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Lecture 23

Physical Chemistry

Chemistry 433

of Surfaces

NC State University

Surface tensionLiquids tend to adopt shapes that minimize their surface area.This places the maximum number of molecules in the bulk.Droplets of liquids tend to be spherical because a sphere isthe shape with the smallest surface-to-volume ratio. The workneeded to change the area by dσ is:

dw = γdσ

The coefficient γ is called the surface tension. It has dimensions of energy per unit area (J/m2). At constant volume and temperature the work of surface formation is equal to the Helmholtz free energy:

dA = γdσ

Since dA < 0 is a spontaneous change surfaces tend to contract.

Curved surfacesA liquid surface is not generally flat. This has two consequences.1. The curvature of the surface affects the vapor pressure.2. The capillary rise or of liquids in narrow tubes results.

We consider bubbles (vapor trapped in liquid) or droplets (liquidin vapor) as spheres of area 4πR2. Here we consider a bubblewhere the pressure inside the cavity is P The outward force iswhere the pressure inside the cavity is Pin. The outward force is 4πR2Pin . The force inwards arises from the external pressurePout and the surface tension. The differential change in area is:

dσ = 4π(R + dR)2 - 4πR2 = 8πRdRSince, dw = 8πγRdR the force is F = 8πγR. Pressure is force perunit area so:

4πR2 Pin = 4πR2 Pout + 8πγRor:

Pin = Pout + 2γ/R

Capillary RiseThe pressure exerted by a column of liquid of density ρ is:

P = ρghAt equilibrium this hydrostatic pressure matches the pressure difference Pin - Pout = ΔP = 2γ/R. The height of a column of li id i ill iliquid in a narrow capillary is obtained from:

2γ/R = ρgh

so that:

h = 2γρgR

Capillary rise problemGiven that the surface tension of water is 0.0728 J/m2

calculate the capillary rise in a glass tube that is 1 mm in radius.

A. 1 mm

B. 1.5 mm

C. 10 mm

D. 15 mm

Capillary rise problemGiven that the surface tension of water is 0.0728 J/m2

calculate the capillary rise in a glass tube that is 1 mm in radius.

A. 1 mm

2 0 0728 J/m2

B. 1.5 mm

C. 10 mm

D. 15 mm

h = 2γρgR =

2 0.0728 J/m2

1000 kg/m3 9.8 m/s2 10– 3 m= 0.015 m = 1.5 cm

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Capillary rise problemGiven that the surface tension of ethanol is 0.032 J/m2

calculate the capillary rise in a glass tube that is 0.1 mm in radius. Assume that density of ethanol is 0.71 g/cm3.

A. 47 mm

B. 65 mm

C. 91 mm

D. 110 mm

Capillary rise problemGiven that the surface tension of ethanol is 0.032 J/m2

calculate the capillary rise in a glass tube that is 0.1 mm in radius. Assume that density of ethanol is 0.71 g/cm3.

A. 47 mm

2 0 032 J/m2

B. 65 mm

C. 91 mm

D. 110 mm

h = 2γρgR =

2 0.032 J/m2

710 kg/m3 9.8 m/s2 10– 4 m= 0.91 m = 9.1 cm = 91 mm

Vapor pressure and surface tensionWhen water or other liquids are finely divided then we must adda term to the free energy expression:

dG = VdP - SdT + γdσ + μdn

The term γdσ is the change in free energy when the surface areais changed. The free energy of bulk water is μH2O. The

dditi l t i li th t th f ill b i i i dadditional term implies that the free energy will be minimized as the total surface area is minimized. This means that the liquidwill form a sphere if no other forces are acting on it. Furthermore,the spheres will tend to combine to form the largest possiblesphere (thereby minimizing the surface-area-to-volume ratio).

This implies that smaller droplets of water that have a higher totalsurface area have a higher vapor pressure than bulk water andtherefore transfer spontaneously into a larger drop.

Gibbs adsorption isothermMolecules at surfaces are attracted to the bulk. This createsa force that tends to minimize surface area. If the surface is stretched, the free energy of the system is increased. Thefree energy per unit surface area is called the surface tension. The SI units of surface tension are millinewtons per meter(mN/m). The surface tension is also the surface free energyper unit area given by:

∂G

where σ is the area. Substances that lower the surface tensionalso lower the surface free energy and therefore they migrateto the surface. The quantitative expression of this phenomenonis called the Gibbs adsorption isotherm:

Γ = adsorption (excess concentration) of solute at surface, mol/m2

γ = surface tension, N/m, a = activity, c = molarity

γ = ∂G∂σ T,P

Γ = – 1RT

dγd ln a ≈ – 1

RTdγ

d ln c

Measuring surface tensionPsurface(nM / m) = γ0 – γ

γ γ0Langmuir balance

Measuring surface tensionThe surface pressure is equal to the difference in the surfacetension of the pure water (γ0) and for the mixture (γ).The surface pressure is usually plotted vs. the area per molecule (in Å2/molecule). This is the reciprocal of the surfaceConcentration (units can be either molecules/ Å2 or moles/m2).As the surface pressure is increased the molecules packmore closely and the area per molecule decreases.

γ γ0Langmuir balance

more closely and the area per molecule decreases.There may be discontinuitiesdue to phase transitions.The maximum surfacepressure is equal to thesurface tension of purewater. At the surface pressure the film collapsesand γ = 0.

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The hydrophobic effectThe free energy change for transfer of hydrocarbons from organicsolvent to water is positive. The dominant contribution to the hydrophobic effect is the entropy. In the organic phase:

μH(H) = μH0(H) + RT ln aH(H)

(H = hydrocarbon, W = water)I t th h i l t ti l iIn water the chemical potential is:

μH(W) = μH0(W) + RT ln aH(W)

Thermodynamic data for the transfer of hydrocarbons from hydrocarbon solvents (H) to water (W) have been obtained by Tanford.

Thermodynamic data for the transfer of hydrocarbons to water

Hydrocarbon μH0(W)-μH

0(H) HH0(W)-HH

0(H) SH0(W)-SH

0(H)(kJ/mol) (kJ/mol) (J/mol-K)

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -C2H6 16.3 -10.5 -88C3H8 20.5 -7.1 -92C4H10 24.7 -3.3 -96C5H12 28.7 -2.1 -105C6H14 32.4 0 -109C6H6 19.2 +2.1 -59

Lipids and SurfactantsBoth lipids and surfactants consist of a hydrophobic tailregion and a hydrophilic head group (amphipathic). Both lipids and surfactants will orient with their head groups pointed towards the water interface and theirtails sequestered from water.

Lipids have the ability to form a bilayer. This propertymakes these molecules the constituents of biologicalmembranes. Bilayers can be gel-like or crystalline. Theycan have a planar phase or form hexagonal phase.

Surfactants can tend to form micelles. Micelles are spherical.The hydrophobic tails form the interior and the charged headgroups are on the surface.

Lipid polymorphism

Biochemistry of Lipids, Lipoproteins and Membranes, Vance & Vance, Elsevier 1996

Lipid structureRepresentative lipids areshown in Figure to the right. There are two acylchains on a glycerol.The third site is a phosphodiester with a number of possiblenumber of possible groups indicated.

The fluid mosaic modelPlasma membrane

Biochemistry of Lipids, Lipoproteins and Membranes, Vance & Vance, Elsevier 1996

Cytosol

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Membrane asymmetryThe inner and outer leaflets of membrane bilayers havedifferent compositions. Erythrocytes are the most studied. The cytosolic side is composed of PE and PS. The PE distribution is ca. 80% in the inner membrane and 20% in the outer membrane. PS is negatively charged and PE is moderately negative in charge. The inner membrane is thus largely negative.g y g

The outer membrane consists of PC, sphingomyelin, and glycolipids. Cholesterol is also important and associates with the membrane to provide added fluidity. Plasma membranes have equimolar quantities of cholesterol. By contrast, the endoplasmic reticulum and mitchondrial membranes have small amounts of cholesterol.

Detergents and Lysophospholipids

Detergents are amphipathic molecules that can have charged or uncharged head groups and single hydrophobic tails. If one of the acyl chains is cleaved from

li id th l h h li id i

S

O-

O O

OO

O

OH

R

a lipid then a lysophospholipid is created.Other associating molecules include aromatic or fused ringcompounds (dyes, purines, pyrimidines) and alicyclic fused ring compounds (bile salts, cholesterol etc.).

SDS

O

Lysophospholipid

Membrane fusionMembrane fusion occurs in fertilization, cell division, exo- andendocytosis, viral infection and intracellular membrane transport.

Model SystemsModel membrane SUVs will undergo fusion when incubatedat or near Tc. For example, continued recycling of sonicatedPC i l th h T 41 C lt i f ti f lPC vesicles through Tc = 41 oC results in formation of largervesicles. For fusion events in vivo calcium is often required.

Biological SystemsIt is difficult to determine the mechanism for membrane fusion.Two proposals that pass through inverted micellar transitionstates are shown on the next slide.

Membrane fusion

Drug DeliveryThe role of lipids in biologymust be understood in orderto enhance drug deliverymethods. Biological membranesare permeable to small neutralmolecules (MW < 1000 Daltons).A number of strategies to enhance drug delivery involvethe use of liposomes. An exampleshown in the Figure consists ofa liposome that is targeted usingsurface-attached monoclonal antibodies that recognize a particularcell-surface antigen.

Transmembrane PotentialThe membrane potential is transmembrane electrical potential

difference expressed inside with respect to outside. Using the outside as the reference potential makes sense as all cells share that outside. This choice of direction (or sign convention) affects other measures of electrical properties, notably transmembrane current, which are all expressed inside with respect to outside. Thus our current convention is that current crossing the membrane from inside to outside is positive. In physics we express the potential as V (unit is the volt).In chemistry the Nernst equation is used to describe the dependence of the oxidation potential on concentration. The symbol is E (or Eo) and the unit is also the volt.

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The Nernst equationThe Nernst equation describes the potential for each half-reaction

a(Ox) + ne- b(Red)The standard potential (i.e. potential at 1 molar concentration) is Eo.The electrode potential is given by:

E = Eo – RT ln [Red]b

where R is the gas constant and F is the Faraday (96,450 J/volt).In an electrochemical cell where two half-reactions are combinedto make a redox reaction, the electromotive force is:

emf = E(+) - E(-)The free energy is G = -nFE. Therefore, the standard free energychange for a redox process is ΔGo = -nF(Eo

ox - Eored) = -nFΔEo.

E = E – nF ln[Ox]a

Application of the Nernst equation to membrane potential

The free energy per mole of solute moved across the membraneΔGconc = -RTln(Co/Ci) where Co is the concentration outside themembrane and Ci is the concentration inside the membrane. Thedifference in charge concentration results in a free energy contributionfrom the voltage difference ΔGvolt = -FΔE (assuming n=1). The balance of forces at equilibrium requires that ΔGvolt = ΔGconc so that the trans-membrane potential is obtained as follows.

ΔGvolt = ΔGconc

– FΔE = – RT ln Co

Ci

ΔE = RTF ln Co

Ci

ProblemAt 25o C the Nernst equation can be simplified to:

Calculate the transmembrane potential if the concentrations of K+

inside and outside the cell are [K+]i = 140 mM and [K+]o = 30 mM, i l

ΔE = 0.059 log Co

Ci

respectively.A. ΔE = -39 mVB. ΔE = -59 mVC. ΔE = -67 mVD. ΔE = -78 mV

ProblemAt 25o C the Nernst equation can be simplified to:

Calculate the transmembrane potential if the concentrations of K+

inside and outside the cell are [K+]i = 140 mM and [K+]o = 30 mM, i l

E = 0.059 log Co

Ci

respectively.A. ΔE = -39 mVB. ΔE = -59 mVC. ΔE = -67 mVD. ΔE = -78 mV

E = 0.059 logK+

o

K+i

= 0.059 log 30140

= – 0.039 V = – 39 mV

ProblemAt 25o C the Nernst equation can be simplified to:

Calculate the transmembrane potential if the concentrations of H+

inside and outside the cell are [H+]i = 0.1 μM and [H+]o = 11 μM, i l

E = 0.059 log Co

Ci

respectively.A. ΔE = 120 mVB. ΔE = 197 mVC. ΔE = 217 mVD. ΔE = 277 mV

ProblemAt 25o C the Nernst equation can be simplified to:

Calculate the transmembrane potential if the concentrations of H+

inside and outside the cell are [H+]i = 0.1 μM and [H+]o = 11 μM, i l

E = 0.059 log Co

Ci

respectively.A. ΔE = 120 mVB. ΔE = 197 mVC. ΔE = 217 mVD. ΔE = 277 mV

E = 0.059 logH+

o

H+i

= 0.059 log 110.1

= 0.120 V = 120 mV

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Coulomb’s law: The force between two charges q1 and q2 is

The force is a vector quantity and can either be attractive

Review of Electrostatics

F = kq1q2

r 2 , k = 14πε0

ε0 permittivity of vacuum 8.854 x 10– 12 C2/N/mThe force is a vector quantity and can either be attractive(negative) or repulsive (positive).

The electric field is the force per unit charge.

+ + - +

Field = Forceq = kqsource

r 2

The Coulomb constant is:

Including the electron charge (so that the charges are nowjust partial charges z1 = eq1 and z2 = eq2) and converting

Electrostatic potential is long range

k = 14πε0

= 8.987 x 109 Nm2/C2

the distance to Å, the potential is:

Two charges at a distance of 10 Å have a potential of 142 kJ/mol.

K = e2

4πε0Å= 2.35 x 10– 18 J = 1420 kJ/mol

U = K z1z2

r(Å)

The electrostatic potential has units of volts and is derivedfrom the field by:

Electrostatic potential is derived from the field

V = Edr = kq2r

To obtain an energy in MKS units volts are multiplied bythe electron charge e = 1.62 x 10-19 coulombs (C).Therefore, 1 eV = 1.62 x 10-19 Joules.The Faraday is just the charge of a mole of electronsNAe = (1.62 x 10-19 C)(6.022 x 1023) = 96,450 C (J/V)Remember Joules = Coulombs x Volts.

The case of a constant electric field, as between charged parallel plate conductors, is a good example of the relationship between work and voltage.

Electric field across parallel plates

+ + + + + + + + + + + + +

dV

The electric field between parallel plates (and across amembrane) is the voltage drop divided by the distance.

- - - - - - - - - - - - - - - - - -

dVField = Voltsdistance

Physiological AspectsTypical values

Expressed with the above convention, all cells have membrane potentials which are inside negative. The values observed vary a little from cell to cell and across different species, but for nerve and muscle cells that are very well studied, a typical value is y yp-80mV (ranging from -90mV to -60mV). (mV = millivolt = 10-3 volt)

Basis The membrane potential is a manifestation of a small separation of charge across the lipid bilayer of the membrane. The lipid is the dielectric of a parallel plate capacitor in which the salty (conductive) solutions separated by the membrane are the plates of the capacitor.

Voltage and Electric FieldAlthough the membrane potentials observed in cells are

<0.1 volt, the electrical field is very large. Changes in this field, brought about through changes in membrane potential can have marked affects on membrane proteins that themselves carry charges. Such influence on membrane proteins can markedly change cell activity.

The field is the voltage divided by the distance of the voltage drop Field = V/d. The units of field relevant to the dimensions of membranes are volts/nm. In the laboratory we often measure the field in volts/cm or volts/m.

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Changes in Transmembrane PotentialAlthough the membrane potentials observed in cells are

<0.1 volt, the electrical field is very large. Changes in this field, brought about through changes in membrane potential can have marked affects on membrane proteins that themselves carry charges. Such influence on membrane proteins can markedly change cell activity.

Any process that changes the separation of charge across the membrane will change membrane potential. These processes can be natural or experimental, and an experimental one serves to illustrate the phenomenon well. It is known as current injection.

Current Injection ExperimentThe microelectrode is a very fine glass tube which can penetrate the membrane, The solution inside the electrode (usually concentrated KCl) will then be connected to the inside of the cell and can be used to carry current into the cell, hence the name current injection. For this to happen the circuit must be completed: the switch must be closed, and the current must go to the reference electrode outside the cell by passing outwards through the membrane. This is shown in the following diagram. The positive charges reaching the inside of the membrane will tend to reduce the inside negativity, and the positive charge carried away from the outer surface will also do so.

Transmembrane Potential

Cell

Microelectrode

Switch + -

ReferenceElectrode- +

Transmembrane Potential

Cell

Microelectrode

Switch + -

ReferenceElectrode- +

TransmembranePotential is reduced

There are several important points to note: Although termed current injection the current is outward from the membrane perspective. There must be a circuit: in through the electrode, out through the cell. The outward current will tend to make the membrane potential become less negative. In other words the positive (outward)

b ill d k h b i ltransmembrane current will tend to make the membrane potential move in a positive direction. If the current could be adjusted in magnitude it could be made large enough to eliminate the original charge separation altogether. This particular case is very important in many attempts to understand membrane phenomena because the membrane potential is temporarily eliminated from influencing ion movements when V = 0.