• 14-1 Rates of Chemical • 14-2 Reaction Rates and...
Transcript of • 14-1 Rates of Chemical • 14-2 Reaction Rates and...
OFB Chapter 14 13/27/2003
Chapter 14Chemical Kinetics
• 14-1 Rates of Chemical Reactions
• 14-2 Reaction Rates and Concentrations
• 14-3 The Dependence of Concentrations on Time
• 14-4 Reaction Mechanisms• 14-5 Reaction Mechanism and
Rate Laws• 14-6 Effect of Temperature on
Reaction Rates• 14-7 Kinetics of Catalysis
OFB Chapter 14 23/27/2003
Chapter 14Chemical Kinetics
• Thus far in the course, we have been using and talking about chemical reactions– Reactants, products, and how much is
involved
• Chemical Kinetics deals with how fast a reaction proceeds– Kinetics = Rates of Chemical Reactions– And how to deduce reaction mechanisms
from observed rates of reactions– Activation Energy– Catalysts
OFB Chapter 14 33/27/2003
∆t∆[X]
tt[X][X] rate average
sLmolsL
mols
mol/L are units
∆t∆[X] rate reaction average
if
if
11
=−−
=
••=•
=
=
−−
To measure rates we could monitor the disappearance of reactants or appearance of products
e.g., NO2 + CO → NO + CO2
OFB Chapter 14 43/27/2003
To measure rates we could monitor the disappearance of reactants or appearance of products
e.g., 2NO2 + F2 → 2NO2F
OFB Chapter 14 53/27/2003
Generalized Reaction
aA + bB → cC + dD
tD
dtC
c
tB
btA
arate
∆∆
∆∆
∆∆
∆∆
][1][1
][1][1
=
+=
−=
−=
OFB Chapter 14 63/27/2003
Order of a Reaction
→ 221
252 2NO OkON + decomposition
→
k[A]rate
k aA n
productse.g.,
=
OFB Chapter 14 73/27/2003
→
][
k aA productse.g.,
nAkrate =
n does not have to be an integer
n = 3/2 “three halves order” reaction
Note that n does not equal the coefficient of the reactant. It is related to the reaction mechanism and determined experimentally.
OFB Chapter 14 83/27/2003
→nm
k
[B]k[A] rate
products bB aA
=
+
OFB Chapter 14 93/27/2003
Chapter 14Chemical Kinetics
• Example 14-2At elevated temperatures, HI reacts according to the chemical equation
2HI → H2 + I2at 443°C, the rate of reaction increases with concentration of HI, as shown in this table.
Data Point 1 2 3[HI] (mol L-1) 0.005 0.01 0.02Rate (mol L-1 s-1) 7.5E-04 3.0E-03 1.2E-02
a) Determine the order of the reaction with respect to HI and write the rate expressionb) Calculate the rate constant and give its unitsc) Calculate the instantaneous rate of reaction for a [HI] = 0.0020M
OFB Chapter 14 103/27/2003
part A toanswer 2
n
4
3
points datatwoanyofratio the take
k[HI] rate
and HI inorder second
0.00500.010
7.5x103.0x10
=
∴
=−
−
2HI → H2 + I2
Data Point 1 2 3[HI] (mol L-1) 0.005 0.01 0.02Rate (mol L-1 s-1) 7.5E-04 3.0E-03 1.2E-02
OFB Chapter 14 113/27/2003
Example 14-2b) Calculate the rate constant and give its units
Example 14-2c) Calculate the instantaneous rate of
reaction for a [HI] = 0.0020M
part A toanswer 2 k[X] rate =
OFB Chapter 14 123/27/2003
Similarly for two or more concentrations
nmp isorder ][][ +== nm BAkrate1)1()1(
arek of units −−−− sLmol pp
C B AExample
→+initial Rate
[A] [B] mol L-1 s-1
1.0E10-4 1.0E10 -4 2.8E10 -6
1.0E10-4 3.0E10 -4 8.4E10 -6
2.0E10-4 3.0E10 -4 3.4E10 -5
OFB Chapter 14 133/27/2003
C B AExampleFor →+ initial Rate
[A] [B] mol L-1 s-1
1.0E10-4 1.0E10 -4 2.8E10 -6
1.0E10-4 3.0E10 -4 8.4E10 -6
2.0E10-4 3.0E10 -4 3.4E10 -5
1st When A is constant (1.0E -4),
B increases X3 and rate increases X3
OFB Chapter 14 143/27/2003
C B AExampleFor →+ initial Rate
[A] [B] mol L-1 s-1
1.0E10-4 1.0E10 -4 2.8E10 -6
1.0E10-4 3.0E10 -4 8.4E10 -6
2.0E10-4 3.0E10 -4 3.4E10 -5
2nd When B is constant (3.0E -4),
A increases X2 and rate increases X4
OFB Chapter 14 153/27/2003
Can now solve for k
1-2- 26
1424
6
12
12
s molL2.8x10
][1x10][1x102.8x10
[B][A]ratek
[B]k[A]rate
=
=
=
=
−−
−
OFB Chapter 14 163/27/2003
14-3 The Dependence of Concentrations on Time
→
][][ k A products
orderfirst for general, in
AktArate =−=∆∆
First Order Reactions
For now, just accept this
important formula
OFB Chapter 14 173/27/2003
If we take the ln of both sides
Recall y = mx + b or
y = b + mxA plot of ln [A] vs t will be a straight line with the
Intercept = ln[A]0
Slope = -k
OFB Chapter 14 183/27/2003
ktAA −= 0]ln[]ln[First Order Reaction
-6
-5
-4-3
-2
-1
00.0E+00 2.0E+04 4.0E+04 6.0E+04
Time (in seconds)
ln [A
} (in
mol
/ L
Slope = -kt
OFB Chapter 14 193/27/2003
First Order Reactions
A useful concept for 1st order reactions (only) is the half=life
i.e., the time for ½ [A]0
(see book for the derivation)
Applies only to First order reactions
TRY
Examples 14-5 and Exercise
OFB Chapter 14 203/27/2003
Second Order Reactions
This is also the equation for a straight line
Y = mx +b
From calculus
→
2
k
ordersecondfor General, In
][][21- rate
products A2
AktA
==∆∆
OFB Chapter 14 213/27/2003
OFB Chapter 14 223/27/2003
Second Order Reaction
0
50
100
150
200
250
0 500 1000
Time (in seconds)
1/[A
] (L
/mol
)oA
kt][
12[A]1
+=
Slope = 2k
Second Order Reactions
Half-life is NOT A useful concept for 2nd order reactions but,
t1/2 = 1/2k[A]0
OFB Chapter 14 233/27/2003
ktAA −= 0]ln[]ln[
First Order Reactions
Second Order Reactions
oAkt
][12
[A]1
+=
OFB Chapter 14 243/27/2003
In the real world, if we don’t know the order of the reaction
[A]n plot both
First Order Reaction
-6-5-4-3-2-10
0.0E+00 2.0E+04 4.0E+04 6.0E+04
Time (in seconds)
ln [A
} (in
mol
/ L
Second Order Reaction
050
100150200250
0 500 1000
Time (in seconds)
1/[A
] (L
/mol
)
OFB Chapter 14 253/27/2003
Chapter 14-4Reaction Mechanisms
• Most reactions proceed not thru a single step but through a series of steps
• Each Step is called an elementary reaction
• Types of elementary Reactions1. Unimolecular (a single reactant)
E.g., A → B + C (a decomposition)
2. Bimolecular (most common type)E.g., A + B → products
3. Termolecular (less likely event)E.g., A + B + C → products
OFB Chapter 14 263/27/2003
Chapter 14Chemical Kinetics
• Reaction Mechanism– is a detailed series of elementary
steps and rates which are combined to yield the overall reaction
– One goal of chemical kinetics is to use the observed rate to chose between several possible reaction mechanisms.
OFB Chapter 14 273/27/2003
(fast) CO NO CO NO 2.)(slow) NONONO NO 1.)
223
322
+→+
+→+
Notice that NO3 is formed and consumed. This is called a reactive intermediate
Notice also that
Step 1 is bimolecular
Step 2 is bimolecular
Reactive intermediate is NO3
OFB Chapter 14 283/27/2003
A direct connection exists between the equilibrium constant of a reaction that takes place in a sequence of steps and the rate constants in each step.
Suppose the reaction proceeds by the following mechanism
][][]][[
]][[][]][][[
22
22
2
2
1
1
2
2
1
121
BADC
BAADCAK
kk
kk
kk
kkKKK
==
=
==
−−−−
The book uses a general reaction to illustrate this principle.
2 A (g) + B (g) ↔ C (g) + D (g)
1.) A (g) + A (g) ↔ A2 (g) rate = k1[A]2
2.) A2 (g) + B (g) ↔ C (g) + D (g) rate = k2[A2][B]
k1
k2
k-1
k-2
rate = k-1[A2]
rate = k-2[C][D]
a.) forward rate = reverse rateb.) Keq = kf / kr
OFB Chapter 14 293/27/2003
14-4 Kinetics and Chemical Equilibrium
E.g., a 3 step mechanism (not in book) similar to the previous general reaction
→← 22
1-k
1 k
ON NO NO 1.) +
→←
OH ON H ON 2.) 222-k
2 k
222 ++
→←
OH N H ON 3.) 223-k
3 k
22 ++
→←
O2H N 2H 2NO 222overall ++
OFB Chapter 14 303/27/2003
→←
O2H N 2H 2NO 222overall ++
From earlier this semester,
at equilibrium
Notice that N2O and N2O2 are reactive intermediates and not in the Equilibrium Constant (K)
Now, let’s derive the same Keq from its elementary steps.
OFB Chapter 14 313/27/2003
A.) at equilibrium
Forward rate = reverse rate
(called “The Principle of Detailed Balance”)
B.) Keq = kf / kr
→← 22
1-k
1 k
ON NO NO 1.) +
]O[Nk[NO]k 2212
1 −=
222
1
1eq1 ][
][KNO
ONkk
==−
OFB Chapter 14 323/27/2003
→←
OH ON H ON 2.) 222-k
2 k
222 ++
O]O][H[Nk]][HO[Nk
222
2222
−=
]][HO[NO]O][H[N
kkK
222
22
2
22 ==
−
eq
A.) at equilibrium
Forward rate = reverse rate
B.) Keq = kf / kr
OFB Chapter 14 333/27/2003
→←
OH N H ON 3.) 223-k
3 k
22 ++
O]][H[Nk]O][H[Nk
223
223
−=
]O][H[NO]][H[N
kkK
22
22
3
3eq3 ==
−
A.) at equilibrium
Forward rate = reverse rate
B.) Keq = kf / kr
OFB Chapter 14 343/27/2003
We added the elementary steps to get the Overall Reaction
O2H N 2H 2NO 222overall +←→
+
=
]O][HNO]][H[N
]][HO[NO]O][H[N
[NO]]O[N
22
22
222
22222
22
22
22
eq ][H[NO]][NO][HK =
OFB Chapter 14 353/27/2003
22
22
22
eq ][H[NO]][NO][HK =
Proven. This is the same K as that written directly from the overall reaction
Notice that the two reactive intermediates N2O2 and N2O cancelled out
OFB Chapter 14 363/27/2003
14-5 Reaction Mechanism & Rate Laws
Typically with a reaction one of several elementary step reaction is the slowest step. This is called the Rate Determining Step (RDS)
Case #1: When the RDS occurs first, the first step is slow and determines the rate of the overall reaction.
→
→
]][F[NOk rateRDS the is 1 Step
(fast) FNO F NO 2.)
(slow) FFNO F NO 1.)
221
2k2
2
2k1
22
=
+
++
OFB Chapter 14 373/27/2003
Reaction Progress
Energy
→
→
]][Fk1[NO rateRDS the is 1 Step
(fast) FNO F NO 2.)
(slow) FFNO F NO 1.)
22
2k2
2
2k1
22
=
+
++
NO2F
F + NO2
NO2+ F2
OFB Chapter 14 383/27/2003
Case #2: When the RDS occurs after one or more Fast steps, mechanisms are often signaled by a reaction order greater than two. The slow step determines the overall rate of the reaction.
But N2O2 is a reactive intermediate
→←
→
]][OO[Nk rate
(slow) 2NO O ON 2.)
(fast) ON NO NO 1.)
2222
2k2
222
221-k
1k
=
+
+
22 2NO O 2NOreactionOverall→+
3 molecule reaction. Is it A Termolecular or Bimolecular reactions? Three way collisions are rare. Try a two step mechanism.
OFB Chapter 14 393/27/2003
Reaction Progress
Energy
2NO + O2
2NO2
→←
→
]][OO[Nk rate
(slow) 2NO O ON 2.)
(fast) ON NO NO 1.)
2222
2k2
222
221-k
1k
=
+
+
N2O2 + O2
OFB Chapter 14 403/27/2003
→←
→
]][OO[Nk rate
(slow) 2NO O ON 2.)
(fast) ON NO NO 1.)
2222
2k2
222
221-k
1k
=
+
+
222
1-
11
221-2
1
[NO]]O[N
kk K
]O[Nk [NO]k
mequilibriuat
==
=
OFB Chapter 14 413/27/2003
2122 [NO]K ]O[N =∴
→←
→
]][OO[Nk rate
(slow) 2NO O ON 2.)
(fast) ON NO NO 1.)
2222
2k2
222
221-k
1k
=
+
+
Substituting for [N2O2] in the rate expression above
OFB Chapter 14 423/27/2003
Reaction Progress
Energy N2O2 + O2
slow
fast
2NO + O2
2NO2
→←
→
]][OO[Nk rate
(slow) 2NO O ON 2.)
(fast) ON NO NO 1.)
2222
2k2
222
221-k
1k
=
+
+
][O[NO]Kk rate 22
12=
OFB Chapter 14 433/27/2003
→←
→←
→ (slow) OI OH HOI OH 3.)
(fast) Cl HOI HOCl I 2.)
(fast)OH HOCl OH OCl 1.)
1-2
k31-
1-
2-k
k21-
1-
1-k
k1
21-
++
++
++
Exercise 14-8 Suppose the mechanism for the reaction given in the Example 14-8 were
What rate law is now predicted?
OFB Chapter 14 443/27/2003
→←
→←
→ (slow) OI OH HOI OH 3.)
(fast) Cl HOI HOCl I 2.)
(fast)OH HOCl OH OCl 1.)
1-2
k31-
1-
2-k
k21-
1-
1-k
k1
21-
++
++
++
Notice that [OH-1] and [HOI] are both reactive intermediates. Need to express them in other species
1-1-1-1 OIClOCl I
solutionaqinreaction Overall
+→+−
OFB Chapter 14 453/27/2003
O]][H[OCl] [HOCl][OH K
21-
1-
1 = [H2O]=1
][HOCl][I][HOI][Cl K 1-
1-
2 =
→←
→←
→ (slow) OI OH HOI OH 3.)
(fast) Cl HOI HOCl I 2.)
(fast)OH HOCl OH OCl 1.)
1-2
k31-
1-
2-k
k21-
1-
1-k
k1
21-
++
++
++
][HOI][OHkrate 13
−=
OFB Chapter 14 463/27/2003
[HOCl]] [OClK ][OH
1-11- =
] [Cl][HOCl] [IK [HOI] 1-
1-2=
][HOI][OHkrate 13
−=Can now substitute [OH-1] and [HOI] in the original rate expression.
=
] [Cl][HOCl] [IK
[HOCl]] [OClKkrate 1-
1-
2
1-
13
OFB Chapter 14 473/27/2003
Reaction Progress
Energy
slowfast
fast
intermediates
→←
→←
→ (slow) OI OH HOI OH 3.)
(fast) Cl HOI HOCl I 2.)
(fast)OH HOCl OH OCl 1.)
1-2
k31-
1-
2-k
k21-
1-
1-k
k1
21-
++
++
++
1-1-1-1 OI Cl OCl Isolution aqin reaction Overall
+→+−
] [Cl] ][I [OClKKkrate 1-
-1-1
213=
[HOI]
[OH-1]
-11 OCl I +−
-1-1 OI Cl +
OFB Chapter 14 483/27/2003
Chain ReactionsA reaction that proceeds through a series of
elementary steps, in polymerization reactions these can be repeated several or many times.
1. Initiation, generation of a reactive intermediate.
• Free Radical
2. Propagation
• A reactive intermediate reacts with starting material or others to generate a new reactive intermediate
• Chain growth in polymers
3. Termination
• Reactive intermediate is “quenched in some manner to give a stable product.
OFB Chapter 14 493/27/2003
→
→ •+•
•
M M R
radical) free ( R Initiator
Initiation
ki
kd
•→+•
•→+•
•→+•
+1)(X kp
x
3 kp
2
2 kp
1
M M M
M M M
M M M
nPropagatio
Coupling M M MnTerminatio
y)(xyx +→•+•
OFB Chapter 14 503/27/2003
• Fiber Optics– UV curable polymers
• Fiber coating• Coloring• Ribbon materials
• Many Polymers– Free-radical polymerization
OFB Chapter 14 513/27/2003
14-6 The Effect of Temperature on Reaction Rates
k of unitsthehasandconstant a isfactor" lexponentia-pre" the is A
andmoleper energy are unitsenergy n Activatiothe is E
where
a
OFB Chapter 14 523/27/2003
Ak RTEa
EquationArrenhius
−=
An Arrenhius Plot
16
16.5
17
17.5
18
18.5
0.0025 0.0027 0.0029 0.0031 0.0033 0.0035
1/T (K-1)
ln k
y = mx + b
Slope = -Ea/R
OFB Chapter 14 533/27/2003
Transition State
OFB Chapter 14 543/27/2003
Reaction Progress
Energy
Ear
Eaf
Transition State
OFB Chapter 14 553/27/2003
OFB Chapter 14 563/27/2003
Reaction Progress
Energy
slowfast
fast
OFB Chapter 14 573/27/2003
Reaction Progress
Energy
slow
fast fast
OFB Chapter 14 583/27/2003
Chapter 14Chemical Kinetics
• Catalyst– provides a lower energy path, but it
does not alter the energy of the starting material and product
– rather it changes the energy of the transition (s), in the reaction
– A catalyst has no effect on the thermodynamics of the overall reaction
• Inhibitor– is a negative catalyst. It slows the rate
of a reaction frequently by barring access to path of low Ea and thereby forcing the reaction to process by a path of higher Ea.
OFB Chapter 14 593/27/2003
OFB Chapter 14 613/27/2003
OFB Chapter 14 623/27/2003
Kinetics of Catalysis• A catalyst has no effect on the
thermodynamics of the overall reaction
• It only provides a lower energy path• Examples
– Pt and Pd are typical catalysts for hydrogenation reactions (e.g., ethylene to ethane conversion)
– Enzymes act as catalysts• Phases
– Homogenous catalysis – the reactants and catalyst are in the same catalyst (gas or liquid phase)
– Heterogeneous catalysis – reaction occurs at the boundary of two different phases (a gas or liquid at the surface of a solid)
OFB Chapter 14 633/27/2003
Chapter 14Chemical Kinetics
• Example / exercise14-1, 14-2, 14-3, 14-4, 14-5, 14-6, 14-7, 14-8, 14-9
• Problems7, 9, 11, 15, 19, 21, 23, 25, 37, 41, 43, 45, 51, 53