Transcript Document

Chemical Kinetics
Chapter 13
Dr. Ali Bumajdad
Chapter 13 Topics
Chemical Kinetics
• Rate of a Reaction
• Reaction Rates and Stoichiometry
• The Rate Law
• Relationship between Reactant Concentration and Time
1) First-order reactions
2) Second-ordered reactions
3) Zero-ordered reactions
• The Effect of Activation Energy and Temperature on Rate
• Collision Theory
• The Arhenius Equation
• Finding Ea from graph and from equation
• Catalysis
Dr. Ali Bumajdad
Chemical Kinetics = Chemistry and Time
•Thermodynamics – does a reaction take place?
•Kinetics – how fast does a reaction proceed?
- Why it is important:
To answer questions like:
1) How quickly a medicine is able to work?
2) How fast the depletion of Ozone?
3) How rapidly food spoils?
4) How fast steel rust?
Rate of a reaction
•Reaction rate is the change in the concentration of a reactant or
a product with time (M/s).
A
B
D[A]
rate = Dt
D[A] = change in concentration of A over
time period Dt
D[B]
rate =
Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is negative.
A
B
time
D[A]
rate = Dt
D[B]
rate =
Dt
•Kinetics can be studied by following a change in property as a
function of time
This property could be:
1) UV absorption
2) Pressure
3) Conductivity
4) Concentration
5) Gas volume
2Br- (aq) + 2H+ (aq) + CO2 (g)
Br2 (aq) + HCOOH (aq)
time
393 nm
light
Detector
D[Br2] a DAbsorption
Br2 (aq) + HCOOH (aq)
2Br- (aq) + 2H+ (aq) + CO2 (g)
slope of
tangent
slope of
tangent
slope of
tangent
[Br2]final – [Br2]initial
D[Br2]
average rate = =Dt
tfinal - tinitial
instantaneous rate = rate for specific instance in time
(slope of a tangent)
rate a [Br2]
rate = k [Br2]
rate
= rate constant
k=
[Br2]
= 3.50 x 10-3 s-1
2H2O2 (aq)
2H2O (l) + O2 (g)
PV = nRT
n
P=
RT = [O2]RT
V
1
[O2] =
P
RT
D[O2]
1 DP
rate =
=
RT Dt
Dt
measure DP over time
Reaction Rates and Stoichiometry
2A
B
Two moles of A disappear and one mole of B that is formed.
•In this case, the rate will be different if we consider the
disappearance of A than if we consider the formation of B
•In order to make it the same we should divide by the coefficient
1 D[A]
rate = =
Dt
2
• In general for:
aA + bB
D[B]
Dt
cC + dD
1 D[A]
1 D[B]
1 D[C]
1 D[D]
rate = ==
=
a Dt
b Dt
c Dt
d Dt
(1)
Q) Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (g)
D[CH4]
D[CO2]
1 D[O2]
1 D[H2O]
rate = =
==
Dt
Dt
Dt
2 Dt
2
Sa. Ex.
a) How is the rate of disappearance of ozone related to the rate of
appearance of oxygen in the following equation:
2O3 (g)
3O2 (g)
b) If at certain time the rate of appearance of O2, D[O2]/Dt =
6.0  10-5 M/s what is the value of the rate of disappearance
of O3, - D[O3]/Dt at the same time.
4.0  10-5 M/s
The Rate Law
•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 that can be determined by experiment
only.
•Rate Law is different than the rate of reaction
1 D[A]
rate = a Dt
Rate = k [A]x[B]y
Reactant not product
aA + bB
cC + dD
Rate = k [A]x[B]y
(2)
reaction is xth order in A
reaction is yth order in B
reaction is (x +y)th order overall
F2 (g) + 2ClO2 (g)
2FClO2 (g)
rate = k [F2]x[ClO2]y
When Double [F2] with [ClO2] constant
Rate doubles
x=1
Quadruple [ClO2] with [F2] constant
y=1
 rate = k [F2][ClO2]
Rate quadruples
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.
F2 (g) + 2ClO2 (g)
2FClO2 (g)
rate = k [F2][ClO2] 1
Q) Determine the rate law and calculate the rate constant
for the following reaction from the following data:
S2O82- (aq) + 3I- (aq)
2SO42- (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
y=1
x=1
rate = k [S2O82-][I-]
Double [I-], rate doubles (experiment 1 & 2)
Double [S2O82-], rate doubles (experiment 2 & 3)
2.2 x 10-4 M/s
rate
k=
=
= 0.08/M•s
2[S2O8 ][I ] (0.08 M)(0.034 M)
Relationship between Reactant Concentration and Time
First-Order Reactions
Reactions with first overall order
• For reaction of the type A
product
D[A] and rate = k [A]
rate = Dt
D[A]
D[A]
and
 = k [A]
= k Dt
Dt
[A]
Using Calculus
ln[A] = ln[A]0 - kt
(3a)
[A] = [A]0exp(-kt) (3b)
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
Unit of first order rate constant
2N2O5
ln[A] = ln[A]0 - kt
Y
= b
(3a)
+ mX
Where m = slope
b = intercept
rate
M/s
= 1/s or s-1
=
k=
M
[A]
4NO2 (g) + O2 (g)
Q) The reaction 2A
B is first order in A with a rate
constant of 2.8 x 10-2 s-1 at 800C. How long will it take for
A to decrease from 0.88 M to 0.14 M ?
ln[A] = ln[A]0 - kt
(3a)
[A]0 = 0.88 M
[A] = 0.14 M
kt = ln[A]0 – ln[A]
ln[A]0 – ln[A]
=
t=
k
ln
[A]0
[A]
k
ln
=
0.88 M
0.14 M
2.8 x
10-2
s-1
= 66 s
First-Order Reactions
• half-life, t½, is the time required for the concentration of a
reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2
[A]0
ln
[A]0/2
ln2
0.693
t½ =
=
=
k
k
k
ln2
0.693
t1/2 =
=
k
k
(4)
Independent of
initial concentration
1) It take the same time for the concentration of reactant to
decrease from 0.1M to 0.05M as it does from 0.05M to 0.025M
2) If you know t1/2 you know k
First-order reaction
A
product
# of
half-lives [A] = [A]0/n
1
2
2
4
3
8
4
16
Q) What is the half-life of N2O5 if it decomposes with a rate
constant of 5.7 x 10-4 s-1?
0.693
t½ = ln2 =
= 1200 s = 20 minutes
-4
-1
k
5.7 x 10 s
How do you know decomposition is first order?
units of k (s-1)
Second-Order Reactions
Reactions with second overall order
• For reactions of the type A
product
D[A]
and rate = k [A]2
rate = Dt
rate
M/s
Unit of second order rate constantk = [A]2 = M2 = 1/M•s
D[A]
and = k [A]2
Dt
Using Calculus
1
1
=
+ kt
[A]
[A]0
(5) where
t½ = t when [A] = [A]0/2
1
t½ =
k[A]0
(6)
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
Zero-Order Reactions
A
product
D[A]
rate = Dt
D[A]
=k
Dt
rate
= M/s
k=
0
[A]
[A] = [A]0 - kt (7)
[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
[A]0
t½ =
2k
rate = k [A]0 = k
(8)
Summary of the Kinetics for reactions of type aA  products
order
Rate law
Zero
Rate = k
First
Rate = k[A]
Second
Rate = k[A]2
ln[A] = -kt +ln[A]o
1
1
 kt 
[A]
[A]o
Integrated rate law
[A] = -kt + [A]o
Plot needed to give
a straight line
[A] versus t
In[A] versus t
1
versus t
[A]
Relationship of k to
the slope of straight
line
Half-life
Slope = -k
Slope = -k
Slope = k
t 12 
[ A]o
2k
t 12 
0.693
k
t 12 
1
k [ A]o
Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order
0
Rate Law
rate = k
1
rate = k [A]
2
[A]2
rate = k
Concentration-Time
Equation
[A] = [A]0 - kt
ln[A] = ln[A]0 - kt
1
1
=
+ kt
[A]
[A]0
Half-Life k Unit
[A]0
t½ =
2k
t½ = ln2
k
1
t½ =
k[A]0
M s-1
s-1
M-1 s-1
(a)
ln[A] = ln[A]0 - kt
ln [A] = - kt
(b,c)
[A]0
(3a)
(3c)
• Plot of ln [A] versus t should be linear for first order reaction
• Since [A] a P (PV=nRT), then
lnP =-kt +ln P0
ln2
0.693
t1/2 =
=
k
k
(4)
1
1
=
+ kt
[A]
[A]0
t½ =
1
k[A]0
(6)
(5)
The Effect of Activation Energy (Ea) and Temperature on Rate
- is the minimum amount of energy
required to initiate a chemical reaction
- Rate increase by decreasing Ea
- Unit: J/mol
-Usually rate increase by increasing T
- Unit: K
Collision Theory
1) How reaction Occurs?
By collisions between molecules or atoms
2) Does all collisions results in a reactions?
No,
a) The collision should be strong enough (more than the
activation energy or the reaction)
b) The collision should be with the right orientation
3) What factor affect the speed (rate) of the reactions?
a) Collision frequency (A)
b) Temperature
c) Activation energy (Ea)
Orientation effect
A+B
Exothermic Reaction
+
AB+
C+D
Endothermic Reaction
The Arhenius Equation
k = A • exp( -Ea / RT )
(9)
(Arrhenius equation)
Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature (K)
ka T
A is the frequency and orientation factor
(10)
Ea 1
lnk = + lnA
R T
Y = m
k a 1/Ea
X + b
1) Finding Ea from graph
Ea 1
lnk = + lnA
R T
Y = m
X + b
2) Finding Ea from Equation
lnk1 =
lnk2 =
lnA
Ea 1
R T1
(1)
lnA
Ea 1
R T2
(2)
Subtracting 2 from 1
lnk1 - lnk2 = - Ea 1 +
R T1
Ea 1
lnk1 - lnk2 =
R T2
E
ln k1 = a 1 - 1
R T2
T1
k2
Ea 1
R T2
1
T1
(11)
Sa. Ex.
Ea 1
lnk = + lnA
R T
Y = m
X + b
(10)
E
ln k1 = a 1 - 1
R T2
T1
k2
(11)
Catalysis
•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
In heterogeneous catalysis, the reactants and the catalysts
are in different phases.
•
Haber synthesis of ammonia
•
Ostwald process for the production of nitric acid
•
Catalytic converters
In homogeneous catalysis, the reactants and the catalysts
are dispersed in a single phase, usually liquid.
•
Acid catalysis
•
Base catalysis
Haber Process
N2 (g) + 3H2 (g)
Fe/Al2O3/K2O
catalyst
2NH3 (g)
Ostwald Process
4NH3 (g) + 5O2 (g)
Pt catalyst
2NO (g) + O2 (g)
2NO2 (g) + H2O (l)
4NO (g) + 6H2O (g)
2NO2 (g)
HNO2 (aq) + HNO3 (aq)
Pt-Rh catalysts used
in Ostwald process
Hot Pt wire
over NH3 solution
Catalytic Converters
CO + Unburned Hydrocarbons + O2
2NO + 2NO2
catalytic
converter
catalytic
converter
CO2 + H2O
2N2 + 3O2
Enzyme Catalysis: very selective biological catalyst
uncatalyzed
D[P]
rate =
Dt
rate = k [ES]
enzyme
catalyzed