Chemistry 433 Surface tension Curved surfaces ...

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solvent to water is positive. The dominant ... The hydrophobic tails form the interior and the charged head .... ε0 permittivity of vacuum 8.854 x 10– 12 C2/N/m.
Chemistry 433 Lecture 23 Physical Chemistry of Surfaces

NC State University

Surface tension Liquids 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 is the shape with the smallest surface-to-volume ratio. The work needed 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 surfaces A 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 (liquid in vapor) as spheres of area 4πR2. Here we consider a bubble where the pressure inside the cavity is Pin. The outward force is 4πR2Pin . The force inwards arises from the external pressure Pout and the surface tension. The differential change in area is: dσ = 4π(R + dR)2 - 4πR2 = 8πRdR Since, dw = 8πγRdR the force is F = 8πγR. Pressure is force per unit area so: 4πR2 Pin = 4πR2 Pout + 8πγR or: Pin = Pout + 2γ/R

Capillary Rise The pressure exerted by a column of liquid of density ρ is: P = ρgh At equilibrium this hydrostatic pressure matches the pressure difference Pin - Pout = ΔP = 2γ/R. The height of a column of li id iin a narrow capillary liquid ill iis obtained from: 2γ/R = ρgh so that:

h=

Capillary rise problem

2γ ρgR

Capillary rise problem

Given 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.

Given 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

A. 1 mm

B. 1.5 mm

B. 1.5 mm

C. 10 mm

C. 10 mm

D. 15 mm

D. 15 mm

h=

2 0.0728 0 0728 J/m 2 2γ = ρgR 1000 kg/m 3 9.8 m/s 2 10 – 3 m

= 0.015 m = 1.5 cm

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Capillary rise problem

Capillary rise problem

Given 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.

Given 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

A. 47 mm

B. 65 mm

B. 65 mm

C. 91 mm

C. 91 mm

D. 110 mm

D. 110 mm

Vapor pressure and surface tension When water or other liquids are finely divided then we must add a term to the free energy expression: dG = VdP - SdT + γdσ + μdn The term γdσ is the change in free energy when the surface area is changed. The free energy of bulk water is μH2O. The additional dditi l tterm iimplies li th thatt th the ffree energy will ill b be minimized i i i d as the total surface area is minimized. This means that the liquid will form a sphere if no other forces are acting on it. Furthermore, the spheres will tend to combine to form the largest possible sphere (thereby minimizing the surface-area-to-volume ratio). This implies that smaller droplets of water that have a higher total surface area have a higher vapor pressure than bulk water and therefore transfer spontaneously into a larger drop.

Measuring surface tension Psurface(nM / m) = γ 0 – γ

γ

γ0 Langmuir balance

h=

2 0.032 0 032 J/m 2 2γ = ρgR 710 kg/m 3 9.8 m/s 2 10 – 4 m

= 0.91 m = 9.1 cm = 91 mm

Gibbs adsorption isotherm Molecules at surfaces are attracted to the bulk. This creates a force that tends to minimize surface area. If the surface is stretched, the free energy of the system is increased. The free 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 energy per unit area given by: γ = ∂G ∂σ

T,P

where σ is the area. Substances that lower the surface tension also lower the surface free energy and therefore they migrate to the surface. The quantitative expression of this phenomenon is called the Gibbs adsorption isotherm:

dγ dγ Γ=– 1 ≈– 1 RT d ln a RT d ln c

Γ = adsorption (excess concentration) of solute at surface, mol/m2 γ = surface tension, N/m, a = activity, c = molarity

Measuring surface tension The surface pressure is equal to the difference in the surface tension 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 surface Concentration (units can be either molecules/ Å2 or moles/m2). As the surface pressure is increased the molecules pack more closely and the area per molecule decreases. There may be discontinuities due to phase transitions. The maximum surface pressure is equal to the surface tension of pure water. At the surface pressure the film collapses γ γ0 and γ = 0. Langmuir balance

2

The hydrophobic effect The free energy change for transfer of hydrocarbons from organic solvent 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) IIn water t the th chemical h i l potential t ti l iis: μH(W) = μH0(W) + RT ln aH(W)

Thermodynamic data for the transfer of hydrocarbons to water μH0(W)-μH0(H) HH0(W)-HH0(H) SH0(W)-SH0(H) (kJ/mol) (kJ/mol) (J/mol-K) ------------------------------------------------C 2H 6 16.3 -10.5 -88 C 3H 8 20.5 -7.1 -92 C4H10 24.7 -3.3 -96 C5H12 28.7 -2.1 -105 C6H14 32.4 0 -109 19.2 +2.1 -59 C 6H 6 Hydrocarbon

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

Lipids and Surfactants

Lipid polymorphism

Both lipids and surfactants consist of a hydrophobic tail region and a hydrophilic head group (amphipathic). Both lipids and surfactants will orient with their head groups pointed towards the water interface and their tails sequestered from water. Lipids have the ability to form a bilayer. This property makes these molecules the constituents of biological membranes. Bilayers can be gel-like or crystalline. They can 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 head groups are on the surface.

Lipid structure Representative lipids are shown in Figure to the right. There are two acyl chains on a glycerol. The third site is a phosphodiester with a number of possible groups indicated.

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

The fluid mosaic model Plasma membrane

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

3

Membrane asymmetry The inner and outer leaflets of membrane bilayers have different 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 g y negative. 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 a lipid li id th then a llysophospholipid h h li id iis created. Other associating molecules include aromatic or fused ring compounds (dyes, purines, pyrimidines) and alicyclic fused ring compounds (bile salts, cholesterol etc.).

R

OO S

OH O O

SDS

Membrane fusion

O

O

Lysophospholipid

Membrane fusion

Membrane fusion occurs in fertilization, cell division, exo- and endocytosis, viral infection and intracellular membrane transport. Model Systems Model membrane SUVs will undergo fusion when incubated at or near Tc. For example, continued recycling of sonicated PC vesicles i l th through h Tc = 41 oC results lt iin fformation ti off larger l vesicles. For fusion events in vivo calcium is often required. Biological Systems It is difficult to determine the mechanism for membrane fusion. Two proposals that pass through inverted micellar transition states are shown on the next slide.

Drug Delivery The role of lipids in biology must be understood in order to enhance drug delivery methods. Biological membranes are permeable to small neutral molecules (MW < 1000 Daltons). A number of strategies to enhance drug delivery involve the use of liposomes. An example shown in the Figure consists of a liposome that is targeted using surface-attached monoclonal antibodies that recognize a particular cell-surface antigen.

Transmembrane Potential The 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.

4

The Nernst equation The Nernst equation describes the potential for each half-reaction a(Ox) + ne-

Application of the Nernst equation to membrane potential The free energy per mole of solute moved across the membrane

b(Red)

The standard potential (i.e. potential at 1 molar concentration) is Eo.

ΔGconc = -RTln(Co/Ci) where Co is the concentration outside the

The electrode potential is given by:

membrane and Ci is the concentration inside the membrane. The difference in charge concentration results in a free energy contribution

b

[Red] E = E o – RT ln nF [Ox] a

from the voltage difference ΔGvolt = -FΔE (assuming n=1). The

where R is the gas constant and F is the Faraday (96,450 J/volt).

balance of forces at equilibrium requires that ΔGvolt = ΔGconc so that

In an electrochemical cell where two half-reactions are combined

the trans-membrane potential is obtained as follows. ΔGvolt = ΔGconc

to make a redox reaction, the electromotive force is: emf = E(+) - E(-)

– FΔE = – RT ln

The free energy is G = -nFE. Therefore, the standard free energy

Co ΔE = RT ln F Ci

change for a redox process is ΔGo = -nF(Eoox - Eored) = -nFΔEo.

Problem

Problem

At 25o C the Nernst equation can be simplified to: ΔE = 0.059 log

Co Ci

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

Co Ci

E = 0.059 log

Co Ci

Calculate the transmembrane potential if the concentrations of K+

Calculate the transmembrane potential if the concentrations of K+

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

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

respectively. i l

respectively. i l

A. ΔE = -39 mV

A. ΔE = -39 mV

B. ΔE = -59 mV

B. ΔE = -59 mV

C. ΔE = -67 mV

C. ΔE = -67 mV

D. ΔE = -78 mV

D. ΔE = -78 mV

Problem

inside and outside the cell are

Co Ci

= 0.1 μM and

o

K+

i

= 0.059 log

30 140

= – 0.039 V = – 39 mV

At 25o C the Nernst equation can be simplified to: E = 0.059 log

Calculate the transmembrane potential if the concentrations of H+ [H+]i

K+

Problem

At 25o C the Nernst equation can be simplified to: E = 0.059 log

E = 0.059 log

[H+]o

= 11 μM,

Co Ci

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,

respectively. i l

respectively. i l

A. ΔE = 120 mV

A. ΔE = 120 mV

B. ΔE = 197 mV

B. ΔE = 197 mV

C. ΔE = 217 mV

C. ΔE = 217 mV

D. ΔE = 277 mV

D. ΔE = 277 mV

E = 0.059 log

H+

o

H+

i

= 0.059 log 11 0.1

= 0.120 V = 120 mV

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Review of Electrostatics

Electrostatic potential is long range

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

qq F = k 12 2 , k = 1 4πε 0 r – 12 2 ε 0 permittivity of vacuum 8.854 x 10 C /N/m The force is a vector quantity and can either be attractive (negative) or repulsive (positive). +

+

-

+

k=

qsource Field = Force q = k r2

1 = 8.987 x 10 9 Nm 2/C 2 4πε 0

Including the electron charge (so that the charges are now just partial charges z1 = eq1 and z2 = eq2) and converting the distance to Å, the potential is: zz U=K 1 2 r(Å)

K=

The electric field is the force per unit charge.

e 2 = 2.35 x 10 – 18 J = 1420 kJ/mol 4πε 0Å

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

Electrostatic potential is derived from the field The electrostatic potential has units of volts and is derived from the field by: V=

The Coulomb constant is:

q Edr = k r2

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

Field = To obtain an energy in MKS units volts are multiplied by the 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 electrons NAe = (1.62 x 10-19 C)(6.022 x 1023) = 96,450 C (J/V)

Volts distance

V

d ------------------

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

Remember Joules = Coulombs x Volts.

Physiological Aspects

Voltage and Electric Field Although the membrane potentials observed in cells are

Typical 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 -80mV (ranging from -90mV to -60mV). (mV = millivolt =

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

yp value is and muscle cells that are veryy well studied, a typical 10-3