Chem.414 - Physical Chemistry II Spring 2012/2013/2014.
-
Upload
rafe-richard -
Category
Documents
-
view
223 -
download
0
Transcript of Chem.414 - Physical Chemistry II Spring 2012/2013/2014.
Chem.414 - Physical Chemistry IIChem.414 - Physical Chemistry II
T ranslational
M echanism s
Collisions
Diffusion Ideal/Dilute
E-Chem (Lab)
Henry/Raoult T herm odynam ics Phase diagram s
Solutions
Boltzm ann Distributions T ransiton State
T heories of Reaction Rates
Rate Law s T -Dependence
Kinetics
Operator Algebra Applications
Quantum Chem istry
Spring 2012/2013/2014
Chemical KineticsChemical Kinetics
0th Order
M echanism s
T -Dependence of Rxn Rates
Steady State Approxim ation Experim ental Techniques
Order & M olecularity
2nd Order nth Order
Concepts of Rxn RatesFirst Order
Study of Chemical KineticsStudy of Chemical Kinetics
• Rate of reaction
• Dependence of concentration of species
• Dependence of temp., pressure, catalyst
• Control of reactions
• Mechanisms [Dominating step (fast vs. slow)]
• Guide to chemical intuition
Reaction Rate and Stoichiometry• For the reaction
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
we know
• In general for
aA + bB cC + dD
Reaction RatesReaction Rates
dt
d
dt
d OHHCClHCRate 9494
dt
d
ddt
d
cdt
d
bdt
d
a
DCBARate
1111
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
Time / s [N2O5] / M ln [N2O5] d[N2O5]/dt(tangential slope)
1 0 1.00
2 200 0.88
3 400 0.78
4 600 0.69
5 800 0.61
6 1000 0.54
7 1200 0.48
8 1400 0.43
9 1600 0.38
10 1800 0.34
11 2000 0.30
4. Consider the following N2O5 reaction: 2 N2O5(soln) ----> 4 NO2(soln) + O2(g)
Let: C = [N2O5]
(a) Using a graph of C vs. t, obtain tangential slopes and plot dC/dt vs. C. Calculate k after fitting with linear regression. (b) Plot ln C vs. t. Calculate k after fitting with linear regression. (c) Plot C vs. t. Fit the data with an appropriate function. Display the equation in standard IRL form with the appropriate variable names for this reaction. (d) Calculate half-live (t2) and life-time (). Compare them to the interpolated values from the plot of C vs. t.
EXCEL
First Order Reactions (to one component)
The Change of Concentration with TimeThe Change of Concentration with Time
0lnln CktC CNCHNCCH Co
39.198
3
Isomeric Transformation of Methyl Isonitrile to Acetonitrile
Differential and Integrated Rate LawsDifferential and Integrated Rate Laws
n-th Order to One Component (Generalized Rate Laws)
Let: C = concentration of reactant A remaining at time tCo = initial concentration of reactant A (i.e. t=0)k = rate constant (units depends on n)
DRL:
IRL:
Differential and Integrated Rate LawsDifferential and Integrated Rate Laws
Rate Law: First Order to One ComponentRate Law: First Order to One Component
Second Order Reactions
The Change of Concentration with TimeThe Change of Concentration with Time
0
11
Ckt
C
)(2
1)()( 2
3002 gOgNOgNO Co
Rate Law: Second Order to One ComponentRate Law: Second Order to One Component
Gas-Phase Decomposition of Nitrogen DioxideGas-Phase Decomposition of Nitrogen Dioxide
)(2
1)()( 2
3002 gOgNOgNO Co
Time / s [NO2] / M
0.0 0.01000
50.0 0.00787
100.0 0.00649
200.0 0.00481
300.0 0.00380
k = 0.543 unit?k = 0.543 unit?Is this reaction first or second order?
Half-Lives, Rate Constants and CHalf-Lives, Rate Constants and Coo
Half-Lives, Rate Constants and CHalf-Lives, Rate Constants and Coo - II - II
Zeroth Order to One Component - CatalysisZeroth Order to One Component - Catalysis
1. Provide the DRL.
2. Determine the IRL.
3. Sketch the IRL: Co=1.00 mol L-1 , k = 5.00x10-3 mol L-1 s-1 .
4. Use Mathcad (or EXCEL) to generate the IRL graph.
5. Obtain the half-life expression.
6. How many half-lives would it take for the reaction to reach equilibrium (i.e. completion)? [ Hint: Solve the IRL for time when C=0. Confirm by graph. ]
Summary of Rate Laws to One-ComponentSummary of Rate Laws to One-Component
First-Order Second-Order Zeroth-Order
DRL
(-dC/dt)kC kC2 k
IRLC = Co·e-kt
ln C = -kt + ln Co
1/C = kt + 1/Co C = -kt + Co
Linear Equation
ln C vs. t 1/C vs. t C vs. t
Linear Plot
Half-Life ln(2)/k 1/kCo Co/2k
Units on k time-1 M-1 time-1 M time-1
m = -k
b = ln Co
m = k
b = 1/Co
m = -k
b = Co
Exponents in the Rate Law• For a general reaction with rate law
we say the reaction is mth order in reactant 1 and nth order in reactant 2.
• The overall order of reaction is m + n + ….• A reaction can be zeroth order if m, n, … are zero.• Note the values of the exponents (orders) have to be determined
experimentally. They are not simply related to stoichiometry.
Concentration and RateConcentration and Rate
nmk ]2reactant []1reactant [Rate
Method of Initial/Comparative RatesMethod of Initial/Comparative Rates
Expt # [NH4+]o / M [NO2
-]o / M (Rate)o / M s-1
1 0.100 0.0050 1.35x10-7
2 0.100 0.0100 2.70x10-7
3 0.200 0.0100 5.40x10-7
)(2)()()( 2224 OHgNaqNOaqNH
Three Component Rate LawThree Component Rate Law
Expt # [BrO3-]o / M [Br-]o / M [H+]o / M (Rate)o / M s-1
1 0.10 0.10 0.10 8.0x10-4
2 0.20 0.10 0.10 1.6x10-3
3 0.20 0.20 0.10 3.2x10-3
4 0.10 0.10 0.20 3.2x10-3
)(3)(3)(6)(5)( 223 OHBraqHaqBraqBrO
Techniques for Multiple Component Rate LawsTechniques for Multiple Component Rate Laws
1. Integration Approach:
Second Order – First Order to each of two components
2. Flooding Technique:
Rate = k [A]x [B]y [C]z
Second Order: First order to each of two components
Consider: A + B ---> Products (D)
ratedCA
dt
dCB
dt
dCD
dt
At t=0; Let: CA = a ; C B = b
At some time t; Let: x (mol/L) of A and B be reacted
CA a x CB b x CD x
dx
dtk a x( ) b x( )
dx
a x( ) b x( )k dt
expands in partial fractions to1
a x( ) b x( )1
b a( ) a x( )1
b a( ) b x( )
by integration, yields1
a x( ) b x( )1
b aln a x( )
1
b aln b x( )
by integration, yields1b a( ) a x( )
1
b a( ) b x( )
1
b aln a x( )
1
b aln b x( )
Re-writing to give the following: 1
a bdx
a xdx
b x
k dt
1
a b0
x
x1
a x1
b x
d k0
t
t1
d
Integrating yields: 1
a bln
b a x( )a b x( )
k t
Z1
a bln
b a x( )a b x( )
Z versus t yields straight line.
Applications of First-Order ProcessesApplications of First-Order Processes
1. Radioactive Decay
2. Bacterial Growth
3. Interest and Exponential Growth
[Credit Card]
4. Loan Balance
Interest and Exponential Growth
P = future valueC = initial depositr = interest rate (expressed as a fraction: eg. 0.06)n = # of times per year interest is compoundedt = number of years invested
C 10000 r 0.06 n 1P t( ) C 1
r
n
n t
Continuous Compound Interestn
1rate
n
n tlim
erate t
n ---> infinity
P 30( ) 57 103
PE t( ) C er t
0 3 6 9 12 15 18 21 24 27 300
8000
1.6 104
2.4 104
3.2 104
4 104
4.8 104
5.6 104
6.4 104
7.2 104
8 104
P t( )
PE t( )
t
PE 30( ) 60 103
PE 30( ) P 30( ) 3061.56
Loan Balance A person initially borrows an amount A and in return agreesto make n repayments per year, each of an amount P.While the person is repaying the loan, interest isaccumulating at an annual percentage rate of r, and thisinterest is compounded n times a year (along with eachpayment). Therefore, the person must continue payingthese installments of amount P unitl the original amountand any accumulated interest is repayed. The equation B(t)gives the amount B that the person still needs to repay aftert years.
B = balance after t yearsA = amount borrowedn = number of payments per yearP = amount paid per paymentr = annual percentage rate (APR)
A 115000 n 12 P 800 r 0.05B t( ) A 1
r
n
n t P
1r
n
n t1
1r
n
1
0 2 4 6 8 10 12 14 16 18 205 10
4
1 104
3 104
7 104
1.1 105
1.5 105
B t( )
t
The Arrhenius Equation• Arrhenius discovered most reaction-rate data obeyed the
Arrhenius equation:
– k is the rate constant, Ea is the activation energy, R is the gas constant (8.3145 J K-1 mol-1) and T is the temperature in K.
– A is called the frequency factor.
– A is a measure of the probability of a favorable collision.
– Both A and Ea are specific to a given reaction.
Temperature and RateTemperature and Rate
RTEa
eAk
Temperature and RateTemperature and Rate
• The balanced chemical equation provides information about the beginning and end of reaction.
• The reaction mechanism gives the path of the reaction.• Mechanisms provide a very detailed picture of which bonds
are broken and formed during the course of a reaction.
Elementary Steps• Elementary step: any process that occurs in a single step.
Reaction MechanismsReaction Mechanisms
Elementary Steps• Molecularity: the number of molecules present in an
elementary step.– Unimolecular: one molecule in the elementary step,
– Bimolecular: two molecules in the elementary step, and
– Termolecular: three molecules in the elementary step.
• It is not common to see termolecular processes (statistically improbable).
Reaction MechanismsReaction Mechanisms
Rate Laws for Elementary Steps• The rate law of an elementary step is determined by its
molecularity:– Unimolecular processes are first order,
– Bimolecular processes are second order, and
– Termolecular processes are third order.
Rate Laws for Multistep Mechanisms• Rate-determining step is the slowest of the elementary
steps. [example]
Reaction MechanismsReaction Mechanisms
Rate Laws for Elementary Steps
Reaction MechanismsReaction MechanismsReaction MechanismsReaction Mechanisms
Rate ExpressionsRate Expressions
yDxCwBvAk
k
1
1
dt
d
ydt
d
xdt
d
wdt
d
v
DCBARate
1111
If elementary steps:
-d[A]/dt = vk1[A]v[B]w – vk-1[C]x[D]y
-d[B]/dt = wk1[A]v[B]w – wk-1[C]x[D]y
d[C]/dt = xk1[A]v[B]w – xk-1[C]x[D]y
d[D]/dt = yk1[A]v[B]w – yk-1[C]x[D]y
Mechanisms with an Initial Fast Step
2NO(g) + Br2(g) 2NOBr(g)
• The experimentally determined rate law can be:
d[NOBr]/dt = kobs[NO]2[Br2] (or) = kobs’[NO][Br2]
• Consider the following mechanism
Reaction MechanismsReaction Mechanisms
NO(g) + Br2(g) NOBr2(g)k1
k-1
NOBr2(g) + NO(g) 2NOBr(g)k2
Step 1:
Step 2:
(fast)
(slow)
NO(g) + Br2(g) NOBr2(g)k1
k-1
NOBr2(g) + NO(g) 2NOBr(g)k2
Step 1:
Step 2:
(fast)
(slow)
Spring 2014
NO(g) + Br2(g) NOBr2(g)k1
k-1
NOBr2(g) + NO(g) 2NOBr(g)k2
Step 1:
Step 2:
(fast)
(slow)
Spring 2014
General MechanismGeneral Mechanism
DCBA Overall Reaction:
Proposed Mechanism:
DBM
CMA
k
k
k
2
1
1Where: D = observable product
M = intermediate
DBM
CMA
k
k
k
2
1
1
DBM
CMA
k
k
k
2
1
1
DBM
CMA
k
k
k
2
1
1
Spring 2014
)(
)(
2
1
1
DBM
CMA
k
k
k
Spring 2014
Hydrogen-Iodine ReactionHydrogen-Iodine Reaction
HIIH 222 Overall Reaction:
Proposed Mechanism:
HIIH
II
k
k
k
22
2
3
1
2
2
2
Where: I• = free radical
?][
dt
HId
HIIH
II
k
k
k
22
2
3
1
2
2
2
HIIH
II
k
k
k
22
2
3
1
2
2
2
HIIH
II
k
k
k
22
2
3
1
2
2
2
Spring 2012
HIIH
II
k
k
k
22
2
3
1
2
2
2
Spring 2012
Rice-Hertzfeld Free Radical Chain Reaction Mechanism
Rice-Hertzfeld Free Radical Chain Reaction Mechanism
)()()( 43 gCOgCHgCHOCH Overall Reaction:
Proposed Mechanism:
ationterchainHCCH
npropagatiochainCHCOCHCHCHOCH
initiationchainCHOCHCHOCH
k
k
k
min2 623
3433
33
3
2
1
?][ 4
dt
CHd
ationterchainHCCH
npropagatiochainCHCOCHCHCHOCH
initiationchainCHOCHCHOCH
k
k
k
min2 623
3433
33
3
2
1
ationterchainHCCH
npropagatiochainCHCOCHCHCHOCH
initiationchainCHOCHCHOCH
k
k
k
min2 623
3433
33
3
2
1
KineticsKinetics
optial rotationabsorption/em issiondielectric constantrefractive indexdilatom etric (Vol.)pressure jum ptem perature jum pelectric fie ldconductivity
Hardw are"Instrum entation"
reaction rates vs. tim econcentrations vs. tim einitia l rates vs. tim e"lives" vs. tim eguess type of ordercom puter fitsflooding/isolationcatalystsm echanism s
Softw are"Brainm entation"
Experim ental T echniquesData to Conclusions
CatalysisCatalysis
Heterogeneous Catalysis• Consider the hydrogenation of ethylene:
C2H4(g) + H2(g) C2H6(g), H = -136 kJ/mol.– The reaction is slow in the absence of a catalyst.– In the presence of a metal catalyst (Ni, Pt or Pd) the reaction occurs
quickly at room temperature.– First the ethylene and hydrogen molecules are adsorbed onto active
sites on the metal surface.– The H-H bond breaks and the H atoms migrate about the metal
surface.
CatalysisCatalysis
CatalysisCatalysis
Enzymes• Enzymes are biological catalysts.• Most enzymes are protein molecules with large molecular
masses (10,000 to 106 amu).
• Enzymes have very specific shapes.• Most enzymes catalyze very specific reactions.• Substrates undergo reaction at the active site of an enzyme.• A substrate locks into an enzyme and a fast reaction occurs.• The products then move away from the enzyme.
CatalysisCatalysis
Enzymes• Only substrates that fit into the enzyme lock can be
involved in the reaction.• If a molecule binds tightly to an enzyme so that another
substrate cannot displace it, then the active site is blocked and the catalyst is inhibited (enzyme inhibitors).
• The number of events (turnover number) catalyzed is large for enzymes (103 - 107 per second).
CatalysisCatalysis
Enzymes
CatalysisCatalysis
][
][][)( 2
SK
SEkMentenMichaelisRate
M
t
Mechanism: Two IntermediatesMechanism: Two Intermediates
ClIOOClIOverall Reaction:
Proposed Mechanism:
rdsslowOIOHHOIOH
fastClHOIHOClI
mequilibriufastveryOHHOClOHOCl
k
k
k
k
2
2
3
2
1
1
][
][][][
OH
OClIk
dt
OId obsExperimentally found:
Show that the proposed mechanism is consistent with the observed RL.
MechanismMechanism
)(3)(2 23 gOgO
Overall Reaction:
Proposed Mechanism:
slowOOO
fastOOO
k
k
k
23
23
22
1
1
][
][][
2
233
O
Ok
dt
Odobs
Observed Rate Law:
slowOOO
fastOOO
k
k
k
23
23
22
1
1
Chemical KineticsChemical Kinetics
0th Order
M echanism s
T -Dependence of Rxn Rates
Steady State Approxim ation Experim ental Techniques
Order & M olecularity
2nd Order nth Order
Concepts of Rxn RatesFirst Order
nCkdt
dCRateDRL tversusCsIRL '
0lnln CktC 0
11
Ckt
C
RTEa
eAk
DBM
CMA
k
k
k
2
1
1