CHE-1H26: Elements of Chemical Physics
Molecular Motion
Dr Yimin Chao
Room 1.45
Email: [email protected]
http://www.uea.ac.uk/~qwn07jsu
1
Outline
•
•
•
•
•
•
•
•
•
1. Transport properties
2. Diffusion and viscosity
2.1 Diffusion process
2.2 Viscosity
3. The relationship of diffusion and viscosity– Einstein-Stokes relationship
4. The effects of temperature on viscosity and diffusion—Arrhenius Law
5. Methods of measurement
6. Rotational diffusion
7. using viscosity and diffusion to measure the shapes and sizes of molecules
References:
• Atkins's Physical Chemistry, 8th Ed, P Atkins and J de Paulo,2002,OUP.
• Elements of Physical Chemistry, 5th Ed, P Atkins and J de Paulo,2009,OUP
• University Physics, 12th Ed, F Sears, MW Zemansky and HD Young, 2000,
Addison and Welsey.
2
Molecular motion-- transport
• Translational and rotational
• Molecular motion in gas and liquid
3
1. Transport properties
• Commonly expressed in ‘phenomenological’
equations that are empirical summaries of
experimental observations
• The rate of migration of a property is measured
by its flux, J, the quantity of that property
passing through a given area (perpendicular to
the direction of flow) in a given time interval
divided by the area and the duration of the
interval.
Flux = amount per SQ M
per Second
Direction of flow
4
Flux
In general the flux J is proportional to the gradient,
X, causing the flux r
r
J = LX
L– proportionality constant
• Both flux and gradient are VECTORS they have
both sign and magnitude.
• The flux acts in a direction such that it reduces
the gradient.
5
Case (1)—Matter: Diffusion-- Fick’s Law
• If matter is flowing (as in diffusion), we speak of a matter
flux of so many molecules per square meter per second
• A positive value of J signifies a flux towards positive z;
A negative value of J signifies a flux towards negative z
• Because matter flows down a
concentration gradient, from high
concentration to low concentration,
c
dc
ddc
c/dz< 0
J is positive if dz is negative, thus
dz
dc
J ( matter ) = − D
dz
-- Fick’s first law of diffusion
J: mol m-2 s-1
•
D– diffusion coefficient
SI units: meter square per second m2S-1
6
Case (2)—Energy: Heat flow-- Fourier’s Law
dT
J (energy) = −κ
dz
J: heat flux,
SI unit: Joule m-2 s-1
κ– thermal conductivity,
SI units: Joules per Kelvin per meter per second J K-1 m-1s-1
7
2. Diffusion and viscosity
2.1 diffusion process:-- Fick’s second law
Fick observed that the diffusion flux was proportional to
the concentration gradient.
Recall
dc
J ( matter ) = − D
dz
- Fick’s first law
8
∆z
Low concentration
Z=z0+ ∆z
Z=z0
High concentration
Direction
of diffusion
z
At z = z0 the amount of matter entering the slab per second is
AJ(z0)
A is the area of the slab, J is the flux in mol m-2 s-1
At z = z0+ ∆z, the amount leaving the slab is
AJ(z0+ ∆z)
The effect of diffusion is to reduce the concentration gradient.
The amounts entering and leaving the slab are not equal.
9
The amount left in the slab will vary with time since the concentration gradient will vary with
time.
The number of mols per second deposited in the slab will be:
dn/dt = AJ(z0) - AJ(z0+ ∆z)
= A[J(z0) - J(z0+ ∆z)]
n is the number of mols. The concentration is the number of mols divided by the volume.
Thus:
The volume of the slab is A ∆z and the concentration in the slab, c, is given by
c = n/ A ∆z
Therefore
dn
dc
A[ J ( z 0 ) − J ( z 0 − ∆ z )]
dt
=
=
dt
A∆z
A∆z
10
if ∆z→ 0, then [J(z0) - J(z0+ ∆z)]→ dJ and ∆z→ dz
therefore
dc/dt = dJ/dz
Recall
J = −D
dc
dz
Fick’s first law
therefore
dc
d dc
( )
= −D
dt
dz dz
thus
dc
d 2c
= −D
dt
dz 2
Fick’s second law
11
Notes to Fick’s Law:
dc
d 2c
= −D
dt
dz 2
• this is a second order differential equation.
• it must be solved using the appropriate
Boundary conditions: the initial conditions of the
system, the boundaries to matter flow, etc
• Analytical solutions are not always possible for
such equations.
12
Two special cases: I
when a thin layer of a diffusant is placed in the middle of a long cylindrical cell,
the variation of concentration with time at a distance z from the centre is given
by
2
c( z, t) =
n0 e
2 π Dt
−(
z
)
4 Dt
n0 is the amount of substance initially present per unit cross section area, mol/m2
A plot of the concentration profiles at different time
13
An example:
A dye solution is placed in a thin layer in the centre of a long cylindrical
cell. If diffusion is allowed to continue for 1 hour and twenty minutes,
calculate the concentration of the dye in mol dm-3 at a distance of 1 cm
from the original position of the thin layer assuming that the diffusion
coefficient of the dye is 0.79x 10-9 m2 s-1 and the initial concentration of
the dye per unit area is 10 mol m–2
The equation for diffusion under these circumstances is
c( z, t ) =
n0
e
2πDt
 z2 

−

4
Dt


z = 1x10-2 m, t = 4800 s.
Therefore
c=
10
−9
2 × 3.142 × 0.79 × 10 × 4800
×e


1×10 −4

− 
−9

 4×0.79×10 × 4800 
= 2.807 mol m-3 = 2.807 x10-3 mol dm-3 = 0.002807 mol dm-3
14
A simulated result:
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16
Two special cases: II
In the case of a thin layer of diffusant placed at the one end of the tube the
profiles appear as follows:
17
2.2 Viscosity
Consider the flow of liquid in a pipe.
•At the wall there is no flow.
•Maximum flow is at the centre.
•The velocity profile is parabolic: at any
point at some distance ry from the centre,
the velocity, u, is given by:
P1 − P2 2 2
u=
ry − r
4Lη
(
r
ry
)
P1 and P2 are the pressures at the start and the end of the pipe.
L is the length of the pipe
η is the viscosity of the liquid in Pa S.
ry is the distance from the centre of the pipe and r is the radius of the pipe
18
The total rate of volume flow–
dry
ry
dV = ∫ u ⋅ 2πry ⋅ dry ⋅ dt
dV
P1 − P 2
=π
⋅ ∫ ry2 − r 2 ⋅ 2ry ⋅ dry
dt
4 Lη r
0
=π
(
)
P1 − P 2 1 4
⋅ [ ry − r 2 ry2 ]0r
4 Lη
2
P1 − P 2 4
=(
)πr = R
8 Lη
Thus,
dV
P1 − P2 4
R=
=(
)πr
dt
8Lη
-- Poiseuille’s Formula
19
An example:
In an experiment a vertical tube of 20 cm length connects two
reservoirs. The rate of flow of the liquid through the tube is 20 cm3s-1.
If the tube has diameter of 0.5 cm calculate the viscosity of the liquid
assuming the pressure difference between the top and bottom of the
tube is 19 Pascals.
Rearranging the equation gives
( P2 − P1 )πr 4
η=
8 LR
P2-P1 = 19 Pa, d = 0.5 cm = 0.5 x 10-2m
L = 20 cm = 20 x 10-2 m, R = 20 cm3s-1= 20 x 10-6m3s-1
Substituting
η = 19 x 3.142 x (0.5 x 10-2)4 /8 x 20 x 10-2 x 20 x 10-6
= 1.166x 10-3 Pascals s
20
Some notes
i.
If the pressure gradient is too great or the flow is too fast
turbulence is observed. There are no simple theories of
turbulence.
ii. For simple liquids η is independent of the rate of flow. For many
materials, e.g. thixotropic paints, blood, tomato ketchup, this is
not the case.
iii. Materials with more complex behaviour are known as nonNewtonian fluids, after Newton who was the first to describe
and analyse viscous flow.
21
Molecular Motion
Dr Yimin Chao
Room 1.45
Email: [email protected]
End of the first lecture
22
The variation of viscosity with concentration
Generally the viscosity of a liquid depends on how easily the molecules of the
liquid can move past one another.
Simple solution
e.g. KOH show a gradual increase in viscosity with concentration.
6
4
3
2
1
9.
9
10
.5
6
11
.2
4
0
0.
08
9
0.
17
9
0.
73
6
1.
32
2
1.
93
8
2.
80
6
4.
21
2
5.
75
6.
85
8.
02
relative viscosity
5
concentration
23
Polymers
At relatively low concentrations (i.e. there is a very large excess of solvent)
the viscous behaviour of polymer solutions can be divided into
two distinct regions: The dilute and semi-dilute regimes
semi dilute
regime
viscosity
c*
dilute
regime
polymer
conce ntratio n
dilute solution
polymers not
interacting
semi dilute solution
polymers entangled
At very low concentrations the viscosity of the solution varies linearly with concentration.
At a critical point called c* there is an abrupt change in slope and the viscosity increases
much more rapidly.
This is due to the entanglement of the polymers.
The polymers are sufficiently highly concentrated that they begin to interact with each
other and become entangled.
This means that the polymers can no longer move independently of one another and the
motion of the liquid is slowed.
24
Outline
•
•
•
•
•
1. Transport properties
2. Diffusion and viscosity
2.1 Diffusion process
2.2 Viscosity
3. The relationship of diffusion and viscosity–
Einstein-Stokes relationship
• 4. The effects of temperature on viscosity and
diffusion—Arrhenius Law
• *5. Methods of measurement
• 6. Rotational diffusion
25
3. The relationship of diffusion and viscosity–
Stokes-Einstein relationship
A random walk
The step-like motion of kinesin molecules in the body
The botanist Brown noticed that when small pollen particles in a liquid
were observed in a microscope they were in constant random motion.
This motion is called Brownian motion.
It reflects the constant bombardment of the pollen grains by the impact
of the molecules of the liquid.
Einstein analysed the motion of these particles by assuming
that they executed a RANDOM WALK.
In a random walk the motion is considered to consist of a series
of steps the direction of each step is assumed to be randomly related
to the direction of the previous step.
26
Mutual and self diffusion
Fick’s law is written for a situation in which a concentration gradient exists.
It is about the rate of change of the concentration gradient.
The diffusion coefficient is thus the MUTUAL diffusion coefficient and is
strictly valid only under defined conditions of concentration gradient. It is the
measure of the rate at which one substance diffuses into another.
Brownian motion shows that diffusion can still go on even when there is no
concentration gradient. SELF DIFFUSION a measure of the random thermal
motion of the molecules in the absence of a concentration gradient.
All diffusion arises because of the constant random thermal motion of
molecules.
A simulation of diffusion due to random motion is shown below.
27
diffusion
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diffusion
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180
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25
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160
350
60
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300
50
120
250
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100
200
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60
20
100
40
5
10
50
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160
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0
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31
Einstein showed that if the particle could be considered as a sphere
in medium of viscosity η. The Mean squared displacement of the particle
could be written as:
kT
z =
t
3πηr
2
z2 is the mean squared displacement of the particle.
This avoids the problem of the displacement being both positive and negative.
k is the Boltzmann constant,
T is the Absolute temperature.
t is time and r is the effective radius of the particle.
r is sometimes known as the Stokes law radius since Stokes derived the relationship between
the radius of a particle and its motion in a viscous medium.
32
The diffusion coefficient may be defined as (by Stokes)
D =
kT
6 πη r
Stokes-Einstein relationship
(Stokes Law)
Thus
z 2 = 2 Dt
Using the Stokes-Einstein relationship,
we can estimate the molecular diameters of molecules.
Example:
Water has a diffusion coefficient of about 2.4x10-9m2s-1 and a viscosity of 1x10-3Pa s.
Rearranging the equation
r=
kT
6πηD
33
kT
r=
6πηD
therefore, r= 1.38x10-23 x 298 /6 x 3.142 x 1 x 10-3 x 2.4 x 10-9 m
=0.908 x10-10 m
=0.0908 nm
•
•
Notes:
In ice the distance between oxygen atoms is 0.276 nm
and the OH bond distance is about 0.09 nm so this value is reasonable.
The radius obtained by the application of Stokes law is an effective radius,
of the order of molecular dimensions but it is affected by interactions
with other molecules in the liquid.
34
• In the case of sucrose and water the simple relationship between
viscosity and diffusion coefficient does not hold. Water diffuses faster
than sucrose because of the effects of the molecular radius but neither
diffusion coefficient has a simple linear relationship to the viscosity.
The deviation is largest at high viscosity (low reciprocal viscosity).
RELATIVE DIFFUSION COEFF
DIFFUSION VS VISCOSITY IN
SUCROSE
18
16
14
12
10
WATER
8
6
4
2
SUCROSE
0
0
0.2
0.4
0.6
0.8
RECIPROCAL VISCOSITY
35
• Local viscosities
If 1% of Agar is dissolved in water, the system turns from
one in which the viscosity is about 1x 10-3 Pa s to one in
which the viscosity is effectively infinite.
i.e. A solid gel is formed.
Gels consist of mobile regions connected by junction
zones.
Gel forming materials are polymers.
They form rigid junction zones connect by flexible regions.
Between the polymers there are cavities filled with solvent.
Junction zone
cavity
In the cavities the local viscosity is the same as that as the solvent. However on the large
scale the viscosity is determined by the larger scale structure.
Diffusion of molecules small compared to the cavity size occurs very easily. When the
molecule gets large enough diffusion is slowed.
When using the Stokes-Einstein relationship it is very important to recognise that
both the viscosity and radius terms must be treated with caution.
36
Outline
•
•
•
•
•
•
•
•
•
1. Transport properties
2. Diffusion and viscosity
2.1 Diffusion process
2.2 Viscosity
3. The relationship of diffusion and viscosity– EinsteinStokes relationship
4. The effects of temperature on viscosity and diffusion—
Arrhenius Law
*5. Methods of measurement
6. Rotational diffusion
7. using viscosity and diffusion to measure the shapes
and sizes of molecules
37
4. The effect of temperature on viscosity and
diffusion
Diffusion is an activated process and often follows an Arrhenius law of the form
D = D0 e
−
Ea
RT
D0 is the diffusion coefficient at infinite temperature,
Ea is the activation energy,
R is the gas constant
and T the absolute temperature
Note: Increasing the temperature increases the numerical value of D.
38
Viscosity decreases with increasing temperature.
Thus:
η = η0e
Ea
RT
Note the change in sign of the exponent.
η0 is the value of viscosity at infinite temperature.
Consider
D =
kT
6 πη r
Substituting the equation for the temperature dependence of viscosity we get
D =
kT
6π r η 0 e
Ea
RT
39
Ea
E
− a
 kT  − RT
e
D = 
= D 0e RT
 6πrη0 
Where
kT
D0 =
6πrη0
Note: For systems obeying the Stokes Einstein equations
the activation energy for viscosity and diffusion are the same.
40
Example:
If the diffusion coefficient of water is 2.34x10-9m2s-1 at 200C what is it at
1000C assuming the activation energy for diffusion is 20 kJmol-1. The gas
constant is 8.3 JMol-1K-1
since
thus
D = D0e
D2
=e
D1
−
−
Ea
RT
Ea  1 1 
 − 
R  T 2 T1 
therefore
D 2 = D1e
E  1 1 
− a − 
R  T 2 T1 
= 2.34 ×10−9 ⋅ e
20×103  1
1 
−
⋅
−

8.31  373 293 
= 1.353 ×10−8 m 2 s −1
41
Outline
•
•
•
•
•
•
•
•
•
1. Transport properties
2. Diffusion and viscosity
2.1 Diffusion process
2.2 Viscosity
3. The relationship of diffusion and viscosity– EinsteinStokes relationship
4. The effects of temperature on viscosity and diffusion—
Arrhenius Law
5. Methods of measurement
6. Rotational diffusion
7. using viscosity and diffusion to measure the shapes
and sizes of molecules
42
5.Method of measurement
5.1 Viscosity. The Ostwald viscometer
Normally calibrate with a liquid of
known viscosity
Since
dV  P1 − P2  4
πr = R
= 
dt  8 Lη 
therefore
η t ρ
= ×
η 0 t0 ρ 0
P1 − P2 = ρgh
V
t
V
ρgh
⋅ πr 4 =
8 Lη
t
R=
(Atkins P666)
43
5.2 Diffusion in liquids
Solution A
Magnetic stirrer
Porous glass
Capillary technique
Solution B
Gel containing
solution A
filling tube
Diaphragm technique
Solution B
stirrer
(Atkins P778)
44
6. Rotational diffusion
Debye was interested in the interactions of dipolar molecules with electric
fields.
When the field is switched on the molecules tend to follow the field:
When the field is switched on the molecule reorients through an angle θ in
a time t.
Electric field switched on
θ
Molecule reorients
through angle θ in
time t
The rate at which the molecules can respond depends on how fast the dipoles
can reorient, this depends on their molecular rotation rates.
The rate of change of angle is dθ/dt. This is a measure of the rate of rotation.
45
In a liquid the rotation of the molecule must be affected by viscosity in
the same way as translational motion, by analogy a rotational diffusion
coefficient D can be defined:
Dr =
kT
8πrη
Note the difference in the numerical factor.
Often rotation is expressed in terms of a correlation time rather than a
rotational diffusion coefficient. The rotational correlation time, τ, can be
defined as:
r2
τ=
2 Dr
τ will decrease as temperature increases.
Thus
τ = τ 0e
Ea
RT
46
We can calculate the expected value of the rotational correlation time of
water as follows:
r2
τ=
2 Dr
and
Therefore by substitution
Example 1:
Dr =
kT
8πrη
4πr 3η
τ=
kT
Water: r =0.09x10-9 m, η= 1x 10-3 Pa s, k=1.38x10-23 jK-1, T =298 K
therefore
4 ⋅ 3 . 14 ⋅ ( 0 . 09 × 10 − 9 ) 3 ⋅ 1 × 10
τ =
1 . 38 × 10 − 23 × 298
−3
=2.2 x 10-12 s.
The measured value is about 1.98 x 10-12 s. So the agreement is very good.
47
Example 2:
In a concentrated sucrose solution the activation energy for viscosity
is about 40 kJmol-1. If the correlation time for rotational motion at 273K
is 150 picoseconds what would you expect it to be at 373K?
τ =τ e
since
thus
0
τ e
= E
τ e
Ea
RT
Ea / RT 2
2
1
a
therefore
τ 2 = 150 ⋅ e
/ RT1
=e
40×103
8.31
Ea  1 1 
 − 
R  T 2 T1 
1 
 1
⋅
−

 373 273 
= 1.35 ps
48
Note 1:
Variation in the correlation time of a probe molecule in sucrose solution
correlation time
picoseconds
200
150
100
50
0
0
2
5
12 15 18 21 24 30 35 40
sucrose concentration %
49
60
50
40
30
20
10
60
50
40
30
21
15
5
0
0
activation energy kj/mol
Note 2:
When the activation energy
for correlation time ( squares) and for viscosity (triangles)
is plotted against concentration for sucrose solutions deviations can be seen.
sucrose concentration %
This is due to the internal structure of the solution.
50
Outline
•
•
•
•
•
•
•
•
•
1. Transport properties
2. Diffusion and viscosity
2.1 Diffusion process
2.2 Viscosity
3. The relationship of diffusion and viscosity– EinsteinStokes relationship
4. The effects of temperature on viscosity and diffusion—
Arrhenius Law
5. Methods of measurement
6. Rotational diffusion
7. using viscosity and diffusion to measure the shapes
and sizes of molecules
51
7. Using Viscosity and Diffusion To Measure
The Shapes And Sizes Of Molecules
Protein molecules can be approximated by three different shapes:
Spheres, prolate spheroid, Oblate spheroid
a
Spheres
c
b
Sphere a=b=c
52
prolate spheroid
c b
a
Prolate b=c<a
If a>>>>>b,c this becomes a rod
53
Oblate spheroid
a
b
c
Oblate a=b>c
If a,b>>>>c this becomes disk
The motion of these shapes in a liquid is different and thus their
effects on viscosity are different. In principal therefore it is possible to
determine the shape of a molecule by measuring its viscosity.
54
For a solution viscosity can be expressed as a function of the solvent viscosity and
concentration.
ηS = η0(1+k1c + k2c2 + k3c3 … )
η0 is the viscosity of the pure solvent and c is the concentration of the solute
If we define
ηrel = ηS/ η0,
Then ηrel = (1+k1c + k2c2 + k3c3 … )
This type of function can express the shape of complex curves
This is an example of a virial expansion.
The data for KOH given above may be fitted to a line of the form
ηrel= 1+0.1054 c +0.003 c2 +.0024c3 –0.002c4
6
relative viscosity
5
4
3
2
1
0
0
5
10
15
concentration
55
then the specific viscosity ηsp is given by
ηsp = ηrel –1 = k1c + k2c2 + k3c3 …
This is a useful definition since it summarises the effects of the solute on the
viscosity.
The intrinsic viscosity, [η], is defined as the ratio ηsp/c as c tends to zero.
Therefore
η sp 
k1c + k 2 c 2 ....
lim   = lim
= k1 =[η],
c →0
c
 c  c →0
by finding the intrinsic viscosity it is possible to obtain the first term in
the virial expansion and so get to the simplest measure of the effects
of the solute on viscosity.
56
Mark-Kuhn-Houwink-Sakurada equation:
[η ] = KM
a
v
Where K and a are constants depend on
the solvent and type of macromolecule;
Mv
- the viscosity average molar mass
57
Example: Using intrinsic viscosity to measure molar mass
The viscosities of a series of solutions of polystyrene in toluene were
measured at 25 oC with the following results:
c/(g dm-3)
η/(10-4kg m-1 s-1)
0
5.58
2
6.15
4
6.74
6
7.35
8
7.98
10
8.64
Calculate the intrinsic viscosity and estimate the molar mass of the
polymer. K= 3.80 × 10-5 dm3 g-1 and a = 0.63.
We draw up the following table:
c/(g dm-3)
0
η/η0
1
100[(η/η0 )-1]/(c/g dm-3)
2
1.102
5.11
4
1.208
5.2
6
1.3117
5.28
8
1.43
5.38
10
1.549
5.49
The points are plotted in the figure right. The extrapolated
intercept at c = 0 is 0.0504, so [η] = 0.0504.
Therefore,
 [η ] 
Mv =  
K 
1
a
= 9.0 × 10 4 g mol −1
58
Generally [η] = ν V/M,
Where V is the hydrated volume and M the mass of the molecule. ν is the
Simha factor. The Simha factor contains information about the shape of
the molecule.
16
14
simha factor
12
10
8
6
4
2
0
10
8
6
5
4
3
2
1
2
3
4
5
6
8
10
axial ratio
prolate
oblate
59
Axial ratios of proteins and viruses estimated by viscosity
[η](10−3m3/kg)
Axial ratio (prolate)
Maximum value
protein
Molecular
weight
Ribonuclease a
13683
3.3
3.9
lysozyme
14211
2.7
3.2
Bushy stunt virus
10700000
3.43
4.0
tropomyosin
93000
52
29
fibrinogen
330000
27
20
60
Sizes of proteins as calculated by Stokes law
Protein
M/Daltons
D/10-11 M2 s-1
r calculated/nm
Cytochrome c
13370
11.4
1.15
Lysozyme
13930
11.2
1.17
Ribonuclease
12640
13.1
1.0
Myoglobin
16890
11.3
1.16
Human serum albumin
68460
6.1
2.15
Alcohol dehydrogenase
73050
6.5
2.01
61