nyb - kinetics -(ppt) -fall 2007
TRANSCRIPT
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202-NYB - KINETICS
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Figure 16.2 The wide range of reaction rates.
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Figure 16.1 Reaction rate: the central focus of chemical kinetics
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Factors That Influence Reaction Rate
Under a specific set of conditions, every reaction has its owncharacteristic rate, which depends upon the chemical nature of the reactants.
Four factors can be controlled during the reaction:
2. Concentration - molecules must collide to react;3. Physical state - molecules must mix to collide;
4. Temperature - molecules must collide with enough energy to react;
5. The use of a catalyst.
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Figure 16.3 The effect of surface area on reaction rate.
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Expressing the Reaction Rate
reaction rate - changes in the concentrations of reactants or products per unit time
reactant concentrations decrease while product concentrationsincrease
rate of reaction = -
for A B
change in concentration of A
change in time= -
conc A 2-conc A 1
t2-t1
(conc A)- t
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Table 16.1 Concentration of O 3 at Various Time in itsReaction with C 2H4 at 303K
C2H4(g ) + O 3(g ) C 2H4 O( g ) + O 2(g )
Time (s) Concentration of O 3 (mol/L)
0.0
20.0
30.0
40.050.0
60.0
10.0
3.20x10 -5
2.42x10 -5
1.95x10 -5
1.63x10 -5
1.40x10 -5
1.23x10 -5
1.10x10 -5
(conc A)- t
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Figure 16.5
The concentrations of O 3 vs. time during its reaction with C 2H4
C2H4(g ) + O 3(g ) C 2H4 O( g ) + O 2(g )
- [C2H4]
t
rate =
- [O3]
t
=
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Figure 16.6 Plots of [C 2H4] and [O 2] vs. time.
Tools of the Laboratory
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In general, for the reaction
aA + bB cC + dD
rate =1
a- = - [A]
t
1
b
[B]
t
1
c
[C]
t= +
1
d
[D]
t= +
The numerical value of the rate depends upon the substance thatserves as the reference. The rest is relative to the balancedchemical equation.
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Sample Problem 16.1
PLAN:
SOLUTION:
Expressing Rate in Terms of Changes inConcentration with Time
PROBLEM: Because it has a nonpolluting product (water vapor), hydrogengas is used for fuel aboard the space shuttle and may be usedby Earth-bound engines in the near future.
2H 2(g) + O 2(g) 2H 2O(g)
(a) Express the rate in terms of changes in [H 2], [O 2], and [H 2O] with time.
(b) When [O2] is decreasing at 0.23 mol/L*s, at what rate is [H
2O]
increasing?
Choose [O 2] as a point of reference since its coefficient is 1. For everymolecule of O 2 which disappears, 2 molecules of H 2 disappear and 2molecules of H 2O appear, so [O 2] is disappearing at half the rate of
change of H 2 and H 2O.-
12
[H2] t
= - [O2]
t= +
[H2O] t
12
0.23mol/L*s = + [H2O]
t12
; = 0.46mol/L*s [H2O]
t
rate =(a)
[O2] t
- = -(b)
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Sample Problem 16.2
SOLUTION:
Determining Reaction Order from Rate Laws
PROBLEM: For each of the following reactions, determine the reaction order
with respect to each reactant and the overall order from thegiven rate law.
(a) 2NO( g) + O 2(g) 2NO 2(g); rate = k [NO]2[O2]
(b) CH 3CHO( g) CH 4(g) + CO( g); rate = k [CH 3CHO] 3/2
(c) H2O 2(aq ) + 3 I-
(aq ) + 2H+
(aq ) I3-
(aq ) + 2H 2O( l); rate = k [H2O 2][I-
]PLAN: Look at the rate law and not the coefficients of the chemical reaction.
(a) The reaction is 2nd order in NO, 1st order in O 2, and 3rd order overall.
(b) The reaction is 3/2 order in CH 3CHO and 3/2 order overall.
(c) The reaction is 1st order in H 2O 2, 1st order in I- and zero order in H +,while being 2nd order overall.
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Table 16.2 Initial Rates for a Series of Experiments in theReaction Between O 2 and NO
Experiment
Initial ReactantConcentrations (mol/L)
Initial Rate(mol/L*s)
1
2
3
4
5
O 2 NO
1.10x10 -2 1.30x10 -2 3.21x10 -3
1.10x10 -2 3.90x10 -2 28.8x10 -3
2.20x10 -2
1.10x10 -2
3.30x10 -2
1.30x10 -2
2.60x10 -2
1.30x10 -2
6.40x10 -3
12.8x10 -3
9.60x10 -3
2NO( g) + O 2(g) 2NO 2(g)
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Sample Problem 16.3
PLAN:
SOLUTION:
Determining Reaction Order from Initial Rate Data
PROBLEM: Many gaseous reactions occur in a car engine and exhaust
system. One of these isNO 2(g ) + CO( g ) NO( g ) + CO 2(g ) rate = k [NO 2]m[CO] n
Use the following data to determine the individual and overall reaction orders.
Experiment Initial Rate(mol/L*s) Initial [NO 2] (mol/L) Initial [CO]
(mol/L)1
2
3
0.00500.0800.0050
0.10
0.100.400.10
0.10
0.20
Solve for each reactant using the general rate law using themethod described previously.
rate = k [NO 2]m[CO] n
First, choose two experiments in which [CO] remainsconstant and the [NO 2] varies.
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Sample Problem 16.3 Determining Reaction Order from Initial Rate Data
continued
0.080
0.0050
rate 2
rate 1
[NO 2] 2[NO 2] 1
m=
k [NO 2]m 2[CO] n2k [NO 2]m 1 [CO] n1
=
0.40
0.10=
m
; 16 = 4m
and m = 2
k [NO 2]m3[CO] n3k [NO 2]m1 [CO] n1
[CO] 3
[CO] 1
n=
rate 3
rate 1=
0.00500.0050
=0.200.10
n ; 1 = 2 n and n = 0
The reaction is2nd order in NO 2.
The reaction iszero order in CO.
rate = k [NO 2]2[CO] 0 = k [NO 2]2
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Table 16.3 Units of the Rate Constant k for Several OverallReaction Orders
Overall Reaction Order Units of k (t in seconds)
0 mol/L*s (or mol L -1 s -1)
1 1/s (or s -1)
2 L/mol*s (or L mol -1 s -1)
3 L2 / mol 2 *s (or L 2 mol -2 s -1)
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Figure 16.8 Graphical determination of the reaction order for thedecomposition of N 2O 5.
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Figure 16.9 A plot of [N 2O 5] vs. time for three half-lives.
t1/2 =
for a first-order process
ln 2
k
0.693
k =
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Sample Problem 16.5
PLAN:
SOLUTION:
Determining the Half-Life of a First-Order Reaction
PROBLEM: Cyclopropane is the smallest cyclic hydrocarbon. Because its60 0 bond angles allow poor orbital overlap, its bonds are weak.As a result, it is thermally unstable and rearranges to propene at1000 0C via the following first-order reaction:
CH 2
H2C CH 2 (g )
H3C CH CH 2 (g )
The rate constant is 9.2s -1, (a) What is the half-life of the reaction? (b) Howlong does it take for the concentration of cyclopropane to reach one-quarter of the initial value?
Use the half-life equation, t 1/2 =0.693
k , to find the half-life.
One-quarter of the initial value means two half-lives have passed.
t1/2 = 0.693/9.2s -1 = 0.075s(a) 2 t 1/2 = 2(0.075s) = 0.150s(b)
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Table 16.4 An Overview of Zero-Order, First-Order, andSimple Second-Order Reactions
Zero Order First Order Second Order
Plot for straight line
Slope, y -intercept
Half-life
Rate law rate = k rate = k [A] rate = k [A]2
Units for k mol/L*s 1/s L/mol*s
Integrated rate law instraight-line form
[A]t =k t + [A]0
ln[A]t =-k t + ln[A] 0
1/[A]t =k t + 1/[A] 0
[A]t vs. t ln[A]t vs. t 1/[A] t = t
k, [A]0 -k, ln[A]0 k, 1/[A]0
[A]0/2k ln 2/ k 1/k [A]0
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Figure 16.10 Dependence of the rate constant on temperature
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The Arrhenius Equation
The Effect of Temperature on Reaction Rate
k = Ae Ea
RT
ln k = ln A - E a /RT
lnk
2
k 1
=E aRT
-1T2
1T1
-
where k is the kinetic rate constant at T
E a is the activation energy
R is the energy gas constantT is the Kelvin temperature
A is the collision frequency factor
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Figure 16.11 Graphical determination of the activation energy
lnk
= -E a /R (1/T) + ln A
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Sample Problem 16.6
PLAN:
SOLUTION:
Determining the Energy of Activation
PROBLEM: The decomposition of hydrogen iodide,
2H I(g) H 2(g) + I2(g)
has rate constants of 9.51x10 -9L/mol*s at 500. K and 1.10x10 -5
L/mol*s at 600. K. Find E a.
Use the modification of the Arrhenius equation to find E a.
lnk 2k 1
=E a
-R
1
T2
1
T1
- E a = - R lnk 2k 1
1
T2
1
T1
--1
1
600K
1
500K-ln
1.10x10 -5L/mol*s
9..51x10 -9L/mol*sE a = - (8.314J/mol*K)
E a = 1.76x10 5 J/mol = 176 kJ/mol
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Figure 16.12Information sequence to determine the kinetic parameters of a reaction.
Series of plots
of concentra-tion vs. time Initial
rates Reactionorders
Rate constant(k ) and actual
rate law
Integratedrate law(half-life,
t1/2 )
Rate constant
and reactionorder
Activationenergy, E a
Plots of concentration
vs. time
Find k atvaried T
Determine slopeof tangent at t 0 for
each plot
Compare initialrates when [A]
changes and [B] isheld constant and
vice versa
Substitute initial rates,orders, and concentrations
into general rate law:rate = k [A] m[B] n
Use direct, ln orinverse plot to
find order
Rearrange tolinear form and
graph
Find k atvaried T
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Figure 16.13
The dependence of possible collisions on the productof reactant concentrations.
AA
AA
BB
BB
AA
AA
BB
BBAA
4 collisions
Add another molecule of A
6 collisions
Add another molecule of B
AA
AA
BB
BB
AA BB
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Figure 16.14
The effect of temperature on the distribution of collision energies
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Figure 16.15 Energy-level diagram for a reaction
REACTANTS
PRODUCTS
ACTIVATED STATE
C o
l l i s i o n
E n e r g y
C o
l l i s i o n
E n e r g y
E a (forward)
E a (reverse)
The forward reaction is exothermic because thereactants have more energy than the products.
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Figure 16.16 An energy-level diagram of the fraction of collisionsexceeding E a .
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Figure 16.17
The importance of molecular orientation to an effective collision.
NO + NO 3 2 NO 2
A is the frequency factor
A = pZ where Z is the collision frequencyp is the orientation probability factor
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Table 16.6 Rate Laws for General Elementary Steps
Elementary Step Molecularity Rate Law
A product
2A product
A + B product
2A + B product
Unimolecular
Bimolecular
Bimolecular
Termolecular
Rate = k [A]
Rate = k [A]2
Rate = k [A][B]
Rate = k [A]2[B]
REACTION MECHANISMS
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