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To fully understand any chemical reaction, there are four aspects of any reaction that To fully understand any chemical reaction, there are four aspects of any reaction that

To fully understand any chemical reaction, there are four aspects of any reaction that - PowerPoint Presentation

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To fully understand any chemical reaction, there are four aspects of any reaction that - PPT Presentation

Stoichiometry identity and relative amounts of reactants and products Spontaneity feasibility of the reaction based on thermodynamics SpeedKINETICS reaction rates Stoppingwhen will the reaction stop Equilibriumnext chapter ID: 934996

reaction rate order kinetics rate reaction kinetics order concentration initial time law integrated reactants mechanism constant kt1 kineticspractice energy

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Slide1

To fully understand any chemical reaction, there are four aspects of any reaction that we study;Stoichiometry (identity and relative amounts of reactants and products).Spontaneity (feasibility of the reaction, based on thermodynamics).Speed…KINETICS (reaction rates).Stopping…when will the reaction stop? (Equilibrium-next chapter)Focus of this chapter (13) is #3!

Kinetics

Slide2

RATES of CHEMICAL REACTIONSAverage Rate Rate = ∆ [A] ∆ tInstantaneous Rate (Rate at any given moment. Equal to the slope of a line tangent to the curve on a conc. vs. time graph.)Initial Rate (Instantaneous rate at the outset of a reaction.)

Kinetics

Slide3

RATE LAWSDifferential Rate Law: relates rate to concentrationIntegrated Rate Law: relates concentration to timeKinetics

Slide4

Differential Rate LawThis is the “default” form of the rate law. If a problem asks for the “rate law” – give this form.FORMAT: Rate = k[A]x[B]y[C]z… (k is the “Rate Constant” which is temperature dependent, A, B & C are reactants, x, y & z are the respective “orders” of those reactants.)METHOD: Method of initial rates (text, pp. 573-576).

Kinetics

Slide5

EXAMPLE: Method of Initial RatesKineticsExp.#[HgCl2], M

[C

2

O

4

2-

], M

Initial Rate (mol/

Lmin

)

10.1050.151.8 x 10-520.1050.307.1 x 10-530.0520.303.5 x 10-540.0520.158.9 x 10-6

2 HgCl

2

+ C

2

O

4

2-

 2

Cl

-

+ 2 CO

2

+ Hg

2

Cl

2

Slide6

KineticsExp.#[HgCl2], M[C2O42-], MInitial Rate (mol/Lmin

)

1

0.105

0.15

1.8 x 10

-5

2

0.105

0.30

7.1 x 10-530.0520.303.5 x 10-540.0520.158.9 x 10-62 HgCl2 + C2O4

2-

 2

Cl

-

+ 2 CO

2

+ Hg

2

Cl

2

Slide7

KineticsRate Laws and Gas Pressure…PV = nRTConcentration = =Since R is a constant, and T will be the same for the given scenario, P is proportional to Concentration.

Slide8

KineticsPractice the Method of Initial Rates, general rate ideas,units for rate constants, etc.,CHANG – pp. 611-612, #13-20ZUMDAHL – pp. 597-599, #9-26

Slide9

INTEGRATED RATE LAWUsed when concentration vs. time data are easier to obtain than initial rates of several trials. Typically only account for single reactant reactions.Determine the order of the single reactant using a graphic method;If [A] vs. time yields a straight line, A is zero order.If ln[A] vs. time yields a straight line, A is first order.If 1/[A] vs. time yields a straight line, A is second order.

Kinetics

Slide10

INTEGRATED RATE LAW EQUATIONSZero Order: [A] = -kt + [A]0First Order: ln[A] = -kt + ln[A]0Second Order: 1/[A] = +

kt

+ 1/[A]

0

Kinetics

Slide11

HALF LIFEThe amount of time required for a sample to reach half of its initial concentration.Integrated rate laws work well for half life, since they relate the concentration of the reactant to the amount of time the reaction has undergone. Here’s the key idea…At t=t1/2, [A] = [A]0/2Kinetics

Slide12

HALF LIFEThe amount of time required for a sample to reach half of its initial concentration.For zero order, since [A] = [A]0/2 …[A]0/2 = -kt1/2 + [A]0Solving for the half-life…kt1/2 = [A]

0

- [A]

0

/2

kt

1/2

= [A]

0

/2

t1/2 = [A]0/2kKinetics

Slide13

HALF LIFEThe amount of time required for a sample to reach half of its initial concentration.For first order, since [A] = [A]0/2 …ln([A]0/2) = -kt1/2 + ln[A]0Solving for the half-life…

kt

1/2

=

ln

[A]

0

-

ln

[A]

0/2= ln([A]0/[A]0/2) kt1/2 = ln(2)=0.693t1/2 = 0.693/kKinetics

Slide14

HALF LIFEThe amount of time required for a sample to reach half of its initial concentration.For second order, since [A] = [A]0/2 …1/([A]0/2) = +kt1/2 + 1/[A]0Solving for the half-life…1/([A]0/2) – 1/[A]

0

= +kt

1/2

kt

1/2

= 2/[A]

0

– 1/[A]

0

= 1/[A]0t1/2 = 1/k[A]0Kinetics

Slide15

% CompletionIf a second order reaction is 65% complete in 12.7 minutes when the initial concentration is 0.200 M, what is the rate constant in L/mol*s?Kinetics

Slide16

PSEUDO-INTEGRATED RATE LAWSIRL’s for reactions with more than one reactant can be improvised by studying the reaction with one concentration MUCH smaller than the rest.Kinetics

Slide17

PSEUDO-INTEGRATED RATE LAWSIf ; A + B + 2 C  productsand [A]0 [B]0 and [C]0, then changes in [B] and [C] will be negligible (they will be vitrually constant).

Rate ≈ k'[A]

n

(where k' = k[B]

x

[C]

y

)

(k' is a “pseudo” rate constant)

Kinetics

Slide18

PSEUDO-INTEGRATED RATE LAWSSince Rate ≈ k'[A]n, a pseudo-integrated rate law can be written based on the order of A under these conditions. (For example, if A is second order, we can approximate that;1/[A] = +k't + 1/[A]0 IF [A]0

[B]

0

and [C]

0

)

Kinetics

Slide19

PSEUDO-INTEGRATED RATE LAWSExample: Zumdahl, p. 601, #44 (Let’s do this one together!)Kinetics

Slide20

KineticsPractice Integrated Rate Laws and time-concentration relationships.CHANG – pp. 612, #21-30ZUMDAHL – pp. 599-601, #27-44

Slide21

COLLISION MODEL of KINETICSAssumes that in order for a reaction to occur……the reactants must collide with each other, …they must collide with a proper orientation, and… the collision must be energetic enough to result in a reaction. (Must overcome the activation energy.)Kinetics

Slide22

COLLISION MODELFactors affecting reaction rate…Concentration – the more crowded the reactants, the more likely a collision is to occur.Temperature – as temperature increases, the molecules move faster, increasing the frequency AND energy of collisions.Catalysis – using a catalyst can provide a different, lower energy pathway for the reaction to proceed.Kinetics

Slide23

COLLISION MODELEvolution of a useful equation…k = zpe-Ea/RT (z = frequency of collisions, p = “steric factor” – which is the fraction of collisions with proper orientation, Ea = activation energy in J/mol) NOT VERY USEFULk = Ae-Ea/RT (z & p are combined to form “Arrhenius factor”. A = frequency of collisions with proper orientation.) STILL NOT VERY USEFUL

ln

k = -Ea/R(1/T) +

ln

A USEFUL

Kinetics

Slide24

KineticsPractice relating rate constant, activation energy & temperature.CHANG – pp. 613-614, #31-47ZUMDAHL – pp. 602, #49-60

Slide25

Reaction MechanismsH2 + 2 NO  N2O + H2OKinetics

KEY VOCABULARY:

Elementary step-

Molecularity

-

Unimolecular

-

Bimolecular-

Termolecular

-

Rate-determining step -

A reaction for which the rate law can be written from its

molecularity

.

The sum of the coefficients of the reactants

One molecule on the reactants side.

Two molecules must collide to react.

Three molecules must collide simultaneously.

The slowest step in the mechanism.

Slide26

Reaction MechanismsWe cannot prove a mechanism, but we can prove that a mechanism is not possible. A plausible mechanism WILL… …have steps that add up to the overall reaction, AND…agree with the experimentally determined rate law. (Its rate determining step will have a rate law, written from its molecularity, which agrees with the observed rate law.Kinetics

Slide27

Reaction MechanismsZumdahl, p. 601, #46Kinetics

Slide28

Reaction MechanismsSPECIAL CONSIDERATION – Mechanisms which have fast, reversible steps and include an intermediate…EX: For the overall reaction – 2 O3  3 O2With observed rate law – Rate = k[O3]2/[O

2

]

The following mechanism is proposed…

O

3

 O

2

+ O

O + O

3  2 O2Is this mechanism plausible?Kinetics

Slide29

2 O3  3 O2Rate = k[O3]2/[O2]O

3

 O

2

+ O

O + O

3

 2 O

2

Kinetics

Slide30

KineticsPractice with Reaction MechanismsCHANG – pp. 614-615, #48-58ZUMDAHL – pp. 601-602, #45-48

Slide31

CATALYSTSHomogeneous Catalysis – catalyst is in the same physical state as the reactants. (Ex. – free radicals in atmospheric chemistry.)Heterogeneous Catalysis – catalyst is in a different physical state than the reactants. (Ex. – using metal plates or other solid surface to catalyze aqueous, liquid or gas-phase reactions, such as in the hydrogenation of unsaturated hydrocarbons.)

Kinetics

Slide32

CATALYSTS4 Steps of Heterogeneous Catalysis…AdsorptionMigrationReactionEscape (More details on p. 590 of Zumdahl)

Kinetics

Slide33

CATALYSTSRelating rate constants to activation energies using different catalysts (or no catalyst).Kineticsln k = -Ea/R*(1/T) + ln A, SO…lnk1 + Ea1/R*(1/T)=lnk2

+ Ea

2

/R*(1/T)

lnk

1

- lnk

2

= Ea

2

/R*(1/T) - Ea1/R*(1/T),SO…

Slide34

KineticsPractice – catalysis and activation energyCHANG – p. 615, #59-66ZUMDAHL – p. 603, #61-64