Chemical Kinetics • Reaction rate • Rate Law • Change of concentration with time • Rate and temperature • Reaction mechanism • Catalysis Kinetics – how fast does a reaction proceed?
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Chemical Kinetics
• Reaction rate
• Rate Law
• Change of concentration with time
• Rate and temperature
• Reaction mechanism
• Catalysis
Kinetics – how fast does a reaction proceed?
I. Reaction Rate
A. Rate is the change in the concentration of a reactant or a product with time (M/s).
- Average rate- Instantaneous rate- Initial rate
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.
A B
rate = –[A]
t
rate = [B]
t
time
Ex: 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 time
From 0.0 s - 400 s average rate = -(0.00296 M-0.0120 M)
(400.0 s - 0.0 s)=2.3 x 10-5 M/s
B. Rate and stoichiometry
2A B
Two moles of A disappear for each mole of B that is formed.
rate = [B]
trate = –
[A]t
12
aA + bB cC + dD
rate = –[A]
t1a
= –[B]
t1b
=[C]
t1c
=[D]
t1d
Q: Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
rate = –[CH4]t
= –[O2]t
12
=[H2O]t
12
=[CO2]t
Q: If the CH4(g) is burning at a rate of 0.44 mol/s, what is the rate of consumption of O2(g)?
0.44 mol/s = –[CH4]t
= –[O2]t
12
–[O2]t = 2 x 0.44 mol/s =0.88 mol/s
C. Rate involving gas
2H2O2 (aq) 2H2O (l) + O2 (g)
PV = nRT
P = RT = [O2]RTnV
[O2] = PRT1
rate = [O2]
t RT1 P
t=
measure P over time
D. Factors affecting reaction rate
• Physical states of reactants
- Homogeneous reaction
- Heterogeneous reaction
• Concentration
• Temperature
• Catalyst
II. 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
aA + bB cC + dD
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
k : rate constant
Rate data for the reaction between F2 and ClO2
F2 (g) + 2ClO2 (g) 2FClO2 (g)
rate = k [F2]x[ClO2]y
Double [F2] with [ClO2] constant
Rate doubles x = 1
Quadruple [ClO2] with [F2] constant
Rate quadruples y = 1
rate = k [F2][ClO2] = 1.2 M-1 s-1k = rate[F2][ClO2]
Overall reaction order : 2
Example:
F2 (g) + 2ClO2 (g) 2FClO2 (g)
rate = k [F2][ClO2]
II. Rate Laws
• Rate laws are always determined experimentally.
• The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.
1
Determine the rate law and calculate the rate constant for the following reaction from the following data: S2O8
2- (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
From experiment 1 & 2: double [I-], rate doubles
Rate1
Rate2
2.2x10-4M/s
1.1x10-4M/s= = 2 =
k(0.08M)x(0.034M)y
k(0.08M)x(0.017M)y= 2y
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
rate = k [S2O82-][I-]
Q:
III. Change in concentration with time (Integrated rate law)
€
rate = − Δ[A ]Δt = k[A]x
Order Rate LawIntegrated rate
law Plot
0
1
2
rate = k
rate = k [A]
rate = k [A]2
ln[A] = ln[A]0 - kt
11[A]
=[A]0
+ kt
[A] = [A]0 - kt [A] vs. t
ln[A] vs. t
1[A]
A Product
A. First-order reaction
€
−[A]Δt
= k[A]
A product
rate = -[A]
t
rate = k [A]
k = rate[A]
= 1/s or s-1M/sM
=
[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
Example: Decomposition of N2O5
1st order reaction : ln[A] = ln[A]0 - kt
kt = ln[A]0 – ln[A]
t =ln[A]0 – ln[A]
k= 74 s
(a) [A]0 = 0.72 M [A] = 0.14 M
ln[A]0
[A]
k=
ln 0.72 M
0.14 M
2.2 x 10-2 s-1=
Q: The reaction 2A –> B is first order in A with a rate constant of 2.2 x 10-2 s-1 at 80oC. (a) How long will it take for A to decrease from 0.72 M to 0.14 M. (b) What is the concentration of A after 20 s if the initial concentration of A is 0.42M?
(b) [A]0 = 0.42 M [A] = ?
t = ?
t = 20 s
ln[A] = ln[A]0 - kt = ln(0.42) - 2.2 x10-2 s-1 x 20 s =-1.3
[A] = exp(-1.3) = e-1.3 =0.27 M
Half-life, t1/2, for first-order reaction
The half-life, t1/2, is the time required for the concentration of a reactant to decrease to half of its initial concentration.
t1/2= t when [A] = [A]0/2
ln[A]0
[A]0/2
k=t1/2
ln2k
=0.693
k=
Q: What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?
t1/2ln2k
=0.693
5.7 x 10-4 s-1= = 1200 s = 20 minutes
How do you know decomposition is first order?
units of k (s-1)
(*indep of concentration)
A product
First-order reaction
# of half-lives [A] = n[A]0
1
2
3
4
1/2
1/4
1/8
1/16
B. Second-order reaction
€
−[A ]Δt = k[A]2
A product
rate = -[A]
t
rate = k [A]2
k = rate[A]2
M/sM2=
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
= 1/M•s
1[A]
=1
[A]0
+ kt
t1/2= t when [A] = [A]0/2
t1/2 =1
k[A]0
1[A]
t
C. Zeroth-order reaction
€
−[A ]Δt = k
A product
rate = -[A]
t
rate = k [A]0= k
k = rate[A]0
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
= M/s t1/2= t when [A] = [A]0/2
[A]
t
[A] = [A]0 - kt
t1/2 =[A]0
2k
Integrated rate law – summary
€
rate = − Δ[A ]Δt = k[A]x
Order Rate LawConcentration-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
t1/2ln2k
=
t/2 =[A]0
2k
t1/2 =1
k[A]0
Determine graphically the rate law for the reaction of
4 PH3(g) -> P4 (g) + 6 H2(g)
Time(s)[PH3]
(M/s)ln[PH3] 1/[PH3]
0
10
30
50
90
150
200
1.000
0.819
0.549
0.368
0.165
0.050
0.018
0.0
-0.2
-0.6
-1.0
-1.8
-3.0
-4.0
1.00
1.22
1.82
2.72
6.06
20.0
55.6
Straight line is obtained whenln[PH3] is plotted against time
Q:
0
0.2
0.4
0.6
0.8
1
0 50 100 150 200
time (s)
[PH
3]
-4
-3
-2
-1
0
0 50 100 150 200
time (s)
ln[PH
3]
0
5
10
15
20
0 50 100 150 200
time (s)
1/[PH
3]
1st order reaction
IV. Reaction Rate and Temperature
A. Collision theory
– Based on the kinetic-molecular theory
– Collisions between reactants leads to reaction
– Requires proper orientation for reaction to occur
– Requires minimum energy (activation energy) for reaction to occur
B. Activation energy, Ea
• Ea: The minimum amount of energy required to initiate a chemical reaction – Transition state, activated complex
• The energy change during the chemical reaction
Exothermic Reaction Endothermic Reaction
Transition state Transition state
C. Temperature dependence of 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
lnk = -Ea
R1T
+ lnA
lnk = -Ea
R1T
+ lnA ln = ( - )Ea
R1T2
k1
k2
1T1
Ea =-1.10 x 8.314 J/K.mol
(-1.09 x 10-4 1/K)
T1 = 25oC = 298 K
ln k1
3k1
Q: The rate constant of a certain reaction triples when the temperature increases from 25oC to 35oC. What is the activation energy of this reaction
ln = ( - )Ea
R1T2
k1
k2
1T1
T2 = 35oC = 308 K k2 = 3 k1 Ea=?
-1.10 =
=1
308 K8.314 J/K.mol
Ea ( - )1
298 K8.314 J/K.mol
Ea
8.314 J/K.mol
Ea (-1.09 x 10-4 1/K )
= 83.9 kJ/mol
Q: Consider the following potential energy profile. (a) Rank the rate of the reaction. (Assume frequency factor is about the same). (b) Which reaction(s) are endothermic?
RP
40 kJ/mol
30 kJ/mol
(1)
R
P
34 kJ/mol 56 kJ/mol
(2)
R P
23 kJ/mol
17 kJ/mol
(3)
Pot
entia
l ene
rgy
reaction progress reaction progress reaction progress
Ea= 40 kJ/mol Ea= 34 kJ/mol Ea= 23 kJ/mol
(a) Slowest reaction to fastest: (1) < (2) < (3) (fastest)
E= 10 kJ/mol E=-22 kJ/mol E= 6 kJ/mol
(b) Endothermic reactions: (1) and (3)
V. Reaction Mechanism
• Detailed molecular process by which reaction occurs• Consists of a series of simple elementary steps or
elementary reactions.
The sequence of elementary steps that leads to product formation is the reaction mechanism.
2NO (g) + O2 (g) 2NO2 (g)
N2O2 is detected during the reaction!
Elementary step: NO + NO N2O2
Elementary step: N2O2 + O2 2NO2
Overall reaction: 2NO + O2 2NO2
+
• Intermediates– Species that appear in a reaction mechanism but not in the
overall balanced equation– Always formed in an early elementary step and consumed
in a later elementary step
Ex. _____ is the intermediate in the above reaction.
Elementary step: NO + NO N2O2
Elementary step: N2O2 + O2 2NO2
Overall reaction: 2NO + O2 2NO2
+
V. Reaction Mechanism
N2O2
• Elementary step: each step in a reaction mechanism
• Molecularity– Number of molecules participate in an elementary step– Unimolecular, bimolecular, termolecular reactions
Rate laws for elementary steps
Unimolecular reaction A products rate = k [A]
Bimolecular reaction A + B products rate = k [A][B]
Bimolecular reaction A + A products rate = k [A]2
exponent in the rate law = coefficient in the elementary step
Ex. Write the rate law for the the elementary step
NO (g) + O3(g) NO2 (g) + O2 (g)
rate = k [NO][O3]
Writing plausible reaction mechanisms
• When the exponent in rate law ≠ coeff in chemical reaction => multistep reaction
• The sum of the elementary steps must give the overall balanced equation for the reaction.
• The rate-determining step should predict the same rate law that is determined experimentally.
aA + bB cC + dD?
The rate-determining step is the slowest step in the sequence of steps leading to product formation.
Q:
Step 1: NO2 + NO2 NO + NO3
Step 2: NO3 + CO NO2 + CO2
What is the equation for the overall reaction?
NO2+ CO NO + CO2
What is the intermediate?
NO3
What can you say about the relative rates of steps 1 and 2?
rate = k[NO2]2 is the rate law for step 1 so step 1 must be slower than step 2
The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps:
Q:
What is the equation for the overall reaction?
2 NO2+ F2 2 NO2F
What is the intermediate?
F
One possible mechanism?
The experimental rate law for the reaction between NO2 and F2 to produce NO2F is rate = k[NO2][F2]. What is the reaction mechanism?
Step 1: NO2 + F2 NO2F + F
Step 2: NO2 + F NO2F
Overall reaction: 2 NO2+ F2 2 NO2F
rate = k[NO2][F2] is the rate law if step 1 is rds
rds
VI. Catalysis
• Change the reaction rate without being used in the reaction
• Homogeneous catalysis
• Heterogeneous catalysis
• Enzyme– Biological catalyst
• 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’
• In heterogeneous catalysis, the reactants and the catalysts are in different phases.
• In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid or gas.
- Haber synthesis of ammonia
- Ostwald process for the production of nitric acid
- Catalytic converters
- Acid catalysis
- Base catalysis
Haber Process
N2 (g) + 3H2 (g) 2NH3 (g)Fe/Al2O3/K2O
catalyst
Enzyme Catalysis
Enzyme Catalysis
uncatalyzedenzyme
catalyzed
rate = [P]
t
rate = k [ES]
E +S ES
ES P + E
Rate law of enzyme catalysis
uncatalyzedenzyme
catalyzed
E +S ES (fast equilibrium)
k1
k-1
ES P + Ek2
k1[E][S]= k-1[ES]
[ES] =k1[E][S]
k-1
rate = = k2 [ES] =[P]
t
k1k2[E][S]k-1
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