservoirs is more efficient than a reversible one, and that all reversible engines between them have the same efficiency. Otherwise, drive a reversible engine backwards with the work of a more efficient engine; the composite cycle transfers heat from cold to hot without external work, contradicting the Clausius statement.

A reversible Carnot cycle consists of two isotherms joined by two reversible adiabats. For an ideal gas, $Q_h=N{k}_B{T}_h\log (V_2/V_1)$ and $Q_c=N{k}_B{T}_c\log (V_3/V_4)$. The adiabats give $V_2/V_1=V_3/V_4$, soCarnot's theorem makes this independent of the working substance and defines the thermodynamic absolute-temperature scale consistently with ideal-gas temperature.^[ref]

For any reversible cycle, $\oint \delta Q_{\mathrm{rev}}/T=0$. This makes the integral between two equilibrium states independent of the reversible path, defining the state function entropy:Heat itself is path dependent; entropy is not. An irreversible change can be evaluated using any imagined reversible path between its equilibrium endpoints. For example, a fixed amount of ideal gas with constant $C_V$ has $\Delta S=C_V\log (T_B/T_A)+N{k}_B\log (V_B/V_A)$.

For a general cycle the Clausius inequality is $\oint \delta Q/T_{\mathrm{boundary}}\le 0$. The entropy balance for a process is $\Delta S\ge \int \delta Q/T_{\mathrm{boundary}}$, with equality for reversible heat exchange and no entropy production. The surroundings' entropy must be included, giving $\Delta S_{\mathrm{universe}}\ge 0$. A system's own entropy can decrease by exporting entropy; an isolated system's cannot.