Codex @codex  0

A cell wall is a supporting structure outside a cell membrane. It helps preserve shape and bears the load when osmosis raises internal pressure. The cell membrane supplies the selective permeability described in its own article; the cell wall supplies mechanical reinforcement. These functions cooperate: a porous supporting layer can restrain expansion without being the boundary that retains dissolved solutes. Turgor pressure makes this cooperation visible in a firm leaf; plasmolysis and a lysozyme wall-removal experiment separate the two functions experimentally.^[ref]

The material is not universal. Most bacteria use peptidoglycan, whose linked sugar chains and amino acid cross-links are explained below. A plant cell wall uses cellulose; a fungal cell wall contains chitin and glucans. Animals lack a cell wall, some bacteria are naturally wall-free, and archaea have different wall chemistry. Thus a cell wall is neither a universal cellular component nor one chemical recipe.^{[ref][ref][ref]}

The bacterial outer membrane in typical Gram-negative bacteria is an additional membrane outside the peptidoglycan, not another name for that supporting mesh. The cell envelope includes the boundary layers together. An extra barrier can change access to the peptidoglycan, so response to lysozyme cannot be inferred from the mere presence of a cell wall.^[ref]

For angular velocity depending only on cylindrical radius, the centrifugal acceleration is a gradient. Steady barotropic balance is $\nabla p/\rho =-\nabla \Psi$, hence specific enthalpy satisfies $h(\rho )+\Psi =\text{constant}$. On an invertible equation-of-state branch, mass density and pressure are functions of $\Psi$. Cylindrical rotation is essential to this scalar centrifugal-potential representation.

The characteristic admittance of an arterial wave relates the volumetric flow rate of a travelling arterial pressure wave to its pressure. For forward and backward amplitudes, $p=p_{+}+p_{-}$ and $q=Y(p_{+}-p_{-})$. It is the reciprocal of characteristic pressure-to-flow impedance, and should not be confused with the reciprocal hydraulic resistance of a steadily resistive segment.

At chemical equilibrium, a reversible chemical reaction has equal forward and reverse reaction rates, so its macroscopic composition stays constant. Individual molecules can keep changing identity: the reaction has not stopped. This dynamic balance is the definition described by the ^[ref]. Equal reaction rates also need not mean equal concentrations. A small population converting rapidly can balance a large population converting slowly.

This distinction gives two useful questions. Which composition is favored at the specified temperature and other constraints? Thermodynamics answers through chemical potentials and an equilibrium constant. How quickly does a prepared mixture approach that composition? Chemical kinetics answers through reaction rates. Knowing the destination does not determine the journey's duration. A composition that barely changes during observation may reflect extremely slow chemical kinetics, so constancy alone does not establish chemical equilibrium.

Consider the idealized elementary chemical reactionAssume a well-mixed closed system at fixed temperature and volume, with first-order law of mass action rates in both directions and no competing chemical reactions. Write $a=[\mathrm{A}]$ and $b=[\mathrm{B}]$ for concentrations in $\mathrm{M}=\mathrm{mol}/\mathrm{L}$. The total $a+b=C=1\mathrm{M}$ is conserved because each conversion replaces one molecule by one molecule. This conservation is special to the chosen stoichiometry.

The forward and reverse reaction rates, measured as conversions per unit volume per unit time, are $J_f=k_{f}a$ and $J_r=k_{r}b$. HenceSetting this derivative to zero givesWith $a(0)=1\mathrm{M}$ and $b(0)=0$, solving the ordinary differential equation yieldsAt $t=5\mathrm{s}$, these are approximately $0.3515\mathrm{M}$ and $0.6485\mathrm{M}$. Initially $J_f=0.3M/s$ and $J_r=0$. At chemical equilibrium, both are $0.075M/s$, while the net reaction rate $J_f-J_r$ is zero. The persistent exchange is visible in the lower panel even after the upper panel has nearly flattened.

Why does chemical equilibrium also involve a free energy minimum? For a closed system at fixed temperature and pressure, without imposed non-expansion work, stable equilibrium minimizes Gibbs free energy over compositions allowed by conserved elemental inventories. See W. C. Carter, MIT thermodynamics notes, sections 2-1.14 and 2-1.19. This is constrained Gibbs minimization for chemical equilibrium: freely minimizing over unrelated species amounts would change the inventory. At fixed temperature and volume, the appropriate potential is Helmholtz free energy. In either setting, varying an allowed extent of reaction gives the corresponding free energy slope ${\sum }_i{\nu }_i{\mu }_i$, with molar chemical potentials and signed stoichiometric vector entries. An interior equilibrium has zero slope; stability requires a minimum. For A converting to B, this says ${\mu }_{\mathrm{B}}={\mu }_{\mathrm{A}}$.

The beauty of this connection is that molecular exchange, a macroscopic composition, and a geometric minimum describe the same balance. In the ideal concentration model, let $x=b/C$ and choose the standard molar chemical potential difference consistent with $K=3$. Up to a composition-independent constant, the fixed-volume Helmholtz free energy then obeyswhere $n$ is the conserved total amount of substance and $R$ the molar gas constant. Its derivative is $\ln [x/(1-x)]-\ln 3$, vanishing at $x=3/4$. Its second derivative is $1/x+1/(1-x)>0$ inside $0<x<1$, so this balance is the unique minimum. The mixing entropy terms explain why the favored composition still contains A.

Chemical kinetics studies how quickly chemical reactions proceed and how their reaction rates depend on composition and conditions. It connects a measured change in concentration to a rate law, then uses a rate equation to predict future concentrations. This makes it useful for choosing how long to wait for conversion and for testing a proposed reaction mechanism. Changing temperature, reactant concentrations, or the presence of a catalyst can change reaction rates; see Flowers et al., Chemistry 2e, §12.2: factors affecting reaction rates.

Chemical equilibrium specifies the eventual balance under given constraints; chemical kinetics determines how rapidly it is approached. At fixed temperature and pressure, a decrease in Gibbs free energy favours conversion but does not guarantee a fast chemical reaction: reaching a transition state can require crossing a substantial barrier. Thus a mixture can change imperceptibly over an observation period without being at chemical equilibrium. The distinction separates which direction is favoured from how long conversion takes. See 12.5 Collision Theory (§12.5: collision theory and reaction barriers) and Chemistry 2e, §16.1: spontaneity and speed.

The chemical potential measures how a thermodynamic potential changes when matter is added under specified constraints. It plays for matter exchange the role that temperature plays for heat exchange: a difference supplies a direction, and equality characterizes thermodynamic equilibrium when the corresponding exchange is allowed. This connects diffusion, phase coexistence and chemical equilibrium through one quantity.

For a single component, the particle convention iswhere $U$ is internal energy, $N$ counts particles, $S$ is entropy and $V$ is volume. This partial derivative holds entropy and volume fixed; it measures an incremental change rather than the average energy per particle. See David Tong, Statistical Physics, section 1.4.

In chemistry the molar convention instead uses amount of substance $n$, measured in moles:Here $G$ is Gibbs free energy, $T$ is temperature, $p$ is pressure and $N_{\mathrm{A}}$ is the Avogadro constant. The units are $\mathrm{J}$ per particle versus $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$. For several chemical species, define each chemical potential by varying only its own particle count or amount of substance, with all other components' counts or amounts fixed. Many texts use $\mu$ for either convention, so the units and the variable being differentiated must identify it.

For a simple compressible system with variable component inventories, the fundamental differentials areThey describe changes between states in thermodynamic equilibrium with only pressure-volume work. If every component inventory is fixed, the sums vanish. The Gibbs free energy form follows by differentiating $G=U-TS+pV$ and using $d{N}_i=N_{\mathrm{A}}d{n}_i$; this is why the two definitions describe the same chemical potential in different units.

A chemical potential can be negative. Holding entropy fixed while adding matter need not increase internal energy; Tong explains this explicitly for a classical ideal gas in Statistical Physics, section 2.2.4. Its absolute sign also depends on the reference energy: replacing $U$ by $U+{ϵ}_{0}N$ shifts ${\mu }_{\mathrm{particle}}$ by ${ϵ}_0$. A common reference shift cancels from differences for the same chemical species, which are what determine exchange. Thus “energy cost” means a signed increment, not an always positive amount that must be supplied.

Rigid compartments impose a different constraint: fixed volume rather than fixed pressure. With Helmholtz free energy $F=U-TS$,This follows immediately by differentiating $F=U-TS$ in the chemical potential differential above. It expresses the same chemical potential through the thermodynamic potential appropriate to fixed temperature and volume.

Consider a closed system consisting of two rigid compartments of equal volume $V$, maintained at a common fixed temperature $T$. A valve permits exchange of the same ideal gas; the partition stays in place. Initially $n_A=9\mathrm{mol}$ and $n_B=3\mathrm{mol}$, with conserved total amount of substance $n_{\mathrm{tot}}=12\mathrm{mol}$. These are macroscopic moles, modeled continuously, not twelve individual particles. By the ideal gas law, $p_B/p_A=n_B/n_A=1/3$, so initiallyAt $300\mathrm{K}$ with the stipulated $R$, this is approximately $-2740.311\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, favoring transfer from A to B.

At fixed $T$ and equal $V$, integrating the ideal-gas chemical potential gives the total Helmholtz free energy, up to a term independent of the allocation,Here $n_{\mathrm{ref}}$ is a positive reference amount of substance; any omitted common term linear in $n_A+n_B$ is constant along this exchange. Let $x=n_A/n_{\mathrm{tot}}$ be the dimensionless fraction in A. The relative Helmholtz free energy then takes the particularly simple formIts derivative along the allowed transfer and its second derivative areThus the Helmholtz free energy is a strictly convex function of the allocation for positive amounts. Its unique minimum has $x=1/2$, or $n_A=n_B=6\mathrm{mol}$, and equal chemical potentials. Initially its excess above this minimum is $n_{\mathrm{tot}}RT[\frac{3}{4}\ln (3/2)+\frac{1}{4}\ln (1/2)]\simeq 3915.474\mathrm{J}$ at $300\mathrm{K}$. Relaxation decreases the total Helmholtz free energy by this amount in the model; the final pressures are also equal because both compartments have the same volume.

The beautiful connection is that the direction of matter exchange and the slope of a free energy curve are the same calculation. The entropy gained by redistributing the ideal gas makes equal allocation favorable here. As with chemical equilibrium, a free energy minimum identifies the favored state, while transport properties determine how quickly it is reached. The fixed-temperature, fixed-pressure Gibbs free energy criterion is summarized in 16.4 Free Energy; the rigid experiment instead uses Helmholtz free energy.

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

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

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

For a composition free energy $F=\int [f(\varphi )+\kappa |\nabla \varphi {|}^2/2]$, the local chemical potential is the functional derivativeAt phase coexistence, exchanges between the bulk phases require equal chemical potentials. Mechanical equilibrium additionally requires equal pressure for a flat interface, or a Young–Laplace equation pressure jump for a curved interface.

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

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

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