chemical kinetics chapter 12. slide 2 reaction rates01 reaction rate: the change in the...
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Chapter 12 Slide 2
Reaction Rates 01Reaction 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 ProductsA B
Rate [A]t
Rate [B]t
2 HI(g) H2(g) + I2(g)
Chapter 12 Slide 3
Reaction Rates and StoichiometryReaction Rates and Stoichiometry
• What is the general rate of the following reaction ?
2 HI(g) H2(g) + I2(g)
Rate = − 12
[HI]t
= [I2]t
Chapter 12 Slide 4
Reaction Rates and StoichiometryReaction Rates and Stoichiometry
• To generalize, for the reaction
aA + bB cC + dD
Rate* = −1a
[A]t
= −1b
[B]t
=1c
[C]t
1d
[D]t
=
*: General rate of reaction
+2 2 2HCO H(aq) + Br (aq) 2H (aq) + 2Br (aq) + CO (g)
red colorless
Which of the expressions below, corresponding to the reaction of bromine with formic acid, is incorrect?
2 2d[CO ] d[Br ]1. =
dt dt
+d[Br ] d[H ]2. =
dt dt
+2d[HCO H] 1 d[H ]
4. = dt 2 dt
2d[Br ] d[Br ]5. = 2
dt dt
2 2d[HCO H] d[Br ]
3. = dt dt
+2 2 2HCO H(aq) + Br (aq) 2H (aq) + 2Br (aq) + CO (g)
red colorless
Which of the expressions below, corresponding to the reaction of bromine with formic acid, is incorrect?
2 2d[CO ] d[Br ]1. =
dt dt
+d[Br ] d[H ]2. =
dt dt
+2d [ H C O H ] 1 d [ H ]
4 . = d t 2 d t
2d[Br ] d[Br ]5. = 2
dt dt
2 2d[HCO H] d[Br ]
3. = dt dt
Chapter 12 Slide 7
How Do we study Rate of a reaction?How Do we study Rate of a reaction?
• Consider the decomposition of N2O5 to give NO2 and O2: 2 N2O5(g) 4 NO2(g) + O2(g)
Brown Colorless
Chapter 12 Slide 8
Reaction Rates: concentration versus time curve 03Reaction Rates: concentration versus time curve 03
Average Rate = Rate between two points in time
•2 N2O5(g) 4 NO2(g) + O2(g)
The slope of each triangle Between two points
Chapter 12 Slide 10
Instantaneous rate: Instantaneous rate: Rate for specific instance in time Slope of the tangent to a concentration versus time curve
Initial Rate
Chapter 12 Slide 11
Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
time
393 nmlight
Detector
[Br2] Absorption3
93 n
m
Br2 (aq)
Chapter 12 Slide 12
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 timeThe slope of a line tangent to the curve at any point is the instantaneous rate at that time
Chapter 12 Slide 14
rate [Br2]
rate = k [Br2]
k = rate[Br2]
= rate constant
The Rate Law; rate = 3.50 x 10-3 s-1 [Br2]
Chapter 12 Slide 16
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.
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
Chapter 12 Slide 17
The Rate Law and Reaction Order are Experimentally DeterminedThe Rate Law and Reaction Order are Experimentally Determined
Chapter 12 Slide 18
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]
The instantaneous rate at the beginning of a reaction is called initial rate
----
Determine the reaction order for:
1 vs 3
1 vs 2
Chapter 12 Slide 19
F2 (g) + 2ClO2 (g) 2FClO2 (g)
rate = k [F2][ClO2]
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.
1
Chapter 12 Slide 20
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
Double [I-], rate doubles (experiment 1 & 2)
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-]
Chapter 12 Slide 21
Determine the Rate Law and Reaction OrderDetermine the Rate Law and Reaction Order
Comparing Experiments 1 and 2, when [NH4+] doubles, the initial
rate doubles.
NH4+(aq) + NO2
−(aq) N2(g) + 2 H2O(l)
Chapter 12 Slide 22
Likewise, comparing Experiments 5 and 6, when [NO2−] doubles,
the initial rate doubles.
NH4+(aq) + NO2
−(aq) N2(g) + 2 H2O(l)
Chapter 12 Slide 23
• This meansRate [NH4
+]
Rate [NO2−]
Rate [NH+] [NO2−]
or
Rate = k [NH4+] [NO2
−]• This equation is called the rate law, and k is the rate constant.
Chapter 12 Slide 24
Rate LawsRate Laws
• The exponents tell the order of the reaction with respect to each reactant.
• This reaction isFirst-order in [NH4
+]
First-order in [NO2−]
Chapter 12 Slide 25
Rate Law & Reaction OrderRate Law & Reaction Order
• The reaction of nitric oxide with hydrogen at
1280°C is: 2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g)
• From the following data determine the rate law and
rate constant.Experiment [NO] [H2] Initial Rate (M/s)
1 5.0 x 10–3 2.0 x10–3 1.3 x 10–5
2 10.0 x 10–3 2.0 x 10–3 5.0 x 10–5
3 10.0 x 10–3 4.0 x 10–3 10.0 x 10–5
k = 1/3(250+250+260) = 250 M-2.s-1Second order in NO, First order in H2
Chapter 12 Slide 26
First-Order ReactionsConcentration and Time Equation
01
First-Order ReactionsConcentration and Time Equation
01• First Order: Reaction rate depends on the reactant concentration raised to first power.
Rate = k[A]
Rate = - A t
[A]t
= k [A]-
A product
Chapter 12 Slide 27
Concentration and Time Equation For A First-Order Reactions
k = rate[A]
= 1/s or s-1M/sM
=
[A]= K Δt-
[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)
[A]-(ln[A] -ln[A]0) = ktln[A] = ln[A]0 - kt
[A] = [A]0exp(-kt)
law decay lexponentiae[A][A] kt0t
ln[A]0
[A]= k t
See next slide for proof of the formula
Chapter 12 Slide 29
First-Order Reactions
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t = 0ln[A] = ln[A]0 - kt
ln[A] = ln[A]0 - kt
ln[A]0
[A]= k t
Chapter 12 Slide 30
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 ?
t = = 66 s
[A]0 = 0.88 M
[A] = 0.14 M
ln[A]0
[A]
k
ln0.88 M
0.14 M
2.8 x 10-2 s-1=
ln[A]0
[A]= k t
t = ?
Chapter 12 Slide 31
What is Half- Life ?The 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
ln[A]0
[A]0/2
k=t½
ln2k
=0.693
k=ln
[A]0
[A]= k t
Half Life For the First Order Reaction
Chapter 12 Slide 32
Reaction OrdersReaction Orders
Units of Rate Constants vs Reaction Orders
Zeroth Order Reaction:
Rate = K [A]0 = K
Chapter 12 Slide 33
What is the order of decomposition of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?What is the half life of decomposition of N2O5 ?
t½ln2k
=0.693
5.7 x 10-4 s-1= = 1200 s = 20 minutes
Therefore, decomposition is first order?
units of k (s-1)
2N2O5(g) 4NO2(g) + O2(g)
Chapter 12 Slide 35
A product
First-order reaction
# of half-lives [A]
1
2
3
4
½ [A]0
1/4 [A]0
1/8 [A]0
1/16 [A]0
[A] = [A]0 x (1/2)n
Chapter 12 Slide 36
First-Order ReactionFirst-Order Reaction
• Show that the decomposition of N2O5 is first order and calculate the rate constant and Half life.
k = 1.7 x 10-3 s-1
2N2O5(g) 4NO2(g) + O2(g)
t1/2 = 408 S
Chapter 12 Slide 37
Second-Order Reactions
A product rate = -[A]t
rate = k [A]2
k = rate[A]2
= 1/M•s or M-1 s-1M/sM2=
[A]t
= k [A]2-
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t = 0
1[A]
=1
[A]0
+ kt
What is Unit of k ?
What is Conc. Vs time equation?
Chapter 12 Slide 38
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 = t1/2
=t1/2
[A]2
[A]0
Chapter 12 Slide 39
= kt +[A]0
1[A]t
1t = t1/2
=t1/2
[A]2
[A]0
Half-life for a second-order reaction
[A]0
1= kt1/2 +[A]0
2=t1/2
k[A]0
1
Chapter 12 Slide 40
Second-Order ReactionsSecond-Order Reactions
=t1/2k[A]0
1
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 42
2 NO2(g) 2NO(g) + O2(g)
a) Is the following reaction first or second order ?b) What is the value of k?
Example:
Chapter 12 Slide 45
Zero Order Reaction:
Rate = k
Example of Zeroth Order Reaction:Decomposition of N2O on hot platinum surface: N2O → N2 + 1/2 O2
Rate [N2O]0 = k[N2O]0 = k d[N2 O]/dt = k
Chapter 12 Slide 46
Reaction Mechanisms 01Reaction Mechanisms 01
• A reaction mechanism
is a sequence of
molecular events, or
reaction steps, that
defines the pathway
from reactants to
products.
Chapter 12 Slide 47
Reaction Mechanisms 02Reaction 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 48
Reaction Mechanisms NO2(g) + CO(g) NO(g) + CO2(g) Reaction Mechanisms NO2(g) + CO(g) NO(g) + CO2(g)
• NO2(g) + NO2(g) NO(g) + NO3(g) Elementary
• NO3(g) + CO(g) NO2(g) + CO2(g) Elementary
• NO2(g) + CO(g) NO(g) + CO2(g) 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.
NO3(g) is called reaction intermediate.
Chapter 12 Slide 49
Reaction Mechanisms 04Reaction Mechanisms 04
• Molecularity: is the number of molecules (or atoms) on the reactant side of the chemical equation.
• Unimolecular: Single reactant molecule.
Chapter 12 Slide 50
Reaction Mechanisms 05Reaction Mechanisms 05
• Bimolecular: Two reactant molecules.
• Termolecular: Three reactant molecules.
Chapter 12 Slide 51
Reaction Mechanisms 06Reaction Mechanisms 06
• Determine individual steps , the reaction intermediates, and the molecularity of each individual step.
2N2O 2N2 + O2
Chapter 12 Slide 52
Rate Laws and Reaction Mechanisms 01Rate Laws and Reaction Mechanisms 01
• Rate law for an overall reaction must be determined experimentally.
• Rate law for elementary step follows from its molecularity.
Chapter 12 Slide 53
Rate Laws and Reaction Mechanisms 02Rate Laws and Reaction Mechanisms 02
• The rate law of each elementary step follows its molecularity.
• The overall reaction is a sequence of elementary steps called the reaction mechanism.
Chapter 12 Slide 54
Rate-Determining StepRate-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 55
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:
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
Chapter 12 Slide 58
The Arrhenius EquationThe Arrhenius Equation
Is it hidden in k?
2N2O5(g) 4NO2(g) + O2(g)
rate = k[N2O5]Where is temperature dependence?
Typically, as the temperature increases, the rate of reaction increases.
Chapter 12 Slide 59
The Arrhenius Equation01The Arrhenius Equation01
• Collision Theory: A bimolecular reaction occurs when two correctly oriented molecules collide with sufficient energy.
• Activation Energy (Ea): The potential energy
barrier that must be surmounted before reactants can be converted to products.
Chapter 12 Slide 60
Temperature dependence of Rate Constat 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]
• The fraction of collisions with an energy equal or more than activation energy( Ea) :
f = e-
Ea/RT
• Raising the temperature speeds up a reaction by providing the energy of activation to more colliding molecules.
Z is a constant related to collision frequency
Chapter 12 Slide 61
• Only the fraction of collisions having proper orientation can result to products.
The Arrhenius Equation02The Arrhenius Equation02
This is called steric factor, p., In the above example p = 0.5
Chapter 12 Slide 62
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 factork = A. f, f = e-Ea/RT
k = Ae-Ea/RT (p.Z )= A
The Arrhenius EquationThe Arrhenius Equation
Chapter 12 Slide 63
K = Ae-Ea/RT
A = pZ
(steric factor)
The Arrhenius EquationThe Arrhenius Equation
Chapter 12 Slide 64
k = A .e-Ea/RT
k = A .e(-Ea/R)(1/T)
Calculating Activation Energy
Ea = -R . (slope)
Ln k
1/T
Chapter 12 Slide 65
2HI(g) + H2(g) I2(g) + H2(g)
Find the activation energy for the following reaction
Chapter 12 Slide 67
Ea = - (8.314 j/K.mol) (-2.24 x 10 4 K)Ea = 190 kj/mol
Slope = -2.24 x 10 4 K
Ea = -R . (slope)
Chapter 12 Slide 68
Effect of Temperature on Fraction of Collisions with Activation energyEffect of Temperature on Fraction of Collisions with Activation energy
f = e-Ea/RT
Effect of TemperatureEffect of Temperature
Collision Theory: As the average kinetic energy increases, the average molecular speed increases, and thus the collision rate increases.
f = e-Ea/RT
Chapter 12 Slide 70
Change of Rate Constant with temperatureChange of Rate Constant with temperature
• If the Ea is known , we can calculate the Rate
Constant when temperature is changed:
lnk2k1
EaR
1T2
1T1
The above formula could be used to determine the rate constant at a different temperature.
Chapter 12 Slide 71
Homework:Determination the Activation Energy
Homework:Determination the Activation Energy
The second-order rate constant for the decomposition of nitrous oxide (N2O) into nitrogen molecule and oxygen atom has been measured at different temperatures:
Determine (graphically)the activation energyfor the reaction.
k (M -1s-1) t (°C)
1.87x10-3 6000.0113 6500.0569 7000.244 750
Ea = 241 KJ/mole
Chapter 12 Slide 73
• A catalyst is a substance that increases the rate of a reaction without being consumed in the reaction.
Catalysis 01Catalysis 01
CatalysisCatalysis
Note that the presence of a catalyst does not affect the energy difference between the reactants and the products
Chapter 12 Slide 75
Catalysis 02Catalysis 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 12 2
Chapter 12 Slide 76
CatalysisCatalysis
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.
H2O2(aq) + I1-(aq) H2O(l) + IO1-(aq)
H2O2(aq) + IO1-(aq) H2O(l) + O2(g) + I1-(aq)
2H2O2(aq) 2H2O(l) + O2(g) overall reaction
rate-determiningstep
fast step
Chapter 12 Slide 77
Catalysis 03Catalysis 03
• Homogeneous Catalyst: Exists in the same phase as the reactants.
• Heterogeneous Catalyst: Exists in different phase to the reactants.
Chapter 12 Slide 79
H H CC CCAA
BB
XX
YYH H
Mechanism of Catalytic Hydrogenation: Mechanism of Catalytic Hydrogenation:
Chapter 12 Slide 80
H
CC CCAA
BB
XX
YY
H H H
Mechanism of Catalytic Hydrogenation: Mechanism of Catalytic Hydrogenation:
Chapter 12 Slide 81
H
H H HCCCC
AABB
XXYY
Mechanism of Catalytic Hydrogenation: Mechanism of Catalytic Hydrogenation:
Chapter 12 Slide 82
H
H H HCCCC
AABB
XXYY
Mechanism of Catalytic Hydrogenation: Mechanism of Catalytic Hydrogenation:
Chapter 12 Slide 83
H H
H
CCCC
AABB
XXYY
H
Mechanism of Catalytic Hydrogenation:Mechanism of Catalytic Hydrogenation: