Transcript Chemical Kinetics - Dr. Parvin Carter
Chemical Kinetics
Chapter 12
Reaction Rates 01
•
Reaction Rate:
The change in the concentration of a reactant or a product with time
(M/s).
• change in concentration divided by the change in time Reactant A B Products Rate [
t A
] 2 HI (
g
) Rate H 2 (
g
) + I Chapter 12 2 (
g
) [
B
t
] Slide 2
Reaction Rates and Stoichiometry
• What is the general rate of the following reaction ?
2 HI (
g
) H 2 (
g
) + I 2 (
g
) Rate = − 1 2 [HI]
t
= [I 2 ]
t
Chapter 12 Slide 3
Reaction Rates and Stoichiometry
• To generalize, for the reaction
a
A +
b
B
c
C +
d
D Rate* = − 1
a
[A]
t
= − 1
b
[B]
t
= 1
c
[C]
t
= 1
d
[D]
t
*: General rate of reaction
Chapter 12 Slide 4
HCO H(aq) + Br (aq) 2 2 + 2H (aq) + 2Br (aq) + CO (g) 2 red colorless Which of the expressions below, corresponding to the reaction of bromine with formic acid, is incorrect?
1. d[CO ] 2 = dt d[Br ] 2 dt 2. dt = + d[H ] dt 3. d[HCO H] 2 = dt d[Br ] 2 dt 4. dt 5. d[Br ] = dt + 1 d[H ] = 2 dt d[Br ] dt 2
HCO H(aq) + Br (aq) 2 2 + 2H (aq) + 2Br (aq) + CO (g) 2 red colorless Which of the expressions below, corresponding to the reaction of bromine with formic acid, is incorrect?
1. d[CO ] 2 = dt d[Br ] 2 dt 2. dt = + d[H ] dt 3. d[HCO H] 2 = dt d[Br ] 2 dt 4 . d [ H C O H ] 2 d t 5. d[Br ] = dt = 2 + 1 d [ H ] d t d[Br ] 2 dt
How Do we study Rate of a reaction?
• Consider the decomposition of N 2 O 5 and O 2 : 2 N 2 O 5
(g)
4 NO 2
(g)
Brown to give NO 2 + O 2
(g)
Colorless Chapter 12 Slide 7
• 2 N 2 O 5
(g)
4 NO 2
(g)
+ O 2
(g)
Reaction Rates:
concentration versus time curve
03
Average Rate = Rate between two points in time
The slope of each triangle Between two points
Chapter 12 Slide 8
Reaction Rates
2N 2 O 5 (
g
) 4NO 2 (
g
) + O 2 (
g
)
Instantaneous rate:
Rate for specific instance in time
Slope of the tangent to a concentration versus time curve Initial Rate Chapter 12 Slide 10
Br 2 (
aq
) + HCOOH (
aq
) 2Br (
aq
) + 2H + (
aq
) + CO 2 (
g
)
time
393 nm light Detector [Br 2 ] a Absorption Chapter 12 Br 2 (
aq
) Slide 11
Br 2 (
aq
) + HCOOH (
aq
) 2Br (
aq
) + 2H + (
aq
) + CO 2 (
g
) slope of tangent slope of tangent slope of tangent
average rate
= [Br 2 ]
t
= [Br 2 ] final
t
final – [Br -
t
2 initial ] initial
instantaneous rate
= rate for specific instance in time
The slope of a line tangent to the curve at any point is the instantaneous rate at that time
Chapter 12 Slide 12
slope of tangent slope of tangent Chapter 12 Slide 13
rate a [Br 2 ] rate =
k
[Br 2 ]
k
rate = [Br 2 ] =
rate constant
The Rate Law; rate = 3.50 x 10 -3 s -1 [Br 2 ] Chapter 12 Slide 14
k = 3.50 x 10 -3 s -1 Chapter 12 Slide 13
The Rate Law and Reaction Order The
rate law
expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.
a
A +
b
B
c
C +
d
D Rate =
k
[A]
x
[B]
y
reaction is
xth order
in A reaction is
yth order
in B reaction is
(x +y)th order overall
Chapter 12 Slide 16
The Rate Law and Reaction Order are Experimentally Determined
Chapter 12 Slide 17
Determine the reaction order for:
F 2 (
g
) + 2ClO 2 (
g
) 2FClO 2 (
g
)
---
rate =
k
[F 2 ]
x
[ClO 2 ]
y
1 vs 3
Double [F 2 ] with [ClO 2 ] constant Rate doubles
x
= 1
1 vs 2
Quadruple [ClO 2 ] with [F 2 ] constant rate =
k
[F 2 ][ClO 2 ] Rate quadruples
y
= 1
The instantaneous rate at the beginning of a reaction is called initial rate
Chapter 12 Slide 18
Rate Laws
• Rate laws are
always
determined experimentally.
• Reaction order is
always
defined in terms of reactant (not product) concentrations.
• The order of a reactant
is not
related to the stoichiometric coefficient of the reactant in the balanced chemical equation.
F 2 (
g
) + 2ClO 2 (
g
) 2FClO 2 (
g
) rate =
k
[F 2 ][ClO 2 ] 1 Chapter 12 Slide 19
Experiment 1 2 3 Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 8 2 (
aq
) + 3I (
aq
) 2SO 4 2 (
aq
) + I 3 (
aq
) [S 2 O 8 2 ] 0.08
0.08
0.16
[I ] 0.034
0.017
0.017
Initial Rate (
M
/s) 2.2 x 10 -4 1.1 x 10 -4 2.2 x 10 -4 rate =
k y = 1 x = 1
rate =
k
[S 2 O 8 2 ]
x
[I ]
y
[S 2 O 8 2 ][I ] Double [I ], rate doubles (experiment 1 & 2) Double [S 2 O 8 2 ], rate doubles (experiment 2 & 3)
k
rate = [S 2 O 8 2 ][I ] 2.2 x 10 -4
M
/s = (0.08
M
)(0.034
M
) = 0.08/
M
• s Chapter 12 Slide 20
Determine the Rate Law and Reaction Order
NH 4 + (
aq
) + NO 2 − (
aq
) N 2 (
g
) + 2 H 2 O (
l
) Comparing Experiments 1 and 2, when [NH 4 + ] doubles, the initial rate doubles.
Chapter 12 Slide 21
NH 4 + (
aq
) + NO 2 − (
aq
) N 2 (
g
) + 2 H 2 O (
l
) Likewise, comparing Experiments 5 and 6, when [NO 2 doubles, the initial rate doubles.
− ] Chapter 12 Slide 22
• • This means Rate Rate Rate [NH 4 + ] [NO 2 − ] [NH + ] [NO 2 − ] Rate =
k
or [NH 4 + ] [NO 2 − ] This equation is called the rate law , and
k
the rate constant .
is Chapter 12 Slide 23
Rate Laws
• The exponents tell the order with respect to each reactant.
of the reaction • This reaction is First-order in [NH 4 + ] First-order in [NO 2 − ] Chapter 12 Slide 24
Rate Law & Reaction Order
• The reaction of nitric oxide with hydrogen at 1280 °C is: 2 NO (g) + 2 H 2
(g)
N 2
(g)
+ 2 H 2 O
(g)
• From the following data determine the rate law and rate constant.
Experiment 1 2 3 [NO] 5.0 x 10 –3 10.0 x 10 –3 10.0 x 10 –3 [H 2 ] 2.0 x10 –3 2.0 x 10 –3 4.0 x 10 –3 Initi al Rate (M/s) 1.3 x 10 –5 5.0 x 10 –5 10.0 x 10 –5 Second order in NO, First order in H 2 k = 1/3(250+250+260) = 250 M Chapter 12 -2 .s
-1 Slide 25
First-Order Reactions
Concentration and Time Equation
01
•
First Order:
Reaction rate depends on the reactant concentration raised to first power.
Rate
= k
[A] A product
Rate = -
A t
[A]
t
=
k
[A] Chapter 12 Slide 26
[A] [A] ln [A] 0 [A] Concentration and Time Equation For A First-Order Reactions = K
Δt
-(ln[A] -ln[A] 0 ) =
kt
ln[A] = ln[A] 0 -
kt
See next slide for proof of the formula
= k t
[A] = [A] 0 exp(
-kt
)
[A] is the concentration of A at any time t
k
= rate [A] =
M
/
s M
= 1/s or s -1
[A] 0 is the concentration of A at time t = 0
[A] = [A] 0 exp(
-kt
) [A] t [A] 0 e kt exponentia l decay law Chapter 12 Slide 27
Integration: Chapter 12 Slide 28
ln [A] 0
= k t
[A] ln[A] = ln[A] 0 -
kt
First-Order Reactions ln[A] = ln[A] 0 -
kt
[A] is the concentration of A at any time
t
[A] 0 is the concentration of A at time
t
= 0 Chapter 12 Slide 29
The reaction 2A B is first order in A with a rate constant of 2.8 x 10 -2 s -1 at 80 0 C. How long will it take for A to decrease from 0.88
M
to 0.14
M
?
t
= ln [A] 0 [A] ln [A] 0 [A]
k
= k t
[A] 0 = 0.88
M
[A] = 0.14
M
t = ?
= 0.88
M
ln 0.14
M 2.8 x 10 -2 s -1
= 66 s Chapter 12 Slide 30
What is Half- Life ?
The
half-life, t ½ ,
is the time required for the concentration of a reactant to decrease to half of its initial concentration.
t ½
=
t
when [A] = [A] 0 /2 ln [A] 0 [A]
= k t
t
½ Half Life For the First Order Reaction [A] 0 ln = [A] 0 /2
k
= ln2
k
= 0.693
k
Chapter 12 Slide 31
Reaction Orders
Zeroth Order Reaction: Rate = K [A] 0 = K
Units of Rate Constants vs Reaction Orders
Chapter 12 Slide 32
What is the order of decomposition of N 2 O 5 if it decomposes with a rate constant of 5.7 x 10 -4 s -1 ?
What is the half life of decomposition of N 2 O 5 ?
2N 2 O 5 (
g
)
4NO 2 (
g
) + O 2 (
g
)
units of
k
(s -1 ) Therefore, decomposition is first order?
t
½ = ln2
k
= 0.693
5.7 x 10 -4 s -1 = 1200 s = 20 minutes Chapter 12 Slide 33
Half life of a First Order Reaction
Chapter 12 Slide 34
Chapter 12 First-order reaction A product # of half-lives 1 2 3 4 [A] ½ [A] 0 1/4 [A] 0 1/8 [A] 0 1/16 [A] 0 [A] = [A] 0 x (1/2)
n
Slide 35
First-Order Reaction
2N 2 O 5 (
g
)
4NO 2 (
g
) + O 2 (
g
)
• Show that the decomposition of N 2 O 5 is first order and calculate the rate constant and Half life.
Chapter 12
k = 1.7 x 10 -3 s -1
t 1/2 = 408 S Slide 36
Second-Order Reactions A product rate = [A]
t
rate =
k
[A] 2 What is Unit of k ?
k
= rate [A] 2 =
M
/
s M 2
= 1/
M
•
s or M -1 s -1
What is Conc. Vs time equation?
[A]
t
=
k
[A] 2 1 [A] = 1 [A] 0 +
kt
[A] is the concentration of A at any time
t
[A] 0 is the concentration of A at time
t
= 0 Chapter 12 Slide 37
Second-Order Reactions
So if a process is second-order in A, a plot of 1/[A] vs.
t
will yield a straight line, and the slope of that line is
k
.
Drive the formula for half life of a second order reaction
t
=
t
1/2
[A]
t
1/2
= [A]
0
2 Chapter 12 Slide 38
Half-life for a second-order reaction 1 [A]
t
=
kt
+ 1 [A]
0
2 [A]
0
=
kt
1/2
+ 1 [A]
0
Chapter 12
t
=
t
1/2
[A]
t
1/2
= [A]
0
2 1
t
1/2
=
k
[A]
0
Slide 39
Second-Order Reactions
1
t
1/2
=
k
[A]
0
For a second-order reaction, the half-life is dependent on the initial concentration.
Each successive half-life is twice as long as the preceding one.
Chapter 12 Slide 40
Example: a) Is the following reaction first or second order ?
b) What is the value of k?
2 NO 2 (g)
2NO
(g)
+ O 2
(g)
Chapter 12 Slide 42
Chapter 12 Slide 43
k = 0.54 M -1 . S -1
Second-Order Reactions Chapter 12 Slide 44
Zero Order Reaction: Rate = k
Example of Zeroth Order Reaction: Decomposition of N 2 O on hot platinum surface: N 2 O → N 2 Rate [N d[N 2 2 O] 0 + 1/2 O = k[N O]/dt = k 2 2 O] 0 = k Chapter 12 Slide 45
Reaction Mechanisms
• A
reaction mechanism
is a sequence of molecular events, or reaction steps, that defines the pathway from reactants to products.
Chapter 12
01
Slide 46
Reaction Mechanisms 02
• Single steps in a mechanism are called elementary steps (reactions).
• An elementary step describes the behavior of individual molecules.
• An overall reaction describes the reaction stoichiometry.
Chapter 12 Slide 47
Reaction Mechanisms NO
2
(
g
) + CO(
g
)
NO(
g
) + CO
2
(
g
)
• NO 2 (
g
) + NO 2 (
g
) NO(
g
) + NO 3 (
g
) • NO 3 (
g
) + CO(
g
) NO 2 (
g
) + CO 2 (
g
) • NO 2 (
g
) + CO(
g
) NO(
g
) + CO 2 (
g
) Elementary Elementary Overall • The chemical equation for an elementary reaction is a description of an individual molecular event that involves the breaking and/or making of chemical bonds.
NO 3 (
g
) is called reaction intermediate.
Chapter 12 Slide 48
Reaction Mechanisms
•
Molecularity:
is the number of molecules (or atoms) on the reactant side of the chemical equation.
•
Unimolecular:
Single reactant molecule.
04
Chapter 12 Slide 49
Reaction Mechanisms
•
Bimolecular:
Two reactant molecules.
05
•
Termolecular:
Three reactant molecules.
Chapter 12 Slide 50
Reaction Mechanisms
• Determine individual steps , the reaction intermediates, and the molecularity of each individual step.
2N 2 O 2N 2 + O 2
06
Chapter 12 Slide 51
Rate Laws and Reaction Mechanisms
• Rate law for an
overall reaction
must be determined experimentally.
• Rate law for
elementary step
follows from its molecularity.
01
Chapter 12 Slide 52
Rate Laws and Reaction Mechanisms
• The
rate law
of each
elementary step
follows its molecularity.
• The
overall reaction
is a sequence of elementary steps called the
reaction mechanism
.
02
Chapter 12 Slide 53
Rate-Determining Step
• The
slowest elementary step
in a multistep reaction is called the
rate-determining step
.
• The overall reaction cannot occur faster than the speed of the rate-determining step.
• The
rate of the overall reaction
is therefore
determined
by the rate of the
rate-determining step
.
Chapter 12 Slide 54
The experimental rate law for the reaction between NO 2 and CO to produce NO and CO 2 is rate =
k
[NO 2 ] 2 . The reaction is believed to occur via two steps: Step 1: Step 2: NO 2 + NO 2 NO + NO 3 NO 3 + CO NO 2 + CO 2 What is the equation for the overall reaction?
NO 2 + CO NO + CO 2 What is the intermediate?
NO 3 What can you say about the relative rates of steps 1 and 2?
rate =
k
[NO 2 ] 2 is the rate law for step 1 so step 1 must be slower than step 2 Chapter 12 Slide 55
Determining Reaction Mechanism From The Rate Law
Activation Energy (
E
a):
Chapter 12 Slide 57
The Arrhenius Equation
Typically, as the temperature increases, the rate of reaction increases.
2N 2 O 5 (
g
) 4NO 2 (
g
) + O 2 (
g
) rate =
k
[N 2 O 5 ] Where is temperature dependence?
Is it hidden in k?
Chapter 12 Slide 58
The Arrhenius Equation 01
•
Collision Theory:
A bimolecular reaction occurs when two correctly oriented molecules collide with sufficient energy.
•
Activation Energy (E a ):
The potential energy barrier that must be surmounted before reactants can be converted to products.
Chapter 12 Slide 59
Temperature dependence of Rate Constat
•
Adding more of the reactants speeds up a reaction by increasing the number of collisions that occur.
•
Collision rate = Z [A][B]
Z is a constant related to collision frequency
The fraction of collisions with an energy equal or more than activation energy( E a ) : f = e -Ea/RT
•
Raising the temperature speeds up a reaction by providing the energy of activation to more colliding molecules.
Chapter 12 Slide 60
The Arrhenius Equation
• Only the fraction of collisions having proper orientation can result to products.
02
This is called steric factor, p., In the above example p = 0.5
Chapter 12 Slide 61
The Arrhenius Equation
Collision rate = Z [A][B] Where Z is a constant, related to the collision frequency . Reaction rate = p.
f
.Z [A][B] Reaction rate = k [A][B] k = p.
f
.Z, Assume p.Z = A , frequency factor A = frequency factor k = A.
f
,
f = e -Ea/RT
k = Ae
-Ea/RT
( p.Z )= A
Chapter 12 Slide 62
The Arrhenius Equation
K = Ae -Ea/RT A = pZ (steric factor) Chapter 12 Slide 63
Calculating Activation Energy
k = A .e
-Ea/RT k = A .e
(-Ea/R)(1/T)
E
a
= -R . (slope)
Chapter 12 Ln k 1/T Slide 64
Find the activation energy for the following reaction 2HI(
g
) + H 2 (
g
)
I 2 (
g
) + H 2 (
g
)
Chapter 12 Slide 65
Calculating Activation Energy
Slope = -2.24 x 10 4 K E a = -R . (slope) E a = - (8.314 j/K.mol) (-2.24 x 10 4 K) E a = 190 kj/mol
Chapter 12 Slide 67
Effect of Temperature on Fraction of Collisions with Activation energy
f = e
-Ea/RT
Chapter 12 Slide 68
Effect of Temperature
f = e
-Ea/RT Collision Theory
: As the average kinetic energy increases, the average molecular speed increases, and thus the collision rate increases.
Change of Rate Constant with temperature
• If the
E a
is known , we can calculate the Rate Constant when temperature is changed: ln
k
2
k
1
E a R
1
T
2 1
T
1 The above formula could be used to determine the rate constant at a different temperature.
Chapter 12 Slide 70
Homework: Determination the Activation Energy
The second-order rate constant for the decomposition of nitrous oxide (N 2 O) into nitrogen molecule and oxygen atom has been measured at different temperatures: k (M -1 s -1 ) t ( °C) Determine (graphically) the activation energy 1.87x10
-3 600 for the reaction.
0.0113
650 0.0569
700 Ea = 241 KJ/mole 0.244
750 Chapter 12 Slide 71
Catalysis 01
Chapter 12 Slide 72
Catalysis 01
• A catalyst is a substance that increases the rate of a reaction without being consumed in the reaction.
Chapter 12 Slide 73
Catalysis
Note that the presence of a catalyst does not affect the energy difference between the reactants and the products
Catalysis 02
• The relative rates of the reaction A + B AB in vessels a –d are 1:2:1:2. Red = A, blue = B, green = third substance C.
(a) What is the order of reaction in A, B, and C?
(b) Write the rate law.
(c) Write a mechanism that agrees with the rate law.
(d) Why doesn’t C appear in the overall reaction?
1 2 Chapter 12 1 2 Slide 75
Catalysis
Catalyst
: A substance that increases the rate of a reaction without itself being consumed in the reaction. A catalyst is used in one step and regenerated in a later step.
H 2 O 2 (
aq
) + I 1 (
aq
) H 2 O 2 (
aq
) + I O 1 (
aq
) 2H 2 O 2 (
aq
) H 2 O(
l
) + I O 1 (
aq
) rate-determining step H 2 O(
l
) + O 2 (
g
) + I 1 (
aq
) fast step 2H 2 O(
l
) + O 2 (
g
) overall reaction Chapter 12 Slide 76
Catalysis 03
•
Homogeneous Catalyst:
Exists in the same phase as the reactants. •
Heterogeneous Catalyst:
to the reactants.
Exists in different phase Chapter 12 Slide 77
Catalysis
•
Catalytic Hydrogenation :
04
Chapter 12 Slide 78
Mechanism of Catalytic Hydrogenation:
B
H H
C A C Y
H H
X Chapter 12 Slide 79
Mechanism of Catalytic Hydrogenation:
B C A
H H
C Y X
H H
Chapter 12 Slide 80
Mechanism of Catalytic Hydrogenation:
H H
A B C C Y X
H H
Chapter 12 Slide 81
Mechanism of Catalytic Hydrogenation:
H H
A B C C Y X
H H
Chapter 12 Slide 82
Mechanism of Catalytic Hydrogenation:
H
A B C
H
C Y X
H H
Chapter 12 Slide 83
Mechanism of Catalytic Hydrogenation:
B A C C Y
H H H
X
H
Chapter 12 Slide 84