Chap 19 Macromolecules

Chapter 19.

Materials 1: macromolecules and aggregates

Atkins / Paula

《 Physical Chemistry, 8th Edition 》

1Chap 19 Macromolecules

Chap 19 Macromolecules

Determination of size and shape19.1 Mean molar masses19.2 Mass spectrometry19.3 Laser light scattering19.4 Ultracentrifugation19.5 Electrophoresis19.6 Viscosity

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Self-assembly19.13 Colloids19.14 Micelles and biological membranes19.15 Surface films

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Structure and dynamics19.7 The different levels of structure19.8 Random coils19.9 The structure and stability of synthetic polymers19.10 The structure of proteins19.11 The structure of nucleic acids19.12 The stability of proteins and nucleic acids

Chap 19 Macromolecules

• There are macromolecules everywhere. inside us and outside us. Some are natural: they include polysaccharides such as cellulose, polypeptides such as protein enzymes, and polynucleotide's Such as deoxyribonucleic acid (DNA). Others are synthetic: they include polymers such as nylon and polystyrene that are manufactured by stringing together and (in some cases) cross-linking smaller building blocks known as monomers.

• Monomers are the individual building blocks, or repeating units, from which polymers are built.

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• Monodisperse: A substance which has a single definite molar mass.

• Polydisperse: A sample in which the mixture of molecules have different molar masses.

• Number-average molar mass :

• where Ni is the number of molecules with molar mass Mi and N is the total number of molecules in the sample. This average is given by osmometry.

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nM

ii

in MNN

M 1

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Chap 19 Macromolecules

• Viscosity-average molar mass :

• where Ni is the number of molecules with molar mass Mi and N is the total number of molecules in the sample. This average is given by viscosity.

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VM

ii

i

ii

i

V MN

MNM

1

Chap 19 Macromolecules

• Weight-average molar mass :

• The second equality shows that the mass-average molar mass is proportional to the mean-square molar mass of macromolecules with molar mass Miand N is the total number of molecules in the sample. The mass-average molar mass is given by light-scattering experiment.

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wM

ii

i

ii

i

ii

iw MN

MNMm

mM

2

1

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• Z-average molar mass :

• The Z-average molar mass is obtained from sedimentation experiment.

• The heterogeneity index is the ration :

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ZM

2

3

ii

i

ii

i

Z MN

MNM

n

W

MM

Chap 19 Macromolecules

• The Flory theta temperature, θ, is the temperature at which the osmotic virial coefficient of a macromolecular solution is zero.

• q solution: A solution that is at its Flory theta temperature is called a θ solution. Such a solution behaves nearly ideally, and so its thermodynamic and structural properties are easier to describe even though its molar concentration need not be low.

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(c) Dynamic light scattering

• Dynamic light scattering, can be used to investigate the diffusion of polymers in solution.

• Consider two polymer molecules being irradiated by a laser beam. Suppose that at a time t the scattered waves from these particles interfere constructively at the detector, leading to a large signal.

• However, as the molecules move through the solution, the scattered waves may interfere destructively at another time t' and result in no signal.

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Chap 19 Macromolecules

• When this behavior is extended to a very large number of molecules in solution, it results in fluctuations in light intensity that depend on the diffusion coefficient, D. which is a measure of the rate of molecular motion and is given by the Stokes-Einstein relation :

• where f is the frictional coefficient, a measure of the forces that retard a molecule's motion. Table 19.2 lists some typical values of D.

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fkTD

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Chap 19 Macromolecules

• For a spherical particle of radius a in a solvent of viscosity '1 (see Section 19.6), the frictional coefficient is given by Stokes's relation:

f = 6π a η• Hence, dynamic light scattering measurements give

the diffusion coefficient and molecular size, in cases where the molecular shape is known.

• If the molecule is not spherical, we use appropriate values off given in Table 19.3.

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Chap 19 Macromolecules

• For dilute monodisperse systems of random coils, it has been found empirically that D is related to the molar mass M of the polymer by:

• The coefficient βD is obtained by determining D at fixed viscosity and temperature for a variety of standard samples with known molar masses.

• As we should expect, bulky polymers of high molar mass migrate more slowly (have a lower diffusion coefficient) through a solvent than polymers of low molar mass.

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60. MD D

Chap 19 Macromolecules

19.4 Ultracentrifugation

• In a gravitational field, heavy particles settle towards the foot of a column of solution by the process called sedimentation. The rate of sedimentation depends on the strength of the field and on the masses and shapes of the particles. Spherical molecules (and compact molecules in general) sediment faster than rod-like and extended molecules.

• When the sample is at equilibrium, the particles are dispersed over a range of heights in accord with the Boltzmann distribution (because the gravitational field competes with the stirring effect of thermal motion).

• The spread of heights depends on the masses of the molecules, so the equilibrium distribution is another way to determine molar mass.

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Chap 19 Macromolecules

• Sedimentation is normally very slow, but it can be accelerated by ultracentrifugation, a technique that replaces the gravitational field with a centrifugal field.

• The effect can be achieved in an ultracentrifuge, which is essentially a cylinder that can be rotated at high speed about its axis with a sample in a cell near its periphery (Fig. 19.6).

• Modern ultracentrifuges can produce accelerations equivalent to about 105 that of gravity (‘105 g’). Initially the sample is uniform, but the 'top' (innermost) boundary of the solute moves outwards as sedimentation proceeds.

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Chap 19 Macromolecules

(a) The rate of sedimentation

• A solute particle of mass m has an effective mass melf = bm on account of the buoyancy of the medium, with

b = l - ρνs

• where ρ is the solution density, νs is the partial specific volume of the solute (νs = (∂V/∂mB), with mB the total mass of solute), and ρνs is the mass of solvent displaced per gram of solute. The solute particles at a distance r from the axis of a rotor spinning at an angular velocity ω experience a centrifugal force of magnitude meff r ω2.

• The acceleration outwards is countered by a frictional force proportional to the speed, s, of the particles through the medium. This force is written fs, where f is the frictional coefficient.

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Chap 19 Macromolecules

(a) The rate of sedimentation

• The particles therefore adopt a drift speed, a constant speed through the medium, which is found by equating the two forces meff r ω2 and fs. The forces are equal when

• The drift speed depends on the angular velocity and the radius, and it is convenient to define the sedimentation constant, S, as

• S may be used to determine either Mn or a. Again, if the molecules are not spherical, we use the appropriate value of fgiven in Table 19.3.

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ff

22 bmrrms eff

A

n

A

n

NMb

NMbbm

rsS

aff 62

Chap 19 Macromolecules

Example 19.4

• Determining a sedimentation constant

• The sedimentation of the protein bovine serum albumin (BSA) was monitored at 25°C. The initial location of the solute surface was at 5.50 cm from the axis of rotation, and during centrifugation at 56850 r.p.m. it receded as follows:

• t/s o 500 1000 2000 3000 4000 5000

• r/ cm 5.50 5.55 5.60 5.70 5.80 5.91 6.01

• Calculate the sedimentation coefficient.

• Method Equation 19.16 can be interpreted as a differential equation for s = dr/dt in terms of r; so integrate it to obtain a formula for r in terms of t. The integrated expression, an expression for r as a function of t, will suggest how to plot the data and obtain from it the sedimentation constant.

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Chap 19 Macromolecules

(a) The rate of sedimentation

• Answer Equation 19.16 may be written

• This equation integrates to

• It follows that a plot of ln(r/ro) against t should be a straight line of lope w2S. Use ω = 2πν, where ν is in cycles per second, and draw up the following table:

• t/s 0 500 1000 2000 3000 4000 5000 • 102 In(r/ro) 0 0.905 1.80 3.57 5.31 7.19 8.87 • The straight-line graph (Fig. 19.7) has slope 1.78 x 1O-5 so

w2S = 1.79 X 10-5 S-I. Because W = 2n x (56 850/60) S-I = 5.95 X 103 S-I, it follows that 5 = 5.02 X 10-13 s. The unit 10-13 s is sometimes called a ' vedberg' and denoted SVj in this case 5= 5.02 Sv.

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Srtr 2

dd

Strro

2ln

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Self-test 19.4 • Calculate the sedimentation constant given the following data

(the other conditions being the same a above):

t /s 0 500 1000 2000 3000 4000 5000

r/cm 5.65 5.68 5.71 5.77 5.84 5.9 5.97

[3.11 Sv]

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