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.

• Gradient of the chemical potential is a force.• Think about gradient of electrical potential energy:

• Extending this to the total chemical potential:

– where f is a frictional coefficient– (1/f) as a mobility coefficient

Mobility and Chemical Potential

qψ = electrical potential energy (q = charge, ψ =electrical potential)

∇qψ =q∂ψ∂x

=qE where E is the electric field and qE is a FORCE

∇μ=∂μ∂x

=Force

u= 1f

⎛ ⎝ ⎜ ⎞

⎠ ⎟ ∂μ∂x

Drift velocity obtained from gradient of the chemical potential

Mobility and Chemical Potential: ExampleE

+

+ -

E =−∂φ∂x

Pote

ntia

l ()

Position (x)

Write down chemical potential as a function of position in this electrophoresis:

If concentration (c) is constant througout:

And the drift velocity is

μ=μ0 +kTln c( )+zeφ

∂μ∂x

=ze∂φ∂x

=qE

u=qEf

=− 1f

⎛ ⎝ ⎜ ⎞

⎠ ⎟ ∂μ∂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 MotionBrownian trajectory

0 1

1

0µm

µm t = 0

t = 0.5 s

• Each vertex represents measurement of position

• Time intervals between measurements constant

• After time (t) molecule moves distance (d)

• 2-dimensional diffusion:

<d2> = <x2> + <y2> = 4Dt

• 3-dimensional diffusion:

<d2> = <x2> + <y2> + <z2> = 6Dt

• In cell membrane, free lipid diffusion:

D ~ 1 µm2/s

“Random walk”

d

A lipid will diffuse around a 10 µm diameter cell:

π ⋅102

⎛ ⎝

⎞ ⎠

2=4Dt⇒ t =61s

Diffusion: Fick’s First Law

0 1

1

0µm

µm

• Jx = Flux in the x direction

• 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:

J x =−D ∂c∂x

⎛ ⎝

⎞ ⎠

Derivation of Fick’s First Law from Entropy of Mixing

μ1 x( ) =μ10 +kTln c1 x( )( ) • Chemical potential of component 1 in mixture.

• Net drift velocity (u) related to gradient of chemical potential by mobility (1/f) where f is frictional coefficient.

• Flux (J1x) is simply concentration times net drift.

• Einstein relation for the diffusion coefficient.

• Entropy is the driving force behind diffusion.

u1 x( ) = 1f

⎛ ⎝ ⎜ ⎞

⎠ ⎟ −∂μ

∂x⎛ ⎝

⎞ ⎠

=− 1f

⎛ ⎝ ⎜ ⎞

⎠ ⎟ kT 1

c1 x( )⎛ ⎝ ⎜ ⎞

⎠ ⎟ ∂c1∂x

J 1x x( ) =c1 x( )u1 x( ) =− 1f

⎛ ⎝ ⎜ ⎞

⎠ ⎟ kT∂c1

∂x

1f

⎛ ⎝ ⎜ ⎞

⎠ ⎟ kT=D

D=μkT where μ≡ mobility (not chemical potential)

Fick’s Second Law: The Diffusion Equation

J x x( ) J x x+dx( )

• 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∂t

⎛ ⎝

⎞ ⎠ = J x x( )−J x x+dx( )( )dy

∂c∂t

⎛ ⎝

⎞ ⎠ =

J x x( )−J x x+dx( )dx

since c=N/ dxdy( )

=− ∂J x∂x

⎛ ⎝ ⎜ ⎞

⎠ ⎟ definition of the derivative

∂c∂t

⎛ ⎝

⎞ ⎠ =D

∂ 2c∂x2

⎛ ⎝ ⎜ ⎞

⎠ ⎟ since J x =−D ∂c

∂x⎛ ⎝

⎞ ⎠

N

Diffusion Example

X

[C] Jx

X

[C] Jx=0

J x =−D ∂c∂x

⎛ ⎝

⎞ ⎠

∂c∂t

=D ∂ 2c∂x2

⎛ ⎝ ⎜ ⎞

⎠ ⎟

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/secD25°C(K+) = 1.9*10-5 cm2/sec

Treat this as a 1D diffusion problem. <x2> = 2Dt

Transport Anisotropy Forces Law MW DependenceProperty (Gradient) (Spherical Particles)

Diffusion Concentration Diffusional D M-1/3

T, Pi Frictional

Sedimentation Centrifugal Centrifugal s M-2/3

Velocity Acceleration Buoyant Frictional

Sedimentation Centrifugal Centrifugal Equilibrium Acceleration, Buoyant

Concentration Diffusional

Viscosity Velocity Shear [] MFrictional

Electrophoresis Electric Field Electrostatic Frictional

Rotary Diffusion Shape RotationalFrictional

dxdcDJ x −=

RTvMD

fNvMs

fvrm

dtdr

A

)1(

)1(

)1(

2

2

22

ρ

ρ

ρω

−=

−=

−=

drdc

rcvRTM app

22 )1( ωρ−

=

dyduAF x

sh =

fZeu ε

=

dΩ(θ, f ,t)dt

=Drot∇2Ω(θ, f,t)

Thermal Motion: Maxwell-Boltzmann Distribution of Velocities

Quantitative description of molecular motion

Average translational molecular kinetic energy:

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:

Both T and m (mass) are in this equation.

Utr =32kT

dP u( )=4π m2πkT

⎛ ⎝

⎞ ⎠

32u2e

−mu2 2kTdu

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– = diameter of spherical molecule (approx. size)

– <u> = mean molecular speed

z= 2π NVσ2 u

u = 8kTπm

⎛ ⎝

⎞ ⎠ =

8RTπM

⎛ ⎝

⎞ ⎠

z=4 π NVσ2 RT

M⎛ ⎝

⎞ ⎠

12

Molecular Collisions

z=4 π NVσ2 RT

M⎛ ⎝

⎞ ⎠

12

Number of collisions per second that one molecule experinces in a gas:

Z = total number of collisions per unit volume per unit time:

Z= NV

⎛ ⎝

⎞ ⎠ z2

Number of molecules

Volume

Prob. Of collision per molecule

2 molecules in each collision

Z=2 π NV

⎛ ⎝

⎞ ⎠

2σ2 RT

M⎛ ⎝

⎞ ⎠

12

Mean Free PathThe mean free path (l) is defined as the average distance a molecule travels between two successive collisions with other molecules.

l = uz

= 12π N V( )σ2

Brownian trajectory

0 1

1

0µm

µm t = 0

t = 0.5 s

“Random walk”

d

Compare to Brownian motion:

Note that l ≠ d, in general.

The vertices observed in Brownian trajectories may involve many collisions.

The apparent “steps” result from the way we collect data.

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 G.– Free energy is a state function, path independent– In general, no rate information– Except for mixing, not reacting systems, where we found:

• 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

−∂μ∂x

= effective force and v=− 1f

⎛ ⎝ ⎜ ⎞

⎠ ⎟ ∂μ∂x

⎛ ⎝

⎞ ⎠

Kinetics vs Equilibrium

• For general chemical reactions we don’t know the exact path or mechanism.

• Nor do we know the precise energy function.

Ener

gy

Reaction Coordinate

Transition statesreaction mechanisms

If we know all the details, physics will tell us the rate.

Details generally not known for chemical reactions.

Kinetics is a quantitative, but largely empirical, study of rates of reactions.

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– Concentration changes with time

• Reactants (decrease)• Products (increase)• Intermediates (increase then decrease)

– Concentration does not change• Catalysts (inhibitor or promoter)• Intermediates in a steady-state process• Components buffered by equilibrium with large reservoir• Solvent

Con

cent

ratio

n

Time

A --> C --> B

Order of a Reaction

• Kinetic order of a reaction describes the way rate depends on concentration: A + B --> C

• {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 Second order overall

– H2 + Br2 -> 2HBr Complex

v=k• A[ ]m B[ ]n P[ ]q

Zero-Order Reactions

• Rate law:

• Seen in some enzymatic reactions such as that of liver alcoholdehydrogenase:

CH3CH2OH + NAD+ CH3CHO + NADH + H+

dcdt

=k0 k0 has units of concentration time-1

LADHethanol acetaldehyde

Con

cent

ratio

n

Time

ethanol

acetaldehyde

v=−d CH3CH2OH[ ]dt

=dCH3CHO[ ]dt

=k0

NAD+ is bufferedEnzyme is saturated

Note: Reaction cannot be strictly zero order at all times.

First Order Reactions

• Rate law:

– Typical of unimolecular reaction mechanism A -> B

– Rearrange and integrate

dcdt

=k1c k1 has units of inverse time

v=−d A[ ]dt

=d B[ ]dt

=k1 A[ ]

d A[ ]A[ ]∫ =−k1 dt∫

ln A[ ]=−k1t+ln A[ ]t=0

ln A[ ]2A[ ]1

⎛ ⎝ ⎜ ⎞

⎠ ⎟ =−k1 t2 −t1( )

Ln[A

]

Time

[A]

Time

[A]0

A[ ]= A[ ]0e−k1t

First Order Reactions: half-life and relaxation time

• Half-life: Time required for half of the initial concentration to react

• 1/2-life of 14C decay is 5770 years

• Relaxation time

Time

[A]0

A[ ]= A[ ]0e−k1t

[A]0/2

Half-life

A[ ]A[ ]0

=12 =e−k1t1 2 t12 =ln2

k1

Time

[A]0

A[ ]= A[ ]0e−k1t

[A]0/e

τ=1k1A[ ]A[ ]0

=e−t τ

Second Order Reactions

• Class I rate law:– Possible bimolecular mechanism A + A --> P– Example: RNA hybridization

– General form

– Separate variables and integrate

v=k2c2 units of k2 are concentration−1 time-1

2 A-A-G-C-U-UA-A-G-C-U-U

U-U-C-G-A-A

k2 v=k2 A2GCU2[ ]2

v=−d A[ ]dt

=k2 A[ ]2

−d A[ ]A[ ]2

=k2dt

1A[ ]− 1

A[ ]0=k2t

1A[ ]0 −x− 1

A[ ]0=k2t

Letting x be the amount of A that has reacted:

Second Order Reactions: Class I

1A[ ]− 1

A[ ]0=k2t

A[ ]= 1k2t+1 A[ ]0 Time

[A]0

A[ ]= 1k2t+1 A[ ]0[A]0/2

Half-life

1A[ ]0 2− 1

A[ ]0=k2t12

t12 = 1k2 A[ ]0

Second Order Reactions: Class II

• Class II reaction rate law for A + B --> P

– 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.

v=k2 A[ ]⋅ B[ ]

Second Order Reactions: Class II

A[ ]=a−x A[ ]0 =aB[ ]=b−x B[ ]0 =b

v=k2 A[ ]⋅ B[ ]General rate law:

Letting x be the concentration of each species that has reacted:

Integrating:

dxa−x( ) b−x( ) =k2dt

1a−blnb a−x( )

a b−x( ) =k2t for a≠b

1A[ ]0 − B[ ]0

ln B[ ]0A[ ]0

+ln A[ ]B[ ]

⎛ ⎝ ⎜ ⎞

⎠ ⎟ =k2t Ln

([A

]/[B

])Time

Slope = ([A]0 - [B]0)·k2

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/secD25°C(K+) = 1.9*10-5 cm2/sec

Treat this as a 1D diffusion problem. <x2> = 2Dt