Chemical Kinetics in Biology Important resources at:

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

[email protected] @SLC2A

Goals Chemical kinetics 1. 2. 3. 4. 5. 6. 7.

Understanding reaction order and rate constants Simple reactions Parallel reactions Sequential reactions Irreversible vs Reversible reactions - the old assumptions are no longer limiting The equilibrium assumption The steady-state assumption

Steady-state Enzyme kinetics 1.

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?

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.

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=

dc dt

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.

€

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

A Continuous-Flow Capillary Mixing Method to Monitor Reactions on the Microsecond Time Scale

M. C. Ramachandra Shastry,* Stanley D. Luck,* and Heinrich Roder*

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

Biophysical Journal Volume 74 May 1998 2714–2721

ATP Regulation of the Human Red Cell Sugar Transporter Anthony Carruthers The Journal of Biological Chemistry vol. 261, pp. 11028-11037,1986

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

Single-channel properties of the reconstituted voltage-regulated Na channel isolated from the electroplax of Electrophorus electricus ROBERT L. ROSENBERG, SALLY A. TOMIKO,

AND

WILLIAM S. AGNEW

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

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

JB Porter,, RD Abeysinghe, L Marshall, RC Hider and S Singh

Blood, Vol 88, No 2 (July 15), 1996: pp 705-713

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

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 238 U = 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 are very different to include processes over such an enormous range of time. However, the tools we use to analyze the acquired time-course data are fundamentally similar.

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= (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.

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.

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

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

v=

dP = kC AmC Bn dt

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

The reaction rate may be given by:

v=

dP = kC A3 C B2 dt

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.

Examples of reaction orders encountered in nature The next fews slides illustrate 4 types of reactions you may observe in the research setting.

• • • •

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.

Download and install GraphPad Prism Step 1 GraphPad Prism version 7 download link(s) and license information for 2016-2017 are:

Prism 7 Windows can be download from:

http://cdn.graphpad.com/downloads/prism/7/InstallPrism7.exe

Your Prism 7 Windows serial number is:

GPS-0346722-LEO5-795B0

Prism 7 Mac OS X can be download from:

http://cdn.graphpad.com/downloads/prism/7/InstallPrism7.dmg

Your Prism 6 Mac serial number is:

GPS-0346722-LEO5-795B0

1. Install the software then run the application for the first time.

2. Enter your serial number at the prompt

3. You will then be asked to register. Complete the form including your email address.

4. You will then receive an email including an activation code. It should look something like this:

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-YYYYYYYY-ZZZZZZZZ-AAAAAAAA-BBBBBBBB This code will activate serial number GPS-0346722-LEO5-795B0 to run on the computer identified by this machine ID: xxxxxxxxxxx. If you have any problems registering Prism, please contact GraphPad technical support at [email protected]

5. Copy the activation code into the Prism Registration Wizard and you should be all set.

Using GraphPad Prism

Support GraphPad 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/

A zero-order reaction. Zero-order reaction Substrate Product

8

zero-order kinetics 100 80

6 v (d[P]/dt)

[Substrate] or [Product]

10

4

60 40 20

2 0

0 0

0

100

200

[S] µM

2

4

6

8

10

TIME

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 data file at: http://inside.umassmed.edu/Global/Kinetics.pzf.zip

Theory of Zero-order Reactions

A zero-order reaction corresponds to the differential rate law

dC dt = k 0 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:

C = C0 + k0 t

or

y = y-intercept + slope * x

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 (yintercept) and k0 (the slope).

16

Zero-order reaction

[Substrate] or [Product]

10

Substrate Product

8 6 4 2 0 0

2

4

6

8

10

TIME

Best-fit values Slope Y-intercept when X=0.0 X-intercept when Y=0.0 Goodness of Fit R squared

Substrate Product Units -1 ± 0 1 ± 0 mols/sec 10 ± 0 0±0 mols 10.00 0 sec 1.000

1.000

18

General rules for zero-order reactions 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 time

3. Zero-order, enzyme catalyzed kinetics are typically observed at saturating [S] There are 2 useful tools for modeling reactions:

• Berkeley Madonna - http://www.berkeleymadonna.com/download.html

• Stella Architect - http://www.iseesystems.com/store/products/stellaarchitect.aspx

I have modeled some reactions using Stella Architect and posted the live models on line.

You can play with a live zero-order simulator at: https://sims.iseesystems.com/anthony-carruthers/

19

A first-order reaction. 1stOrder 5 4

100

[Substrate] [Product]

2

v (d[P]/dt)

[A] or [B]

80

3

60 40

first-order kinetics 20 0

1 0 0

0

100

200

[S] µM

2

4

6

8

10

TIME

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?

20

Theory of First-order Reactions

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

dC dt = k 1 C 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. The reaction:

A

k1

B

has the rate law:

v=-

d[A] d[B] = =k1[A] dt dt

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).

First Order reactions - loss of substrate Theory

21

Integrated rate law

[A] = [A] 0 e -k t

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

1

Defining [A]o as [A] at [A] at zero-time and

integrating between A at time 0 and time t gives

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

y

=

slope x + intercept

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

ln2 0.693 t 1/2 = k = k 1 1 and because τ = 1/k1,

t1/2 = 0.693 τ

First Order reactions - product formation Integrated rate law

[B] = [B] ∞ {1 - e -k t} 1

t1/2

time

Half-life Defining [B] at t1/2 as [B]∞/2

ln2 0.693 t 1/2 = k = k 1 1 and because τ = 1/k1,

t1/2 = 0.693 τ

22

23

Returning to our example of a first order reaction, 1stOrder 5

[A] or [B]

4 3

[Substrate] [Product]

2 1 0 0

2

4

6

8

10

TIME

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 1stOrder 10

[Substrate]

[A]

1

0.1

0.01 0

2

4

6

8

10

TIME

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

24

1st Order 5

[Substrate]

4 3 2 1.4 sec

1 0 0

1.4 sec 1.4 sec

1

2

3

4

5

6

7

8

9

10

TIME

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.

25

Non-linear regression analysis To 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] 0 e -k t 1

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

Every nonlinear regression method follows these steps:

26

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

27

One phase decay fit for [Substrate]

One phase associa2on fit for [Product]

Y=(Y0 - Plateau)*exp(-k*t) + Plateau Best-fit values Y0 Plateau k Half Life Tau = 1/k Goodness of Fit Degrees of Freedom R square

5.000 0

mM 0.5000

mM per sec

Y=Y0 + (Plateau-Y0)*(1-exp(-K*t))



1.386 sec 2.000 sec 1.000

147

Best-fit values Y0 Plateau k Half Life Tau = 1/k Goodness of Fit Degrees of Freedom R square

5.000

0

mM 0.5000

mM per sec



1.386 sec 2.000 sec 1.000

147

General rules for 1st order reactions

28

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 = -k

3. 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 = -k

5. 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 regression

8. 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

k1 k2

B

The equations are unchanged but now

k = k1 + k2

Let’s explore this at: https://sims.iseesystems.com/ anthony-carruthers/

Conclusion

1. Irreversible first-order reactions have explicit solutions

2. Reversible first-order reactions have explicit solutions but may also be solved numerically.

Second Order Reactions 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.

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.

[Proflavin]2

2 proflavin

]2

[

2

A–A–G–C–U–U 2 A–A–G–C–U–U U–U–C–G–A–A

[

2

2

2

]2

2nd Order class 1

5

[A] or [B]

4

[Substrate] [Product]

3 2 1 0 0

2

4

6

8

10

TIME

[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.

Nonlin fit of 2ndOrderIrrev:Residuals 0.4 0.3

[Substrate] [Product]

0.2 0.1 0.0 -0.1 -0.2

0

1

2

3

4

5 TIME

6

7

8

9

10

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

Theory of Class 1 Second-order Reactions Defining [A] at zero-time = [A]0, it can be shown that

1 1 − = k2t [ A] [ A]0

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

multiply both sides by [A]0

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

Thus one expects a linear relation between the reciprocal of the reactant concentration and time. Class 1, 2nd order Transform of data

8

Linear regression analysis Best-fit values Slope 0.66 ± 0 Y-intercept when X=0.0 X-intercept when Y=0.0 Goodness of Fit R square1.000

slope == [A] [A]00 kk22 slope

4

1st order data 2nd order data

2

1.0 ± 0 -1.515

2nd order data

0 0

2

4

6

8

10

TIME

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

half-time vs [A]

How starting [A] affects rate of 2nd order reaction 15

8

6

1 2 3 4 5 6 7 8 9 10

[A]0/[A]

10

4

Increasing [A]0

5 2

0 0

2

4

6

8

0 0

10

2

1.5

1.0

4

6 TIME

[A]

[A]0 k per sec

t1/2 sec

[A]0/[A]

6

A0k (slope) vs A0 second order Best-fit values Slope Y-intercept when X=0.0 X-intercept when Y=0.0 1/slope

[A]0 k per sec 0.1320 ± 2.842e-009 -3.974e-009 ± 1.763e-008 3.010e-008 7.576

0.5

0.0 0

2

4

6 [A]o

8

10

8

10

General rules for 2nd order reactions (Class 1)

1. Standard 1st order analysis does not work 2. Plotting [A]0/[A] vs time produces a straight line with slope [A]0 k 3. 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)

A+A

k1 k -1

B

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

Let’s explore at: https://sims.iseesystems.com/anthony-carruthers/

1.0 A

0.5

1/A

[A] or [B]

15

5

time

1.000 ± 1.028e-008 1.000 ± 5.950e-008 -1.000 1.000

5

A B

0.0 0

10

Best-fit values Slope Y-intercept when X=0.0 X-intercept when Y=0.0 1/slope

10

0 0

5

time

10

k1

A+A

k -1

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

B

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

One-phase association Best-fit values Y0 Plateau K Tau Half-time Span

A

B

0.9896 0.5007 1.740 0.5747 0.3984 -0.4888

0.005212 0.2496 1.740 0.5747 0.3984 0.2444

200 150

1/(At-0.5)

[A] or [B]

1.0

0.5 A B

100 50 0

0

5 time

0.0

0

5

10

time

[B] eq k K eq = k 1 = -1 [A] neq

Conclusion

1. Irreversible second-order reactions have explicit solutions

2. Reversible second-order reactions do not have useful explicit solutions and must be solved numerically.

10

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.

For example, in the reaction

k1

E+S

ES

k2

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). kf

R+L

kr

LR

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.

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].

Pseudo 1st Order

[L]

0.0010

10 5.995

0.0008

3.594

[LR] µM

2.154

0.0006

1.292 .774

0.0004

.464 .278

0.0002

.167 .1

0.0000

0

5

10

TIME The data were fitted with the one-phase association 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].

Explore at: https://sims.iseesystems.com/anthony-carruthers/

1. 2.

We obtain k or kobs from the one-phase association equation fits. If we then plot kobs versus [L], you can see that the plot is linear with 2 constants - slope and y-intercept. kobs vs L

25

kobs per sec

kobs per sec

20 15

Best-fit values Slope Y-intercept when X=0.0 X-intercept when Y=0.0 1/slope

1.999 ± 0.0001861 0.5012 ± 0.0007352 -0.2507 0.5002

10 5 0 0

2

4

6

8

10

[L] µM

We will show below that: 1.

The slope is kf

2. 3.

The y-intercept is kr The x-intercept is -kr/kf

Derivation of psedo-first order equation for second order reaction when one species (e) is fixed and low.

k2 !!! ⇀ e + S↽ !! ! eS k1

d [ eS ] = k2 [ e][ S ] − k1[ eS ] dt [ e] = [ et ] − [ eS ]

d [ eS ] = k2 {[ et ] − [ eS ]}[ S ] − k1[ eS ] dt 1 d [ eS ] = dt k2 {[ et ] − [ eS ]}[ S ] − k1[ eS ] 1

∫ k2 {[ e ] − [ eS ]}[ S ] − k1[ eS ] , d [ eS ] = t

−log(k1[ eS ] − [ et ] k2[S] + [ eS ] k2[S]) k1 + k2[S]

∫ dt,t = t −log(k1[ eS ] − [ e ] k2[S] + [ eS ] k2[S]) =t t

k1 + k2[S]

log(k1[ eS ] − [ et ] k2[S] + [ eS ] k2[S]) = −t(k1 + k2[S]) [eS](k1 + k2[S]) − [ et ] k2[S] = e−t (k1+ k 2[S ]) [eS] = [eS] =

[ e ] k2[S] + e

−t (k1+ k 2[S ])

t

(k1 + k2[S])

[ e ] k2[S] t

(k1 + k2[S])

(1 − +e−t (k1+ k 2[S ]) )

Theory for pseudo first-order reactions For our reaction

The rate of LR formation is given by:

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

1.

The time dependent component of this expression is e-t(kr+kf[L])= e-t.kobs

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/kf

5.

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.

Note 1. 2.

We obtain plateau from the one-phase association equation fits. We then plot plateau versus [L], you can see that the plot shows saturation kinetics with a Bmax and Kd

Where Kd = 0.25 µM = kr/kf indicating that the 2 analyses are internally consistent.

What is the difference between a first order reaction and a secondorder reaction that behaves like a first order reaction? •

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

•

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]).

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

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 1” 2nd order reaction is k[S]0 and when [S]0 is 0, kobs = 0

•

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

•

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

Parallel Reactions

Parallel reactions are of the type: k1

B

A k2

C

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

d[A] = k1 [A]+ k2 [A] = (k1 + k2 )[A] dt d[B] = k1 [A] dt d[C] = k2 [A] dt

−

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

€

[A] = −(k1 + k2 )t ln [A]0 [A] = [A]0 e−(k1 +k 2 )t

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

d[B] = k1 [A] = k1[A]0 e−( k1 +k2 )t dt d[C] = k2 [A] = k2 [A]0 e−( k1 +k2 )t dt Separating variables and integrating and assuming [B] = [C] = 0 o o

k1[A]0 (1− e−(k1 +k 2 )t ) k1 € + k2 k [A] [C] = 2 0 (1− e−(k1 +k 2 )t ) k1 + k2 [B] =

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

Time 0 1 2 3 4.62 5 7 9 11 13 15 20

12

0.1 0.05 10 A 10.00 8.61 7.41 6.38 5.00 4.72 3.50 2.59 1.92 1.42 1.05 0.50

10 B 0.00 0.93 1.73 2.42 3.33 3.52 4.33 4.94 5.39 5.72 5.96 6.33

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

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

C 0.00 0.46 0.86 1.21 1.67 1.76 2.17 2.47 2.69 2.86 2.98 3.17

[A], [B] or [C]

k1 k2 [A]o

A B C

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8

t1/2 = 4.62 s thus k=0.693/ 4.62 = 0.15 s-1

6 4 2 0 0

5

10

time

15

20

25

All reactions are limited by the fastest rate constant

Let’s explore at: https://sims.iseesystems.com/anthony-carruthers/

Now B peaks then declines. Changes in all 3 species governed by separate rate constants

Now C peaks then declines.

Conclusion

1. Irreversible parallel reactions have explicit solutions

2. Reversible parallel reactions do not have useful explicit solutions and must be solved numerically

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

k1

A These are just as difficult to solve for:

v1 = −

k2

B

C

d[ A] = k1[ A] dt

[ A] = [ A]0 e d[B] = k1[ A] − k2 [B] dt

− k1t

= k1[ A]0 e

− k1t

− k2 [B]

Integrating - assuming that [B]0 = 0

[B] =

v2 =

k1[ A]0 − k1t −k t {e − e 2 } k2 − k1

d[C] = k2 [B] dt

Integrating - assuming that [C]0 = 0

[C] = [ A]0 [1 −

=

k2 k1[ A]0 − k1t −k t {e − e 2 } k2 − k1

1 −k t −k t {k2 e 1 − k1e 2 }] k2 − k1

Ao k1 k2

10 5 0.05

t 0 0.1 0.25 0.5 1 2 3 4 5 7 9 13 15 20 25

A 10 6.07 2.87 0.82 0.07 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

B 0 3.92 7.08 9.02 9.54 9.14 8.69 8.27 7.87 7.12 6.44 5.27 4.77 3.72 2.89

C 0 0.01 0.05 0.16 0.39 0.86 1.31 1.73 2.13 2.88 3.56 4.73 5.23 6.28 7.11

A

A B C

[A], [B] or [C]

8

B

k2

C

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 firstorder conversion to C. If the appearance of C is our measure of velocity, when k1 >> k2 its appearance follows simple first order kinetics.

12 10

k1

6 4 2 0 0

5

10

15

20

25

-2 time min

k1=5 k-1=0 k2=0.05 k-2=0 Introduce reversibility k1=5 k-1=2 k2=0.05 k-2=0

k1=5 k-1=0 k2=0.05 k-2=0.05

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.

Lets explore at: https://sims.iseesystems.com/anthony-carruthers/ 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 2phase decay. The appearance of C follows simple first order kinetics.

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)

Ao k1 k2

10 0.05 0.2

t 0 1 2 3 4 5 6.93 9 13 15 20 25

A 10 9.51 9.05 8.61 8.19 7.79 7.07 6.38 5.22 4.72 3.68 2.87

B 0 0.44 0.78 1.04 1.23 1.37 1.52 1.57 1.49 1.41 1.17 0.93

C 0 0.05 0.17 0.35 0.58 0.84 1.40 2.05 3.29 3.87 5.16 6.20

A

[A], [B] or [C]

B

k2

C

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.

12

A B C

10

k1

8 6 4 2 0 0

5

10

15

20

25

time min

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

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

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

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

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.

Conclusion

1. Irreversible sequential reactions have explicit solutions

2. Reversible sequential reactions do not have useful explicit solutions and must be solved numerically

How can we use what we have learned to analyze more complex reactions?

• Simple, reversible reactions have explicit algebraic solutions that have specific predictive utility.

• Reversible, multi-step reactions do not have useful explicit solutions but can be solved numerically.

• By breaking multi-step reversible reactions into connected but discreet reversible reactions, we may simplify their analysis.

Equilibria and Kinetics

All reactions approach equilibrium. For every forward step there is a reverse step. 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

k1

A The rate of disappearance of A is

B

k-1

−d[ A] = k1[ A] − k−1[B] dt

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

k1 [B]eq = =K eq k−1 [ A]

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 at equilibrium, all nodes within a multi-node path must also be at equilibrium.

A

B

Thus the reaction

is not possible.

C

Rather, each step in the reaction must be reversible. Thus the mechanism is:

k1

A k3

k-1 k-3 k-2

B k2

C The relation to thermodynamics requires further that:

k1 k 2k 3 [B]eq K= = = [ A]eq k 1 k 2 k3 thus k1 k2 k3 = k 1 k 2 k

3

(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.

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

k1

A+B

k-1

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

X

k2

X

k-2

v2 =X k2

P+Q

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

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

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

thus

[X] =

k1 [A] [B] k-1

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

v=

d [Q] d [P] = = k2 [X] dt dt

substituting from above we see that:

v = k2 k1 [A] [B] k-1 Thus v can be expressed in terms of reactant concentrations only. The criterion for the prior equilibration approximation is v v