1 chemical kinetics: rates of reactions chapter 13 svante a. arrhenius 1859-1927.* developed concept...

Post on 13-Jan-2016

219 Views

Category:

Documents

1 Downloads

Preview:

Click to see full reader

TRANSCRIPT

1

Chemical Kinetics:Chemical Kinetics:Rates of ReactionsRates of Reactions

Chapter 13Chapter 13

Svante A. Arrhenius1859-1927.*Developed concept ofactivation energy; assertedsolutions of salts containedions.

2

Chemical KineticsChemical Kinetics• Kinetics is the study of how fast chemical reactions

occur.• There are 3 important factors which affect rates of

reactions:– reactant concentration [molarity or pressure]

– temperature

– action of catalysts

• Goal: to understand chemical reactions at the molecular level.

3

Chemical KineticsChemical KineticsReaction RatesReaction Rates• Speed of a reaction is measured by the change in

concentration with time [brackets means concentration, units of molarity, M]

• For a reaction A B

At the beginning of the reaction (t = 0), let us assume that

[A] = 1.0 M and [B] = 0.

Average rate = change in conc. of B change in time

=Δ[B]

Δ t

4

Reaction RatesReaction Rates A B

Con

c (M

)

[ ]

[ ]

Concentration of reactant and product as function of time

5

Reaction ratesReaction rates

The rate for this reaction depends on the time. At beginning of reaction, rate is fast, and the rate slows down as the reaction proceeds.

At any time, the rate is given by either: (a) change in [A] with change in time, or… (b) change in [B] with change in time.

There are two ways of calculating rates: (a) average rate (rough and imprecise) (b) instantaneous rate (more exact)

Average rate = change in conc. of B change in time

=Δ[B]

Δ t

- Δ[A]

Δ t=

6

Average rate is rate between two time intervals

At t=10 min, [B]=0.26 M; at t=30 min, [B]=0.60 MAverage rate = [B]/ t = (0.60-0.26)/(30-10) = (.34 M / 20 min) = 0.017 M/min

con

c(M

)

[ ]

[ ]

What is rate between 10 min and 30 min?

7

The instantaneous rate is more precise than the average rate.It is the rate at some specific instance in time; the concept is derived from the calculus.It is the slope of a curve at a specific instance in time. This is obtained by constructing a tangent to the curve at the indicated point.

“rise” = Δy

“run” = Δx

ΔyΔx

M

time

Slope =

8

0.6 M

38 min

What is instantaneous rate at t = 20 min?

Con

c (M

)

= 0.96 M/sRate =Δ[B] Δ t =

0.6 M38 min 0.016 M/min=

9

Reaction RatesReaction Rates A B

Con

c (M

)

[ ]

[ ]

When do we use (+) and (-) signs for rate?

Slopes for reactants (A) are always negative (-)Slopes for products (B) are always positive (+)But, all rates are positive (+). Therefore….If slope is (-), then Rate = - slope = (-)(-) = +If slope is (+), then Rate = +slope = (+)(+) = +

10

Reaction Rates and StoichiometryReaction Rates and Stoichiometry• For the reaction: N2 + 3 H2 2 NH3

we can expression reaction rate as:

(a)

(b)

Δt

]Δ[NRate 2 (N2 is a reactant)

Δt

]Δ[HRate 2 (H2 is a reactant)

(c) Δt

]Δ[NHRate 3

(NH3 is a product)

But these rates are not equal!NH3 is produced twice as fast as N2 is reacted. So,

Δt

]Δ[N2

Δt

]Δ[NH 23

11

The Dependence of Rate on ConcentrationThe Dependence of Rate on Concentration• In general rates increase as concentrations increase.

NH4+(aq) + NO2

-(aq) N2(g) + 2H2O(l)

x2 same x2

x2same x2

Let’s isolate these four experiments (1,2,5,6)

12

Initial [NH4+] Initial [NO2

-] Rate (M/s) 1. 0.0100 0.200 5.4 x 10-7

2. 0.0200 0.200 10.8 x 10-7

5. 0.200 0.0202 10.8 x 10-7

6. 0.200 0.0404 21.6 x 10-7

X2

same X2

X2

X2

same

as [NH4+] doubles (with [NO2

-] constant), the rate doubles,

so: Rate [NH4+]

Rate = k[NH4+][ NO2

-] where k is the rate constantThis reaction is 1st order with respect to NH4

+ and

1st order with respect to NO2- and is

1+1=2 or 2nd order overall

as [NO2-] doubles (with [NH4

+] constant), the rate doubles,

so: Rate [NO2-]

We conclude: Rate [NH4+][NO2

-]

13

The Dependence of Rate on ConcentrationThe Dependence of Rate on ConcentrationDetermine rate expression and rate constant, k, for following:

___[A]___ ___[B]____ __Rate(M/s)____ 0.2 0.6 3.5 0.4 0.6 7.0 0.4 1.2 14.0

___[A]___ ___[B]____ __Rate____ 0.4 0.7 2.1 0.8 0.7 8.4 0.8 1.0 8.4

___[A]___ ___[B]____ __Rate____ 0.1 0.4 1.8 0.1 0.8 7.2 0.2 0.4 3.6

14

The Change of Concentration with TimeThe Change of Concentration with TimeFirst-Order ReactionsFirst-Order Reactions

•A plot of ln[A]t versus t is a straight line with slope -k and intercept ln[A]0.

•In the above we use the natural logarithm, ln, which is log to the base e.

A products

Rate = k[A]Δt

Δ[A] When integrated, it gives the

1st order rate equation:

kt[A]

[A]ln

t

o OR: ot ln[A]ktln[A]

15

The Change of Concentration with TimeThe Change of Concentration with TimeFirst-Order ReactionsFirst-Order Reactions

0AlnAln ktt

16

The Change of Concentration with TimeThe Change of Concentration with TimeHalf-LifeHalf-Life• Half-life is the time taken for the concentration of a reactant to

drop to half its original value.

• That is, half life, t1/2 is the time taken for [A]0 to reach ½[A]0.

• Mathematically,

21kt

[A]

[A]ln

o21

o OR:

k

0.693ln2

k

1t

21

17

The Change of Concentration with TimeThe Change of Concentration with TimeHalf-LifeHalf-Life

18

Data are for first order reaction of CH3NC.Calculate (a) rate constant and (b) half-life

ln(A0/At) = ktln(502/95.5) = k(8000 s)ln(5.26) = k(8000 s)1.66 = k (8000 s)k = 2.07 x 10-4 s-1

t1/2 = 0.693/k = 0.693/2.07 x 10-4 s-1

= 3340 sec

Sample problem

19

The Change of Concentration with TimeThe Change of Concentration with TimeSecond-Order ReactionsSecond-Order Reactions• For a second order reaction with just one reactant:

0A1

A1 kt

t

•A plot of 1/[A]t versus t is a straight line with slope k and intercept 1/[A]0

•For a second order reaction, a plot of ln[A]t vs. t is not linear.

20

The Change of Concentration with TimeThe Change of Concentration with TimeSecond-Order ReactionsSecond-Order Reactions

21

The Change of Concentration with TimeThe Change of Concentration with TimeZero-Order ReactionsZero-Order Reactions• For a zero order reaction with just one reactant:

A0 – At = kt

• A plot of [A]t versus t is a straight line with slope -k and intercept [A]0

• Zero-order reactions are used in heterogeneous catalysis reactions, such as gasoline production in refineries and catalytic hydrogenation.• For zero-order reactions, the rate does not change.

22

Temperature and RateTemperature and Rate

23

Temperature and RateTemperature and Rate• As temperature increases, the rate increases.• Since the rate law has no temperature term in it, the

rate constant must depend on temperature.

Activation EnergyActivation Energy•Arrhenius: molecules must possess a minimum amount of energy to react. Why?

–In order to form products, bonds must be broken in the reactants.

–Bond breakage requires energy.

•Activation energy, Ea, is the minimum energy required

to initiate a chemical reaction.

24

Temperature and RateTemperature and RateActivation EnergyActivation Energy

Ea

25

Temperature and RateTemperature and RateActivation EnergyActivation Energy

26

Temperature and RateTemperature and RateActivation EnergyActivation Energy• The change in energy for the reaction is the difference

in energy between reactants and products.

• The activation energy, Ea , is the difference in energy between reactants and transition state.

• The rate depends on Ea.

• Notice that if a forward reaction is exothermic, then the reverse reaction is endothermic.

27

28

Temperature and RateTemperature and RateThe Arrhenius EquationThe Arrhenius Equation• Arrhenius discovered most reaction-rate data obeyed the Arrhenius equation:

• If we have a lot of data, we can determine Ea and A graphically by rearranging the Arrhenius equation:

• If we do not have a lot of data, then we can use

RTaEAek

–Both A and Ea are specific to a given reaction.

–k is the rate constant

–Ea is the activation energy

–R is the gas constant (8.314 J/K-mol)

–T is the temperature in K.

–A is the frequency factor.

29

Temperature and RateTemperature and RateThe Arrhenius EquationThe Arrhenius Equation• If we have a lot of data, we can determine Ea and A

graphically by rearranging the Arrhenius equation:

ART

Ek a lnln

122

1 11ln

TTR

E

kk a

•If we do not have a lot of data, then we can use

30

Data for a reaction.Use two points to calculate Ea

11

ln318R

E

0.0521

0.332k a288

31

Reaction MechanismsReaction Mechanisms• First-order reactions are generally unimolecular, i.e., they depend

only on the spontaneous reaction of one molecule.• Second-order reactions are generally bimolecular, i.e., they depend

upon the collision of one molecule with another to exchange atoms.• Zero-order reactions depend upon a catalytic bottle-neck, such as

catalytic hydrogenation or cracking of petroleum to form gasoline.• Third, or fractional, order reactions are possible and involve very

complex mechanisms.

32

CatalysisCatalysis• A catalyst changes the rate of a chemical reaction, by

lowering the activation energy Ea.

33

Catalysis – catalytic converterCatalysis – catalytic converterA platinum surface in the exhaust system of an automobile:Promotes the complete oxidation of fuel.Converts carbon monoxide to carbon dioxide.Converts nitrogen oxides to N2 and O2.

Catalysis – catalytic hydrogenationCatalysis – catalytic hydrogenationAdds H2 to C=C bonds of oils to convertunsaturated compounds to saturated compounds(fats are the typical examples).The most common catalyst is nickel.

34

CatalysisCatalysis Hydrogenation of ethylene, C2H4

C2H4(g) + H2(g) C2H6(g), H = -136 kJ/mol.

35

Catalysis - EnzymesCatalysis - EnzymesAn enzyme is a protein molecule, with a specific shape

which catalyzes a specific reaction. The substance undergoing a reaction is called a substrate, which locks into an enzyme, resulting in a fast reaction (“lock and key” model).

top related