Chemical Kinetics in

Biology

http://glutxi.umassmed.edu/index.html

http://glutxi.umassmed.edu/grad.html

Important resources at:

1

Goals

Chemical kinetics

1.! Understanding reaction order and rate constants

2.! Analysis of reaction rates

Steady-state Enzyme kinetics

3.! How to model steady-state kinetics

a.! The King-Altman method

b.! The method of Cha

How do we analyze time course data and then what do we do with it?

2

Chemical Kinetics in Biology

What is “Chemical Kinetics”? The study of reaction rates.

Why do we study Chemical Kinetics? This method, in combination with “steady-state

kinetic analysis” reveals fundamental reaction pathways.

What is Relationship between Chemical Kinetics and thermodynamics? “Thermodynamics” tells us whether a reaction can

proceed spontaneously but does not inform us about the rate at which the reaction will proceed. This information has to be obtained experimentally.

3

Chemical Kinetics in Biology - studying rates of Biological reactions

The methods of reaction rate analysis were developed for studying relatively simple systems encountered by chemists. These approaches are also valuable in analyzing more complex biological processes because, oftentimes, one or a few steps control the rate of an extensive chain of reactions.

All steps involved in metabolism, replication, cell division, muscle contraction etc. are subject to the same basic principles as the elementary reactions of the chemist.

The rate or velocity, v, of a reaction or process describes how fast it occurs. Usually, the velocity is expressed as a change in concentration per unit time,

€

v = dcdt

but it may also express the change in a population of cells with time, the increase or decrease in the pressure of gas with time or the change in absorption of light by a colored solution with time.

4

Primary reactions of sensory rhodopsinsI. Lutz†, A. Sieg†, A. A. Wegener‡, M. Engelhard‡, I. Boche§, M. Otsuka§, D. Oesterhelt§, J. Wachtveitl†¶, and W. Zinth†i

962–967 PNAS January 30, 2001 vol. 98 no. 3

A Continuous-Flow Capillary Mixing Method to Monitor Reactions on theMicrosecond Time ScaleM. C. Ramachandra Shastry,* Stanley D. Luck,* and Heinrich Roder*

Biophysical Journal Volume 74 May 1998 2714–2721

ROBERT L. ROSENBERG, SALLY A. TOMIKO, AND WILLIAM S. AGNEW

Single-channel properties of the reconstituted voltage-regulated Na channel isolated from the electroplax of Electrophorus electricus

Proc. Natl. Acad. Sci. USA Vol. 81, pp. 5594-5598, 1984

5

Kinetics of removal and reappearance of non-transferrin-bound plasma iron with deferoxamine therapy

Blood, Vol 88, No 2 (July 15), 1996: pp 705-713JB Porter,, RD Abeysinghe, L Marshall, RC Hider and S Singh

The Journal of Biological Chemistryvol. 261, pp. 11028-11037,1986

ATP Regulation of the Human Red Cell Sugar TransporterAnthony Carruthers

Kinetics of Corneal Epithelium Turnover In VivoRichard J. Cenedella and Charles R. Fleschner

Investigative Ophthalmology & Visual Science, Vol. 31 p 1957, (1990)

6

Chemical kinetics is the study of the rates of reactions.

Some reactions (e.g. 2H2 + O2 ⇆ 2H2O) proceed so slowly as to be unmeasurable.

Radioisotopes of some nuclei have very long lifetimes (e.g. τ for 238U = 2.3 x 1017 s (4.47 billion years)).

Other reactions, such as the growth of bacterial cells, are slow (τ = 1 x 104 s) but measurable.

The biological reactions depicted on the last 2 slides range from picoseconds (10-12 s) to days (105 s).

Clearly, the methods of observation must be very different to include processes over such an enormous range of time.

7

Quick review of Rate Law

The rate of a process depends in some way on the concentrations or amounts involved. The reaction rate is a function of concentration:

v=ƒ (concentrations)

Substances that influence v can be grouped into 2 categories:

1) Those whose [] changes with time: e.g. A. reactants decrease with time B. products increase with time C. intermediates increase then decrease during a reaction e.g. consider C in the following reaction

A ⇌ C ⇌ B

2) Those whose [] do not change with time: e.g. A. catalysts (promoters/inhibitors) including enzymes and active surfaces B. Intermediates in a steady-state process including reactions under flow C. Components buffered by means of equilibrium with large reservoirs D. Solvents and the environment in general These concentrations do not change during a single run but may be changed from one experiment to the next. The concentrations of these components frequently do influence the rates of reactions.

8

Characteristics of a Reaction

Let’s consider 3 aspects of a reaction:the stoichiometry

the mechanism the order

The stoichiometry of the reaction tells us how many moles of each reactant are needed to form each mole of product: H2 + 0.5 O2 = H2O

or 2H2 + O2= 2H2O

(both are correct stoichiometric expressions).

The mechanism of the reaction tells us how the molecules react to form products. For the above reaction, the mechanism is thought to involve H, O and OH radicals:

H2 ⇄ 2H

H + O2 ⇄ OH + O

OH + H2 ⇄ H2O + H

O + H2 ⇄ OH + H

each reaction is an elementary reaction; the 4 reactions describe the proposed mechanism.

9

The order of the reaction describes how the velocity of the reaction depends upon the concentration of reactants.

Consider the following reaction:A + B ⇄ P

Because the velocity of the reaction may depend on the concentrations of several species, we must distinguish between order with respect to a particular component and the overall order which is the sum of exponents of all components.

For example, if the reaction is 3A + 2B ⇄ P

v =

dPdt

= kCA3CB

2The reaction rate may be given by:

For this reaction, the rate law is of the form:

where the concentrations CA, CB are raised to powers m and n that are usually

integers or zero (C0A = constant). The order of the reaction with respect to a

particular component (A or B) is just the exponent of the concentration term.

v =

dPdt

= kCAmCB

n

10

The next several slides illustrate 4 types of reactions you may observe in the research setting.

Examples of reaction orders encountered in nature

• Zero-order kinetics• First order kinetics• True Second order kinetics• Second order kinetics characterized by pseudo-first order behavior.

In order to analyze time course data, you need a good software tool.

11

GraphPad Prism version 6 download link(s) and license information for 2014-2015 are:Prism 6 Windows

http://cdn.graphpad.com/downloads/prism/6/InstallPrism6.exePrism 6 Windows serial number: GPW6-200512-LEM5-16772

 Prism 6 Mac OSX http://cdn.graphpad.com/downloads/prism/6/InstallPrism6.dmg

Prism 6 Mac serial number: GPM6-200513-LEM5-F3EF2

Step 1

Download and install GraphPad Prism

12

Thank you for registering GraphPad Prism. To activate Prism on your computer, copy the code below and paste it into the Prism registration wizard:

XXXXX-XXXXXXXX-XXXXXXXX-XXXXXXXX-XXXXXXXX (the email will contain the actual code)

This code will activate serial number (the email will insert the Windows or Mac serial # here) to run on the computer identified by this machine ID: YYYYYYYYYYY (the email will identify your computer’s true ID).

If you have any problems registering Prism, please contact GraphPad technical support at support@graphpad.com

Step 2When you launch Prism you will asked for your email address and will subsequently receive an email stating:

13

Using GraphPad Prism

SupportGraphPad Prism is very easy to learn and to use but extensive support is available through:• The built in help system• http://www.graphpad.com/scientific-software/prism/#learn• https://twitter.com/graphpad

http://www.graphpad.com/scientific-software/prism/

14

Zero-order reaction

0 2 4 6 8 100

2

4

6

8

10

TIME

[Sub

stra

te] o

r [P

rodu

ct]

SubstrateProduct

A zero-order reaction.

Note that [substrate] decreases linearly with time and [product] increases linearly with time. This observation suggests that we should perform “Linear Regression” analysis of the data to obtain constants (slopes) for substrate loss and product formation.

You can download this file at:

http://inside.umassmed.edu/Global/Kinetics.pzf.zip

zero-order kinetics

0 100 2000

20

40

60

80

100

[S] µM

v (

d[P

]/dt)

15

Theory of Zero-order ReactionsA zero-order reaction corresponds to the differential rate law

The units of k0 are molarity per sec. This is a “zero-order reaction because there is no concentration term in the right hand of the equation

Defining C0 as the concentration at zero time and C as the concentration at any other time, the integrated rate law is:

or y = y-intercept + slope * x

dtdC k0=

C C k t0 0= +

This is the equation for a linear relation between the independent (time) and dependent (concentration) variables.

We can therefore subject the raw data to linear regression analysis to obtain C0 (y-intercept) and k0 (the slope).

16

16

Zero-order reaction

0 2 4 6 8 100

2

4

6

8

10

TIME

[Sub

stra

te] o

r [P

rodu

ct]

SubstrateProduct

Best-­‐fit  values ! ! ! ! Substrate!! Product! ! Units          Slope! ! ! ! ! -­‐1  ±  0  ! ! 1  ±  0       mols/sec          Y-­‐intercept  when  X=0.0! ! 10  ±  0! ! 0 ± 0!! ! mols          X-­‐intercept  when  Y=0.0! ! 10.00!! ! 0! ! ! sec      Goodness  of  Fit! !          R  squared! ! ! ! 1.000!! ! 1.000

17

18

1. Plot of St or Pt vs time produces a straight line with slope = -k (for St) or k (for Pt)

2. k has units of mols produced or consumed per unit time3. Zero-order, enzyme catalyzed kinetics are typically observed at

saturating [S]

General rules for zero-order reactions

18

A first-order reaction.

Note here that [substrate] decreases in a curvilinear fashion with time and [product] increases in a curvilinear manner with time. This observation indicates that the reaction is NOT zero-order. How can we analyze this further?

1stOrder

0 2 4 6 8 100

1

2

3

4

5

TIME

[A] o

r [B] [Substrate]

[Product]

first-order kinetics0 100 200

0

20

40

60

80

100

[S] µM

v (

d[P

]/d

t)

19

19

Theory of First-order Reactions

A first-order reaction corresponds to the differential rate-law:

The units of k1 are time-1 (e.g. s-1). There are no concentration units in k1 so we do not need to know absolute concentrations - only relative concentrations are needed.

dtdC k C1=

The reaction:

A Bk1

where k1 is the rate constant for this reaction.

The velocity may be expressed in terms of either the rate of disappearance of reactant (-d[A]/dt) or the rate of appearance of product (d[P]/dt).

has the rate law:

v dtd[A]

dtd[B] k A][1=- = =

20

20

First Order reactions - loss of substrate

Theory Integrated rate law

Defining [A]o as [A] at [A] at zero-time and integrating between A at time 0 and time t gives

ln [A] k t ln[A]1 0=- +

y = slope x + intercept

[A] [A] e0 k t1= -

Defining [A] at t1/2 as [A]0/2

t kln2

k0.693

1/21 1

= =

and because τ = 1/k1, t1/2 = 0.693 τ

21

-d[A] = k1 [A]0 dt

21

First Order reactions - product formation

Half-life

[B] [B] e{ }1�k t1= - -

Integrated rate law

Defining [B] at t1/2 as [B]∞/2

t kln2

k0.693

1/21 1

= =

and because τ = 1/k1, t1/2 = 0.693 τ

22

22

23Returning to our example of a first order reaction,1stOrder

0 2 4 6 8 100

1

2

3

4

5

TIME

[A] o

r [B] [Substrate]

[Product]

1stOrder

0 2 4 6 8 100.01

0.1

1

10

TIME

[A]

[Substrate]

The data suggest that [substrate] falls from 5 mM to an equilibrium value of 0 mM.If we plot the log [substrate] vs time (or show the y-axis data on a log scale), we obtain

This produces a linear plot which is consistent with 1st order kinetics!

23

A second clue comes from the measurement of half-times. As [Substrate] declines from 5 - 2.5 mM, from 2.5 - 1.25 mM and from 1.25 to 0.625 mM, the time required for each 50% reduction is unchanged at ≈1.4 sec.

This is characteristic of “first-order decay” as observed with radioactive decay.

Constant decay times and the linear relationship between log {[S]t - [S]∞} vs time indicate a first order process. Let us check this by applying a first-order analysis to the data.

1st Order

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

TIME

[Sub

stra

te]

1.4 sec

1.4 sec

1.4 sec

24

24

25Non-linear regression analysisTo do this we subject the data to nonlinear regression (the plot is nonlinear) using an appropriate equation for first-order reactions.

The integrated rate law for first-order substrate loss is

[A] [A] e0 k t1= -

Nonlinear regression finds the values of those parameters of the equation (k1 and [A]0) that generate a curve that comes closest to the data. The result is the best possible estimate of the values of those parameters.To use nonlinear regression, therefore, you must choose a model or enter one.  GraphPad Prism offers a model for first-order reactions called “One-Phase Decay”

The equation is: Y=(Y0 - Plateau)*exp(-k*X) + Plateau

In which the parameters are defined as:1. Y0 is the Y value when X (time) is zero or [A]0 in this case.

2. Plateau is the Y value at infinite time (0 for our data set).

3. k is the rate constant k1 (per unit time).

4. Span is the difference between Y0 and Plateau

25

URL: http://www.graphpad.com/help/Prism5/Prism5Help.html?how_regression_works2.htm

1. Start with initial estimated values for each parameter in the equation.

2. Generate the curve defined by the initial values. Calculate the sum-of-squares - the sum of the squares of the vertical distances of the points from the curve.

3. Adjust the parameters to make the curve come closer to the data points - to reduce the sum-of-squares. There are several algorithms for adjusting the parameters - Prism uses the Marquardt algorithm.

4. Adjust the parameters again so that the curve comes even closer to the points. Repeat.

5. Stop the calculations when the adjustments make virtually no difference in the sum-of-squares.

6. Report the best-fit results. The precise values you obtain will depend in part on the initial values chosen in step 1 and the stopping criteria of step 5. This means that repeat analyses of the same data will not always give exactly the same results.

Every nonlinear regression method follows these steps:26

26

One  phase  decay!Perfect  fit! ! [Substrate]! [Product]! ! UnitsBest-­‐fit  values ! !          Y0!! ! ! ! ! 5.000!! ! 0       mM          Plateau! ! ! ! ! 0! ! ! 5.000       mM          k! ! ! ! ! ! 0.5000! ! 0.5000     per  sec          Half  Life!! ! ! ! 1.386  ! ! 1.386       sec          Tau  =  1/k! ! ! ! 2.000!! ! 2.000       sec          Goodness  of  Fit! !          Degrees  of  Freedom! ! 48! ! ! 48          R  square!! ! ! ! 1.000!! ! 1.000

Y=(Y0 - Plateau)*exp(-k*t) + Plateau

1stOrder

0 2 4 6 8 100

1

2

3

4

5

TIME

[Sub

stra

te] o

r [Pr

oduc

t]

[Substrate][Product]

27

27

1. First-order enzyme catalyzed kinetics are typically observed at subsaturating [S] 2. Plot of log (St-S∞) vs time produces a straight line with slope = -k3. The half-time (t1/2) and k are invariant of the starting value of St chosen.4. Plot of log (P∞-Pt) vs time produces a straight line with slope = -k5. t1/2 = 0.693/k

6. k has units of time-1 (e.g. s-1). There are no concentration units in k so we need not know absolute concentrations - only relative concentrations are needed.

7. k may be obtained by direct curve fitting procedures using nonlinear regression8. The full equation for loss of substrate is

[S]t = {[S]0 - [S]∞} e-(k.t) + [S]∞9. The full equation for product formation is

[P]t = [P]∞ (1 - e-(k.t))10. When a first order reaction is reversible (as most are), e.g.

A Bk

k

2

1

The equations are unchanged but now k = k1 + k2

General rules for 1st order reactions 28

28

0 5 100.0

0.5

1.0

time

[A] o

r [B]

AB

One-phase associationBest-fit values

Y0PlateauKTauHalf-timeSpan

A

1.0004.699e-0100.50002.0001.386-1.000

B

-5.029e-0091.0000.50002.0001.3861.000

A → B k1 = 0.5 s-1k1

29

0 5 100.0

0.5

1.0

Time sec

[A] o

r [B]

One-phase associationBest-fit values

Y0PlateauKTauHalf-timeSpan

Perfect fit

1.0000.50001.0001.0000.6931-0.5000

-1.618e-0090.50001.0001.0000.69310.5000

k1 = k-1 = 0.5 s-1k1

[ ][ ]

K kk

AB

eqeqn

eq

1

1= =-

k k kobs 1 1= + -

A B?k-1

AB

30

http://www.berkeleymadonna.com/jmadonna/jmadrelease.html#!

Berkeley Madonna is an extremely fast, general purpose differential equation solver. It runs on both Windows and Mac OS. Developed on the Berkeley campus under the sponsorship of NSF and NIH, it is currently used by academic and commercial institutions for constructing mathematical models for research and teaching

http://www.berkeleymadonna.com/features.html

Check out its features at:

31

Conclusion1. Irreversible first-order reactions have explicit

solutions2. Reversible first-order reactions have explicit

solutions but may also be solved numerically.

32

Second Order Reactions

Class 1 (A+A ⇌ P)v=k2[A]2

Although one or more reactants may be involved, the rate law for many reactions depends only on the second power of a single component. e.g.

Fall into 2 main categories depending on whether the rate law depends: 1) upon the second power of a single reactant species, or 2) the product of the concentrations of two different reagents.

2 proflavin [Proflavin]2

2[ ]2

2 A–A–G–C–U–UA–A–G–C–U–U

U–U–C–G–A–A

2[ 2 2]2

33

2nd Order class 1

0 2 4 6 8 100

1

2

3

4

5

TIME

[A] o

r [B]

[Substrate]

[Product]

[substrate] decreases in a curvilinear fashion with time and [product] increases in a curvilinear manner with time. This observation indicates that the reaction is NOT zero-order. How can we analyze this further?

The curves drawn through the points were made by nonlinear regression assuming first order kinetics (one-phase decay equation). Note the systematic deviations from the fit. This strongly suggests that this reaction does not follow first order kinetics.

We can investigate this further by plotting the residuals of the fit (how each point deviates from the calculated fit) vs time.

34

Nonlin fit of 2ndOrderIrrev:Residuals

0 1 2 3 4 5 6 7 8 9 10-0.2

-0.1

0.0

0.1

0.2

0.3

0.4

TIME

[Substrate][Product]

This confirms the poor fit and that we should consider either an error in data sampling or another model for the data.

35

Class 1, 2nd order Transform of data

0 2 4 6 8 100

2

4

6

8

TIME

[A] 0/

[A]

2nd order data

Linear  regression  analysisBest-­‐fit  values !          Slope! 0.66  ±  0          Y-­‐intercept  when  X=0.0! 1.0  ±  0          X-­‐intercept  when  Y=0.0! -­‐1.515        Goodness  of  Fit!          R  square!1.000

slope = [A]0 k2

Thus one expects a linear relation between the reciprocal of the reactant concentration and time.

Defining [A] at zero-time = [A]0, it can be shown that

1[A]

−1

[A]0

= k2t

1[A]

= k2t +1

[A]0

multiply both sidesby [A]0

[A]0

[A]= [A]0 k2t +1

Theory of Class 1 Second-order Reactions

[A]0/[A] versus time for normalized 1st and 2nd order kinetics with identical t½

Class 1, 2nd order Transform of data

0 2 4 6 8 100

2

4

6

8

TIME

[A] 0/

[A]

1st order data

2nd order data

slope = [A]0 k2

36

A0k (slope) vs A0 second order

0 2 4 6 8 100.0

0.5

1.0

1.5

[A]o

[A] 0

k pe

r sec

Best-fit valuesSlopeY-intercept when X=0.0X-intercept when Y=0.01/slope

[A]0 k per sec

0.1320 ± 2.842e-009-3.974e-009 ± 1.763e-0083.010e-0087.576

half-time vs [A]

0 2 4 6 8 100

2

4

6

8

[A]

t1/2

sec

How starting [A] affects rate of 2nd order reaction

0 2 4 6 8 100

5

10

15

TIME

[A] 0/

[A]

12345678910

Increasing[A]0

37

General rules for 2nd order reactions (Class 1)

1. Standard 1st order analysis does not work2. Plotting [A]0/[A] vs time produces a straight line with

slope [A]0 k3. Plotting slope vs [A]0 produces a straight line with

slope k and y-intercept 0.4. The half-time (t1/2) falls with increasing [A].5. The units of k are concentration-1.time-1.6. This analysis breaks down when the reaction is

reversible (i.e. when kr ≥ kf/10)

38

0 5 100

5

10

15

time1/

A

Best-fit valuesSlopeY-intercept when X=0.0X-intercept when Y=0.01/slope

A

1.000 ± 1.028e-0081.000 ± 5.950e-008-1.0001.000

A A Bk

k

1

1

+-

k1 = 1 M-1. s-1; k-1 = 0

0 5 100.0

0.5

1.0

time

[A] o

r [B

]

AB

39

0 5 100.0

0.5

1.0

time

[A] o

r [B]

AB

One-phase associationBest-fit values

Y0PlateauKTauHalf-timeSpan

A

0.98960.50071.7400.57470.3984-0.4888

B

0.0052120.24961.7400.57470.39840.2444

0 5 100

50

100

150

200

time

1/(At-0.5)

A A Bk

k

1

1

+-

k1 = k-1 = 0.5 M-1. s-1

[ ][ ]

K kk

AB

eqeqn

eq

1

1= =-

k1= 0.5 M-1.s-1; k-1 = 0.5 s-1

40

Conclusion1. Irreversible second-order reactions have

explicit solutions2. Reversible second-order reactions do not

have useful explicit solutions and must be solved numerically.

41

2nd order reactions Class 2 (pseudo-first order; A+B⇌P)

A reaction that is 2nd order overall may be first order with respect to each of the two reactants.

E+S ESk1

k2

For example, in the reaction

If the enzyme E were maintained at a constant low [] (e.g. [E] < [S]/100) and the substrate were varied, the reaction could be written as:

v= [E]k1[S]

Let us review this by examining ligand (L) binding to a receptor (R).

Upon rapid mixing of R and L, the receptor may undergo a fluorescence change allowing measurement of ligand binding. Alternatively, it may be possible to measure ligand binding by use of radiolabeled Ligand and filter-bound receptor. Either way, the time course of ligand binding may be examined to determine whether it displays first or second order kinetics.

R L LRkr

kf+

42

The data were fitted with the one-phase decay equation and the fit is excellent in each case (the residuals < [LR]/100)

You can also see that the reaction becomes faster at higher [L] - k increases and t1/2 falls with increasing [L].

Pseudo 1st Order

0 5 100.0000

0.0002

0.0004

0.0006

0.0008

0.0010

TIME

[LR

] µM

.167

.278

.1

.464

.774

1.292

2.154

3.594

5.995

10

[L]

At zero-time, various concentrations of L (µM) were mixed with 1 nM R. The time course of LR formation was monitored at each [L].

43

We will show below that:

1. The slope is kf2. The y-intercept is kr3. The x-intercept is -kr/kf

kobs vs L

0 2 4 6 8 100

5

10

15

20

25

k obs

per s

ec

[L] µM

Best-fit valuesSlopeY-intercept when X=0.0X-intercept when Y=0.01/slope

kobs per sec

1.999 ± 0.00018610.5012 ± 0.0007352-0.25070.5002

1. We obtain k or kobs from the one-phase decay equation fits. 2. If we then plot kobs versus [L], you can see that the plot

is linear with 2 constants - slope and y-intercept.

44

For our reaction

The rate of LR formation is given by:

1. The time dependent component of this expression is e-t(kr+kf[L]).2. Thus kobs = (kr+kf[L])3. In a plot of kobs versus [L], kobs increases linearly with [L] (slope = kf) and

the y-intercept = kr.4. The x-intercept (when kobs = 0) = -kr/kf5. Hence, analysis of the time course of L binding to R at varying [L] permits

computation of kf, kr and kf/kr = Keq for the reaction.6. This is ONLY true when [L] >> [R]. Under these conditions first-order

kinetics are observed ([L] does not change significantly). If [L] ≈ [R] the system will behave like a class 1 second order reaction.

Theory for pseudo first-order reactions

Defining [R]0 as the amount of receptor at t= 0, it can be shown that:

45

What is the difference between a first order reaction and a second-order reaction that behaves like a first order reaction?

• A second-order reaction that behaves like a first order reaction is called a pseudo-first-order reaction. Its rate constant, kobs, increases linearly with [S] (i.e. kobs = kr+kf[S]).

• A true first-order reaction is characterized by a rate-constant, k, that is independent of [substrate] or [product].

What is the difference between a class 1 second order reaction and a class 2 second-order (pseudo-first-order) reaction?

• kobs for a “class 1” 2nd order reaction is k[S]0 and when [S]0 is 0, kobs = 0

• A “class 1” 2nd order reaction is not described accurately by first order equations but when 1/[S] is plotted vs time, the plot is linear.

• kobs for a “class 2” 2nd order reaction is kf[S]0 + kr and when [S]0 is 0, kobs = kr.

• A “class 2” 2nd order reaction is described accurately by first order equations.

46

Parallel ReactionsParallel reactions are of the type:

A

B

C

k1

k2

€

−d[A]dt

= k1[A]+ k2[A] = (k1 + k2 )[A]

d[B]dt

= k1[A]

d[C]dt

= k2[A]

€

ln [A][A]0

=−(k1 + k2 )t

[A]= [A]0e−(k1+k2 ) t

Two separate routes of A breakdown exist. The rate expressions are:

The solution to the first eqn has the form of a first-order rate law

47

To find out how [B] and [C] change with time, we substitute for [A] from the last eqn.

€

d[B]dt

= k1[A] = k1[A]0e−(k1+k2 ) t

d[C]dt

= k2[A] = k2[A]0e−(k1+k2 ) t

Separating variables and integrating and assuming [B]o = [C]o = 0

€

[B]= k1[A]0k1 + k2

(1− e−(k1+k2 ) t )

[C]= k2[A]0k1 + k2

(1− e−(k1+k2 ) t )

Thus in parallel reactions, if one step is much faster than the others, it dominates the reactions.

t1/2�  =�  4.62�  sthus�  

k=0.693/4.62

=�  0.15�  s-­1

k1 0.1k2 0.05[A]o 10

Time A B C0 10.00 0.00 0.001 8.61 0.93 0.462 7.41 1.73 0.863 6.38 2.42 1.21

4.62 5.00 3.33 1.675 4.72 3.52 1.767 3.50 4.33 2.179 2.59 4.94 2.4711 1.92 5.39 2.6913 1.42 5.72 2.8615 1.05 5.96 2.9820 0.50 6.33 3.17

0

2

4

6

8

10

12

0 5 10 15 20 25

time

[A],

[B]

or

[C]

A

B

C

48

k1=0.1k-1=0

k2=0.05k-2=0

k1=0.1k-1=0.1k2=0.05k-2=0

k1=0.1k-1=0

k2=0.05k-2=0.05

Now C peaks then declines.

Now B peaks then declines. Changes

in all 3 species governed by separate rate

constants

All reactions are limited by the

fastest rate constant

Introduce reversibility

49

Conclusion1. Irreversible parallel reactions have explicit

solutions2. Reversible parallel reactions do not have

useful explicit solutions and must be solved numerically

50

Series Reactions (first order)Some reactions are of the type:

A B Ck1 k2

v1 = −

d[A]dt

= k1[A]

[A] = [A]0 e−k1t

d[B]dt

= k1[A]− k2[B] = k1[A]0 e−k1t − k2[B]

Integrating - assuming that [B]0 = 0

[B] =

k1[A]0

k2 − k1

{e−k1t − e−k2t}

v2 =

d[C]dt

= k2[B] =

k2k1[A]0

k2 − k1

{e−k1t − e−k2t}

[C] = [A]0[1−

1k2 − k1

{k2e−k1t − k1e

−k2t}]Integrating - assuming that [C]0 = 0

These are difficult to solve for:

51

When k1 >> k2, the second reaction is the rate-determining step. A will be rapidly converted to B and, during most of the reaction, B undergoes a first-order conversion to C. If the appearance of C is our measure of velocity, when k1 >> k2 its appearance follows simple first order kinetics.

-2

0

2

4

6

8

10

12

0 5 10 15 20 25

time min

[A],

[B

] or

[C

]

ABC

Ao 10

k1 5

k2 0.05

t A B C

0 10 0 0

0.1 6.07 3.92 0.01

0.25 2.87 7.08 0.05

0.5 0.82 9.02 0.16

1 0.07 9.54 0.39

2 0.00 9.14 0.86

3 0.00 8.69 1.31

4 0.00 8.27 1.73

5 0.00 7.87 2.13

7 0.00 7.12 2.88

9 0.00 6.44 3.56

13 0.00 5.27 4.73

15 0.00 4.77 5.23

20 0.00 3.72 6.28

25 0.00 2.89 7.11

A B Ck1 k2

52

k1=5k-1=0

k2=0.05k-2=0

When k1 >> k2, the second reaction is the rate-determining step. A is rapidly converted to B and B undergoes a first-order conversion to C. If the appearance of C is our measure of velocity, when k1 >> k2 its appearance follows simple first order kinetics.

k1=5k-1=2

k2=0.05k-2=0

Introduce reversibility

When the first reaction is made reversible, the second reaction remains the rate-determining step but now the peak of intermediate B is lower and A declines as a 2-phase decay. The appearance of C follows simple first order kinetics.

k1=5k-1=0

k2=0.05k-2=0.05

When the second reaction is made reversible, the second reaction remains the rate-determining step, the peak of intermediate B is unchanged. The appearance of C follows simple first order kinetics, is faster (kobs =k2+k-2) and B and C achieve equilibrium (Keq=k2/k-2)

53

When k2 >> k1, the reaction begins with the very slow conversion of A to B followed by a very rapid conversion of B to C. In this case, [B] remains low throughout and C appears as A disappears. The time course of C formation indicates a lag phase.

Ao 10k1 0.05k2 0.2

t A B C0 10 0 01 9.51 0.44 0.052 9.05 0.78 0.173 8.61 1.04 0.354 8.19 1.23 0.585 7.79 1.37 0.846.93 7.07 1.52 1.409 6.38 1.57 2.0513 5.22 1.49 3.2915 4.72 1.41 3.8720 3.68 1.17 5.1625 2.87 0.93 6.20

0

2

4

6

8

10

12

0 5 10 15 20 25

time�  min

[A],�  [B]�  or�  [C]

ABC

A B Ck1 k2

54

k1=0.05k-1=0k2=0.2k-2=0

When k2 >> k1, the reaction begins with the very slow conversion of A to B followed by a very rapid conversion of B to C. In this case, [B] remains low throughout and C appears as A disappears. The time course of C formation indicates a lag phase.

k1=.05k-1=0k2=0.2k-2=0.2

When the second reaction is made reversible, the first reaction remains the rate-determining step, B does not peak over this time course, A declines as in the original condition but C increases more slowly.

k1=.05k-1=.05k2=0.2k-2=0

When the first reaction is made reversible, it remains the rate-determining step and the differences between this condition and irreversibility in the 1st step are subtle.

Introduce reversibility

55

Conclusion1. Irreversible sequential reactions have explicit

solutions2. Reversible sequential reactions do not have

useful explicit solutions and must be solved numerically

56

Equilibrium and KineticsAll reactions approach equilibrium. For every forward step there is a reverse reaction. In practice we sometimes ignore the reverse step because the concentrations of products are kept very small. However, you have seen how the reverse reaction influences time courses. Furthermore, it is important to know the relationship between kinetic rate constants (k) and the thermodynamic equilibrium constant (K).

For the elementary first order reaction

A Bk1

k-1

Often, there is more than one path for the reaction of A to form B. To be consistent with the principles of equilibrium thermodynamics, we MUST apply the principles of microscopic reversibility. This states that if A can react to form B by 2 or more paths, we cannot have a mechanism by which A→B only by one path and B→A by another.

Thus the reaction is not possible.A

C

B

The rate of disappearance of A is

−d[A]dt

= k1[A]− k−1[B]

At equilibrium -d[A]/dt=0, therefore

[B]eq

[A]eq =k1

k−1

= K

57

Each step in the reaction must be reversible. Thus the mechanism is:

The relation to thermodynamics requires further that:

(note the product of rate constants in one direction = the product of all rate constants in the opposite direction). Thus the 6 rate constants are not independent.

A

C

Bk1

k-1

k-2k2k3

k-3

K =[B]eq

[A]eq =k1

k−1

=k2k3

k−2k−3

thusk1k−2k−3 = k−1k2k3

58

Complex reactionsEnzyme mediated reactions involve a series of reversible steps. For example:

The exact solution to the rate equations in this case is very complex. Because the elementary reactions are bimolecular and 5 molecular species are involved, it is useful to learn some approximations that may be applied.

A + B

X

Xk1

k-1

P + Qk2

k-2

Prior- or Rapid-equilibrium approximation

Here, we assume steps k1 and k-1 are rapid relative to k2. A, B and X rapidly attain a state of quasi-equilibrium, such that v1 = v-1

k1[A][B] = k-1[X]

Thus we obtain the equilibrium expression

v1 =A B k1; v-1 =X k-1

v2 =X k2

K = k-1k1 = [A] [B]

[X]

59

Thus v can be expressed in terms of reactant concentrations only. The criterion for the prior equilibration approximation is

v ≈ v2<<v1 ≈ v-1

which may be read as:

“The overall velocity of the reaction is limited by the slow step 2 and the velocity is much slower than the forward and reverse reactions of step 1 which are essentially in equilibrium”

Step 2 is the rate-limiting step and the rate of product formation is given by

substituting from above we see that:

[ ][ ][ ]K k

kA BX

1

1= =-

[ ] [ ][ ]X kk A B

1

1=-

thus

[ ] [ ] [ ]v dtd P

dtd Q k X2= = =

[ ][ ]v kk k A B

1

2 1=-

60

Steady-state approximationSometimes an intermediate is formed that is highly reactive. Thus it never truly builds up to an appreciable level.

To a first approximation, X reacts as rapidly as it is formed.v1 = v2

k1 [A][B] = k2 [X][D]

hence

The significance of this is not that [X] is constant throughout the reaction. Such is never the case. However, it is true that the slope d[X]/dt of the curve of [X] versus time is much smaller than that of the other reactants or products.

A + B

X + D

Xk1

Pk2

(slow)

(fast)

€

v = d[P]dt

= k1[A][B]

Another approach is to consider all steps involving the formation and disappearance of the reactive intermediate and to set the sum of their rates to zero. e.g.

€

d[X]dt

= k1[A][B]− k2[X][D] ≅ 0

61

Steady state approximation in chemical kineticsThis approximation consists of assuming that the concentration of a reactive intermediate in a reaction mechanism is constant, i. e. its derivative is zero.Its use helps us to solve the differential equations that arise from rate equations, which lack an analytical solution for most mechanisms beyond the most simple ones. The steady state approximation is applied, for example in Michaelis-Menten kinetics.

There are two parts to this reaction:

1) Formation of ES

2) ES breakdown to product P and free enzyme E

k2E+S ES E+P

k1 k3

62

S

P

E

ES

k1 10k2 1INIT E 0.01INIT ES 0INIT S 1k3 1k4 2INIT P 0

E S ES E Pk

k

k

k

2

1

4

3+ +

63

Defining the rate of product formation, v as

! ! ! ! v = k3 [ES]! (1

If k3[ES] << k1 [E][S] ≈ k2[ES] (i.e. we make the rapid equilibrium assumption)

+d[ES]/dt = k1 [E][S] (2

Rate of breakdown of [ES]

-d[ES]/dt = k2[ES] (3

64

In the "steady state" the concentrations of intermediates (e.g. ES) are unchanged, whereas [S] + [P] can change. If we limit measurements of v to early stages, [ES] does not change (there is no reverse reaction)! ! !

d[ES]/dt = 0 (4

! ! i.e.k1 [E] [S] = k2[ES] (5

hence

(6[ES] = [E][S]k1k2

(7[ES] = [E][S]k2

k1

65

The following steps are algebraic tricks

[Et] = [E] + [ES] (8

where Et is total enzyme

if we divide the velocity equation (1 by Et we obtain

v[Et ]

=k3[ES][E] + [ES] ! (9

66

We can rearrange this to

v[Et ]k3

=[ES]

[E] + [ES](10

then substitute for [ES] from equation (6 to give

v[Et ]k3

=

k1k2[E][S]

[E]+ k1k2[E][S]

=

k1k2[S]

1+ k1k2[S]

(11=

[S]k2k1

+ [S]

67

Define ! Vm as:! ! [Et]k3 = Vm! ! (12

vVm

=

[S]Km

1+ [S]Km

(14

Km =k2 + k3k1

(13

v =Vm[S]Km + [S] (15or

Km =k2k1

= KS

68

Steady-state assumption• The rate constants describing a reversible reaction are not independent.• The law of Microscopic Reversibility states that when the reaction is

drawn as a “King Altman” diagram, the product of rate constants for the forward reaction is identical to the product of rate constants for the reverse reaction.

• When d[S]/dt or d[P]/dt are constant, the concentration of intermediates in a reaction are said to be in a steady-state or an internal equilibrium.

• When the concentration of intermediates in a reaction are in a steady-state, this means that the rate of their formation and breakdown are identical.

• Knowing this and the inter-relationship between forward and reverse rate constants permits definition of the intermediates in terms of [reactant], [product] and rate constants.

• The next class will build on this.

69