Title: Chapter 13 - Rates of Reaction Chemical Kinetics
1Chapter 13 - Rates of ReactionChemical Kinetics
- End-of-Chapter Problems Pages 577-589
- Sapling Homework, Chapter 13, due 11/25/2013
2End-of-Chapter Problems - Chapter 13 pages 567-579
- 1 - 12 16 17 18 20 23 25 26 33
34 - 35 36 37 39 45 49 51 53 55 57
- 59 60 65 67 68 71 75 83
84 -
- 87 89 117 125
3Lecture Outline
- I. Background on Rates Mechanisms
- II. Rates
- III. Rate Law (Variables T, Mech M)
- IV. Order
- Determining order a) Varying M, measuring
rates - Determining order b) Integrated rate
expression graph - 1st Order
- 2nd Order
- 0th Order
- Order Summary
- V. Temperature Arrhenius Equation
- VI. Mechanisms
- Introduction, examples and catalysts
4I. 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)
5I. 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
6I. 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 an Insoluble Reactant
7II. 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 (- since want a
rate) -
- In order to obtain rate, we need a way to measure
?M of a reactant or product with respect to time.
8II. 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
ie rate at t0.0 (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.
9II. Rates
10II. Rates
11II. Rates
- Calculate the Average Rate 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 - - ave Rate 1.1 x 10-4 M/s
12II. 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 given ? (wavelength) This is Beers
Law from Ch 121 (know) - 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.
13 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.
14III. 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
15III. 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 NO2n F2m k NO21 F21
-
16III. 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)
17III. 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 told that it is a one step in
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
18IV. 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 - If 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 (told it is elementary) - No need for experimentation order comes from
balancing coefficients - rate kO32
19IV. 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.
20Order 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.
21IV. 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
22IV. 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
23IV. 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
24- 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.
25IV. 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.
26IV. 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
27IV. 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
28Example 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.
29IV. 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 of both sides INVERSE ln
N2O5 N2O5 - INVERSE ln -4.500 e -4.500
0.0111 -
- N2O5 0.0111 M
30IV. 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
31IV. 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.
(Note Reaction must be 1st order examine units
for k) - Calculate
- 1) t1/2
- 2) How much will be left after one half-life?
- 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.
32IV. 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.
33IV. Order Integrated Rate Law - Second Order
Rxns Example
Plot of lnNO2 vs t is not linear not 1st
order.
34IV. 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
35IV. Order Integrated Rate Law - Summary
? Rate when double M None
Double Quadruple
36IV. Order Integrated Rate Law - Summary
A? B ?A/?t kAn
Note slope 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
37V. 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.
38V. Temperature Reaction Rate A)
Potential Energy Diagram for an Elementary
Reaction
?H
(Exothermic)
What is Ea for reverse reaction?
39V. 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 559 contains a form of the Arrhenius
equation which may be useful for some online HW
questions.
40V. 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)
41VI. 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.
42VI. 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 - 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
43VI. 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 2Io (fast equilibrium)
- 2) 2Io 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
44IV. 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.
45VI. Mechanisms C) Catalysts
A catalyst speeds up the rxn by lowering the Ea
provides a different mechanism with a lower Ea
NewEa
46VI. Mechanisms C) Catalysts
47Final ExamTable
- Chapter 13 Equations
- Integrated Rate Equation Half-Life
- At - kt Ao Ao/2k
- lnAt - kt lnAo 0.693/k
- 1/At kt 1/Ao 1/kAo
-
- R 8.31 J/Km 0.0821 La/Km k A e-Ea/RT
-
- Water
- ?Hvap 40.7 kJ/m ?Hfus 6.01 kJ/m
- S water 4.18 J/goC Kb 1.86 oC/m Kf
0.512 oC/m - Selected values for ?Hof and ?Hbond energy
in kJ/mole