1 16-1 Chemical Kinetics: Rates and Mechanisms of Chemical Reactions 16-2 Learning Objectives: Reaction Rate Expressing the Reaction Rate **The Rate Law and Its Components **Integrated Rate Laws: Concentration Changes over Time Catalysis: Speeding Up a Reaction Theories of Chemical Kinetics *Reaction Mechanisms: The Steps from Reactant to Product 16-3 The wide range of reaction rates • Explosion • Ripening of fruits/vegetables • Digestion • Fossilization from ancient vegetation to fossil fuel 16-4 A faster reaction (top) and a slower reaction (bottom): How fast the product is formed or reactant is consumed?
Transcript
1
16-1
Chemical Kinetics:
Rates and Mechanisms of Chemical Reactions
16-2
Learning Objectives:
Reaction Rate
Expressing the Reaction Rate
**The Rate Law and Its Components
**Integrated Rate Laws: Concentration Changes over Time
Catalysis: Speeding Up a Reaction
Theories of Chemical Kinetics
*Reaction Mechanisms: The Steps from Reactant to Product
16-3
The wide range of reaction rates
• Explosion• Ripening of fruits/vegetables• Digestion• Fossilization from ancient vegetation to fossil fuel
16-4
A faster reaction (top) and a slower reaction (bottom): How fast the product is formed or reactant is consumed?
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16-5
Factors That Influence Reaction Rate
Particles must collide in order to react: Greater collision rate, faster rate.
Higher the concentration of reactants, reaction rate↑– Greater number of collisions. Household bleach (CloroxTM).
Physical state of the reactants influences reaction rate. - Substances must mix in order for particles to collide.
Example: Surface area (powder form vs. chunky form)
Higher the temperature, reaction rate↑– At higher temperatures particles have more energy and therefore
collide more often and more effectively. Milk spoils faster at higher temperature.
16-6
The effect of surface area on reaction rate
A hot steel nail glows feebly when placed in O2.
The same mass of steel wool (more surface area) bursts into flame.
16-7
Sufficient collision energy is required for a reaction to occur.
16-8
Expressing the Reaction Rate
Reaction rate: the changes in concentrations of reactants or products per unit time.
For the general reaction A → B, we measure the concentration of A at t1 and at t2:
“-” : because the concentration of A is decreasing. This gives the rate a positive value.
Δ[A]ΔtRate =
change in concentration of Achange in time
conc A2 - conc A1
t2 - t1= - = -
“[xxx]” = concentration in moles per liter.
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16-9
Average rate
Instantaneous rate = slope of tangent line at t
Initial rate = instantaneous rate at t = 0
C
A
B
16-10
Question: point A or B?Which has the higher [O3]?
Which has higher reaction rate?
How does the rate change over time, increase or decrease?
A
BKey: A; A; decrease
16-11
Plots of [reactant] and [product] vs. time.
C2H4 + O3 → C2H4O + O2
[O2] increases just as fast as [C2H4] decreases.
Rate = -Δ[C2H4]
Δt= -
Δ[O3]Δt
Δ[C2H4O]Δt
Δ[O2]Δt
= =
Quick Question: a or b?
Which has higher Δ[O2]?
Which has higher reaction rate? Aa
b
Key: b; a16-12
Plots of [reactant] and [product] vs. time.H2 + I2 → 2HI
[HI] increases twice as fast as [H2] decreases.
Rate = -Δ[H2]
Δt= -
Δ[I2]Δt
=Δ[HI]
Δt12
Δ[I2]Δt
Rate =Δ[H2]
ΔtΔ[IH]
Δt= -2 = -2
The expression for the rate of a reaction and its numerical value depend on which substance serves as the reference.
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16-13
In general, for the reactionaA + bB → cC + dD
where a, b, c, and d are the coefficients for the balanced equation, the rate in terms of each species is expressed as:
The Reaction Rates depends on the stoichiometry of Reaction
Analogy: For the same person, heart rate and breathing rate is not the same at the same moment.
_________________________________________
16-14
Example: The balanced equation for the reaction of bromate ion with bromide in acidic solution is given by
BrO3- + 5Br- + 6H+ → 3Br2 + 3H2O
At a particular instant in time, the value of -△[Br-]/△t is 1.0 × 10-2
mol/L ∙ s. What is the value of △[Br2]/△t in the same unit?
6.0 × 10-3 mol/L ∙ s
Practice: What is the value of -△[H+]/△t in the same unit?
16-15
The Rate Law: How Rate Depends on Concentrations
For any general reaction occurring at a fixed temperature
aA + bB → cC + dD
Rate = k[A]m[B]n
k = Rate Constantm and n : Reaction Orders, determined by experiment.
k depends on temperature and catalyst, not affected by concentrations.
Order: zero order; first order; second order
16-16
Individual and Overall Reaction Orders
For the reaction 2NO(g) + 2H2(g) → N2(g) + 2H2O(g):
The rate law is rate = k[NO]2[H2]
The above reaction is 2nd order for NO, 1st order with respect to H2; 3rd order overall.
If the concentration of a reactant does NOT affect the rate, the order of this reactant is ______ order.
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16-17
Plots of Reactant concentration, [A], vs. Time
16-18
Plots of Rate vs. Reactant Concentration [A]
16-19
Determining Reaction Orders: A. Initial Rate Method
For the general reaction aA + bB → dC + dD, rate law as Rate = k[A]m[B]n
To determine the values of m and n, we run a series of experiments and find the order ONE by ONE
To find n, only ___ is changed, measure new initial rate3.
Set first trial ([A]1 and [B]1 measure initial rate1 at time = 0).
To find m, only [A] is changed while [B] and [C] unchanged, measure new initial rate2 .
16-20
Initial Rate Method to Determine the order
ExperimentInitial Rate(mol/L·s)
Initial [A](mol/L)
Initial [B](mol/L)
1 1.75x10-3 2.50x10-1 3.00x10-1
2 3.50x10-3 5.00x10-1 3.00x10-1
3 3.95x10-3 2.50x10-1 4.50x10-1
The mathematical way to find the order:
order = ______________
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16-21
Finding the order with respect to A
compare Trial# 1 and ___, where [B] is kept constant but [A] changes:
m =
Rate = k[A]m[B]n
TrialInitial Rate(mol/L·s)
Initial [A](mol/L)
Initial [B](mol/L)
1 1.75x10-3 2.50x10-1 3.00x10-1
2 3.50x10-3 5.00x10-1 3.00x10-1
3 3.95x10-3 2.50x10-1 4.50x10-1
16-22
Practice: Finding the rate constant k
Choose any trial to determine the rate constant k
k =__________
Trial 1: k = 0.0778 mol-2 L-2 s-1
Rate = k[A]1[B]2
TrialInitial Rate(mol/L·s)
Initial [A](mol/L)
Initial [B](mol/L)
1 1.75x10-3 2.50x10-1 3.00x10-1
2 3.53x10-3 5.00x10-1 3.00x10-1
3 3.92x10-3 2.50x10-1 4.50x10-1
Your k = __________________
16-23
Easier Way to Find Reaction Orders? Double the concentration(s)
For the simple reaction A → products:
If the rate doubles when [A] doubles, rate ∝ [A]1, reaction is _____ order with respect to A.
If the rate quadruples when [A] doubles, rate ∝ [A]2, reaction is __________ order with respect to [A].
If the rate does not change when [A] doubles, rate ∝ [A]0, reaction is _______ order with respect to A.
16-24
Units of the Rate Constant k for Common Reaction Orders
Overall Reaction Order
Units of k(t in seconds)
0 mol/L·s (or mol L-1s-1)
1 1/s (or s-1)
2 L/mol·s (or L mol-1s-1)
3 L2/mol2·s (or L2 mol-2s-1)
General formula:
Lmol
unit of t
order-1
Units of k =
The value of k is easily determined from experimental rate data. The units of k depend on the overall reaction order.
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16-25
Information sequence to determine the kinetic parameters of a reaction.
Series of plots of concentration vs.
time
Initial rates
Reaction orders
Rate constant (k) and actual rate law
Determine slope of tangent at t0 for each plot.
Compare initial rates when [A] changes and [B] is held constant (and vice versa).
Substitute initial rates, orders, and concentrations into rate = k[A]m[B]n, and solve for k.
16-26
Integrated Rate LawsAn integrated rate law includes time as a variable.
First-order rate equation:rate = - Δ[A]
Δt= k [A]
Second-order rate equation:
rate = - Δ[A]Δt
= k [A]2
Zero-order rate equation:
rate = - Δ[A]Δt
= k [A]0 = k
Remember both rate and concentration depends on time
Integrated Rate Law:
_____________________
_____________________
_____________________
16-27
Graphical presentation of 1st order reaction
First-order reaction
ln[A]0[A]t
= kt
integrated rate law
ln[A]t = _______ + _____
straight-line form
A plot of ln[A] vs. time gives a straight line for a first-order reaction.Rate constant k = - slope
16-28
Graphical presentation of 2nd order reaction
Second-order reaction integrated rate law1
[A]t
1
[A]0- = kt
straight-line form1
[A]t= kt +
Plot of 1/[A] vs. time: linear. Rate constant k = slope
y = mx + b
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16-29
Graphical presentation of zero order reaction
Zero-order reaction
Plot of [A] vs. time gives a straight line for a first-order reaction.Rate constant k = - slope
integrated rate law[A]t - [A]0 = - kt
straight-line form[A]t = _____ + ____
16-30
Example: Determining the Reactant Concentration after a Given Time
PROBLEM: At 1000oC, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 87 s-1, to two molecules of ethylene (C2H4).
(a) If the initial C4H8 concentration is 2.00 M, what is the concentration after 0.010 s?
a. 0.83 mol/L
Given: k = 87 s-1, t = 0.010 s, [A]0 = 2.00 M, ln([A]t/[A]0) = - kt
Find: [A]t
16-31
Example: Apply Integrated rate law in Problem solving
PROBLEM: At a certain temperature, cyclobutane (C4H8) decomposes in a first-order reaction, with a rate constant of 1.0 s-1.
(b) What fraction of C4H8 has decomposed after 3.0 sec?
0.95
Given: k = 1.0 s-1, t = 3.0 s, ln([A]t/[A]0) = - kt
Find: 1 - [A]t/[A]0
16-32
Example: Apply Integrated rate law in Problem solving
At a certain temperature, cyclobutane (C4H8) decomposes in a first-order reaction. After 10.0 seconds, 20.0% of cyclobutane decomposed. Determine the rate constant at this temperature.
0.0223 s-1
Given: t = 10.0 s, 1 - [A]t/[A]0) = 0.200
Find: k
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16-33
Graphical determination of Reaction order: Decomposition of N2O5
Three ways to graph:A. [A] vs. timeB. ln[A] vs. timeC. 1/[A] vs. timeSince the plot of ln[N2O5] vs. time gives a straight line, the reaction is first order.
16-34
Reaction Half-life
The half-life (t1/2) for a reaction is the time taken for the concentration of a reactant to drop to half its initial value.
For a first-order reaction, t1/2 does not depend on the starting concentration.
t1/2 =ln 2k
= 0.693k
The half-life for a first-order reaction is a constant, independent on the concentration of reactant
16-35
[N2O5] vs. time for three reaction half-lives.
t1/2 =
for a first-order process
ln 2k
0.693k
=
16-36
Half-life Equations
For a first-order reaction, t1/2 does not depend on the initial concentration.
For a second-order reaction, t1/2 is inversely proportional to the initial concentration:
1k[A]0
t1/2 =
For a zero-order reaction, t1/2 is directly proportional to the initial concentration:
[A]02k0t1/2 =
10
16-37
Practice: Determining the Half-Life of a First-Order ReactionAt 1000°C, the rearrangement reaction of cyclopropane has rate constant of 9.2 s-1, (a) What is the half-life of the reaction? (b) How long does it take for the concentration of cyclopropane to reach one-tenth of the initial value?
a. Half life t1/2 = 0.693/k = 0.075 s b. 0.25 s16-38
An Overview of Zero-Order, First-Order, and Simple Second-Order Reactions
Zero Order First Order Second Order
Rate law rate = k rate = k[A] rate = k[A]2
Units for k mol/L·s 1/s L/mol·sHalf-life
Integrated rate lawin straight-line form
[A]t = -kt + [A]0 ln[A]t = -kt + ln[A]0
Plot for straight line [A]t vs. t ln[A]t vs. t vs. t
Slope, y intercept -k, [A]0 -k, ln[A]0 k,
[A]02k
ln 2k
1k[A]01
[A]t= kt +
1[A]0
1[A]t
1[A]0
16-39
Collision Theory and Concentration
The basic principle of collision theory is that particles must collide in order to react.
An increase in the concentration of a reactant leads to higher frequency of collisions, hence increasing reaction rate.
The number of collisions depends on the product of the numbers of reactant particles, not their sum.Concentrations are multiplied in the rate law, not added.
16-40
Temperature Affects Rate Constant
Temperature has a dramatic effect on reaction rate.For many reactions, an increase of 10°C will double or triple the rate.
Experimental data shows that k increases exponentially as T increases: the Arrhenius equation
k = Ae-Ea/RTk = rate constantA = frequency factorEa = activation energy
Higher T increased ratelarger k
k = Ae-Ea/RTk = rate constantA = frequency factorEa = activation energy
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16-41
Increase of the rate constant with temperature for the hydrolysis of an ester.
Reaction rate and k increase exponentially as T increases.
16-42
Activation Energy
In order to be effective, collisions between particles must exceed a certain energy threshold.
The lower the activation energy, the faster the reaction.
When particles collide effectively, they reach an activated state. The energy difference between the reactants and the activated state is the activation energy (Ea) for the reaction.
Smaller Ea increased ratelarger f larger k
16-43
Energy-level diagram for a reaction.
Collisions must occur with sufficient energy to reach an activated state.
This particular reaction is reversible and is exothermic in the forward direction.
16-44
Temperature and Collision Energy
An increase in temperature causes an increase in the kinetic energy of the particles. This leads to more frequent collisions
Also, at a higher temperature, higher fraction of molecules has the kinetic energy above the activation energy Ea
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16-45
The Effect of Ea and T on the Fraction (f) of Collisions with Sufficient Energy to Allow Reaction
Ea (kJ/mol) f (at T = 298 K)
50 1.70x10-9
75 7.03x10-14
100 2.90x10-18
T f (at Ea = 50 kJ/mol)25°C (298 K) 1.70x10-9
35°C (308 K) 3.29x10-9
45°C (318 K) 6.12x10-9
16-46
Calculating Activation Energy
k = Ae -Ea/RT
Arrhenius equation: Exponential form vs. Logarithm form
ln k = ln A -Ea 1R T
Use rates at two different temperatures:
At T1, ln k1 = ln A -Ea 1R T1
At T2, ln k2 = ln A -Ea 1R T2
Ea= ___________________
16-47
Graphical determination of the Activation energy:lnk (as y) vs. 1/T (as x)
ln k = ln A -Ea 1R T
Ea = (-8.314 J/mol) x slopeUnit: Joule
A = exp(y intercept)Unit: same unit as rate constant
A = exp(y intercept)y = -5.64x103 + 16.6R² = 0.9999
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
00.0031 0.0032 0.0033 0.0034 0.0035
lnk
1/T (K-1)
46.9 kJ/mol 1.6 x 107 L/mol s
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16-49
Practice: The decomposition of hydrogen iodide, 2HI(g) → H2(g) + I2(g), has rate constants of 9.51x10-9 L/mol·s at 500. K and 1.10x10-5 L/mol·s at 600. K. Find Ea.
Ea = 1.76x105 J/mol = 1.76x102 kJ/mol 16-50
Molecular Structure and Reaction Rate
For a collision between particles to be effective, it must have both sufficient energy and the appropriate relative orientation between the reacting particles.
k = Ae -Ea/RT
A in the Arrhenius equation is the frequency factor for the reaction.
A = pZ p = orientation probability factorZ = collision frequency
p depends on the specific reaction and is related to the structural complexity of the reactants.
16-51
Molecular orientation for an Effective Collision
NO(g) + NO3(g) → 2NO2(g)
Only one relative orientation of these two molecules, Effective Orientation, that leads to an effective collision.
Think: Which bonds are broken? Which bonds are formed?
16-52
Transition State Theory
Effective collision between particles leads to the formation of a transition state or activated complex.
The transition state is an unstable species that contains partial bonds. It is a transitional stage between reactants and products.
Transition states cannot be isolated.
The transition state exists at the point of maximumpotential energy. The energy required to form the transition state is the activation energy.
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16-53
Transition state: Reaction between BrCH3 and OH-
The transition state contains partial bonds (dashed) between C and Br and between C and O. It has a trigonal bypyramidal shape.
BrCH3 + OH- → Br- + CH3OH
16-54
Quantum mechanical Study of the reaction between BrCH3 and OH-.
16-55
Reaction Energy Diagram: Endothermic vs. Exothermic Reaction
16-56
Example Drawing Reaction Energy Diagrams and Transition States
For reaction: O3(g) + O(g) → 2O2(g)
The Ea(fwd) is 19 kJ, and the ΔHrxn for the reaction as written is -392 kJ. Draw a reaction energy diagram, label the reactant, product, activation energy, reaction enthalpy change, and transition state. Predict a structure for the transition state, and calculate Ea(rev).
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16-57
Reaction Mechanism, Elementary Step
Two Step Mechanism:
Decomposition of Ozone
Unlike a balanced equation, many chemical reactions do not occur with all the reactant molecules together turning into final products.
The mechanism of a reaction is the sequence of single reaction steps (elementary steps) that make up the overall equation.
16-58
Reaction intermediates
Reaction intermediate is the species (molecules, atoms, ions, etc.) produced in the initial elementary steps.
Reaction intermediates typically are unstable and reactive, consumed in the subsequent elementary steps.
16-59
Molecularity of Elementary Steps
Each elementary step is characterized by its molecularity, the number of particles involved in the reaction: Unimolecular; Bimolecular; Termolecular(three molecules)
16-60
Elementary Step Molecularity Rate Law
A product
2A product
A + B product
2A + B product
Unimolecular
Bimolecular
Bimolecular
Termolecular
Rate = [A]
Rate = k[A]2
Rate = k[A][B]
Rate = k[A]2[B]
The rate law for an elementary step can be deduced from the reaction stoichiometry – reaction order equals molecularity for an elementary step only.
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16-61
Example: Reaction mechanism, Rate law for Elementary Steps, Reaction intermediate
PROBLEM: The following elementary steps are proposed for a reaction mechanism:(1) NO2Cl(g) → NO2(g) + Cl(g)(2) NO2Cl(g) + Cl(g) → NO2(g) + Cl2(g)
(a) Write the overall balanced equation.
(b) Write the rate law for each step.
(c) Identify the reaction intermediate.
16-62
The Rate-Determining Step of a Reaction
The slowest elementary step in a reaction is the rate-determining or rate-limiting step. Analogy: Traffic between Orange County and Riverside
The rate law for the rate-determining step becomes the rate law for the overall reaction.
The reaction NO2(g) + CO(g) → NO(g) + CO2(g) has been proposed to occur by a two-step mechanism:
The elementary steps must add up to the overall balanced equation.
The elementary steps must be reasonable.
The mechanism must correlate with the observed rate law. If first elementary step is the rate-determining, its rate law should match the experimental rate law
A mechanism is a hypothesis –we cannot prove it is correct, but if it is consistent with the data, and can be used to predict results accurately, it is a useful model for the reaction.
16-64
Example: Propose a Mechanism with a Slow Initial Step
The overall reaction 2NO2(g) + F2(g) →2NO2F(g) has an experimental rate law Rate = k[NO2][F2].In the first elementary step is rate-determining step, the
mechanism would be:
For step 1 (the slow step), rate1 = k1 x ______ which matches the experimental rate law
Step 1:
Step 2: The second NO2 molecule needs to be reactant so that the overall reaction has two NO2 molecules
