Transcript No Slide Title

Mobility and Chemical Potential
• Molecular motion and transport disucssed in TSWP Ch. 6. We can
address this using chemical potentials.
• Consider electrophoresis:
– We apply an electric field to a charged molecule in water.
– The molecule experiences a force (F = qE)
– It moves with a constant velocity (drift velocity = u)
• It obtains a speed such that the drag exactly opposes the electrophoretic force. No
acceleration = steady motion.
– u = qE·1/f where f is a frictional coefficient with units of kg s-1.
• qE is a force with units of kg m s-2 giving u with the expected ms-1 velocity units.
– Think of (1/f) as a mobility coefficient, sometimes written as µ.
– u = mobility · force
– We can determine the mobility by applying known force (qE) and
measuring the drift velocity, u.
Mobility and Chemical Potential
• Gradient of the chemical potential is a force.
• Think about gradient of electrical potential energy:
q 
electrical potential energy ( q = charge,  = electrical potential)
q  q

 qE where E is the electric field and qE is a FORCE
x
• Extending this to the total chemical potential:

 Force
x
1 
u   
Drift velocity obtained from gradient of the chemical potential
 f x
 
– where f is a frictional coefficient
– (1/f) as a mobility coefficient
Mobility and Chemical Potential: Example
+
    kTln c  ze
0
If concentration (c) is constant
througout:


 ze  qE
x
x
And the drift velocity is
u
1 
qE
  
f
 f x
+
Potential ()
Write down chemical potential
as a function of position in this
electrophoresis:
-
E
E

x
Position (x)
What if c is not constant?
Can the entropy term give rise to an
effective force that drives motion?
This is diffusion, and we can derive
Fick’s Law (TSWP p. 269) from
chemical potentials in this way.
Brownian Motion
Brownian trajectory
1
• Each vertex represents measurement of position
“Random walk”
• Time intervals between measurements constant
µm
• After time (t) molecule moves distance (d)
t=0
• 2-dimensional diffusion:
<d2> = <x2> + <y2> = 4Dt
d
• 3-dimensional diffusion:
t = 0.5 s
0
0
µm
A lipid will diffuse around a 10 µm
diameter cell:
 10 
 4Dt  t  61s
 2 
2
1
<d2> = <x2> + <y2> + <z2> = 6Dt
• In cell membrane, free lipid diffusion:
D ~ 1 µm2/s
Diffusion: Fick’s First Law
1
• Jx = Flux in the x direction
µm
• Flux has units of #molecules / area
(e.g. mol/cm2)
• Brownian motion can lead to a net flux of
molecules in a given direction of the concentration
is not constant.
• Ficks First Law:
0
0
µm
1
c 
J x  D
x 
Derivation of Fick’s First Law from
Entropy of Mixing
1 x   10  kTln c1 x 
• Chemical potential of component 1 in mixture.
1   
u1  x     
 f  x 
• Net drift velocity (u) related to gradient of
chemical potential by mobility (1/f) where f is
frictional coefficient.
1   1 c1

   kT
 f  c1  x   x
1  c1
J1x  x   c1  x u1  x     kT
 f  x
1 
 kT  D
 f 
• Flux (J1x) is simply concentration times net drift.
• Einstein relation for the diffusion coefficient.
• Entropy is the driving force behind diffusion.
D  kT
where   mobility (not chemical potential)
Fick’s Second Law: The Diffusion Equation
• Consider a small region of space (volume for 3D,
area for 2D)
• Jx(x) molecules flow in and Jx(x+dx) molecules
flow out (per unit area or distance per unit time).
N 
 J x  x   J x  x  dx dy
 t 
N
J x  x
c  J x  x   J x  x  dx 

t 
dx
J x  x  dx 
J 
   x 
 x 
 2 c 
c 
 D 2 
t 
x 
since c  N / dxdy
definition of the derivative
c 
since J x   D
x 
Diffusion Example
X
[C]
Jx
c 
J x  D
x 
 2 c 
c
 D 2 
t
x 
X
[C]
Jx=0
A pore (100 nm2) opens in a cell membrane that separates
the cell interior (containing micromolar protein
concentrations) from the protein free exterior. How fast
do the proteins leak out? (Assume D ~ 10-6 cm2/s)
Diffusion Question
Thought problem:
Axons of a nerve cell are long processes that can extend more than 1 meter for
nerve cells that connect to muscles or glands. If an action potential starting in the
cell body of the neuron proceeds by diffusion of Na+ and K+ to the synapse, how
long does it take the signal to travel 1m?
D25°C(Na+) = 1.5*10-5 cm2/sec
D25°C(K+) = 1.9*10-5 cm2/sec
Treat this as a 1D diffusion problem.
<x2> = 2Dt
Transport
Property
Anisotropy Forces
(Gradient)
Diffusion
Concentration Diffusional
T, Pi
Frictional
Sedimentation
Velocity
Centrifugal
Acceleration
Centrifugal 
Buoyant 
Frictional 
Law
J x  D
Centrifugal
Centrifugal 
Acceleration, Buoyant 
Concentration Diffusional 
Viscosity
Velocity
Shear 
Frictional 
Electrophoresis Electric Field
Electrostatic 
Frictional 
Rotary Diffusion Shape
Rotational
Frictional
dc
dx
dr m 2 r (1  v2  )

dt
f
M (1  v2  )
s
fN A

Sedimentation
Equilibrium
MW Dependence
(Spherical Particles)
D  M-1/3
s  M-2/3
MD(1  v2  )
RT
M app 
RT
dc
2
(1  v2  ) rc dr
Fsh  A
u
dux
dy
Ze
f
d( , f ,t)
2
 Drot  (, f ,t )
dt
[] M
Thermal Motion: Maxwell-Boltzmann
Distribution of Velocities
Quantitative description of molecular motion
Average translational molecular kinetic energy:
Utr 
3
kT
2
k = Boltzmann constant = R/N0
= 1.38 x 10-23 JK-1 molecule-1
Probibility of a molecule in a dilute gas having
speed between u and u + du:
3
 m  2 2 mu
dPu   4
u e
2kT 
2
2kT du
Both T and m (mass) are in this equation.
Molecular Collisions
• Speeds of molecules in gas phase can be very large.
– ~ 500 m/s for O2 at 20 °C
– Molecules do not travel far before colliding (at atm pressures)
• z = number of collisions per second that one molecule
experinces in a gas
– N = molecules; V = total volume
– s = diameter of spherical molecule (approx. size)
z  2
N 2
s u
V
u 
– <u> = mean molecular speed
1
N RT 
z  4  s2
 M 
V
2
8kT  8RT 

m   M 
Molecular Collisions
Number of collisions per second that one molecule experinces in a gas:
1
N RT 
z  4  s2
 M 
V
2
Z = total number of collisions per unit volume per unit time:
Number of molecules
Prob. Of collision per molecule
Z
N  z
V 2
2 molecules in each collision
Volume
1
2
N  2 RT 
Z2 
s
V 
 M 
2
Mean Free Path
The mean free path (l) is defined as the average distance a molecule
travels between two successive collisions with other molecules.
l
Brownian trajectory
1
Compare to Brownian motion:
µm
“Random walk”
Note that l ≠ d, in general.
The vertices observed in Brownian
trajectories may involve many collisions.
t=0
The apparent “steps” result from the way we
collect data.
d
t = 0.5 s
0
0
u
1

z
2  N V s 2
µm
1
Mean free paths will generally be much
shorter in condensed systems.
Molecular Collisions and Reaction Kinetics
• Chemical reactions that create a bond generally require
molecular collisions.
– Kinetic rate of these reactions should depend on the rate and
energy of molecular collisions.
– Collisions bring the reactants together
– Kinetic energy from molecular motion enables reaction to cross
over activation barrier.
• This is not just translational motion, but also includes molecular vibrations,
which are more important in large molecules like proteins.
Kinetics vs Equilibrium
• Equilibrium configuration depends only on DG.
– Free energy is a state function, path independent
– In general, no rate information
– Except for mixing, not reacting systems, where we found:


 effective force
x
and
1  
v    
 f x 
• v is the drift velocity and f is the drag coefficient
• We can solve this problem because we know the exact path
• And we know the energy at each point along this path
Kinetics vs Equilibrium
• For general chemical reactions we don’t know the exact
path or mechanism.
• Nor do we know the precise energy function.
Transition states
reaction mechanisms
If we know all the details, physics will tell
us the rate.
Energy
Details generally not known for chemical
reactions.
Kinetics is a quantitative, but largely
empirical, study of rates of reactions.
Reaction Coordinate
It is useful because kinetic behavior reveals
information about reaction mechanism.
(e.g. signatures of life beyond Earth)
Rate Law & Definitions
• Reaction velocity: v = dc/dt
• Rate Law: Substances that influence the rate of a reaction
grouped into two catagories
• Reactants (decrease)
• Products (increase)
• Intermediates (increase then decrease)
– Concentration does not change
•
•
•
•
Concentration
– Concentration changes with time
A --> C --> B
Time
Catalysts (inhibitor or promoter)
Intermediates in a steady-state process
Components buffered by equilibrium with large reservoir
Solvent
Order of a Reaction
• Kinetic order of a reaction describes the way rate depends
on concentration: A + B --> C
v  k   Am Bn  Pq
• {m,n,q} are usually integers but not always
• Order with respect to A is m etc.
• Overall order of reaction is sum: m + n + q
– Kinetic order depends on reaction mechanism, it is not determined
by stoichiometry.
– H2 + I2 -> 2HI
– H2 + Br2 -> 2HBr
Second order overall
Complex
Zero-Order Reactions
• Rate law:
dc
 k0
dt
k0 has units of concentration time
-1
• Seen in some enzymatic reactions such as that of liver
alcoholdehydrogenase:
ethanol
Concentration
CH3CH2OH + NAD+
LADH
acetaldehyde
CH3CHO + NADH + H+
v
ethanol
acetaldehyde
d CH3CH2 OH  dCH3 CHO

 k0
dt
dt
NAD+ is buffered
Enzyme is saturated
Time
Note: Reaction cannot be strictly zero
order at all times.
First Order Reactions
• Rate law:
dc
 k1c
dt
k1 has units of inverse time
– Typical of unimolecular reaction mechanism
– Rearrange and integrate
d A
  A  k1  dt
ln  A  k1t  ln At 0
 A 
ln  2   k1 t2  t1 
A1 
Ln[A]
d A dB

 k1 A
dt
dt
[A]0
[A]
v
A -> B
Time
 A  A0 e k1t
Time
First Order Reactions:
half-life and relaxation time
• Half-life: Time required for half of the initial
concentration to react
[A]0
 A 1 k1 t1 2
 e
 A0 2
t1 2 
ln 2
k1
[A]0/2
Half-life
• 1/2-life of 14C decay is 5770 years
• Relaxation time
t  1 k1
 A
 e t t
 A0
 A  A0 e k1t
Time
[A]0
 A  A0 e k1t
[A]0/e
t
Time
Second Order Reactions
• Class I rate law:
v  k2 c 2
units of k 2 are concentration
1
time -1
– Possible bimolecular mechanism A + A --> P
– Example: RNA hybridization
k2
2 A-A-G-C-U-U
A-A-G-C-U-U
v  k2 A2 GCU2 
2
U-U-C-G-A-A
– General form
v
d A
 k2 A2
dt
– Separate variables and integrate

d A
 A
2
 k2 dt
1
1

 k2t
A
A
   0
Letting x be the amount of A
that has reacted:
1
1
k t
 A0  x  A0 2

Second Order Reactions: Class I
1
1

 k2t
 A A0
 A 
1
1
k 2 t 1  A0
1
k t
 A0 2  A0 2 1 2

t1 2 
1
k2  A0
[A]0
 A 
[A]0/2
Half-life
1
k2 t 1  A0
Time
Second Order Reactions: Class II
• Class II reaction rate law for
A + B --> P
v  k2 A B
– First order with respect to either A or B
– Second order overall
• Some examples:
NO(g) + O3(g) --> NO2(g) + O2(g)
v = k2[NO][O3]
H2O2 + 2Fe2+ + 2H+(excess) --> 2H2O + 2Fe3+
v = k2[H2O2][Fe2+ ]
– Reactions of different stoichiometric ratios can exhibit class II
kinetics.
Second Order Reactions: Class II
General rate law:
v  k2 A B
Letting x be the concentration of each species that has reacted:
 A  a  x
A0  a
B  b  x
B0  b
dx
 k2 dt
a  x b  x 
1
ba  x 
ln
k t
a  b ab  x  2
for a  b
 B0
1
 A
ln
  k2 t
 ln
 A0   B0  A0
B
Slope = ([A]0 - [B]0)·k2
Ln([A]/[B])
Integrating:
Time
Last Homework
From TSWP:
Ch6: 1, 3,13 + 12 & 24 (which cover sedimentation)
Ch7: 1, 4, 10 (more problems may be added next week)
From Class:
A pore (100 nm2) opens in a cell membrane that separates
the cell interior (containing micromolar protein
concentrations) from the protein free exterior. How fast
do the proteins leak out? (Assume D ~ 10-6 cm2/s)
If an action potential starting in the cell body of the neuron proceeds by diffusion
of Na+ and K+ to the synapse, how long does it take the signal to travel 1m?
D25°C(Na+) = 1.5*10-5 cm2/sec
D25°C(K+) = 1.9*10-5 cm2/sec
Treat this as a 1D diffusion problem.
<x2> = 2Dt