Chemical Kinetics Thermodynamics – does a reaction take place? Kinetics – how fast does a...
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Transcript of Chemical Kinetics Thermodynamics – does a reaction take place? Kinetics – how fast does a...
Chemical Kinetics
Thermodynamics – does a reaction take place?
Kinetics – how fast does a reaction proceed?
Reaction rate is the change in the concentration of a reactant or a product with time (M/s).
A B
rate = -[A]t
rate = [B]t
[A] = change in concentration of A over time period t
[B] = change in concentration of B over time period t
Because [A] decreases with time, [A] is negative.
13.11GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
A B
13.1
rate = -[A]
t
rate = [B]
t
time
2GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
Reaction rate = change in Reaction rate = change in concentration of a reactant or concentration of a reactant or product with time.product with time.
Three “types” of ratesThree “types” of rates initial rateinitial rateaverage rateaverage rateinstantaneous rateinstantaneous rate
Reaction Rates Reaction Rates
3GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
average rate = -[Br2]
t= -
[Br2]final – [Br2]initial
tfinal - tinitial
slope oftangent
slope oftangent slope of
tangent
instantaneous rate = rate for specific instance in time13.14GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
2NO22NO+O2
5GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
6GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
13.3
Concentrations & RatesConcentrations & Rates
0.3 M HCl 6 M HCl
Mg(s) + 2 HCl(aq) ---> MgClMg(s) + 2 HCl(aq) ---> MgCl22(aq) + H(aq) + H22(g)(g)
8GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
Reaction OrderWhat is the rate expression for aA + bB → cC + dD
Rate = k[A]x[B]y
where x=1 and y=2.5?Rate = k[A][B]2.5
What is the reaction order?First in A, 2.5 in B
Overall reaction order?2.5 +1 = 3.5
9GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
Interpreting Rate Laws Interpreting Rate Laws Rate = k [A]Rate = k [A]mm[B][B]nn[C][C]pp
If m = 1, rxn. is 1st order in AIf m = 1, rxn. is 1st order in ARate = k [A]Rate = k [A]11
If [A] doubles, then rate goes up by factor of __ If [A] doubles, then rate goes up by factor of __ If m = 2, rxn. is 2nd order in A.If m = 2, rxn. is 2nd order in A.
Rate = k [A]Rate = k [A]22
Doubling [A] increases rate by ________Doubling [A] increases rate by ________ If m = 0, rxn. is zero order.If m = 0, rxn. is zero order.
Rate = k [A]Rate = k [A]00
If [A] doubles, rate ________If [A] doubles, rate ________
10GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
Determine the rate law and calculate the rate constant for the following reaction from the following data:
S2O82- (aq) + 3I- (aq) 2SO4
2- (aq) + I3- (aq)
Experiment [S2O82-] [I-]
Initial Rate (M/s)
1 0.08 0.034 2.2 x 10-4
2 0.08 0.017 1.1 x 10-4
3 0.16 0.017 2.2 x 10-4
rate = k [S2O82-]x[I-]y
Double [I-], rate doubles (experiment 1 & 2)
y = 1
Double [S2O82-], rate doubles (experiment 2 & 3)
x = 1
k = rate
[S2O82-][I-]
=2.2 x 10-4 M/s
(0.08 M)(0.034 M)= 0.08/M•s
13.2
rate = k [S2O82-][I-]
11GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
Deriving Rate LawsDeriving Rate Laws
Expt. Expt. [CH[CH33CHO]CHO] Disappear of CHDisappear of CH33CHOCHO(mol/L)(mol/L) (mol/L•sec)(mol/L•sec)
11 0.100.10 0.0200.020
22 0.200.20 0.0810.081
33 0.300.30 0.1820.182
44 0.400.40 0.3180.318
Derive Derive rate law rate law and and kk for for CHCH33CHO(g) --> CHCHO(g) --> CH44(g) + CO(g)(g) + CO(g)
from experimental data for rate of disappearance from experimental data for rate of disappearance of CHof CH33CHOCHO
12GENERAL CHEMISTRY(KINETIKS)
Rate of rxn = k [CHRate of rxn = k [CH33CHO]CHO]22
Deriving Rate LawsDeriving Rate Laws
Rate of rxn = k [CHRate of rxn = k [CH33CHO]CHO]22
Here the rate goes up by ______ when initial conc. Here the rate goes up by ______ when initial conc. doubles. Therefore, we say this reaction is doubles. Therefore, we say this reaction is _________________ order._________________ order.
Now determine the value of k. Use expt. #3 data—Now determine the value of k. Use expt. #3 data—0.182 mol/L•s = k (0.30 mol/L)0.182 mol/L•s = k (0.30 mol/L)22
k = 2.0 (L / mol•s)k = 2.0 (L / mol•s)Using k you can calc. rate at other values of [CHUsing k you can calc. rate at other values of [CH33CHO] CHO]
at same T.at same T.
13GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 14–14
Expession of R &KIodide ion is oxidized in acidic solution to
triiodide ion, I3- , by hydrogen peroxide.
– A series of four experiments was run at different concentrations, and the initial rates of I3
- formation were determined. – From the following data, obtain the reaction orders with respect to
H2O2, I-, and H+.– Calculate the numerical value of the rate constant.
)(2)()(2)(3)( 2322 lOHaqIaqHaqIaqOH
Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 14–15
A Problem to ConsiderInitial Concentrations (mol/L)
H2O2 I- H+ Initial Rate [mol/(L.s)]Exp. 1 0.010 0.010 0.00050 1.15 x 10-6
Exp. 2 0.020 0.010 0.00050 2.30 x 10-6
Exp. 3 0.010 0.020 0.00050 2.30 x 10-6
Exp. 4 0.010 0.010 0.00100 1.15 x 10-6
– Comparing Experiment 1 and Experiment 2, you see that when the– H2O2 concentration doubles (with other concentrations constant),– the rate doubles.– This implies a first-order dependence with respect to H2O2.
Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 14–16
A Problem to ConsiderInitial Concentrations (mol/L)
H2O2 I- H+ Initial Rate [mol/(L.s)]Exp. 1 0.010 0.010 0.00050 1.15 x 10-6
Exp. 2 0.020 0.010 0.00050 2.30 x 10-6
Exp. 3 0.010 0.020 0.00050 2.30 x 10-6
Exp. 4 0.010 0.010 0.00100 1.15 x 10-6
– Comparing Experiment 1 and Experiment 3, you see that when the– I- concentration doubles (with other concentrations constant),– the rate doubles.– This implies a first-order dependence with respect to I-.
Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 14–17
A Problem to ConsiderInitial Concentrations (mol/L)
H2O2 I- H+ Initial Rate [mol/(L.s)]Exp. 1 0.010 0.010 0.00050 1.15 x 10-6
Exp. 2 0.020 0.010 0.00050 2.30 x 10-6
Exp. 3 0.010 0.020 0.00050 2.30 x 10-6
Exp. 4 0.010 0.010 0.00100 1.15 x 10-6
– Comparing Experiment 1 and Experiment 4, you see that when – the H+ concentration doubles (with other concentrations constant),– the rate is unchanged.– This implies a zero-order dependence with respect to H+.
Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 14–18
A Problem to Consider
– Because [H+]0 = 1, the rate law is:
]I][OH[kRate 22
– The reaction orders with respect to H2O2, I-, and H+, are 1, 1, and 0, respectively.
Initial Concentrations (mol/L)H2O2 I- H+ Initial Rate [mol/(L.s)]
Exp. 1 0.010 0.010 0.00050 1.15 x 10-6
Exp. 2 0.020 0.010 0.00050 2.30 x 10-6
Exp. 3 0.010 0.020 0.00050 2.30 x 10-6
Exp. 4 0.010 0.010 0.00100 1.15 x 10-6
Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 14–19
A Problem to ConsiderInitial Concentrations (mol/L)
H2O2 I- H+ Initial Rate [mol/(L.s)]Exp. 1 0.010 0.010 0.00050 1.15 x 10-6
Exp. 2 0.020 0.010 0.00050 2.30 x 10-6
Exp. 3 0.010 0.020 0.00050 2.30 x 10-6
Exp. 4 0.010 0.010 0.00100 1.15 x 10-6
– You can now calculate the rate constant by substituting values from– any of the experiments. Using Experiment 1 you obtain:
Lmol
010.0Lmol
010.0ksL
mol1015.1 6
Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 14–20
A Problem to ConsiderInitial Concentrations (mol/L)
H2O2 I- H+ Initial Rate [mol/(L.s)]Exp. 1 0.010 0.010 0.00050 1.15 x 10-6
Exp. 2 0.020 0.010 0.00050 2.30 x 10-6
Exp. 3 0.010 0.020 0.00050 2.30 x 10-6
Exp. 4 0.010 0.010 0.00100 1.15 x 10-6
– You can now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1 you obtain:
)/(102.1/010.0010.0
1015.1 216
smolLLmol
sk
Zero-Order Reactions
13.3
A product rate = -[A]
trate = k [A]0 = k
k = rate
[A]0=RATE= M/s
[A]
t= k-
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
t½ = t when [A] = [A]0/2
t½ =[A]0
2k
[A] = [A]0 - kt
ZERO ORDER REACTION
0
2
4
6
8
10
12
0 2 4 6 8 10 12
TIME
[A]
First-Order Reactions
13.3
A product rate = k [A]
k = rate
[A]= 1/s or s-1
M/s
M=
[A]
t= k [A]-
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
[A] = [A]0exp(-kt)ln[A] = ln[A]0 - kt
t½ =0.693
k
22GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
Second-Order Reactions
13.3
A product rate = -[A]
trate = k [A]2
k = rate
[A]2= 1/M•s
M/s
M2=
[A]
t= k [A]2-
[A] is the concentration of A at any time t[A]0 is the concentration of A at time t=01
[A]=
1
[A]0
+ kt
t½ = t when [A] = [A]0/2
t½ =1
k[A]0
SECOND ORDER REACTION
0
2
4
6
8
10
12
0 1 2 3 4 5
TIME
[A]
SECOND ORDER REACTION
0
0.1
0.2
0.3
0.4
0.5
0.6
0 1 2 3 4 5
TIME
1/[A
]
First step in evaluating rate data is to graphically interpret the order of rxn
Zeroth order: rate does not change with lower concentration
First, second orders:Rate changes as a function of concentration
Summary of the Kinetics of Zero-Order, First-Orderand Second-Order Reactions
Order Rate Law
Concentration-Time Equation Half-Life
0
1
2
rate = k
rate = k [A]
rate = k [A]2
ln[A] = ln[A]0 - kt
1
[A]=
1
[A]0
+ kt
[A] = [A]0 - kt
t½
ln2
k=
t½ =[A]0
2k
t½ =1
k[A]0
13.3
Collision TheoryCollision TheoryCollision TheoryCollision TheoryOO33(g) + NO(g) ---> O(g) + NO(g) ---> O22(g) + NO(g) + NO22(g)(g)
10 31 COLLISINS/LIT.SReactions require Reactions require (A)(A) geometry geometry(B) (B) Activation energy Activation energy
26GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
Three possible collision orientations-- a) & b) produce reactions,while c) does not.
Collision TheoryCollision Theory
TEMPERATURE10 c0 T 100-300% RATE25 c0 T 35 c0 2% COLLISINEFFECTIVE COLLISIONS
Friday, April 21, 2023 GENERAL CHEMISTRY(KINETIKS) 28
29GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
① ① with sufficient energywith sufficient energy
The colliding molecules must have a total kinetic energy equal to or greater than the activation energy, Ea. The activation energy is the mini-mum energy of collision required for two molecules to initiate a chemical reaction. It can be thought of as the hill in below. Regardless of whether the elevation(平面 ) of the ground on the other side is lower than the original position,there must be enough energy imparted to the golf ball to get it over the hill.
30GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
The Collision Theory of Chemical Kinetics
Activation Energy (Ea)- the minimum amount of energy required to initiate a chemical reaction.
Activated Complex (Transition State)- a temporary species formed by the reactant molecules as a result of the collision before they form the product.
31GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
32GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
A + B C + D
Exothermic Reaction Endothermic Reaction
The activation energy (Ea) is the minimum amount of energy required to initiate a chemical reaction.
13.433GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
Temperature Dependence of the Rate Constant
k = A • exp( -Ea/RT )
Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature
A is the frequency factor
(Arrhenius equation)
13.434GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
2.05
2.1
2.15
2.2
2.25
2.3
2.35
2.4
2.45
0 200 400 600 800 1000 1200
TEMPERATURE
K
Arrhenius Equation
• k = A • exp( -Ea/RT )
• Can be arranged in the
form of a straight line
ln k = ln A-Ea/RT
Plot ln k vs. 1/T
slope = -Ea/R
Friday, April 21, 2023 GENERAL CHEMISTRY(KINETIKS) 35
0.74
0.76
0.78
0.8
0.82
0.84
0.86
0.88
0.9
0 0.0005 0.001 0.0015 0.002 0.0025
1/T
Ln
K
•k = A • e( -Ea/RT )
ln k = ln A-Ea/RT=
(-Ea/R)(1/T) + ln A
Ea/RTdecreases
- Ea/RTincreases
e- Ea/RT
increases kincreases
REACTI ONSPEEDS UP
I f Tincreases
36GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
300400C0 Rate20000 Times400500C0 Rate400 Times
Ea=60KJ/mol300310C0 Rate2 TimesEa=260KJ/mol300310C0 Rate25 Times
37GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
K in different T
Friday, April 21, 2023 GENERAL CHEMISTRY(KINETIKS) 38
Log(K2/K1)=-Ea/2.303R(1/T2-1/T1)
Log(K2/K1) =Ea/2.303R(T2-T1)/T1T2
Ln(K2/K1)=-Ea/R(1/T2-1/T1)
MechanismsMechanismsOO33 + NO reaction occurs in a single + NO reaction occurs in a single
ELEMENTARYELEMENTARY step. Most others involve a step. Most others involve a sequence of elementary steps.sequence of elementary steps.
Adding elementary steps gives NET reaction.Adding elementary steps gives NET reaction.
39GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
MechanismsMechanisms
Most rxns. involve a sequence of elementary Most rxns. involve a sequence of elementary steps.steps.
2 I2 I-- + H + H22OO22 + 2 H + 2 H++ ---> I ---> I22 + 2 H + 2 H22OO
Rate = k [IRate = k [I--] [H] [H22OO22]]NOTENOTE1.1. Rate law comes from experimentRate law comes from experiment2.2. Order and stoichiometric coefficients Order and stoichiometric coefficients
not necessarily the same!not necessarily the same!3.3. Rate law reflects all chemistry Rate law reflects all chemistry
including including the slowest step the slowest step in multistep in multistep reaction.reaction.
40GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
MechanismsMechanisms
Proposed MechanismProposed Mechanism
Step 1 — slowStep 1 — slow HOOH + IHOOH + I-- --> HOI + OH --> HOI + OH--
Step 2 — fastStep 2 — fast HOI + IHOI + I-- --> I --> I22 + OH + OH--
Step 3 — fastStep 3 — fast 2 OH2 OH- - + 2 H + 2 H++ --> 2 H --> 2 H22OO
Rate of the reaction controlled by slow step —Rate of the reaction controlled by slow step —
RATE DETERMINING STEPRATE DETERMINING STEP, rds., rds.
Rate can be no faster than rds!Rate can be no faster than rds!
The species HOI and OHThe species HOI and OH-- are are reaction reaction intermediates.intermediates.
Proposed MechanismProposed Mechanism
Step 1 — slowStep 1 — slow HOOH + IHOOH + I-- --> HOI + OH --> HOI + OH--
Step 2 — fastStep 2 — fast HOI + IHOI + I-- --> I --> I22 + OH + OH--
Step 3 — fastStep 3 — fast 2 OH2 OH- - + 2 H + 2 H++ --> 2 H --> 2 H22OO
Rate of the reaction controlled by slow step —Rate of the reaction controlled by slow step —
RATE DETERMINING STEPRATE DETERMINING STEP, rds., rds.
Rate can be no faster than rds!Rate can be no faster than rds!
The species HOI and OHThe species HOI and OH-- are are reaction reaction intermediates.intermediates.
Most rxns. involve a sequence of elementary steps.Most rxns. involve a sequence of elementary steps. 2 I2 I-- + H + H22OO22 + 2 H + 2 H++ ---> I ---> I22 + 2 H + 2 H22OO
Rate = k [IRate = k [I--] [H] [H22OO22]]
41GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
MechanismsMechanisms 2NO+F22NOF R=K[NO][F2]
1=NO+F2NOF+F R1=K1[NO][F2]
Rate limiting step(RDS)2=NO+F NOF R2=K2[NO][F]
Friday, April 21, 2023 GENERAL CHEMISTRY(KINETIKS) 42
SN1
OH- + (CH3)3CBr(CH3)3COH + Br-
R=K[(CH3)3CBr]
1)(CH3)3CBr(CH3)3C+ +Br- R1=K1[(CH3)3CBr ]2) (CH3)3C+ +OH- (CH3)3COH R2=K2[(CH3)3C+] +[OH-]
43GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
SN2
OH- +CH3Br CH3OH + Br-
R=K[CH3Br][OH-]
44GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
45GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
Rate Laws Rate Laws and and MechanismsMechanisms
NO2 + CO reaction: Rate = k[NO2]2
Single step
Two possible Two possible mechanismsmechanisms Two steps:
46GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.
k = A • exp( -Ea/RT ) Ea k
uncatalyzed catalyzed
ratecatalyzed > rateuncatalyzed
Ea < Ea‘ 13.647GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
For the decomposition of hydrogen peroxide:
A catalyst(Br-) speeds up a reaction by lowering the activation energy. It does this by providing a different mechanism (机制 ) by which the reac-tion can occur. The energy profiles for both the catalyzed and uncatalyzed reactions of decomposition of hydrogen peroxide are shown in Figure below.
Br-
48GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
Homogeneous CatalysisN2O (g) (g) N2 (g) + O2 (g)
1== N2O (g) N2 (g) + O (g)
2==O (g) + N2O (g) N2(g)+ O2 (g) Ea=240KJ/mol Cl2
Cl2(g)2Cl(g)N2O(g)+Cl(g)N2(g)+ClO(g) 2ClOCl2(g)+O2(g)Ea=140KJ/mol
49GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
C2H4 + H2 C2H6
with a metal catalyst
a. reactantsb. adsorptionc. migration/reactiond. desorption
Heterogeneous Catalysis
50GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
Enzyme Catalysis
13.651GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
Catalytic Converters
13.6
CO + Unburned Hydrocarbons + O2 CO2 + H2Ocatalytic
converter
2NO + 2NO2 2N2 + 3O2
catalyticconverter
52GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023
The End!!!!!!!The End!!!!!!!
53GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023