Rate Law: equation describing the relationship between the reaction rate and concentration of a reactant or reactants. Rate = k[A]m[B]n where k is called the rate ...
Chemical Kinetics: Rates and Mechanisms of Chemical Reactions C.P. Huang University of Delaware 1
Content 1. 2. 3. 4. 5. 6.
Introduction Reaction rate Rate laws Analysis of rate equations Rate theories Reaction mechanisms 1. 2. 3.
Complex Catalysis Chain
7. Reactions in solution 8. Reactions at interface 2
1. INTRODUCTION
3
Objectives
• Chemical Kinetics: study of rates of chemical reactions and mechanisms by which they occur • A reaction may be spontaneous but does not occur at measurable rates
4
Why kinetics • How fast a reaction can take place? • What steps or pathways are involved in any chemical reaction? • How complete is the chemical reaction? • What are the two major factors controlling the outcome of chemical reactions? • Chemical thermodynamics • Chemical kinetics
5
2. REACTION RATE
6
Definition of reaction rate • Rate – change in some variable per unit time Rate
1/time
• Reaction rate – change in concentration per unit time; M/s or mol/(L-s) • Rates are determined by monitoring concentration as a function of time • Rates are negative for reactants and positive quantities for products
7
Types of reaction rate • Instantaneous rate – rate at a specific time • Average rate – ∆[A] over a specific time interval • Initial rate – instantaneous rate at t = 0 • Note: Rates and rate laws are not based on stoichiometry!! They must be determined experimentally.
8
Factors affecting reaction rate • Kinetics are very difficult to describe from first principles – Structure, elements, behavior • Rate of reaction describes how fast reactants are used up and products are formed • There are 4 basic factors that affect reaction rates – Nature of reactants – Effective concentrations – Temperature – Presence of catalysts – Number of steps
9
Nature of reactants: particle size • The degree of intimacy among particles obviously depends on the physical nature of the particles. • Particles in the liquid state are closer than in the solid state. • Likewise, particles in a finely divided solid will be closer than in a chunk of the solid • In both situations, there is a larger surface area available for the reaction to take place • This leads to an increase in rate. The smaller the particles the faster the reaction rates
10
Nature of reactants: bonding • Ions react rapidly: Ag+ + Cl- AgCl(s) Very fast • Reactions which involve bond breaking are slower: NH4+ + OCN- OC(NH2)2 • Redox reactions in solutions are slow • Transfer of electrons are faster than those of atoms. • Reactions between covalently bonded molecules are slow: 2 HI(g) H2(g) + I2(g)
11
Concentration • For every reaction the particles must come into intimate contact with each other. • High concentrations by definition implies that particles are closer together (than dilute solutions). • So rate increases with concentration. • Surface area – larger surface area increases reaction • Mixing increases interaction • Need to minimized precipitation or colloid formation
12
Temperature • Temperature affects rate by affecting the number and energy of collisions • So an increase in temperature will have the effect of increasing reaction rate
13
Reaction rates and stoichiometry • Rate has units of moles per liter per unit time - M s-1, M h-1 • Consider the hypothetical reaction aA + bB cC + dD 1 [ A] 1 [ B ] • We can write r
a t b t 1 [C ] 1 [ D] c t d t
14
Example At a given time, the rate of C2H4 reaction is 0.23 M/s. What are the rates of the other reaction components? C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(g) 0.23 M/s
?
?
?
15
3. RATE LAW
16
Rate law • Consider the following reaction aA + bB products • Rate Law: equation describing the relationship between the reaction rate and concentration of a reactant or reactants Rate = k[A]m[B]n where k is called the rate constant
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Concentration and rate aA + bB → cC + dD • General form of rate law: rate = k[A]m[B]n [A], [B] – concentration, in M or P k – rate constant; units vary m, n – reaction orders • Reaction orders and, thus, rate laws must be determined EXPERIMENTALLY!!! – Note: m ≠ a and n ≠ b – Overall order = sum of individual orders 18
Reaction order • Rate = k[A][B]0 m = 1 and n = 0 - reaction is first order in A and zero order in B - overall order = 1 + 0 = 1 - usually written: Rate = k[A] • The values of the reaction order must be determined experimentally; they cannot be found by looking at the equation, i.e., the stoichiometry of the reaction
19
Rate law • m, n are called reaction orders - they indicate the sensitivity of the rate to concentration changes of each reactant • NOTE: the orders have nothing to do with the stoichiometric coefficients in the balanced overall equation • An exponent of 0 means the reaction is zero order in that reactant - rate does not depend on the concentration of that reactant
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Rate Law • An exponent of 1, rate is directly proportional to the concentration of that reactant - if concentration is doubled, rate doubles - reaction is first order in that reactant • An exponent of 2, rate is proportional to the square of concentration of that reactant – if concentration is double, rate is quadrupled – reaction is second order in that reactant
• The overall reaction order is the sum of all the orders
21
Reaction orders For the reaction: A →B, the rate law is: rate = k[A]m Order (m)
∆[A] by a factor of
Rate increases by
Zero (0)
2, 4, 15, ½, etc.
None
2 3 2 3 ½
2X 3X 4X 9X ¼X
1st
(1)
2nd (2)
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Example 2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g) rate = k[NO]2[H2] 1. 2. 3. 4. 5. 6.
What is the order with respect to NO? What is the order with respect to H2? 3 What is the overall order?
2 1
If [NO] is doubled, what is the effect on the reaction rate? quadrupled If [H2] is halved, what is the effect on the reaction rate? halved What are the units of k? M-2-s-2 23
Example PtCl2(NH3)2 + H2O PtCl(H2O)(NH3)2 + Cl-
Calculate the rate of reaction when the concentration of PtCl2(NH3)2 is 2.0x10-2 M. Rate = 9x10-3 (h-1) x 0.02 (M) = 1.8x10-5 (M-h-1)
What is the rate of Cl- production under these conditions? d[Cl]/dt = rate = 1.8x10-5 (M-h-1) 24
Example 2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g) rate = k[NO]2[H2] k = 6.0 x 104 M-2s-1 @1000K
Calculate rate when [NO] = 0.025 M and [H2] = 0.015 M. Rate = 6x10-4 (M-2-s-1)x0.025 M x 0.015 M =225 (M-s-1)
25
Measurement of reaction rate Rate may be measured in three ways • Average reaction rate: a measure of the change in concentration with time • Instantaneous rate: rate of change of concentration at any particular instant during the reaction • Initial rate: instantaneous rate at t = 0; that is, when the reactants are first mixed
26
Measurements of reaction rate C C
C Initial rates nc
Instantaneous rate
nc
nt
C1 C2 C3 t
t
t1
t2
t3
t
Here nc is true order, with respect to concentration and nt is order with respect to time. When nt < nc, reaction is inhibitory and when nt >nc reaction is autocatalytic. 27
C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq) Time, s
[C4H9Cl], M
0
0.1000
50
0.0905
100
0.0820
150
0.0741
200
0.0671
300
0.0549
400
0.0448
500
0.0368
800
0.0200
10,000
0
In this reaction, the concentration of butyl chloride, C4H9Cl, is measured at various times.
28
Time, s
[C4H9Cl], M
Average rate, M/s
0
0.1000
50
0.0905
1.9E-4
100
0.0820
1.7E-4
150
0.0741
1.6E-4
200
0.0671
1.4E-4
300
0.0549
1.22E-4
400
0.0448
1.01E-4
500
0.0368
0.8E-4
800
0.0200
0.56E-4
10,000
0
29
C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq)
Initial rate
Instantaneous rate at 500 s
30
Determination of rate law The method of initial rate • Measuring the initial rates as a function of the initial concentrations • Avoids problems of reversible reactions • Initially there are no products so they cannot affect the measured rate • This method is chosen to check the effect of a single reactant on the rate
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Bench-Top Reactors
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Atlas Reactors
http://www.syrris.com/batch-products/atlas-parallel-system
33
Stopped-Flow Technique
http://www.hi-techsci.com/techniques/stoppedflow 34
Temperature Jump
http://www.hi-techsci.com/techniques/stoppedflow 35
Quench-Flow Technique
http://www.hi-techsci.com/techniques/stoppedflow
36
Summary of Experimental Methods Method
Range of half-life, s
Conventional
102 – 108
Flow
10-3 - 102
Relaxation
10-10 -1
pressure jump
10-6 - 1
temperature jump
10-7 - 1
field pulse
10-10 – 10-3
shock tubes
10-9 – 10-3
Kinetic spectroscopy
10-15 – 10-10
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4. ANALYSIS OF RATE LAW
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Types of rate laws • Differential rate law or rate law: Shows how the reaction rate changes with concentration • Integrated rate law: Shows how concentration changes with time
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First order Log [A]
R
1 d[ A] k (T ,I )[ A ] ν A dt 1.0
[ A ] [ A ] o e k ( T ,I )t k ( T ,I ) log[ A ] log[ A ] o t 2.303
0.1
0.01
0.001
t 1
2
3 40
Second order
R
1 d[ A ] k ( T )[ A ] 2 A dt
1 1 ν i k ( T )t [A] [ A ]0
[ A ]o A [B]o ln B [A]
[B]o ln A [ A ]o
B [B]o A [ A ]o k(T )t
41
Third order 1 d[ A ] R k(T )[ A ][B ][C ] A dt
A B C 1 [ A ]o [B]o [C]o
1 1 2k(T )t 2 2 [A] [ A ]o
42
Summary Rate Equations Rate = k[A]n; [A] = concentration at time t, [Ao] = initial concentration, [X] = product conc. 0 order
[A0]-[A] = kt, [X] = kt
1st order ln[A ] - ln[A] = kt, 0 2nd order
3rd order
k = M/s ln[A0] - ln([Ao] - [X]) = kt
k = 1/s
k = M-1-s-1
k = M-2-s-1
43
Half-life • The half-life, t1/2, is defined as the time it takes for the reactant concentration to drop to half its initial value
• Note: the half-life for a first order reaction does not depend on the initial concentration • The value of the half-life is constant
44
Half-Life • Half-life – A =Aoe-t – = ln2/t1/2 – If a rate half life is known, fraction reacted or remaining can be calculated
(CH3)2N2(g) N2(g) + C2H6(g) Time Pressure (torr) 0 36.2 30 46.5 t 0
A 36.2 30 36.2(1-x)
B 0 36.2x
C 0 36.2x
46.5 = 36.2(1-x+2x)=36.2(1+x)=36.2+36.2x
Ct =Coe-kt
46.5-36.2 = 36.2x x = 0.285
kt0.5=ln(2) to.5 = - [ln(2) t]/[ln(Ct/Co)]
A = 36.5(1-0.285) = 25.9
to.5 = - [ln(2) (30)]/[ln(25.9/36.5)]=62.1 (min) 45
t0.5 2t0.5 t0.53 t0.5
t0.5
t0.5
46
5. RATE THEORY
47
ARRHENIUS’ EQUATION • In 1885, Hood proposed the following equation: A' logk B T • In 1884, vant’s Hoff-Arrhenius proposed the following equation: d ln K c E dT RT 2
Kc is the equilibrium constant 48
ARRHENIUS’ EQUATION k1 k 1 A B CD
k1[ A][ B] k 1[C ][ D] Kc
k1 [C ][ D] k 1 [ A][ B]
Ea d ln k dT RT 2 Ea ln k C RT
d ln k1 E 12 1 dt RT d ln k 1 E 12 1 E E1 E 1 dt RT d ln k1 d ln k 1 E dt dt RT 2
C = constant
Activated state; X*
E2 E1 Final state E=Ea
Initial state
49
Arrhenius Equation • Svante Arrhenius developed an equation for the mathematical relationship between k and Ea.
• A is the frequency factor, which represents the number of effective collisions. E • e RT : Boltzmann expression for the fraction of system having energy in excess of the value, Ea, so that is may be identified with the fraction of reactant molecules that are activated complexes a
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Arrhenius Equation Or log k = log A –Ea/(2.303 R) (1/T) Slope = Ea/2.303 R = Ea/4.57 Cal
Y=mx+b
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To find slope, m Ln k
m=
k2 ln k 1 1 1 T T1 2
Ea/3R
1/T
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Arrhenius’ Equation • This is Arrhenius’ Equation • Can be arranged in the form of a straight line • ln k = (-Ea/R)(1/T) + ln A • Plot ln k vs. 1/T slope = -Ea/R Log [k]
EA/2.303 = dlog K/d(1/T)
1.0
Ea < 5 kcal/mol: diffusion control reaction Ea > 5 kcal/mol; reactive control
0.1
Diffusion regime
Reaction regime
0.01
0.001
1/T 1
2
3
53
Activation Energy, Ea • Energy barrier (hump) that must be overcome for a chemical reaction to proceed • Activated complex or transition state – arrangement of atoms at the top of the barrier
54
Temperature Effects Maxwell-Boltzmann Distribution
• At higher temperatures, more molecules will have adequate energy to react. • This increases the reaction rate. 55
Example A reaction which rate constant doubles when temperature is raised by 10oK from 300 oK. What is its Ea value? Given k310 = 2k300 or k1 = 2k0 Given k310 = 2k300 or k1 = 2k0 Ea
k k T1To 310)(300 (8.34) ln o 53594(J / mol) ln 0 2k o T o T1 k 1 (300 310)
56
Summary of A and Ea Reaction
A
Ea(kJ/mol)
N2ON2+O N2O5 2NO+O2 N2+ON+NO OH+H22H2O+H
8x1011 6x1014 1x1011 1x1011
251 88 315 42
1st 1st 2nd 2nd
CO2+OH-HCO3- 1x1011
315
2nd
57
Example It is given for a specific reaction, k =0.00123 /ms at 290oK and 0.0394/ms at 350oK. What is its Ea and k at 308oK? ln k 1 ln A
Ea RT1
(350)(298) 0.0394 Ea 57811( J / mol) 57.8(kJ / mol) (8.34 ) 350 298 0 . 00123
E 1 k 1 ln 2 a k1 R T2 T1 TT k E a 1 2 R ln 2 T2 T1 k1
ln k ln(0.00123 )
57811 1 1 8.70 0.578 5.94 8.324 308 298
k e 5.94 0.00263 2.63x10 3 (1/ ms)
58
Collision Theory • COLLISION THEORY: a reaction results when reactant molecules, which are properly oriented and have the appropriate energy, collide • The necessary energy is the activation energy, Ea
M. Silberberg
59
Molecular Orientation and Effective Collisions • Not all collisions leads to a reaction • For effective collisions proper orientation of the molecules must be possible
60
Molecular Collisions
Before collision
Effective collision
Before collision No-effective collision 61
Collision Theory two identical gas molecules colliding with each other at a velocity, v (molecules/cm3-s) [N.C. Leuis (1918) and Eyring (1935)]
v z AA e z AA
E RT
zAA = # collision/(s-cm3)
1 2 d 2 cn 2 2 c
molecules/(cm3-s)
8κT = average velocity of each molecule πm
m: mass of each molecules k = Boltzmann constant
1 8 T T 2d2n2 2n2 d2 2 m m 1 E m A mB 2 RT 2 v n A nB d AB 8T e dAB: average distance, sum of radii m A mB z AA
62
Collision Theory 1 2
v n A nB d
k = k’Na (cm3/mol-s)
2 AB
E m A mB RT e 8T m A mB
1 2
E m m v B RT k' d 2AB 8T A e m A mB n A nB 1 2
E m m B RT k Na d 2AB 8T A e m A mB 1 2
m A mB 2 A Na d AB 8T Z m A mB 1
k BT e 2
E RT
B
k PZe
E RT
Z T
63
Example Estimate he rate constant for the decomposition of HI(g) molecules at 321.4oC according to the collision theory. Given for the HI(g) molecules the following prosperities: 2HI(g) = H2(g) +I2(g) = 3.5 Å = 3.5x10-8 cm; E = 44,000 cal/g-mol; mA = mB = MHI = 128 g; R = Na = 1.38x10-16x6.023x1023 = 8.3x107 (erg/oK-mol) = 1.98 (cal/oK-g-mol) T = 273 + 321.4 = 594.6 oK
64
Example
1 2
2 44000 (8.34 x107 )(594.6 ) 7 k' (3.5 x10 ) 88.3x10 594.6 e 128 1.70x10 10 e 37.4 8 2
9.72x10 27 (cm 3 / molecules s)
k =9.72x10-27 (cm3/moleculesx)(6.023x1023)(molecules/mol) = 5.86x10-3 (cm3/mol-s) =5.86x10-3x(1000) =5.85x10-6 (L/mol-s)
65
Example Collision of two water molecules at room temperature, 298 oK, given: d = 0.30 nm, T=298oK; mA=mB=18(g-mol-1)/6.023x1023 =2.99x10-26 kg; =1.38x10-23 J/oK
Z 0.3x10 9 2
81.38x10
23
298
26 2.99x10 / 6x10 2.37x10 10 (cm 3 / molecules s) 26 2
1 2
2.37x10 16 (m 3 / molecules s)
66
Example
Reaction
k’ (cm3-molecules-1-s-1)
k (L-mol-1-s-1)
CH4+OHCH3 + H2O
8.14x10-15
4.9x1012
CH3+O2 CH2O+OH
3.76x10-36
2.26x10-9
CH3O+O CH2O+OH
1.7x10-11
1.02x1016
k (L-mol-1-s-1) = 1000 x6.023x1023 k’ (cm3-molecules-1-s-1)
67
Transition State Theory • During a chemical reaction, reactants do not suddenly convert to products • The formation of products is a continuous process of bonding breaking and forming • At some point, a transitional species is formed containing “partial” bonds • This species is called the transition state or activated complex
68
Transition State Theory • The transition state is the configuration of atoms at the maximum of the reaction energy diagram • The activation energy is therefore the energy needed to reach the transition state • Note also that the transition state can go on to form products or break apart to reform the reactants
rA
uA R
b uBC
rB
69
Transition State Theory A + B = AB* C r = [AB*]; (s-1); [AB*](molecules/cm3) a *AB *AB [ AB*] K a A aB A B [ A][ B] *
*AB r K * [ A][ B ] A B
K* e
G*
K = 1.38x10-16 erg/oK
RT
T
= 6.624x10-27 erg/s
70
Transiton Theory T *AB r A B
S * / R H * / RT e
T *AB S */ R e A A B T *AB S */ R H */ RT e r e A B
T
f e S */ R e H */ RT
TZe H */ RT
E =H* 71
Summary of Rate Theory
Rate Theory Arrhenius Collision Activated Complex (Transition)
Equation k Ae 1 2
k BT e
Ea
A term RT
E RT
T S * R H* RT k e e
A BT1/2
T S * R e
Exponential term Ea RT Ea RT H * RT
72
6. REACTION MECHANISMS
73
Reaction Mechanisms • Reactions occur in a series of elementary steps collectively called a mechanism. • Determining the reaction mechanism is the overall goal of kinetic studies. • One step, the rate-determining step (RDS), is much slower than the other. • Usually, an intermediate (isolable) or a transition state (non-isolable) is formed at some point during the reaction. • molecularity – the number of molecules that participate in a reaction
74
Reaction Mechanism • MECHANISM: the step-by-step pathway by which a reaction occurs • Each step is called an elementary step – NO2(g) + CO(g) NO(g) + CO2(g) • Mechanism: – NO2(g) + NO2(g) NO(g) + NO3(g) – NO3(g) + CO(g) NO2(g) + CO2(g) • NO3 is a reaction intermediate
75
Reaction Mechanism • The slow step is called the ratedetermining step (RDS) or rate-limiting step (RLS) • A reaction can never occur faster than its slowest step • Overall reaction = sum of all elementary steps • The mechanism proposed must be consistent with the rate law
76
Elementary Steps and Molecularity Molecularity
Elementary Reaction
Rate Law
Unimolecular
A products
Rate = k[A]
Bimolecular
A + A products
Rate = k[A]2
Bimolecular
A + B products
Rate = k[A][B]
Trimolecular
A + A + A products
Rate = k[A]3
Trimolecular
A + A + products
Rate = k[A]2[B]
Trimolecular
A + B + C products
Rate = k[A][B][C]
Molecularity is the number of molecules reacting.
77
Steady-state Approach • When reaction mechanism has several steps of comparable rates, the rate-limiting step is often not obvious. There are intermediates in some steps. • SSA: a method used to derive a rate law based on the assumption that one intermediate is consumed as quickly as it is generated. 2 N2O5 4NO2 + O2
N2O5 NO2 + NO3 NO3 + NO2 NO+ NO2 + O2 NO3 + NO 2NO
(1) (2) (3)
78
2N2O5 4NO + O2 N2O5
k1 k-1
NO2 + NO3
NO3 + NO NO3 + NO
k2 k3
NO + NO2 + O2 2NO
d [ NO ] k 2 [ NO3 ][ NO ] k3 [ NO3 ][ NO ] 0 dt
k 2 [ NO3 ][ NO2 ] [ NO ] k3 [ NO3 ] d [ NO3 ] k1[ N 2O5 ] k 2 [ NO3 ][ NO2 ] k3 [ NO3 ][ NO ] k 1[ NO2 ][ NO3 ] 0 dt
[ NO3 ]
k1[ N 2O5 ] k 2 [ NO3 ][ NO2 ] k3[ NO3 ][ NO ] k 1[ NO2 ][ NO3 ]
79
[ NO3 ]
k1[ N 2O5 ] k 2 [ NO2 ] k3 [ NO] k 1[ NO2 ]
k1[ N 2O5 ] k (k 2 k 1 )[ NO2 ] k3 2 [ NO2 ] k3
k1[ N 2O5 ] (2k 2 k 1 )[ NO2 ]
d [O2 ] k 2 [ NO3 ][ NO2 ] dt k k k [ N O ][ NO2 ] 1 2 3 2 5 (2k 2 k 1 )[ NO2 ]
r
k1k 2 k3 [ N 2O5 ] k '[ N 2O5 ] (2k 2 k 1 ) 80
Catalysis • Catalyst – increases the rate of a reaction without being consumed or changing chemically • Accomplishes this by lowering the activation energy by changing the reaction mechanism. • Heterogeneous vs. homogeneous catalysis • Examples: – Catalytic converter – Enzymes in the body – Ozone depletion
81
Catalysis • Reaction rates are also affected by catalysts • Catalyst: a substance that increases the rate of a reaction without being consumed in the reaction • Catalysts work by providing alternative pathways that have lower activation energies • A catalyst may be homogeneous or heterogeneous • Homogeneous: catalyst and reactants are in the same phase 82
Energy
Ea
Uncatalyzed pathway Catalyzed pathway
Ea
products
E Reactants
Reaction process
83
• Heterogeneous: catalyst in a different phase • Typically: a solid in a liquid • An important example: catalytic converters in automobile - convert pollutants to CO2 H2O, O2, N2 - usually Pt, Pd, V2O5, Cr2O3, CuO • Cars must use unleaded fuels – lead poisons the catalytic bed 84
Enzyme Kinetics Michaelis-Menten mechanism for enzyme kinetics is:
85
Catalytic efficiency, = kcat/KM = k2/[(k-1+k2)/k1] max, = k1; if k2 >>k-1 k1 = rate of formation of ES Diffusion limit ~ 108 – 109 M-1-s-1 For enzyme sized molecules at room temperature Decomposition of hydrogen peroxide = 4x108 M-1-s-1
Turnover number or catalytic constant = number of catalytic cycles performed by the activate Site in a given time intervals divided by the duration of that
86
Lineweaver Burk Plot
1 KM 1 1 v vmax S S
vmax [ S ] v K M [S ]
S K M v
vmax
S vmax
v/[S]
[S]/v
1/v
v vmax v S K M K M
1/vmax
KM/vmax
1/KM vmax/KM KM/vmax
1/vmax 1/[S]
[S]
v 87
Hypothetical
KM = 20
Vmax = 2
88
Enzyme Inhibition Reversible: enzyme activity can be regenerated by removing the inhibitors. Irreversible: complete loss of enzyme activity after a period of time
89
Competitive Inhibition E+S +
k1 k-1
k2 ES E + P
EI
k3
KM
E I K I EI
I KI
E S k1 k2 ES k1
E+Q
k-3
d [ ES ] k1[ E ][ S ] k 1[ ES ] k 2 [ ES ] 0 dt d [ EI ] k3 [ E ][ I ] k 3[ EI ] k 4 [ EI ] 0 dt
Eo [ E ] [ ES ] [ EI ] Eo [ E ]
[ E ][ S ] [ E ][ I ] KM KI
[S ] [I ] Eo [ E ]1 K K M I
[E]
Eo [S ] [ I ] 1 KM KI 90
1/v
[I]
d [ P] k 2 [ ES ] dt
KM [I ] 1 vmax K I
d [ P] [ E ][ S ] k2 dt KM
1/vmax [S]
d [ P] k 2 Eo [S ] [ S ] [ I ] dt KM 1 KM KI
v
d [ P] dt
k 2 Eo [ S ] [I ] K M 1 [ S ] KI
vmax k 2 Eo
1 1 KM v vmax vmax
vmax [ S ] v [I ] [ S ] K M 1 KI
[I ] 1 1 K I [S ]
Inhibitor is replaced from the active sites by substrate at high [S] 91
Non-competitive inhabitation Inhibitor binds to some other binding sites; Inhibitor combines with both free E and ES 2 possible mechanisms: Ternary complex is a dead-end complex and does not breakdown to yield products Ternary complex breaks down at a slow rate than he ES complex KM ES E + P +
E+S + I
I
KI EI + S
KM
KI EIS E + P
92
E+S +
k1 k-1
I
1/v k2 ES E + P + I
KI EI + S
k3 k-3
KI
[S ] K M 1 K M
EIS E + P
v k 2 [ ES ]
1/vmax-app
[S ] v KM vmax 1 [ S ] [ I ] [ I ][ S ] KM K I K IKM
1 vmax app
1 vmax
[I ] vmax K I
1/vmax-app
Vmax=k2E0 v vmax
1/(vmaxKI)
Henri-Michaelis-Menton Eq
[S ] [I ] [S ] [ S ]1 K M 1 K K M I
1/[S]
1/vmax [I]
93
Uncompetitive inhibition No binding site for inhibitor until substrate is bound to the enzyme, so only ternary complex is possible
94
Uncompetitive Inhibition KM E+S ES E + P +
1/v
I
KM vmax
KI EIS E + Q
1
[I ] 1 vmax K I
v k 2 [ ES ] [S ] v KM vmax 1 [ S ] [ I ][ S ] KM K IKM
1/[S]
1 1 [I ] v' vmax K I 1/v’
Vmax=k2E0
1 1 [ I ] 1 KM 1 v vmax [ S ] vmax K I
1/KI 1/vmax [I]
95
E nS ES n P K
[ ES n ] [ E ][ S ]n
Yield
Eo [ E ] K [ E ][ S ]n [E]
Hill eq.
[ ES n ] K [ E ][ S ]n Y [ E ] [ ES n ] [ E ] K [ E ][ S ]n
Y K [ S ]n 1 Y
KEo [ S ]n 1 K [ S ]n
n: Hill coefficient
Y log n log[S ] log K 1 Y
v k 2 [ ES ]
[ ES ]
Eo 1 K [ S ]n
k 2 E o K [ S ]n v k 2 [ ES n ] 1 K [ S ]n 1 1 K n v vmax [ S ] vmax
Eo [ E ] [ ES ] 96
97
Chain Reactiosn Chain reactions usually involve free radicals H2+Br2=2HBr The experimental rate law is 1 2
d [ HBr ] k '[ H 2 ][ Br2 ] [ HBr ] dt 1 k'' [ Br2 ]
98
Initiation
Br2 Br Br
Propagation
ka
kb Br H 2 HBr H
kc H Br2 HBr Br
Inhibition
Termination
kd H HBr H 2 Br
ke Br Br Br2
99
100
101
C2H6 = C2H4 + H2
102
103
The Lindemann Mechanism N2O5 = NO2 + NO3.
k1 >> k-2 [N2O5] Rate k2[N2O5]2 k1 bAPA
bA PAbB P B R ks (1 bA PA bB PB ) 2 bA PAbB P B bB P B ' PB R ks ks ks 2 (bA PA ) bA PA PA
bBPB
