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Rates of ReactionChemical Kinetics

I. Background on Rates Mechanisms

- Chemists study reactions. Some of what we study

- - Major Minor Products
- - Reactants
- - Completion
- - Effects of Catalysts
- - Separation and Purification of Products
- - Effects of Variables on Rxn Speed (rate) on

Products - - Rates of the Reaction
- - Mechanism of the Reaction
- This chapter deals with rates mechanisms of

reactions. - Mechanism Step-by-step progress of the chemical

reaction. - Rate How fast the reaction proceeds (usually

? M / Time)

I. Background on Rates Mechanisms

- Study of rates is useful since the results will
- 1) Indicate how to manipulate factors to

control the reaction - 2) Lead to the mechanism of the reaction
- 3) Indicate time needed to get a given amount

of product - 4) Indicate amount of product in a given amount

of time - Study of mechanisms important. With the

mechanism we can - 1) Predict products of similar reactions
- 2) Better understand the reaction
- 3) Accurately manipulate the reaction for a

desired result - 4) Organize and simplify the study of organic

chemistry - OH

I - Example CH3-CH-CH3 H I-

-------) CH3-CH-CH3 H2O

I. Background on Rates Mechanisms

- Main Factors which influence reaction rate
- Concentrations of Reactants - Rates usually

increase as reactant concentrations increase. - Reaction Temperature - An increase in temperature

increases the rate of a reaction. - Presence of a Catalyst (not all rxns have

catalysts) - A catalyst is a substance which increases the

rate of a reaction without being consumed in the

overall reaction. - The concentration of the catalyst or its surface

area (if insoluble) are variables which influence

the rate. - Some catalysts are incredibly complex - like

enzymes and others are quite simple H

H2O CH2 CH2 ------) CH3-CH2-OH H - Type of Reactants
- Surface Area of Insoluble Reactant

II. Rates

- Reaction Rate either the increase in M of

product per unit time or the decrease in M of

reactant per unit time ?M / ?T - Note X moles X / Liter
- Example H Catalyst
- Sucrose H2O --------------) Glucose

Fructose - Rate rate of formation of either product.
- Rate ? M of glucose / ?sec ?glucose

/ ?sec - or
- Rate rate of disappearance of either

reactant. - Rate - ?sucrose / ?sec
- In order to obtain rate, we need a way to measure

?M of any reactant or product with respect to

time.

II. Rates

- Example 2 N2O5 -----) 4 NO2 1 O2
- If we want to equalize the rates then
- Rate ?O2 1/4 ?NO2 - 1/2

?N2O5 - ?t ?t ?t
- - divide by balancing coefficients when we

equalize rates. - Various Rates can be determined 1)

instantaneous rate at a given time 2) average

rate over a long period of time or 3) the

initial rate rate at the beginning of the rxn

(this is used the most). - On next slide the ?O2 versus time is plotted

for a reaction. - Note 1) how the rate changes with time 2) that

rate is the tangent at a given point on the curve.

II. Rates

II. Rates

II. Rates

- Example of average rate calculation for
- I- ClO- -----) Cl- IO-
- Given the I- and time in seconds, then what is

the average rate? - Time (s) M I-
- 2.00 0.00169
- 8.00 0.00101
- - Rate - ?M / ?T
- - Rate - (0.00101 - 0.00169) M / (8.00 - 2.00)

s 6.8x10-4 M / 6.00s - - Rate 1.1 x 10-4 M/s

II. Rates

- Need to obtain the change in M of a given reagent

per change in time could follow any parameter

related to concentration. - Examples of what one might follow to obtain

rates - - a change in pressure (if gas produced or

consumed in the rxn) - - a change in pH (if acidity changes in the rxn)
- - a change in absorbance of electromagnetic

radiation (EMR) - Usually measure absorbance of Visible or UV EMR

at a given ? - caused by a change in reactant or

product concentration. - A ?bc at a given ? (wavelength) This is

Beers Law - A absorbance use spectrophotometer to

measure has no units. - ? molar absorptivity a constant _at_ given ?

has units of M-1cm-1 - b pathlength of EMR through sample usually

1.00 cm cuvette used. - c concentration in M
- A plot of A versus M _at_ given ? will yield a

straight line and the equation - A Ebc intercept If follow ?A then

can convert to ?M get rate.

III. Rate Law rate k x Am x Bn

for A B

- Rate Law relates the rate to temperature

concentration. - Rate law is given in terms of REACTANTS only

(convention). - k rate constant handles the temperature

variable. - The exponents are the order handle the

concentration variables. - General form of the rate law for a A b

B c C d D - rate k x Am x Bn
- - Order for A is m order for B is n

Overall order is m n - - m n are determined experimentally.
- - k, also determined experimentally units

depend upon overall order.

III. Rate Law - Rate Constant Units

- Note Assume time is in seconds (s). Rate k

Ax - Solve for k plug in units k Rate / Ax
- (this may be useful for one of the online HW

problems) - Overall Rxn Order, x Units for k
- zero Ms-1
- first s-1

- second M-1s-1
- third M-2s-1

III. Rate Law

- Example 2 NO2 1 F2 ----) 2 NO2F

Rate k NO2n x F2m - In the laboratory, the overall rate was found to

be second order. - n m 2 Possibilities n2 m0

n0 m2 n1 m1 - Experiments demonstrated that n1 and m1 How?

By running the reaction at least three times

1 getting rate at certain initial

concentrations of NO2 F2 2 getting rate

when keeping NO2 the same doubling F2 3

getting rate when keeping F2 the same

doubling NO2. - They found that doubling NO2 doubled the rate

doubling F2 doubled the rate so, both

coefficients had to be 1. - Rate k NO21 F21

III. Rate Law - Rate Constant

- Determination of the rate constant, k. You are

given - 1) aA bB ? cC dD
- 2) Rate kA0B1 (0 1 were determined

experimentally) - 3) M of A B 2.00 moles/L Rate 2.50 x

10-2 M/s - Determine the value for k give complete rate

expression. - Rate kA0B1 k Rate
- A0B1
- k Rate 2.50 x 10-2 M/s 2.50 x

10-2 M/s - A0B1 2.00 M0 x 2.00 M1

1 x 2.00 M1 - k 1.25 x 10-2 s-1
- Rate (1.25 x 10-2 s-1) A0 B1

(completed rate expression)

III. Rate Law

- Rate Law for a reaction is found experimentally

except for a single step in a mechanism

(elementary reaction). Assume rxn is NOT

elementary unless it is one step in a mechanism. - Given 1NO2 1CO -----) 1NO 1CO2
- By experiment, the rate law was found to be
- rate kNO22CO0 or rate kNO22

( Note CO0 1 ) - The order WRT each reactant the overall order

are - 2nd order WRT NO2 0th order WRT CO

2nd order overall

IV. Order

- The concentration variables are handled by the

exponents - the order. - The orders are determined experimentally except

for one case An - elementary reaction.
- Elementary reactions are one step reactions which

are the individual steps in a mechanism. (For

an elementary reaction only the balancing

coefficients determine the order.) - Important - Example 1 for a multistep reaction 1 CH2Br2

2 KI ---) 1 CH2I2 2 KBr - Experimentation found that m n were both

first order then - rate k CH2Br2 1 KI1
- Example 2 for an elementary reaction 2 O3

---) 3 O2 (assume elementary) - No need for experimentation order comes from

balancing coefficients - rate kO32

IV. Determination of Order

- Order - from units of k If you are given the

units of the rate constant for a reaction, then

you will know the overall order (slide 14). Not

too common. - Order by Method 1 - from altering M Measure

initial rates keeping one reactant constant and

change the concentration of another observe the

rates calculate order as illustrated in the next

few slides. - Order by Method 2 - from integrated rate

expression Use calculus integrate the rate

expression between the limits of time 0 time

t. By plotting out the variables of these

integrated rate expressions you can determine the

order. This will be shown in the lecture, and

you will be doing this in the kinetics lab.

Order by Method 1 - from altering M Measure

initial rates keeping one reactant constant and

change the concentration of another observe the

rates calculate order as illustrated in the next

few slides.

IV. Determination of Order by varying M

- Example 1 Determine the order for rate

expression for - 2N2O5 ---) 4NO2 1O2 rate k

N2O5m - Exp 1 Rate 4.8x10-6 Ms-1 at 1.0x10-2 M

N2O5 - Exp 2 Rate 9.6x10-6 Ms-1 at 2.0x10-2 M

N2O5 - Order Note that when N2O5 doubles, the rate

doubles. Since - rate a N2O5m rate doubles when N2O5

doubles, the value of - must be 1 the order is 1.
- - rate a N2O5m rate doubles when

N2O5 doubles, then - go from 1m 1 to 2m 2 m has to be

1 - Rate law rate k x N2O51

IV. Determination of Order by varying M

- Summary
- Effects of doubling reagent M while keeping

others constant - Rate remains the same 0th order M0
- Rate doubles 1st order M1
- Rate quadruples 2nd order M2
- Rate increases eightfold 3rd order M3

IV. Determination of Order by varying M

- Example 2 2 NO Cl2 -----) 2 NOCl -

Calculate order of Rxn - Exp Initial NO Initial Cl2 Initial Rate,

Ms-1 - 1 0.0125 0.0255 2.27x10-5
- 2 0.0125 0.0510 4.57x10-5
- 3 0.0250 0.0255 9.08x10-5
- Rate kNOmCl2n
- a) calculate n From 1 2 - double Cl2 keep

NO constant - rate increases by factor of 2.01 n 1
- b) calculate m From 1 3 - double NO keep

Cl2 constant - rate increases by factor of 4.00 m 2
- Rate kNO2Cl21
- 2nd order wrt NO 1st order wrt Cl2

3rd order overall

- Order by Method 2 - from integrated rate

expression - Integrate the rate expression between the limits

of time 0 and time t. By plotting out the

variables of these integrated rate expressions

you can determine the order. You will be doing

this in the kinetics lab.

IV. Determination of Order by Integrated Rate

Expression

- Summary on use of logarithms
- Log involves s to the base 10 Log 10x

x - Ln Natural log uses s to the base e Ln ex

x - Ln A/B Ln A - Ln B (or Log)
- Ln A x B Ln A Ln B (or Log)
- Ln Ab b Ln A (or Log)
- To obtain either log or ln use the appropriate

calculator function. - Log 2.1x10-4 - 3.68 (note significant figure

change - see below) - (-4.0000000. 0.32 -3.68 cut off at

first doubtful digit) - To remove Ln Log use the inverse ex 10x

functions on cal. - Inverse log 3.00 or 103.00 1.0 x 103
- Inverse ln 3.00 or e3.00 20.

IV. Order Integrated Rate Law - First Order Rxns

- 1) 1st Order Reactions aA -----) Products

If 1st order, then - -?A/?t kA1 (rate expression)
- This plot for first order data only gives minimal

information

A

Time

IV. Order Integrated Rate Law - First Order Rxns

- 1) 1st Order Reactions aA -----) bB -?A/?t

kA1 - if we integrate from time t to 0, we get the

following - Y mX b
- lnAt - kt lnAo or

lnAt/Ao -kt - where At M of A at time t Ao M

of A at t 0 - - A plot of lnAt versus t gives a straight line

(Y mX b) - b Slope (m) - k
- Note Only linear for 1st order

lnAt

Time, t

Example of an integrated rate plot for a 1st

order reaction

Slope -k rise/run

Must be 1st order since plot of lnN2O5 vs t is

linear. Can get k from the slope.

IV. Order Example using 1st order integrated

equation

- Example 2N2O5 ---) 4NO2 O2 rate k

N2O51 (1st order) - Given k 4.80x10-4s-1 N2O5to 1.65x10-2

M what is N2O5 at 825 s? - ln At - k x

t ln Ao - ln N2O5 - 4.80x10-4s-1 x 825 s

ln 1.65x10-2 - ln N2O5 - 0.396

- 4.104 - ln N2O5 - 4.500
- Take inverse ln or anti ln of both sides get
- inverse ln -4.500 or e -4.500

0.0111 - N2O5 0.0111 M

IV. Order Integrated Rate Law - First Order Rxns

- Half-Life (t1/2) of 1st Order Reaction
- t1/2 time it takes for Ao to decrease to 1/2

initial M ½Ao - ln At / Ao -kt ln 1/2Ao / Ao

-kt1/2 - ln 1/2 -kt1/2 -0.693 -kt1/2 t1/2

0.693 / k - Note 1) Time for 1/2 to disappear is

independent of A for 1st order reaction. 2)

This is an easy way to calculate 1st order rate

constant, k. - Example If t1/2 189 sec for 1st order

decomposition of 1.0 mole of H2O2, then how much

H2O2 will be left after 378 sec? - Note 378/189 2 Goes through

two half lives - 1.0 mol ---) 0.50 mol ---) 0.25 mol

IV. Order Integrated Rate Law - First Order Rxns

- Example Given a) k 3.66x10-3s-1 for

decomposition of H2O2 and b) H2O2o 0.882 M.

Calculate - 1) t1/2
- 2) How much will be left after one half-life?
- (Note Reaction must be 1st order examine units

for k) - 1) t1/2 0.693/k
- t1/2 0.693 / 3.66x10-3s-1 189 s
- 2) M of H2O2 cut in half in one half-life

(t1/2) will go from 0.882 to 0.441 M in 189 s.

IV. Order Integrated Rate Law - Second Order Rxns

- 2) Second Order Reactions
- - Assume that aA -----) Products is 2nd

order - Rate - ?A / ? t k A2
- Integrate rate expression from time t to 0 get

following - 1/At k t 1/Ao So, a plot of

1/At vs t should give a straight line with

slope k and y intercept 1/Ao - t1/2 1 Note Now t1/2 depends on

initial M - k x A0
- Note can tell if reaction is 2nd order from

1/A vs t plot.

IV. Order Integrated Rate Law - Second Order

Rxns Example

Plot of lnNO2 vs t is not linear not 1st

order.

IV. Order Integrated Rate Law - 0th Order

- 3) 0th Order Reactions Assume A ---) B

is 0th order - Rate -kA0
- Rate -k
- - Integrated Rate Equation for a 0th order

reaction - At -k x t A0
- - a plot of At versus t will give a straight

line - - Again, if you let At 1/2 Ao then t

t1/2 - - t1/2 A0 / 2k

IV. Order Integrated Rate Law - Summary

? Rate when double M None

Double Quadruple

IV. Order Integrated Rate Law - Summary

A? B ?A/?t kAn

Note slope -k or k in each case

At

0th Order n0 At - kt Ao

Time, t

1st Order n1 lnAt - kt lnAo

1/At

2th Order n2 1/At kt 1/Ao

Time, t

V. Temperature

- A collision needs to occur before a reaction can

take place, the rate constant ( rate) of the

reaction depends upon the - 1) collision frequency (temperature)
- 2) number of collisions having enough energy for

rxn (Ea) - 3) orientation of particles upon collision
- Ea energy of activation minimum energy of

collision in order for the reaction to take

place. - Ea ?H can be represented by Potential Energy

Diagram can draw for one step or for several

steps in a mechanism.

V. Temperature Reaction Rate A)

Potential Energy Diagram for an Elementary

Reaction

?H

(Exothermic)

What is Ea for reverse reaction?

V. Temperature Reaction Rate Arrhenius

Equation

- Arrhenius Equation relates rate constant (k),

temperature (T), energy of activation (Ea in

J/mole), orientation factor. - k A e-Ea/RT R gas constant use R

8.31 J/(Kmole) - Take ln of both sides ln k -(Ea/R) 1/T

ln A - Y m X b
- Measure k at several temperatures and make plot

of ln k - versus 1/T. Slope of the curve - Ea/R

(will give Ea) - Note A is a constant includes orientation

factor. - Note Page 549 contains a form of the Arrhenius

equation which is useful for some online HW

questions.

V. Temperature Reaction Rate Arrhenius

EquationData below from 4 experiments - detn of

rate constant, k, at 4 temperatures for a rxn

ln A

o

Slope -Ea/R Can now determine Ea

ln k

o

o

o

1/T (in K-1)

ln k -(Ea/R) 1/T ln A

From k Ae-Ea/RT Y m X

b Use R 8.31J/(K.mol)

VI. Mechanisms A) Introduction

- Mechanism step by step progress of chemical

reaction. - Most likely mechanism is determined

experimentally from a study of rate data. - The mechanism consists of one or more elementary

reactions which add up to give you the overall

reaction. - Species which is generated then consumed in the

mechanism is called an intermediate Species

which is added, consumed then regenerated is a

catalyst. - Step with largest Ea (slowest step) is called the

rate determining step governs overall reaction

rate.

VI. Mechanisms B) Example 1 - Information

- Overall Rxn O3 2NO2 -----) O2 N2O5
- Suggested two-step mechanism (from

experimentation) - Step 1) O3 NO2 -----) NO3 O2

(slow) - Step 2) NO3 NO2 -----) N2O5
- rate k O31NO21 - all from slow first step
- Notes a) Two elementary reactions (NOTE

balancing coefficients orders in an

elementary rxn) - b) Steps add up to give overall rxn
- c) NO3 is an intermediate (produced used

up) - d) There is no catalyst
- e) Slowest step governs the overall rate
- mechanism is useful will give us a) practical

data, b) rate law, c) theoretical data, d)

understanding of reaction

VI. Mechanisms B) Example 2 - Calculate Rate

Expression

- Determine a) general rate expression b)

complete rate law from the following mechanism

Note can directly get the order for an

elementary rxn from the balancing coefficients. - 1) 1I2 2I (fast equilibrium)
- 2) 2I 1H2 2HI (slow)
- a) Overall Rxn from addn of steps 1I2 1H2

2HI - General rate law rate kI2xH2y
- b) Complete rate expression from mechanism
- From step 2 (rxn rate slow step rate)

rate k2 Io2 x H21 - From step 1 Keq Io2/I21 Io2

KeqI21 Substitute into above - rate k2 Keq I21 H21
- rate k I21 H21

IV. Mechanisms C) Catalysts

- Catalyst A chemical which speeds up a reaction

without being consumed in the reaction. - - They operate by lowering the Ea for the rate

determining step. - - One example is Pt which speeds up the

following rxn - CO 1/2 O2 -----) CO2
- - Pt can be used in catalytic converter for

your car exhaust. - Most famous catalysts are proteins called

enzymes. - - Enzymes extremely specific biochemical

catalysts that allow complex reactions to take

place in living systems under mild conditions. - - Enzymes are very complex, well designed, and

usually have molecular weights in the tens of

thousands. - - Their mode of operation uncovered only 60

years ago.

VI. Mechanisms C) Catalysts

A catalyst speeds up the rxn by lowering the Ea

provides a different mechanism with a lower Ea

NewEa

VI. Mechanisms C) Catalysts