Kuo Kan Liang

RCAS, AS and BST, NTU

Rates of Reactions

Empirical StudiesReaction rate (progress with time) measurements:

The progress of the reactions are expressed as the change of the concentration / pressure of reactant(s) / product(s) / intermediate(s)

Measured by monitoring physical quantities proportional to the concentration or pressure. For example, absorption of radiation.

Photon-matter interactions Frequency of light: Wavelength: Angular frequency: Photon energy:

Spectroscopic Methods

⇒

D*

D’

D

Absorption

Light scatteringD + h�

⌫� = c/⌫

! = 2⇡⌫ = 2⇡c/�

E = h⌫ = ~! = hc/�

D⇤

D0 + h⌫0

D + h⌫

Planck’s constant: h = 6.626 × 10−34 J s ħ = h / 2π = 1.055 × 10−34 J s

Organic / biological molecules: UV-Vis (Ultra-violet and visible) IR (Infrared)

Fundamental principle of absorption spectroscopy: Beer-Lambert law

A

Absorption Spectroscopy

The light beam is characterized by the intensity (energy per unit area per unit time) at the frequency of interest:

Beam cross-section = A

Molecule concentration: [D]

Part of the incident energy is absorbed by the molecule ⇒ I(ω) / I0(ω) < 1

Light path length L

I (�)

L

I0(!)I(!)

Part of the incident energy is absorbed by the molecule ⇒ I(ω) / I0(ω) < 1

Light path length L

Absorbance (optical density):

Extinction coefficient:

Beer-Lambert Law

↵(!) = � log10I(!)

I0(!)

A

Beam cross-section = A

Molecule concentration: [D]

I (�)

L

I0(!)I(!)

"(!) =↵(!)

[D]L

Beer-Lambert LawConsider a thin layer of the sample (Thickness = dL) When one molecule absorbs one photon at frequency ω, the energy of the light beam reduces by ħω Number of incident photon per unit time: I(ω)A/ħω Number of photons absorbed per unit time: −A dI/ħω. ( dI < 0 ) The probability per unit time that a photon is absorbed when there is one mole of D in the volume: p(ω) ⇒

�AdI

~! = p(!)A [D] dLAI(!)

~! I I + dI

dL

Beer-Lambert LawLinear dependence of the absorbance on molecule concentration and light-path:

Exponential dependence of the intensity on absorbance, molecule concentration, and light-path:

I(!) = I0(!) 10�↵(!) = I0(!) 10

�"(!)[D]L

�Z L

L0=0d(log10 I) = � log10

I

I0

= ↵(!) =p(!)A [D]

ln 10

Z L

L0=0dL0

=

p(!)A

ln 10

[D]L = "(!) [D]L

� AdI

~� =

p(�)A [D]AI(�) dL

~� )

1

ln 10

dI

I= d(log10 I) = �p(�)A [D] dL

ln 10

Composition of 2-Component Rxn Mixture

Reaction: A ⇄ B

The absorbance of the reaction mixture is measured at two different frequencies:

α

α1

α2

(↵(!1) = "A1 [A]L+ "B1 [B]L

↵(!2) = "A2 [A]L+ "B2 [B]L

[A] =

����↵(!1) /L "B1

↵(!2) /L "B2

��������"A1 "B1

"A2 "B2

����, [B] =

����"A1 ↵(!1) /L"A2 ↵(!2) /L

��������"A1 "B1

"A2 "B2

����

Isosbestic PointIf the extinction coefficients of the two components A and B are the same at a frequency ωi, that is, εA(ωi) = εB(ωi) = ε(ωi), this frequency ωi is called the isosbestic point. Since for our model reaction we have [A] + [B] = [A]0 + [B]0 at any time during the reaction,

↵(!i) = "(!i) [A]L+ " (!i) [B]L

= " (!i) ([A] + [B])L

= constant

Isosbestic PointThe existence of isosbestic point is the evidence of having two species in the reaction mixture in chemical equilibrium without intermediate.

Definition of Reaction RateReaction rate: rate of change of the amount (partial pressure, concentration, ... not activity) of certain reactant or product species.

Rate is always taken as positive

Production rate(s) of product(s), consumption rate(s) of reactant(s) ...

Rate of finite process:

Rate of infinitesimal process: (Differential or instantaneous rate)

v = |� [J ] /�t|

v = |d [J ] /dt|

Stoichiometrically Normalized Reaction Rates

Consider the species based on which the reaction rate is defined, and define a rate which applies to all the species in a reaction Examples:

A → B : -d[A]/dt = +d[B]/dt (NH2)2CO(aq) + 2 H2O(l) → 2 NH4+(aq) + CO32–(aq) :d[NH4+] / dt = 2 d[CO32–] / dt = –2 d[(NH2)2CO] / dt

Stoichiometrically normalized rate: νA A + νB B → νC C + νD D

v ⌘ d [D] /dt

⌫D=

d [C] /dt

⌫C=

d [B] /dt

�⌫B=

d [A] /dt

�⌫A

Measuring Fast ReactionsReal-time measurements with specialized instrumentations

Flow methods (continuous flow, stopped flow, …)

Time-resolved spectroscopic methods

Relaxation methods(ways to apply the above methods)

Batch measurements

Quenching methods(When real-time methods do not have proper dynamic range)

Flow Method

Steady flow with velocity v

S

Distance ↔ time ( t = S / v )

Continuous-Flow Method

t = t0 + S / v

t: time since initiation

t0: mixing dead time

S

Continuous-Flow MethodDisadvantages:

Must be liquid-phase reactions

Require a large amount of reactants

To achieve higher time-resolution for faster reaction, even more reactants are required

Time resolution: ∆t = ∆S / v (∆S = spatial resolution) Fixed spatial resolution, then higher v ⇒ better time resolution

Stopped-Flow Method

Stopped-Flow MethodProcedure:

Mixing with the flow stopped

Start the flow

Stop the flow

Measurement with time-resolved spectroscopic methods (CD, FTIR, …)

Dead time: td = t0 + S/v

Time-resolution: depends on the spectroscopic method

Relaxation MethodManfred Eigen, Nobel Prize in Chemistry 1967

If a parameter such as temperature, pressure, or denaturant concentration is switched to a new value so rapidly that the chemical system cannot respond during the switching process, the subsequent relaxation towards equilibrium can be monitored.

Protein Folding KineticsProtein unfolding can be induced by denaturant, temperature change, or pH change, etc. Protein unfolding/refolding kinetics can be measured by relaxing the unfolding agent

Protein + high [D] + solvent

Protein + low [D] + solvent

Solvent

Sudden change

Obs

erve

d ki

netic

s

Refolding Kinetics

Sudden change

Obs

erve

d ki

netic

s

Quenching MethodChemical reaction is stopped some time after it is initiated

Advantage: composition of the reaction mixture can be studied with slow methods in an off-line manner

Chemical quench flow: species that can quench the reaction is added into the flow

Freeze quench: temperature is lowered to an extend that no more reaction will continue

Freeze quench example: cryo-em

Nigel Unwin (Cambridge): open structure of nicotinic acetylcholine receptor (nAchR)

Tube formed by nAchR →

FIGURE 5 Mature tubes obtained after incubation for 4-5 wk at 17°C. Ribbons of paired receptors are visible in a "defective" region along the length of the tube in b and in the small rounded end regions of the tubes in c and d. Note also the zones of densest staining on either side of the middle portions of the tubes (see Fig. 2). Bar, 0.1 /~m. x 123,000.

I 2 0 6 THE JOURNAL OF CELL BIOLOGY - VOLUME 99, 1984

on February 22, 2011jcb.rupress.org

Downloaded from

Published October 1, 1984

Free quench example: cryo-em

Nigel Unwin (Cambridge): open structure of nicotinic acetylcholine receptor (nAchR)

Flash Photolysis (Pump Probe) The absorption spectra of the reactant(s) is different from that of the product(s) in a (photo-induced) reaction

Example: Photo-isomerization of rhodopsin

Polli et al, Nature 467, 440-443 (2010)

probe time – pump time

Red

inte

nsity

⽂文字

Flash Photolysis (Pump Probe)

Sampleretror

eflector

Probe pulse

Pump pulse

The sample originally absorbs the pumping radiation but not the probing radiation. After it absorbs the probe pulse, however, it will absorb the probing radiation

probe time – pump time

Red

inte

nsity

⽂文字

Flash Photolysis (Pump Probe)

Probe pulse

Pump pulse

Even if the probe pulse arrives at the same time as the pump pulse, the system may not be able to absorb it, yet.

probe time – pump time

Red

inte

nsity

⽂文字

Flash Photolysis (Pump Probe)

Probe pulse

Pump pulse

If the probe pulse arrives shortly after the arrival of the pump pulse, the system may a b s o r b s i t q u i t e strongly, resulting in a very weak transmitted probe pulse.

probe time – pump time

Red

inte

nsity

⽂文字

Flash Photolysis (Pump Probe)

Probe pulse

Pump pulse

If the probe pulse arrives much later, the ‘pumpe d ’ mo lecu le s would have already relaxed (reacted) so that the probe is not absorbed that much anymore.

Flash Photolysis (Pump Probe)ÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊ

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Ê

Ê

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ÊÊ

1. 1.5 2. 2.5 3. 3.5 4. 4.5 5.

0.2

0.4

0.6

0.8

1.0

Relative

inte

nsity

of

tran

smitt

ed p

robe

pul

se

Delay time

The difference in the probe beam intensity

is the induced absorption

Flash Photolysis (Pump Probe) Example: Photo-isomerization of rhodopsin

Polli et al, Nature 467, 440-443 (2010)

Empirical Studies of Reaction Kinetics

Rate Laws / Rate ConstantsThe differential rate generally depends on the instantaneous configuration. Under given physical condition and configuration, the rate is usually a constant.

A + B → P, typically:(power law) εA, εB are not necessarily integers

Example of more complicated rate law:

Michaelis-Menten law: ← Not a power law

v / [A]"A [B]"B

v =vmax

[S]

Km + [S]

Rate Laws / Rate ConstantsIf the rate law is a power law, for example, the order of the reaction is εA + εB.

The constant k is the rate constant. There is no definite relation between the stoichiometry coefficients { νA, νB } and the exponents { εA, εB }.

The dimension / unit of the rate constant depends on the rate law

Example: first-order reaction A → B, dimension of k: time–1

v = k [A]"A [B]"B

Determination of Rate LawsRate law = fundamental physical model of the reaction.

Rate law = expressing the instantaneous rate as a function of time, physical conditions, and system composition.

Empirical studies: determine function form of the rate law directly from experimental data.

Isolation method

Initial rate method

Isolation MethodExample: A + B + C + … → P

Generally

Goal: Find k, εA, εB, and so on.

To find εA, prepare B, C, … in large excess ⇒

[B], [C], … can be approximated as constants [B]0, [C]0, …. ⇒(pseudo-εA-order)

v = k [A]"A [B]"B [C]"C . . .

v = k ([B]"B [C]"C . . .) [A]"A

Isolation MethodMeasure rate and [A] as functions of time simultaneously.

Plot the logarithm of rate as function of logarithm of [A]:

: effective rate constantln v = ln k0A + "A ln [A]

k0A

ln v

ln [A]slope =

εA

Initial Rate MethodIsolation method is often combined with the initial rate method.

Different terms in a composite rate law are determined one by one (divide-and-conquer method).

In the isolation method, the “isolation” approximation is worse and worse when more and more A is consumed.

Initial Rate MethodIf, for different initial concentration of A, [A]0, the progress of the reaction is monitored for a short time, even [A] can be treated as a constant. → Initial rate v0 Measure v0 for different [A]0, then, like in the isolation method, we have ln v0 = ln k’A + εA ln [A]0

ln v0

ln [A]0

slope = εA

Initial Rate MethodFor power law reaction rate law, linear regression (linear least-square) method can be used to analyze the experimental data.

Example: Binding of glucose (G) to the enzyme hexokinase (H). Assume that the rate law is Determine k, εH, and εG according to the experimental data.

v0 = k [H]"H0 [G]

"G0 ) log10 v0 = log10 (k [H]

"H0 ) + "Glog10 [G]0

Example

log10(v0)log10([G]0)

-3 -2.81 -2.51 -2.4-2.87 0.699 0.881 1.19 1.3-2.52 0.845 1.04 1.36 1.49-2 1.32 1.53 1.85 1.98log 1

0([H

] 0)

log10(v0)log10([G]0)

-3 -2.81 -2.51 -2.4-2.87 0.699 0.881 1.19 1.3-2.52 0.845 1.04 1.36 1.49-2 1.32 1.53 1.85 1.98

<x> = -2.68, <x2> = 7.24, σx2 = 0.0566

Examplelo

g 10([H

] 0)

Example

log10(v0)log10([G]0)

-3 -2.81 -2.51 -2.4 log10k’-2.87 0.699 0.881 1.19 1.3 3.72-2.52 0.845 1.04 1.36 1.49-2 1.32 1.53 1.85 1.98

<x> = -2.68, <x2> = 7.24, σx2 = 0.0566 <y>=1.018, <xy>=-2.67, Cov(x,y)=0.0570

εG = 1.066, log10k’ = 3.72

log 1

0([H

] 0)k0 ⌘ k [H]

"H0

Example

log10(v0)log10([G]0)

-3 -2.81 -2.51 -2.4 log10k’-2.87 0.699 0.881 1.19 1.3 3.72-2.52 0.845 1.04 1.36 1.49 4.06-2 1.32 1.53 1.85 1.98

<x> = -2.68, <x2> = 7.24, σx2 = 0.0566 <y>=1.184, <xy>=-3.11, Cov(x,y)=0.0608

εG = 1.072, log10k’ = 4.06

log 1

0([H

] 0)k0 ⌘ k [H]

"H0

Example

log10(v0)log10([G]0)

-3 -2.81 -2.51 -2.4 log10k’-2.87 0.699 0.881 1.19 1.3 3.72-2.52 0.845 1.04 1.36 1.49 4.06-2 1.32 1.53 1.85 1.98 4.6

<x> = -2.68, <x2> = 7.24, σx2 = 0.0566 <y>=1.670, <xy>=-4.41, Cov(x,y)=0.0619

εG = 1.0927, log10k’ = 4.60

log 1

0([H

] 0)k0 ⌘ k [H]

"H0

Example

log10(v0)log10([G]0) εG = 1

-3 -2.81 -2.51 -2.4 log10k’-2.87 0.699 0.881 1.19 1.3 3.72-2.52 0.845 1.04 1.36 1.49 4.06-2 1.32 1.53 1.85 1.98 4.6

<x>=-2.46, <x2>=6.20, σx2=0.128

εH = 1.0169, log10k = 6.63 <y>=4.124, <xy>=-10.023, Cov(xy)=0.1299

log 1

0([H

] 0)k0 ⌘ k [H]

"H0

v0 = k[G]0[H]0, k = 4.26 × 106 M-1s-1

Integrated Rate LawFrom the differential (instantaneous) rate law we find

Often, we may be more concerned about [A](t)

1st-order rate law:

v =1

⌫A

����d [A]

dt

���� = k [A]"A

v = � 1

⌫A

d [A]

dt= k [A] ) �k⌫Adt =

d [A]

[A]= d ln[A]

Z [A]

[A]0

d ln [A]0 = �k⌫A

Z t

0d⌧ ) ln

[A]

[A]0= �k⌫At

1st-Order Rate LawHalf-life t1/2 : time when [A] = [A]0 / 2. Let νA = 1 and

For a 1st-order reaction, half-life is independent of [A]0.

Lifetime (time constant): τ = 1 / k

When t = τ, [A] = [A]0 e-1.

Integrated rate law: [A] = [A]0 e-kt = [A]0 e-t/τ

ln[A]0 /2

[A]0= �k⌫At1/2 ) t1/2 =

ln 2

k

Meaning of LifetimeAt time t = 0, in one liter of the reaction mixture, there are [A]0 moles or NA × [A]0 molecules inside At time t, there are NA × [A] molecules in the reaction mixture, where [A] = [A]0 e-kt. During time t and t + dt, the number of A molecules which react and disappear is k × NA × [A] × dt. The time that these molecules remain as A is t. The average time that each molecule A remain as A, that is, their lifetime as A, is

⌧ =

R1t=0 kNA [A]0 e

�kt

t dt

NA [A]0=

1

k

Z 1

x=0x e

�x

dx =1

k

1st-Order Rate Law

0

1

0

1

0

1

0

1

t1/2

t1/2

1/2

1/4

1/81/16

Increa

sing k

Time

[A]/

[A] 0

1st-Order Rate LawIf ln [A] is plotted against t, the result should be a straight line for 1st-order reaction. If it is not a straight line, the reaction may not be a 1st-order reaction.t / min. C / μM ln C

30 699 6.55

60 622 6.43

120 413 6.02

150 292 5.68

240 152 5.02

360 60 4.09

480 24 3.180 100 200 300 400 500

3.0

3.5

4.0

4.5

5.0

5.5

6.0

6.5

Time / min.

ln (

C / μM

)

2nd-Order ReactionCase 1, A → P: d[A]/dt = -k[A]2

Half-life:

�d [A]

[A]2= d

✓1

[A]

◆= k dt

) 1

[A]=

1

[A]0+ kt

[A]0 = [A] (1 + kt [A]0) ) [A] =[A]0

1 + kt [A]02

[A]0=

1

[A]0+ kt1/2 ) kt1/2 =

1

[A]0) t1/2 =

1

k [A]0

2nd-Order ReactionCase 2, A + B → P: d[A]/dt = d[B]/dt = -k[A][B]

If [A]0 = [B]0, X = [A]0 - x and dX/dt = -kX2. The function form of the result is the same as in case 1.

A + B ! P[A]0 [B]0 ?�x �x ?

[A]0 � x [B]0 � x ?

9>>=

>>;) dx

dt

= k ([A]0 � x) ([B]0 � x)

2nd-Order Reaction

0

1

0

1

0

1

0

1

Time

[A]/

[A] 0

2nd-order

1st-orderInc

reasin

g k

2nd-Order ReactionCase 2, A + B → P: d[A]/dt = d[B]/dt = -k[A][B]

If [A]0 ≠ [B]0, [B]-[A]=[B]0-x-[A]0+x=[B]0-[A]0, so

1

[A]� 1

[B]=

[B]� [A]

[A] [B]=

[B]0 � [A]0[A] [B]

) d [A]

[A] [B]= �k dt

) �k dt =1

[B]0 � [A]0

✓d [A]

[A]� d [A]

[B]

◆

=1

[B]0 � [A]0

✓d [A]

[A]� d [B]

[B]

◆

) �kt =1

[B]0 � [A]0

✓ln

[A]

[A]0� ln

[B]

[B]0

◆