Download - Chapter 13 Chemical Kinetics
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Chapter 13
Chemical Kinetics
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The study of reaction rate is called chemical kinetics.
Reaction rate is measured by the change of
concentration (molarity) of reactants or products
per unit time.
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Δt
reactantΔreactantreactantr
if
if
tt
Unit:
Molarity of A: [A], e.g. [NO2]: molarity of NO2
Δt
productΔproductproductr OR
if
if
tt
reactants products,
− −
[reactant]↓ and [product]↑
mol·L−1·s−1 ≡ M·s−1
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5 122
NO 0.0079 M 0.0100 Mr NO 4.2 10 M s
50 s 0 s-
t
0 s → 50 s:
5 122
NO 0.0065 M 0.0079 Mr NO 2.8 10 M s
100 s 50 s-
t
50 s → 100 s:
2NO2 (g) 2NO (g) + O2 (g)
− = r(NO2)
rate is a function of time
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−
5 1NO 0.0035 M 0.0021 Mr NO 2.8 10 M s
100 s 50 s-
t
5 122
O 0.0018 M 0.0011 Mr O 1.4 10 M s
100 s 50 s-
t
50 s → 100 s:
−
2NO2 (g) 2NO (g) + O2 (g)
50 s → 100 s:
= r(NO2)
rate is a function of time
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1
2
Δn
Δn
V
nΔ
V
nΔ
]Δ[O
]Δ[NO
Δt]Δ[O
Δt]Δ[NO
)r(O
)r(NO
2
2
2
2
O
NO
O
NO
2
2
2
2
2
2
2NO2(g) 2NO(g) + O2(g)
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a A + b B c C + d D
r(A) a
r(B) b
r(A) a
r(C) c
d
r(D)
c
r(C)
b
r(B)
a
r(A) r =
r does not depend upon the choice of species
r(A) r(B) ,
a b
r(A) r(C)
a c
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r(N2O5) = 4.2 x 10−7 M·s−1
What are the rates of appearance of NO2 and O2 ?
2N2O5(g) 4NO2(g) + O2(g)
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Consider the following balanced chemical equation:
H2O2(aq) + 3 I–(aq) + 2 H+(aq) I3–(aq) + 2 H2O(l)
In the first 10.0 seconds of the reaction, the concentration of I– dropped from 1.000 M to 0.868 M.
(a)Calculate the average rate of this reaction in this time interval.
(b) Predict the rate of change in the concentration of H+ (that is, [H+]/t) during this time interval.
Example 13.1. page 567
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Consider the general reaction aA + bB cC and thefollowing average rate data over some time period Δt:
1Δ[A]0.0080 M s
Δt 1Δ[B]
0.0120 M sΔt
1Δ[C]0.0160 M s
Δt
Determine a set of possible coefficients to balance thisgeneral reaction.
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−
All the rates in this table are average rates.
− = r(NO2)
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0 mileBarnesville
16 miles Griffin
56 miles Atlanta
2:00 pm 2:16 pm 3:16 pm
Average Speed from B to G = 16 miles ÷ 16 min = 1.0 mile/min
Average Speed from G to A = 40 miles ÷ 60 min = 0.7 mile/min
Δl
Δt
0limt
l d l
t d t
Instantaneous speed at green spot
Instantaneous speed contains more information
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l
t
Δl
Δt
t
l
Instantaneous speed at the red point = slope of the red solid line =
Barnesville0 mile 0 min
Atlanta56 miles
Griffin20 miles
16 min 76 min
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Reaction rate is a function of time
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Factors that affect reaction rates
State of the reactants
Concentrations of the reactants
Temperature
Catalyst
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a A + b B c C + d D
r = k [A]m [B]n
Differential rate law: how r depends on concentrations
m, n: reaction order, mth order for A, nth order for B(m+n): overall reaction order
m and n must be measured from experiments. They canbe different from the stoichiometry.
k: rate constant: depends on temperature, but not concentrations
1 Δ[A] 1 d[A]
a Δt a dt
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(1) 2N2O5(g) 4NO2(g) + O2(g) r = k[N2O5]
(2) CHCl3(g) + Cl2(g) CCl4(g) + HCl(g) r = k[CHCl3][Cl2]1/2
(3) H2(g) + I2(g) 2HI(g) r = k[H2][I2]
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aA + bB cC +dD
r = k [A]m [B]n
overall reaction order (m+n) unit of k ⇌
Units
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(1) 2N2O5(g) 4NO2(g) + O2(g) r = k[N2O5]
(2) CHCl3(g) + Cl2(g) CCl4(g) + HCl(g) r = k[CHCl3][Cl2]1/2
(3) H2(g) + I2(g) 2HI(g) r = k[H2][I2]
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Experiment Number
Initial [NH4+] (M) Initial [NO2
−] (M) Initial Rate (M·s−1)
1 0.100 0.0050 1.35 x 10−7
2 0.100 0.010 2.70 x 10−7
3 0.200 0.010 5.40 x 10−7
r = k [NH4+]m [NO2
−]n
NH4+ (aq) + NO2
− (aq) N2(g) + 2H2O(l)
How to find the rate law by experiment: method of initial rates
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A very common method to investigate how each factoraffects the whole system:
Change one thing at a time while keep the others constant.
z = f (x,y)
x
z
x
z
y
How does the change of x affect z?
y
z
y
z
x
How does the change of y affect z?
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Experiment Number
Initial [NH4+] (M) Initial [NO2
−] (M) Initial Rate (M·s−1)
1 0.100 0.0050 1.35 x 10−7
2 0.100 0.010 2.70 x 10−7
3 0.200 0.010 5.40 x 10−7
r = k [NH4+]m [NO2
−]n
NH4+ (aq) + NO2
− (aq) N2(g) + 2H2O(l)
How to find the rate law by experiment: method of initial rates
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(a) Determine the differential rate law
(b) Calculate the rate constant
(c) Calculate the rate when [NO] = 0.050 M and [H2] = 0.150 M
2NO(g) + 2H2(g) N2(g) + 2H2O(g)
Experiment Number
Initial [NO] (M) Initial [H2] (M) Initial Rate (M·s−1)
1 0.10 0.10 1.23 x 10-3
2 0.10 0.20 2.46 x 10-3
3 0.20 0.10 4.92 x 10-3
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(a) Determine the differential rate law
(b) Calculate the rate constant
(c) Calculate the rate when [NO] = 0.050 M and [H2] = 0.150 M
2NO(g) + 2H2(g) N2(g) + 2H2O(g)
Experiment Number
Initial [NO] (M) Initial [H2] (M) Initial Rate (M·s−1)
1 0.10 0.10 1.23 x 10-3
2 0.10 0.30 3.69 x 10-3
3 0.30 0.10 1.11 x 10-2
again
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Experiment Number
Initial [A] (M) Initial [B] (M) Initial Rate (M·s−1)
1 0.100 0.100 4.0 x 10−5
2 0.100 0.200 4.0 x 10−5
3 0.200 0.100 1.6 x 10−4
A + B C
(a) Determine the differential rate law
(b) Calculate the rate constant
(c) Calculate the rate when [A] = 0.050 M and [B] = 0.100 M
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From the data, determine:
(a) the rate law for the reaction
(b) the rate constant (k) for the reaction
EXAMPLE 13.2 Determining the Order and Rate Constant of a Reaction
NO2(g) + CO(g) NO(g) + CO2(g)
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Use the data in table to determine
1) The orders for all three reactants 2) The overall reaction order3) The value of the rate constant
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r = k [BrO3−]m [Br−]n [H+]p
r1 = k (0.10 M)m (0.10 M)n (0.10 M)p = 8.0 x 10−4 M · s−1
r2 = k (0.20 M)m (0.10 M)n (0.10 M)p = 1.6 x 10−3 M · s−1
r3 = k (0.20 M)m (0.20 M)n (0.10 M)p = 3.2 x 10−3 M · s−1
r4 = k (0.10 M)m (0.10 M)n (0.20 M)p = 3.2 x 10−3 M · s−1
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one quiz after lab
Relationship among reaction rates as expressedby different species.
d
r(D)
c
r(C)
b
r(B)
a
r(A) r =
overall reaction order (m+n) unit of k ⇌
Method of initial rates:
table of experimental data rate order, k, rate at otherconcentrations.
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a A + b B c C + d D
r = k [A]m [B]n
Differential rate law: how r depends on concentrations
Differential rate law: differential equation
[ ] [ ][ ] [ ]m nA d A
r k A Ba t a dt
How concentration changes as a function of time integrated rate law
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A Products
Δ[A] d[A]r = r(A) k[A]
Δt dt
First order reaction differential rate law:
First order reaction integrated rate law:
0ln[A]ktln[A] kt[A]
[A]ln 0 or
integrated rate law: how concentration changes as a function of time.
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0ln[A] kt ln[A]
Plot ln[A] vs. t gives a straight line
Slope = −k, intercept = ln[A]0
First order reaction integrated rate law:
y = mx + b
[A] is the molarity of A at t
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[N2O5] (M) Time (s)
0.1000 0
0.0707 50
0.0500 100
0.0250 200
0.0125 300
0.00625 400
N2O5(g) 2NO2(g) + ½ O2(g)
Use these data, verify that the rate law is first order in N2O5,and calculate the rate constant.
k = 6.93 x 10−3 s−1
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Read a similar Example 13.3 on page 575
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Using the data given in the previous example, calculate [N2O5]
at 150 s after the start of the reaction.
[N2O5] (M) Time (s)
0.1000 0
0.0707 50
0.0500 100
0.0250 200
0.0125 300
0.00625 400
0.0354 M
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Practice on Example 13.4 on page 576 and check your answer
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The half-life of a reaction, t1/2, is the time required for a
reactant to reach one-half of its initial concentration.
[N2O5] (M) Time (s)
0.1000 0
0.0707 50
0.0500 100
0.0250 200
0.0125 300
0.00625 400
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The half-life for first order reaction:
kt
2ln2/1
The half-life for first order reaction does NOT depend onconcentration.
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[N2O5] (M) Time (s)
0.1000 0
0.0707 50
0.0500 100
0.0250 200
0.0125 300
0.00625 400
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A Plot of [N2O5] versus Time for the Decomposition Reaction of N2O5
kt2 5 2 5 0[N O ] [N O ] e
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A certain first order reaction has a half-life of 20.0 s.
a)Calculate the rate constant for this reaction.
b)How much time is required for this reaction to be 75 %
complete?
kt
2ln2/1 kt
[A]
[A]ln 0
a) k = 0.0347 s−1 b) k = 40.0 s
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Try Example 13.6 and For Practice 13.6on page 579 and check your answers
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A Products
2k[A]dt
d[A]
Δt
Δ[A]r(A)
Second order reaction differential rate law:
Second order reaction integrated rate law:
0[A]
1kt
[A]
1
integrated rate law: how concentration changes as a function of time.
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Plot 1/[A] vs. t gives a straight line
Slope = k, intercept = 1/[A]0
Second order reaction integrated rate law:
0[A]
1kt
[A]
1
y = mx + b
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The half-life for second order reaction:
02/1 ]A[
1
kt
The half-life for second order reaction depends on initial concentration.
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A certain reaction has the following general form: A BAt a particular temperature [A]0 = 2.80 x 10−3 M, concentration versus time data were collected for this reaction,and a plot of 1/[A] versus time resulted in a straight line with aslope value of 3.60 x 10−2 M−1·s−1
a)Determine the (differential) rate law, the integrated rate law, and the value of the rate constant.
b) Calculate the half-life for this reaction.
c) How much time is required for the concentration of A to decrease to 7.00 x 10−4 M ?
0[A]
1kt
[A]
1
02/1 ]A[
1
kt
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For first order reaction, show that
from the integrated rate law
kt
2ln2/1
0ln[A] kt ln[A]
(show your work, do not copy the question)
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A Products
kk[A]dt
d[A]
Δt
Δ[A]r(A) 0
Zero order reaction differential rate law:
Zero order reaction integrated rate law:
0[A]kt[A]
integrated rate law: how concentration changes as a function of time.
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Plot [A] vs. t gives a straight line
Slope = −k, intercept = [A]0
Zero order reaction integrated rate law:
0[A]kt[A]
y = mx + b
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The half-life for zero order reaction:
kt
2
]A[ 02/1
The half-life for zero order reaction depends on initial concentration.
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The Decomposition Reaction 2N2O(g) 2N2 (g) + O2 (g) takes place on a Platinum Surface
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The decomposition of ethanol on alumina surface C2H5OH(g) C2H4(g) + H2O(g) was studied at 600 K.Concentration versus time data were collected for this reaction, and a plot of [C2H5OH] versus time resulted in a straight line with a slope value of −4.00 x 10−5 M·s−1
a)Determine the (differential) rate law, the integrated rate law, and the value of the rate constant.
b) If the initial concentration of C2H5OH was 1.25 x 10−2 M, calculate the half-life of this reaction.
c) How much time is required for all the 1.25 x 10−2 M C2H5OH to decompose ? 0[A]kt[A]
kt
2
]A[ 02/1
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Factors that affect reaction rates
State of the reactants
Concentrations of the reactants
Temperature
Catalyst
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r = k [A]m [B]n
k: rate constant: depends on temperature, but not concentrations
aA + bB cC +dD
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A Plot Showing the Exponential Dependence of the Rate Constant on Absolute Temperature
Collision model: molecules must collide to react.
T ↑ v ↑ kinetic energy = ½ mv2 ↑
Not all collisions lead to products
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Prentice Hall © 2003 Chapter 14
RTEaef /Fraction of molecules whose Ek > Ea is
T ↑ f ↑ Ea ↑ f ↓
activation energy
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RTEazezfk /
Another factor needs to be taken into account
molecular orientation
exptkk
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Several Possible Orientations for a Collision Between Two BrNO Molecules
BrNO + BrNO 2NO +Br2
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212
1 11ln
TTR
E
k
k a
Arrhenius equation
RTEaAek /
RTEapzek / steric factor p ≤ 1
A: frequency factor How k depends on T
T ↑ k ↑ Ea ↑ k ↓
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At 550 °C the rate constant for
CH4(g) + 2S2(g) CS2(g) + 2H2S(g)
is 1.1 M·s−1, and at 625 °C the rate constant is
6.4 M·s−1. Using these value, calculate Ea for this reaction.
212
1 11ln
TTR
E
k
k a
Ea = 1.4 x 105 J/mol
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Try Example 13.8 on page 585 and check your answers
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The ball can climb over the hill only if its kinetic energy is greaterthan Ep = mgh, where m is the mass of the ball, h is the heightof the hill, and g is gravitational acceleration.
h
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(a) The Change in Potential as a Function of Reaction Progress (b) A Molecular Representation of the Reaction
BrNO + BrNO 2NO +Br2
O N Br
exothermic reaction
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Chemical Equation
Reactants Products
Reaction Mechanism
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NO2 + CO NO + CO2
Step 1: NO2 + NO2 NO3 + NO Step 2: NO3 + CO NO2 + CO2 +
NO2 + CO NO + CO2
NO3: intermediate, does notappear in overall reaction
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Whole process is called the reaction mechanism.
Each single step is called an elementary reaction/step.An elementary reaction is a single collision.
NO2 NO2 NO3 NO
NO3 CO NO2 CO2
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For an elementary reaction
aA + bB cC + dD
r = k[A]a[B]b
Not for overall reaction!
The number of molecules that react in an elementary reaction is called the molecularity of that that elementary reaction (not applicable to overall reaction).
Overall reaction: r = k[A]m[B]n , find m and n by experiments
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1 ― unimolecular2 ― bimolecular3 ― termolecular
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What determines the rate of an overall reaction?
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Team A
Team B
Slowest step: rate determining step
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Ep
Reaction progress
Reactants
Products
∆E
intermediate
Which step is the rate determining step?
exothermic orendothermic?
Ea, 2
Ea, 1
2nd
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Factors that affect reaction rates
State of the reactants
Concentrations of the reactants
Temperature
Catalyst
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Catalyst is a substance that speeds up a reaction without
being consumed itself.
Catalyst changes the reaction mechanism through a lower
activation energy pathway.
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Energy Plots for a Given Reaction
Ea
Ea
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catalyst
homogeneous
heterogeneous
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Homogeneous catalyst: same phase as reactants.
3O2(g) 2O3(g)
2NO(g) + O2(g) 2NO2(g)
2NO2(g) 2NO(g) + 2O(g)
2O2(g) + 2O(g) 2O3(g)
light
+
3O2(g) 2O3(g)
NO(g): catalyst; NO2(g), O(g): intermediates
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The Decomposition Reaction 2N2O(g) 2N2 (g) + O2 (g) takes place on a Platinum Surface
Heterogeneous catalyst: different phase from reactants.
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These Cookies Contain Partially Hydrogenated Vegetable Oil
C = C + H2 − C − C −
− −−−
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milk sugar = lactose
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