nyb - kinetics -(ppt) -fall 2007

8/14/2019 NYB - Kinetics -(PPT) -Fall 2007

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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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nyb - kinetics -(ppt) -fall 2007

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