1

Rate laws, Reaction OrdersRate laws, Reaction OrdersThe rate or velocity of a chemical reaction isloss of reactant or appearance of product inconcentration units, per unit time

!

d[P]

dt= "

d[S]

dt

The rate law for a reaction is of the form

!

Rate =d[P]

dt= k[A]

n a [B]nb [C]

n c ...

The order in each reactant is its exponent, niThe overall order of the reaction is the sum of the ni

Reaction OrderReaction OrderMolecularityMolecularity

• The order of a reaction is determined by thenumber of participants that must cometogether before the rate-limiting-step (RLS)of the reaction

• It is determined by the mechanism, notstoichiometry of a reaction

• It must be determined (measured)experimentally

Determining Reaction OrderDetermining Reaction OrderInitial Rate Method

• Choose initial concentrations of reactants• Measure “Initial Rate”, before concentrations of reactants

change appreciably (tangents)

Initial rate Course of reaction

Determining Reaction OrderDetermining Reaction OrderInitial Rate Method

• Choose initial concentrations of reactants• Measure “Initial Rate”, before concentrations of reactants

change appreciably• Change initial concentrations, and measure initial rate

again•Example: For A + B C, assume Rate = k[A]na[B]nb

[A]o [B]o Rate if then1 1 R1 R2= R1 na = 02 1 R2 R2= 2R1 na = 11 2 R3 R2= 4R1 na = 2

R3= R1 nb = 0, etc.•Once you know the ni, plug in and solve for k

Determining Reaction OrderDetermining Reaction OrderIntegrated Rate Equation Method

• Allow the rx to proceed and measure change inconcentration of reactant or product.

• By trial and error, determine which equation fits the data

Determining Reaction OrderDetermining Reaction OrderIntegrated Rate Equation, Zero Order

• Choose initial concentrations of reactants• Measure Rate over a substantial portion of the reaction• The product and reactant concentrations change

“automatically”• Fit to theory

!

[A]t" [A]

0= "kt

!

d[B]

dt= "

d[A]

dt= k[A]

0= k

•Example: zero order reaction A B•The rate law is

!

d[A][A ]

o

[A ]

" = # k

0

t

" dt

•Integrating from [A]0 at t = 0 to [A]t at t = t

•So, a plot of [A]t - [A]0 vs t is linear, with a slope of -k

2

Integrated Rate EquationFirst Order Reaction

A B• The rate law is

• Rewrite as

• Integrate from [A]0 at t = 0 to [A]t at t = t

• A plot of ln (A0/At) vs t is linear, with a slope of k• If so, the reaction is 1st Order in A and zero order in all

other components

!

d[B]

dt= "

d[A]

dt= k[A]

!

d[A]

[A][A ]

o

[A ]

" = # k

0

t

" dt

!

ln[A]t " ln[A]0 = ln(At

A0

) = "kt!

d[A]

[A]= "kdt

!

ln(A0

At

) = ktor

Integrated Rate EquationSecond Order Reaction

A B• The rate law is

• Rewrite as

• Integrate from [A]0 at t = 0 to [A]t at t = t

• So, a plot of (1/[A]t) vs t is linear, with a slope of k• If so, the reaction is 2nd Order in A and zero order in all

other components

!

d[B]

dt= "

d[A]

dt= k[A]

2

!

[A]"2d[A]

[A ]o

[A ]

# = " k

0

t

# dt

!

1

[A]"1

[A]0

= kt!

d[A]

[A]2

= "kdt

!

1

[A]=1

[A]0

+ ktor

Determining Reaction OrderDetermining Reaction OrderMultiple reactants, first order in each

A + B → C

• Use stoichiometry to relate all reactant concentrations totheir initial values and to [A]

• If [A]0 = [B]0 and the stoichiometry is 1, then theintegrated rate equation is the same as for 2nd order in A

• If [A]0 ≠ [B]0 and Stoichiometry is n (A + nB → C)Then substitute [B] = [B]0 -2([A]0 -[A]) before integrating

• Remember, stoichiometry doesn’t determine reactionorder, but it may be useful in guessing what the ordermight be

!

d[C]

dt= "

d[A]

dt= "

d[B]

dt= k[A][B]

Units of Rate ConstantsUnits of Rate Constants

If rate = , it must have units of

And the products on the right side of the equation musthave the same units, so the units of k must be different

!

d[A]

dt

!

(concentration)

(time)

Zero order reactions Rate (M•min-1) = k (M•min-1)

First order reactions Rate (M•min-1) = k (min-1) [A](M)

Second order reactions Rate (M•min-1) = k (M-1•min-1) [A]2(M2)

Third order reactions Rate (M•min-1) = k (M-2•min-1) [A]3(M3)

Sometimes initial rate and integratedSometimes initial rate and integratedrate measurements disagree - Why?rate measurements disagree - Why?

The integrated rate equation gives the concentrations ofcomponents whose concentrations change during the reaction

consider: A + B → C Rate = k2[A][B]The reaction is second order overall, first order in each reactantRate decreases as [A] decreases and as [B] decreases.What if [B] remains constant during a run?

because it is present in large excess (e.g., solvent), because its concentration is buffered (e.g., H+, OH-), orbecause it is a catalyst, and is regenerated.

Then the reaction is pseudo-zero order in B, and pesudo-first order overall:

Rate = kapparent[A], where kapparent = k2[B]

Model an Enzyme Reaction

The enzyme and substrate(s) get togetherTransformations occurThe product(s) is (are) released, regenerating enzyme

!

E + Sk1" # "

k$1% " " E • S

k2" # "

k$2% " " E + P

We have to worry about four concentrations[E], [S], [E•S], and [P]

and four reactions, some probably 1st order, some 2nd

k1[E][S], k2[E•S], k-1[E•S] and k-2[E][P]Too complicated!

3

Simplifying Assumptions1. Measure initial rates V0 (<10% of reaction)

• then [P] = 0• so we can neglect the reverse reaction, k-2[E][P]• and we can ignore product inhibition (later)

• and [S]0 = [S] + [E•S]2. Use catalytic amounts of enzyme, [E] << [S]

• Then [S]0 = [S] + 03. Enzyme exists in only two forms, E and E•S, which sum to ET4. [E•S] reaches a constant level rapidly and remains constant

during the measurement• Could happen in two ways• k-1 >> k2, rapid equilibrium• k-1 ≤ k2, steady state

5. V0 = k2[E•S]

k1

k-1

k2E + S E•S E + P

Simplifying Assumptionssteady state k1

k-1

k2E + S E•S E + P

!

d[E •S]

dt= 0 = k

1[E][S]" k

-1[E • S]" k

2[E • S]

!

0 = k1[E][S]" [E • S]{k-1 + k2}or

But we also know that [S] = [S]0 and ET = [E]+[E•S], so that [E] = ET - [E•S], andV0 = k2[E•S] (assumes the catalytic step is the RLS)

So, let’s substitute for free enzyme [E] and solve for [E•S]

!

0 = k1{ET - [E • S]}[S]0 " [E • S]{k-1 + k2}

Collect terms in [E•S] (next slide)

Steady Statek1

k-1

k2E + S E•S E + P

!

0 = k1[S]0ET - [E • S]{k1[S]0 + (k-1 + k2)}

!

V0 =k2[S]0ET

[S]0 +(k-1 + k2)

k1

!

V0=Vmax[S]

0

KM

+ [S]0

!

KM =(k-1 + k2)

k1

!

Vmax

= k2ET

!

[E • S] =k1[S]0ET

k1[S]0 + (k-1 + k2)solve for [E•S]

!

V0 = k2[E • S]= k2k1[S]0ET

k1[S]0 + (k-1 + k2)

multiply by k2

Simplifying Assumptionsrapid equilibrium k1

k-1

k2E + S E•S E + P

[E•S] is controlled by the equilibrium and is slowly bledoff toward product

!

KS

=[E][S]

0

[E • S]

!

[E • S] ={E

T- [E •S]}[S]

0

KS

!

[E • S] =[E][S]

0

KS

!

[E • S]{1+[S]

0

KS

} =ET[S]

0

KS

!

[E • S] =

ET[S]

0

KS

1+[S]

0

KS

=ET[S]

0

KS+[S]

0

!

V0= k

2[E • S]=

k2ET[S]

0

KS+[S]

0

=Vmax[S]

0

KS+[S]

0

The same equation

Comparisonsteady state vs rapid equilibrium

So, they really only differ in therelative magnitude of k-1 and k2

Steady State

units:

!

KM

=k"1 + k

2

k1

!

time"1

time-1conc

"1= concentration

!

KS

=k"1

k1

!

k1[E][S]

0= k"1[E • S]

!

k"1

k1

=[E][S]

0

[E • S]

Rapid Equilibrium

units:

Because at equilibrium

so

!

time"1

time-1conc

"1= concentration

k1

k-1

k2E + S E•S E + P

Michaelis-Menten Michaelis-Menten EquationEquation

0

500

1000

0 20 40 60 80 100 120 140 160

[S]o

Vo

Vmax (units of velocity)

KM (units of concentration)

4

If [S]0 is too high poor estimation of KM - looks zero order

If [S]0 is too lowpoor estimation of KM -no evidence of saturationpoor estimation of Vmax - looks first order

Need data with [S]0 in the vicinity of KM, Need data with [S]0 both above and below KM

What Data are Required?What Data are Required?

So… how can it be bothfirst order and zero order?0

500

1000

0 20 40 60 80 100 120 140 160

[S]o

Vo

Order of the ReactionOrder of the Reaction

0

500

1000

0 20 40 60 80 100 120 140 160

[S]o

Vo !

V0

=Vmax[S]

0

KM

+ [S]0

"Vmax[S]

0

[S]0

=Vmax

!

V0

=Vmax[S]

0

KM

+ [S]0

"Vmax[S]

0

KM

=Vmax

KM

[S]0

Mixed order zero order

first order

!

[S]0

>>KM

!

[S]0

<<KM X

X

More OrderMore Order

!

V0

=d[P]

dt= "

d[S]

dt=V

max

!

V0

= "d[S]

dt=Vmax

KM

[S]

Zero Order Region

!

[S]0

>>KM

First Order Region

!

[S]0

<<KM

!

" d[S]

[S ]0

[S ]t

# =Vmax

dt

0

t

#

!

[S]0" [S] =V

maxt

So, a plot of [S]0 -[S] or [P] vs t is linear, with slope = Vmax

!

"d[S]

[S][S ]0

[S ]

# =Vmax

KM

dt

0

t

#

!

ln[S]

0

[S]=Vmax

KM

t

So, the semilog plot vs t is linear, with slope = Vmax/KM

In Between -can’t simplify … invert before integrating:

!

Vmaxt = KMln[S]0

[S]+ {[S]" [S]0} The sum of the limiting cases

VVoo and [S] and [S]oo Relations Relations

!

V0

=Vmax[S]

0

KM

+ [S]0

"V0

Vmax

=[S]

0

KM

+ [S]0

!

V0

Vmax

= 0.2 =[S]

0

KM

+ [S]0

So, if V0 is 20% of Vmax,

and so, S0 = 0.25KM

!

[S]0

KM

=0.2

1" 0.2= 0.25

Or, if [S]0 = 5KM, how close is the rate to Vmax?

!

V0

Vmax

=5K

M

KM

+ 5KM

= 0.83

!

V0

Vmax

=10K

M

KM

+10KM

= 0.91How about [S]0 = 10KM?

Complications -Complications -Substrate InhibitionSubstrate Inhibition

Vo vs [S]o

0

250

500

750

1000

0 2 4 6 8 10 12

[S]o

Vo

It’s actually hard to get to VmaxAt [S]0 = 10KM, you observe only 91% of VmaxThe example actually had Vmax = 1000

And, sometimes substrate binds incorrectly and inhibitsNever reaches VmaxCan’t get KM

Lineweaver-BurkLineweaver-Burk

!

1

V0

=KM

+ [S]0

Vmax[S]

0

=KM

Vmax

1

[S]0

+1

Vmax

Invert both sides:

!

x - intercept =-1

KM

1[S]0

1V0

!

slope =KM

Vmax

!

y - intercept =1

Vmax

5

Lineweaver-BurkLineweaver-Burk“Double-reciprocal” plot1/V0 vs 1/[S]0 has linear region, even with substrate inhibition

Eadie-Hofstee Eadie-Hofstee PlotPlotalternative linear transformalternative linear transform

!

V0=Vmax[S]

0

KM+[S]

0

!

V0

[S]0

!

V0

!

Vmax

!

-1

KM

!

-Vmax

KM

!

V0KM+V

0[S]

0= V

max[S]

0

!

V0

[S]0

KM+V

0= V

max

!

V0

[S]0

+V0

KM

=Vmax

KM

!

V0

[S]0

=Vmax

KM

"1

KM

V0

Non-Michaelis-Menten Non-Michaelis-Menten BehaviorBehaviorcooperativitycooperativity

Negative Cooperativity Michaelis-Menten

Positive Cooperativity

Vo

[So]

Induced FitInduced FitEnzymes and Ideas are Flexible

Conformational ChangeParadigm change

AllosteryEnergy Coupling (e.g., F1F0)

Hexokinase HexokinaseBound Glc

CooperativityCooperativity

+ S

S S

S

+ SKd1 Kd2

S

Higher [S] always shifts the equilibrium to the right(from cubes to cylinders)

If Kd1 > Kd2, cooperativity is positive.If Kd1 < Kd2, cooperativity is negative.

How Cooperative How Cooperative isis it? it?

Koshland’s Cooperativity factor:

!

RS

=[S]0 givingV = 0.9Vmax

[S]0 givingV = 0.1Vmax

For normal MM kinetics

!

V

Vmax

= 0.9 =[S]

0

KM

+ [S]0

[S]0

= 9KM

!

V

Vmax

= 0.1=[S]

0

KM

+ [S]0

[S]0

= 1

9KM

!

RS

=9

1/9= 81

For positive cooperativity, Rs < 81For negative cooperativity, Rs > 81

6

Dependence of VDependence of Voo on E on ETT

Vmax = k2ET, and V0 ∝ Vmax, alwaysAt low [S]0, V0 = Vmax[S]0/KMAt high [S]0, V0 = VmaxSo, V0 ∝ Vmax, always, always, always

ExceptIf an irreversible inhibitor is presentIf a second substrate is limitingIf the assay system becomes limitingIf an Enzyme-cofactor or other equilibrium is involved

Dependence of VDependence of Voo on E on ETTexceptions - irreversible inhibitorexceptions - irreversible inhibitor

The irreversible inhibitor inactivates the enzymestoichiometricallyWhen you have added more enzyme than there isinhibitor, the reaction rate is proportional to ET

V0

ETEnzyme is inactivited until irreversible inhibitor is used up

normal behavior

Dependence of VDependence of Voo on E on ETTexceptions - second substrate becomes limitingexceptions - second substrate becomes limiting

Example: glucose oxidaseβ-D-glucose + O2 + H2O δ-gluconolactone + H2O2But O2 is not very soluble in water (2.4 x 10-4M @ 25°C)So, rate can’t continue to increase with ET forever.

V0

ET

[O2] starts to become limiting, here

normal behavior

Dependence of VDependence of Voo on E on ETTexceptions - coupled enzyme assay becomes limitingexceptions - coupled enzyme assay becomes limiting

A common way to assay an enzyme is to use a second enzyme thatuses the product of the first enzyme to make something that’s easyto detect, such as NADH, which absorbs at 340nm

V0

ET

Coupling enzyme can’t keep up

normal behavior

A B CE1 E2

NAD+ NADH + H+

More than one SubstrateMore than one Substrate

Except for isomerases, most enzymes have more than one substrateHow do we deal with this fact?A + E E•A + B… is awkward notationA better notation is due to W. W. Cleland (Madison,WI):Rewrite normal MM sequence E + S E•S E + P as

E E•S E•P E

S

P

Then, more complex reactions can be written simplyNote also that

!

V0

=Vmax[S]

KM

+ [S]=

Vmax

KM

[S]+1

ModelsModelstwo substrates on, two products releasedtwo substrates on, two products released

E E•A E•A•B E•P•Q EQ E

A

P

E + A + B P + Q + E

Substrates could add together, one after the other

B

Q

This is an ordered bi bi reactionKinetic equation can be generated as beforeWe’ll assume rapid equilibrium

7

Ordered bi biOrdered bi biassume rapid equilibriumassume rapid equilibrium

!

KA

=[A][E]

[EA]

!

KB

=[B][EA]

[EAB]

E + A EA+B

EAB E + P + QKB

KA

kcat

V0 = kcat[EAB]

ET = [E] + [EA] + [EAB]

1. Substitute [E] = ET - [EA] - [EAB] into eq for KA2. Solve for [EA]; substitute that into eq for KB3. Solve for [EAB]4. Multiply by kcat:

!

V0

=Vmax

1+K

B

[B]+K

AK

B

[A][B]

Random bi biRandom bi bieither substrate can add first, either producteither substrate can add first, either productcan leave firstcan leave first

A

P

B

Q

A

B

Q

P

E

E•A

E•B E•A•B E•P•Q E•P

E•Q

E

Binding of A and B may or may not be independent

Random bi biRandom bi bi

V0 = kcat[EAB]

ET = [E] + [EA] + [EB] + [EAB]

!

V0

=Vmax

1+"K

A

[A]+"K

B

[B]+"K

AKB

[A][B]

E + A EA+B

EAB E + P + QKB

KA

kcat

+B

EB + A

αKBαKA

!

KB

=[B][E]

[EB]!

"KA

=[A][EB]

[EAB]

!

KA

=[A][E]

[EA]

!

"KB

=[B][EA]

[EAB]Same steps - just more of themsolvingsubstitutingmultiply by kcat

Ping-PongPing-PongOne product leaves before second substrate addsOne product leaves before second substrate adds

E E•A E•P F F•B EQ E

A

P

B

Q

!

V0

=Vmax

1+K

B

[B]+K

A

[A]

Double reciprocal plotsDouble reciprocal plotsVary one substrate at different fixed levels of other substrate

[B]!

1

V0

=KA

Vmax

KB

[B]

"

# $

%

& ' 1

[A]+

1

Vmax

1+KB

[B]

"

# $

%

& '

Double reciprocal plotsDouble reciprocal plotsordered bi bi, continuedordered bi bi, continued

Note that plot of samedata vs second substratelooks different(rearrange to see why)

!

1

V0

=KA

Vmax

KB

[B]

"

# $

%

& ' 1

[A]+

1

Vmax

1+KB

[B]

"

# $

%

& '

[A]

8

Random biRandom bi

α = 1

[B]!

1

V0

="K

A

Vmax

1+KB

[B]

#

$ %

&

' ( 1

[A]+

1

Vmax

1+"K

B

[B]

#

$ %

&

' (

α = .5

[B]

Double reciprocal plotsDouble reciprocal plotsping pongping pong

!

1

V0

=KA

Vmax

1

[A]+

1

Vmax

1+KB

[B]

"

# $

%

& '

Constant slope

Decreasing intercept

[B]

Getting KGetting KIIslope or intercept slope or intercept replotsreplotsIf slope changes (competitive or noncompetitive)

If 1/V-intercept changes, (noncompetitive or uncompetitive)

For one inhibitor concentration, solve the approprite equationFor more than one [I], plot the parameter (slope or intercept) vs [I]

To get KI,divide the intercept of the replot by the slope of the replot

!

Slope =Slope[I]= 0(1+[I]

KI

)

!

Slope =Slope[I]= 0 +Slope[I]= 0

KI

[I]

!

Intercept = Intercept[I]= 0(1+[I]

KI

)

!

Intercept = Intercept[I]= 0 +Intercept[I]= 0

KI

[I]

Y =

Slo

pe o

r 1/V

-inte

rcep

t

[I]

!

Slope =Y[I]= 0

KI

!

Intercept = Y[I]= 0

!

KI =Replot Intercept

Replot Slope

Reversible InhibitionReversible Inhibitionmixed typemixed type

E + S ES E + P

EI + S ESI

+ IKI

X

αKI+ I

KM k2

αKM

Substrate & inhibitor binding are neither independent normutually exclusive. Binding of one alters K of the other bya factor of α.

!

Vo =V

max[S]

0

KM(1+

[I]

KI

) + [S]0(1+

[I]

"KI

)!

ET

= [E]+ [ES]+ [EI]+ [ESI]!

[EI]=[E][I]

KI

!

[ESI]=[ES][I]

KI

Two inhibitory forms of the enzymeTwo dissociation constants to account for them

Double Reciprocal PlotsDouble Reciprocal Plots

[I]

-1/KM

1/Vmax

Mixed TypeInhibition

1/V0

1/[S]0

α = 2Vmax = 100KM = 2

!

1

[S]0

= "1

#KM

!

1

V0

=1

Vmax

1"1

#

$

% &

'

( )

!

slope =KM

Vmax1+[I]

KI

"

# $

%

& '

!

intercept =1

Vmax1+

[I]

"KI

#

$ %

&

' (