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Chapter Twelve:

CHEMICAL KINETICS

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Contents p526

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Two gases can coexist indefinitely at 25℃.

To be useful, reactions must occur at a reasonable rate.

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2H2(g) + Cl2(g) → 2 HCl(g)

2H2(g) + O2(g) → 2H2O(l)

N2(g) + 3H2(g) → 2NH3(g)

The area of chemistry that concerns reaction rates is

called chemical kinetics.

One of the main goals of chemical kinetics is to

understand the steps by which a reaction rakes place.

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12-1 Reaction Ratesp527

2NO2(g) → 2NO(g) + O2(g)

Change in concentration (conc.) of a reactant

or product per unit time.

Rate =conc. of A at time conc. of Aat time2 1

2 1

t t

t t

--

At

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Calculate the average rate at which the concentration

of NO2 changes over the first 50 seconds of the

reaction using the data given in Table 12.1.

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Rate LawsRate = k [NO2]n

k = rate constant

n = rate order

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12-2 Rate Laws: An Introductionp532

Rate LawsRate = k[NO2]n k = rate constant

n = rate order

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Differential rate law

Integrated rate law

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Reaction Rate and Concentration

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Rate Laws: A Summary p534

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12-3 Determining the Form of theRate Law

Figure 12.3

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Method of Initial Rates p535

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P537Ex 12.1 Determining a Rate Law

The reaction between bromate ions and bromide ions

in acidic aqueous solution is given by the equation

Table 12.5 gives the results from four experiments.

Using these data, determine the orders for all three

reactant, the overall reaction order, and the value of

the rate constant.

)(OH3)(Br3)(H6)(Br5BrO 223 llaqaqaq

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Solution: p537

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12-4 The Integrated Rate Law p538

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p539First-Order Rate Laws

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(12.2)

The integrated first-order law is

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p539Integrated First-Order Rate Law

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P539Ex 12.2 First-Order Rate Laws IThe decomposition of N2O5 in the gas phase was

studied at constant temperature.)()(4)(2 2252 gOgNOgON

Using these data, very that the rate law is first order in[N2O5], and calculate the value of the rate constant, wherethe rate = -Δ[N2O5]/Δt.

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Solution p540

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P540

Ex 12.3 First-Order Rate Laws IIUsing the data giving in Sample Exercise 12.2. calculate

[N2O5] at 150 s after the start of the reaction.Solution:

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Since the reaction is first order, the slop of the line

equals –k, where

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Half-Life of a First-Order Reaction p541

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P542Ex 12.4 Half-Life for First-Order ReactionA certain first-order reaction has a half-life of 20.0

minutes.

a. Calculate the rate constant for this reaction.

b. How much time is required for this reaction to be75% complete?

Solution

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Second-Order Rate Lawsp543

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Half-life of the second-order reaction p543

(12-6)

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Ex 12.5 Determining Rate Laws P544

Butsdiene reacts to form its dimer according to theequation )()(2 12864 gHCgHC

The following data were collected for thisreaction at a given temperature:

a. Is this reaction first order or secondorder?

b. What is the value of the rate constantfor the reaction?

c. What is the half-life for the reactionunder the conditions of thisexperiment?

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Solution

a. The data necessary to make these plots are as follows:

b. For a second-order reactions, a plot of 1/[C4H6]versus t produces a straight line of slope k.

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c. The expression for the half-life of a second reaction is

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Zero-Order Rate Laws p546

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Zero-Order Reactionp546

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Integrated Rate Laws for reaction withMore Than One Reactant

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The decomposition reaction of N2O iszero order

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pseudo-first-order rate lawp547

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Rate Laws

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Half-Life of Reactions

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Summaryp548

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Important points:p548

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p548Table 12.6 Summary of the kinetic for reactions of the typeaA → products that are zero, fist, or second order in [A]

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12-6 Reactions Mechanisms p549

Gaseous NO3 is an intermediate, a species that isneither a reactant nor a product but that is formed andconsumed during the reaction sequence.

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A Molecular Representation of the

Elementary Steps in the Reaction of NO2

and CO

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Figure 12.9

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Decomposition of N2O5

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Oscillating Reactions

)(2)()(2 222 gFNOgFgNO

FFNOFNO K 2221

P551Ex 12.6 Reaction MechanismsThe balanced equation for the reaction of the gases nitrogendioxide and fluorine is

The experimentally determined rate law is

Rate = k [NO2][F2]

A suggested mechanism for this reaction is

Is this an acceptable mechanism. That is, does it satisfy thetwo requirements?

FNONOF K22

2

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Solution:

The second requirement is that the mechanism must agreewith the experimentally determined rate law. Since theproposed mechanism states that first rate determining, theoverall reaction rate must be that of the first step. The firststep is bimolecular, so the rate law is

Rate = k1 [NO2] [F2]

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12-7 A Model for Chemical Kineticsp552

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The exponential dependence of rate constant onabsolute temperature

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Change in Potential Energy

Figure 12.11

(a) The change in potential energy as function of a reaction progressfor the reaction 2BrNO → 2NO + Br2. The activation energy E1represents the energy needed to disrupt the BrNO molecules so thatthey can form products. The quantity △E represents the net changeun energy in going from reactant to products. (b) A molecularrepresentation of the reaction.

p553Transition state

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The number of collisionsp554

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Possible orientations for a collisionp554

Arrhenius equation p554

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P555Ex 12.7 Determining Activation Energy I

)()(4)(2 2222 gOgNOgON

The reaction was studied

at several temperatures,

and the following values

of k were obtained :

Calculate the value ofEa for this reaction. Figure 12.14

Solution

To obtain the value of Ea, we need to construct a plot of In(k )versus 1/T. First we must calculate values of In(k ) and 1/T, asbelow:

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Ex 12.8 Determining Activation Energy II

sL/mole

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The gas-phase reaction between methane and diatomic

sulfur is given by the equation

At 550 ℃ the rate constant for this reaction is 1.1 .

Using these values, calculate Ea for this reaction.

)(SH2)(CS)(S2)(CH 2224 gggg

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Solutionp557

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The Gas Phase Reaction of NO and Cl2

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Transition States and Activation Energy

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12-8 Catalysis p558

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Heterogeneous Catalysis p558

Effect of a Catalyst on the Numberof Reaction-Producing Collisions

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Figure 12.16 Effect of a catalyst on the number of a reaction-

producing collisions. Because a catalyst provides a reaction pathway

with a lower activation energy, a much greater fraction of the collisions

is effective for the catalyzed pathway (b) than for the uncatalyzed

pathway (a) (at a given temperature). This allows reactants to become

products at a much higher rate, even through there is no temperature.

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Involve four steps in aheterogeneous catalysis reaction

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Automobiles: Air Purifiers?p560

Chlorine atoms can catalyze thedecomposition of ozoneFreon-12(CCl2F2) is used as refrigerants and propellants in aerosol

cans. The chemical inertness of Freons makes them valuable but

also creates a problem, since they remain in the environment a

long time. Eventually, they migrate into the upper atmosphere to

be decomposed by high-energy light. Among the decomposition

products are chlorine atoms:

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The ozone destruction in the

atmosphere

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Homogeneous Catalysis