Title: Chemical Kinetics http://www.chem1.com/acad/webtext/dynamics/dynamics-3.html http://ibchem.com/IB/ibnotes/brief/kin-sl.htm
1Chemical Kinetics http//www.chem1.com/acad/we
btext/dynamics/dynamics-3.html
http//ibchem.com/IB/ibnotes/brief/kin-sl.htm
http//www.docbrown.info/page03/3_31rates.htmThee
ffectofaCatalysthttp//www.practicalchemistry.or
g/experiments/the-rate-of-reaction-of-magnesium-wi
th-hydrochloric-acid,100,EX.html
1.
2What do we mean by kinetics?
- Kinetics is the study of
- the rate at which chemical reactions occur.
- The reaction mechanism or pathway through which a
reaction proceeds.
2.
3How an Airbag Works http//www.youtube.com/wa
tch?vdZfLOnXoVOQ
- When sodium azide, NaN3, is ignited by a spark,
it releases nitrogen gas which can instantly
inflate an airbag.(one 25th of a second) - SiO2
- 10NaN3 (s) 2KNO3 (s)gt K2O (s) 5 Na2 O(s)
16N2 (g)
4Reaction Rate
- The change in the concentration of a reactant or
a product with time (M/s). - Reactant ? Products
- A ? B
- Since reactants go away with time
- Rate - ?A/?t ?B/?t mol
dm-3 s-1 - Simulation
- http//www.chm.davidson.edu/vce/kinetics/ReactionR
ates.html
5A gt B Rate - ?A/?t
?B/?t until completion
6http//www.avogadro.co.uk/kinetics/rate_equatio
n.htm
- When the rate does not depend on the
concentration
7Reaction Mechanism
- Most chemical reactions consist of a sequence of
two or more simpler reactions. - For example, recent evidence indicates that the
reaction - 2O3 ---gt 3O2
- occurs in three steps after intense ultraviolet
radiation from the sun liberates chlorine atoms
from certain compounds in Earths stratosphere.
8For example
- 1. Chlorine atoms decompose ozone according to
the equation - Cl O3 --- gt O2 ClO
- 2. Ultraviolet radiation causes the decomposition
reaction - O3 --- gt O2 O
- 3. ClO produced in the reaction in step 1 reacts
with O produced in step 2 according to the
equation - ClO O --- gt Cl O2
9- A complex reaction is one that consists of two or
more elementary steps. The complete sequence of
elementary steps that make up a complex reaction
is called a reaction mechanism.
10Rate Determining Step
- Chemical reactions, too, have a weakest link in
that a complex reaction can proceed no faster
than the slowest of its elementary steps. In
other words, the slowest elementary step in a
reaction mechanism limits the instantaneous rate
of the overall reaction.
11Reaction Rates
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)
- The reaction slows down with time because the
concentration of the reactant decreases.
11.
12Collision Theory
- http//www.saskschools.ca/curr_content/chem30_05/2
_kinetics/kinetics2_1.htm - For a reaction to occur Particles must collide
with each other and these collisions must be
effective - Sufficient energy(Activation energy) and proper
orientation.
13Factors That Affect Reaction Rateshttp//www.doc
brown.info/page03/3_31rates.htmhttp//www.blackgo
ld.ab.ca/ict/Divison3/reactionrates/reactionrates.
htm
- The Nature of the Reactants
- Chemical compounds vary considerably in their
chemical reactivities. - Concentration of Reactants
- As the concentration of reactants increases, so
does the likelihood that reactant molecules will
collide. - Temperature
- At higher temperatures, reactant molecules have
more kinetic energy, move faster, and collide
more often and with greater energy. - Catalysts
- Change the rate of a reaction by changing the
mechanism. - Surface Area
- http//www.physchem.co.za/OB12-che/rates.htm
.
14(No Transcript)
15Maxwell Boltzman
- As you increase the temperature the rate of
reaction increases. As a rough approximation, for
many reactions happening at around room
temperature, the rate of reaction doubles for
every 10C rise in temperature.
16Catalyst and Ea
- A catalyst provides an alternative route for the
reaction. That alternative route has a lower
activation energy.
17http//www.gcsescience.com/rc4-sodium-thiosulfa
te-hydrochloric.htm
- investigate the effect that the temperature of
water has on the reaction rate of antacid
tablets. - investigate the change in reaction rate when
varying sizes of zinc particles are reacted with
sulfuric acid. - explore the relationship between concentration
and reaction rate. - observing how quickly reactants are used up or
how quickly products are forming. This can be
calculated in three ways of precipitation - When
the products of the reaction is a precipitate,
which will cloud the solution. Observing a marker
through the solution and timing how long it takes
for this marker to disappear will determine the
rate of reaction. - Hydrochloric Acid and Sodium Thiosulphate
- HCl(aq) Na2S2O3(aq)
NaCl(aq) SO2(g) S(s)
H2O(l)
18The graph lines W, X, original, Y and Z on the
left diagram are typical of when a gaseous
product is being collected
- X ,a catalyst was added, forming the same amount
of product, but faster. - Z could represent taking half the amount of
reactants or half a concentration. The reaction
is slower and only half as much gas is formed - W might represent taking double the quantity of
reactants, forming twice as much gas e.g. same
volume of reactant solution but doubling the
concentration, so producing twice as much gas,
initially at double the speed (gradient twice as
steep). - W gt X gt original gt Y gt Z
- See SG graphs
19Measuring the rate of a reaction
- The change in concentration can be measured by
using any property that changes during conversion
of reactants to products - CaCO3(s) HCl(aq) gt CaCl2(aq) H2O(l)
CO2(g) - 1. mass and volume changes for gaseous reactions
- 2. change in pH for acids and bases
- 3. change in conductivity for metals
- 4. colorimetry for color changes
20Measuring the volume of gas produced with time
21Mass Loss Method
22Graphs
- The graph below shows the volume of carbon
dioxide gas produced against time when excess
calcium carbonate is added to x cm3 of 2.0 mol
dm-3 hydrochloric acid. - (i) Write a balanced equation for the
reaction. - (ii) State and explain the change in the
rate of reaction with time. Outline how you
would determine the rate of the reaction at a
particular time. - (iii) Sketch the above graph on an answer
sheet. On the same graph, draw the curves you
would expect if - I. the same volume (x cm3) of 1.0 mol dm-3
HCl is used. - II. double the volume (2x cm3) of 1.0 mol
dm-3 HCl is used. - Label the curves and explain your
answer in each case.
23Graph mass x time
24SG page 36
25Example 1 Reaction Rates
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)
C4H9Cl M
- In this reaction, the concentration of butyl
chloride, C4H9Cl, was measured at various times,
t.
?C4H9Cl ?t
Rate
25.
26Reaction Rates Calculation
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl (aq)
Average Rate, M/s
- The average rate of the reaction over each
interval is the change in concentration divided
by the change in time
26.
27Reaction Rate Determination
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)
- Note that the average rate decreases as the
reaction proceeds. - This is because as the reaction goes forward,
there are fewer collisions between the reacting
molecules.
27.
28Reaction Rates
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)
- A plot of concentration vs. time for this
reaction yields a curve like this. - The slope of a line tangent to the curve at any
point is the instantaneous rate at that time.
28.
29Reaction Rates
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)
- The reaction slows down with time because the
concentration of the reactants decreases.
29.
30Reaction Rates and Stoichiometry
C4H9Cl(aq) H2O(l) ??? C4H9OH(aq) HCl(aq)
- In this reaction, the ratio of C4H9Cl to C4H9OH
is 11. - Thus, the rate of disappearance of C4H9Cl is the
same as the rate of appearance of C4H9OH.
30.
31Reaction Rates Stoichiometry
- Suppose that the mole ratio is not 11?
Example H2(g) I2(g) ??? 2 HI(g)
2 moles of HI are produced for each mole of H2
used.
The rate at which H2 disappears is only half of
the rate at which HI is generated
31.
32Concentration and Rate
- Each reaction has its own equation that gives its
rate as a function of reactant concentrations. - This is called its Rate Law
- The general form of the rate law is
- Rate kAxBy
- Where k is the rate constant, A and B are the
concentrations of the reactants. X and y are
exponents known as rate orders that must be
determined experimentally - To determine the rate law we measure the rate at
different starting concentrations.
32.
33Concentration and Rate
NH4 (aq) NO2- (aq) ? N2 (g) 2H2O (l)
- Compare Experiments 1 and 2when NH4 doubles,
the initial rate doubles.
33.
34Concentration and Rate
NH4 (aq) NO2- (aq) ? N2 (g) 2H2O (l)
- Likewise, compare Experiments 5 and 6 when
NO2- doubles, the initial rate doubles.
34.
35Concentration and Rate
NH4 (aq) NO2- (aq) ? N2 (g) 2H2O (l)
This equation is called the rate law, and k is
the rate constant.
35.
36The Rate Law
- A rate law shows the relationship between the
reaction rate and the concentrations of
reactants. - For gas-phase reactants use PA instead of A.
- k is a constant that has a specific value for
each reaction. - The value of k is determined experimentally.
- Rate K NH4 NO2-
- Constant is relative here-
- the rate constant k is unique for each reaction
- and the value of k changes with temperature
36.
37The Rate Law
- Exponents tell the order of the reaction with
respect to each reactant. - This reaction is
- First-order in NH4
- First-order in NO2-
- The overall reaction order can be found by adding
the exponents on the reactants in the rate law. - This reaction is second-order overall.
- Rate K NH4 1NO2- 1
37.
38Determining the Rate constant and Order
- The following data was collected for the
reaction of substances A and B to produce
products C and D. - Deduce the order of this reaction with
respect to A and to B. Write an expression for
the rate law in this reaction and calculate the
value of the rate constant.
38.
39Graphical Representation
- http//www.chem.purdue.edu/gchelp/howtosolveit/Kin
etics/IntegratedRateLaws.html - A ? B
40Zero Order
- For a zero order reaction
- A versus t (linear for a zero order
reaction) - rate k (k - slope of
line)
41First Order
- For a 1st order reaction
- rate kA (k - slope of line)
42First-Order Processes
- Consider the process in which methyl isonitrile
is converted to acetonitrile.
How do we know this is a first order reaction?
42.
43First-Order Processes
- This data was collected for this reaction at
198.9C.
Does ratekCH3NC for all time intervals?
43.
44First-Order Processes
- When Ln P is plotted as a function of time, a
straight line results. - The process is first-order.
- k is the negative slope 5.1 ? 10-5 s-1.
44.
45Half-Life of a Reaction
- Half-life is defined as the time required for
one-half of a reactant to react. - Because A at t1/2 is one-half of the original
A, - At 0.5 A0.
45.
46Half-Life of a First Order Reaction
- For a first-order process, set At0.5 A0 in
integrated rate equation
NOTE For a first-order process, the half-life
does not depend on the initial concentration,
A0.
46.
47First Order Rate Calculation
Example 1 The decomposition of compound A is
first order. If the initial A0 0.80 mol
dm-3. and the rate constant is 0.010 s-1, what
is the concentration of A after 90 seconds?
47.
48First Order Rate Calculation
Example 1 The decomposition of compound A is
first order. If the initial A0 0.80 mol
dm-3. and the rate constant is 0.010 s-1, what
is the concentration of A after 90 seconds?
LnAt LnAo -kt
LnAt Ln0.80 - (0.010 s-1 )(90 s) LnAt
- (0.010 s-1 )(90 s) Ln0.80 LnAt
-0.90 - 0.2231 LnAt -1.1231 At
0.325 mol dm-3
48.
49First Order Rate Calculations
Example 2 A certain first order chemical
reaction required 120 seconds for the
concentration of the reactant to drop from 2.00 M
to 1.00 M. Find the rate constant and the
concentration of reactant A after 80 seconds.
49.
50First Order Rate Calculations
Example 2 A certain first order chemical
reaction required 120 seconds for the
concentration of the reactant to drop from 2.00 M
to 1.00 M. Find the rate constant and the
concentration of reactant A after 80 seconds.
Solution k 0.693/t1/2 0.693/120s 0.005775
s-1 LnA Ln(2.00) -0.005775 s-1 (80 s)
-0.462 Ln A - 0.462 0.693 0.231 A
1.26 mol dm-3
50.
51First Order Rate Calculations
Example 3 Radioactive decay is also a first
order process. Strontium 90 is a radioactive
isotope with a half-life of 28.8 years. If some
strontium 90 were accidentally released, how long
would it take for its concentration to fall to 1
of its original concentration?
51.
52First Order Rate Calculations
Example 3 Radioactive decay is also a first
order process. Strontium 90 is a radioactive
isotope with a half-life of 28.8 years. If some
strontium 90 were accidentally released, how long
would it take for its concentration to fall to 1
of its original concentration?
Solution k 0.693/t1/2 0.693/28.8 yr 0.02406
yr-1 Ln1 Ln(100) - (0.02406 yr-1)t -
4.065 t - 4.062 . - 0.02406
yr-1 t 168.8 years
52.
53- A versus t (linear for a zero order
reaction) - ln A versus t (linear for a 1st order
reaction) - 1 / A versus t (linear for a 2nd order
reaction)
54Second Order
- For a 2nd order reaction
- rate kA2
(k slope of line)
55Determining Reaction OrderDistinguishing Between
1st and 2nd Order
The decomposition of NO2 at 300C is described by
the equation
A experiment with this reaction yields this data
Time (s) NO2, M
0.0 0.01000
50.0 0.00787
100.0 0.00649
200.0 0.00481
300.0 0.00380
55.
56Determining Reaction OrderDistinguishing Between
1st and 2nd Order
Graphing ln NO2 vs. t yields
- The graph is not a straight line, so this process
cannot be first-order in A.
Time (s) NO2, M ln NO2
0.0 0.01000 -4.610
50.0 0.00787 -4.845
100.0 0.00649 -5.038
200.0 0.00481 -5.337
300.0 0.00380 -5.573
56.
57Second-Order Reaction Kinetics
A graph of 1/NO2 vs. t gives this plot.
Time (s) NO2, M 1/NO2
0.0 0.01000 100
50.0 0.00787 127
100.0 0.00649 154
200.0 0.00481 208
300.0 0.00380 263
- This is a straight line. Therefore, the process
is second-order in NO2. - The slope of the line is the rate constant, k.
57.
58Half-Life for 2nd Order Reactions
- For a second-order process, set
- At0.5 A0 in 2nd order equation.
In this case the half-life depends on the initial
concentration of the reactant A.
58.
59Sample Problem 1 Second Order
- Acetaldehyde, CH3CHO, decomposes by second-order
kinetics with a rate constant of 0.334
mol-1dm3s-1 at 500oC. Calculate the amount of
time it would take for 80 of the acetaldehyde
to decompose in a sample that has an initial
concentration of 0.00750 M.
The final concentration will be 20 of the
original 0.00750 M or 0.00150
1 . .00150
1 . .00750
0.334 mol-1dm3s-1 t
666.7 0.334 t 133.33 0.334 t 533.4
t 1600 seconds
59.
60Sample Problem 2 Second Order
- Acetaldehyde, CH3CHO, decomposes by second-order
kinetics with a rate constant of 0.334
mol-1dm3s-1 at 500oC. If the initial
concentration of acetaldehyde is 0.00200 M. Find
the concentration after 20 minutes (1200 seconds)
Solution
1 . At
1 . 0.00200 mol dm-3
0.334 mol-1dm3s-1 (1200s)
1 . At
0.334 mol-1dm3 s-1 (1200s) 500 mol-1dm3
900.8 mol-1dm3
1 _____. 900.8 mol-1dm3
At
0.00111 mol dm-3
60.
61Summary of Kinetics Equations
First order Second order Second order
Rate Laws
Integrated Rate Laws complicated
Half-life complicated
61.
62Temperature and Rate
- Generally speaking, the reaction rate increases
as the temperature increases. - This is because k is temperature dependent.
- As a rule of thumb a reaction rate increases
about 10 fold for each 10oC rise in temperature
62.
63The Collision Model
- In a chemical reaction, bonds are broken and new
bonds are formed. - Molecules can only react if they collide with
each other. - These collisions must occur with sufficient
energy and at the appropriate orientation.
63.
64The Collision Model
- Furthermore, molecules must collide with the
correct orientation and with enough energy to
cause bonds to break and new bonds to form
64.
65Activation Energy
- In other words, there is a minimum amount of
energy required for reaction the activation
energy, Ea. - Just as a ball cannot get over a hill if it does
not roll up the hill with enough energy, a
reaction cannot occur unless the molecules
possess sufficient energy to get over the
activation energy barrier.
65.
66Reaction Coordinate Diagrams
- It is helpful to visualize energy changes
throughout a process on a reaction coordinate
diagram like this one for the rearrangement of
methyl isonitrile.
66.
67Reaction Coordinate Diagrams
- It shows the energy of the reactants and products
(and, therefore, ?E). - The high point on the diagram is the transition
state.
- The species present at the transition state is
called the activated complex. - The energy gap between the reactants and the
activated complex is the activation energy
barrier.
67.
68MaxwellBoltzmann Distributions
- Temperature is defined as a measure of the
average kinetic energy of the molecules in a
sample.
- At any temperature there is a wide distribution
of kinetic energies.
68.
69MaxwellBoltzmann Distributions
- As the temperature increases, the curve flattens
and broadens. - Thus at higher temperatures, a larger population
of molecules has higher energy.
69.
70MaxwellBoltzmann Distributions
- If the dotted line represents the activation
energy, as the temperature increases, so does the
fraction of molecules that can overcome the
activation energy barrier.
- As a result, the reaction rate increases.
70.
71MaxwellBoltzmann Distributions
- This fraction of molecules can be found through
the expression - where R is the gas constant and T is the
temperature in Kelvin .
71.
72Arrhenius Equation
- Svante Arrhenius developed a mathematical
relationship between k and Ea -
- where A is the frequency factor, a number that
represents the likelihood that collisions would
occur with the proper orientation for reaction.
Ea is the activation energy. T is the Kelvin
temperature and R is the universal thermodynamics
(gas) constant. - R 8.314 J mol-1 K-1 or 8.314 x 10-3 J mol-1
K-1
72.
73Arrhenius Equation
- Taking the natural logarithm of both sides, the
equation becomes
y mx b
When k is determined experimentally at several
temperatures, Ea can be calculated from the slope
of a plot of ln k vs. 1/T.
73.
74Arrhenius Equation for 2 Temperatures
When measurements are taken for two different
temperatures the Arrhenius equation can be
symplified as follows
Write the above equation twice, once for each of
the two Temperatures and then subtract the lower
temperature conditions from the higher
temperature. The equation then becomes
74.
75Arrhenius Equation Sample Problem 1
The rate constant for the decomposition of
hydrogen iodide was determined at two different
temperatures 2HI ? H2 I2. At 650 K, k1
2.15 x 10-8 dm3 mol-1s-1 At 700 K, k2
2.39 x 10-7 dm3 mol-1s-1 Find the activation
energy for this reaction.
2.39 x 10-7 Ea Ln ---------------- -
------------------------ x 2.15 x 10-8
(8.314 J mol-1 K-1) Ea 180,000
J mol-1 180 kJ mol-1
1 1 ------ -- ------ 700K 650K
75.
76Overview of Kinetics Equations
First order Second order Second order
Rate Laws
Integrated Rate Laws complicated
Half-life complicated
Rate and Temp (T)
76.
77Reaction Mechanisms
- The sequence of events that describes the actual
process by which reactants become products is
called the reaction mechanism.
77.
78Reaction Mechanisms
- Reactions may occur all at once or through
several discrete steps. - Each of these processes is known as an elementary
reaction or elementary process.
78.
79Reaction Mechanisms
- The molecularity of a process tells how many
molecules are involved in the process. - The rate law for an elementary step is written
directly from that step.
79.
80Multistep Mechanisms
- In a multistep process, one of the steps will be
slower than all others. - The overall reaction cannot occur faster than
this slowest, rate-determining step.
80.
81Slow Initial Step
NO2 (g) CO (g) ??? NO (g) CO2 (g)
- The rate law for this reaction is found
experimentally to be - Rate k NO22
- CO is necessary for this reaction to occur, but
the rate of the reaction does not depend on its
concentration. - This suggests the reaction occurs in two steps.
81.
82Slow Initial Step
- A proposed mechanism for this reaction is
- Step 1 NO2 NO2 ??? NO3 NO (slow)
- Step 2 NO3 CO ??? NO2 CO2 (fast)
- The NO3 intermediate is consumed in the second
step. - As CO is not involved in the slow,
rate-determining step, it does not appear in the
rate law.
82.
83Fast Initial Step
- The rate law for this reaction is found
(experimentally) to be - Because termolecular ( trimolecular) processes
are rare, this rate law suggests a two-step
mechanism.
83.
84Fast Initial Step
Step 1 is an equilibrium- it includes the
forward and reverse reactions.
84.
85Fast Initial Step
- The rate of the overall reaction depends upon the
rate of the slow step. - The rate law for that step would be
- But how can we find NOBr2?
85.
86Fast Initial Step
- NOBr2 can react two ways
- With NO to form NOBr
- By decomposition to reform NO and Br2
- The reactants and products of the first step are
in equilibrium with each other. - Therefore,
- Ratef Rater
86.
87Fast Initial Step
- Because Ratef Rater ,
- k1 NO Br2 k-1 NOBr2
- Solving for NOBr2 gives us
87.
88Fast Initial Step
- Substituting this expression for NOBr2 in the
rate law for the rate-determining step gives
88.
89Catalysts
- Catalysts increase the rate of a reaction by
decreasing the activation energy of the reaction. - Catalysts change the mechanism by which the
process occurs. - Some catalysts also make atoms line up in the
correct orientation so as to enhance the reaction
rate
89.
90Catalysts
- Catalysts may be either homogeneous or
heterogeneous - A homogeneous catalyst is in the same phase as
the substances reacting. - A heterogeneous catalyst is in a different phase
90.
91Catalysts
- One way a catalyst can speed up a reaction is by
holding the reactants together and helping bonds
to break. - Heterogeneous catalysts often act in this way
91.
92Catalysts
- Some catalysts help to lower the energy for
formation for the activated complex or provide a
new activated complex with a lower activation
energy
AlCl3 Cl2 ? Cl AlCl4- Cl C6H6 ? C6H5Cl
H H AlCl4- ? HCl AlCl3 Overall
reaction C6H6 Cl2 ? C6H5Cl HCl
92.
93Catalysts Stratospheric Ozone
In the stratosphere, oxygen molecules absorb
ultraviolet light and break into individual
oxygen atoms known as free radicals The
oxygen radicals can then combine with ordinary
oxygen molecules to make ozone. Ozone can also
be split up again into ordinary oxygen and an
oxygen radical by absorbing ultraviolet light.
93.
94Catalysts Stratospheric Ozone
The presence of chlorofluorcarbons in the
stratosphere can catalyze the destruction of
ozone. UV light causes a Chlorine free radical
to be released
The chlorine free radical attacks ozone and
converts it Back to oxygen. It is then
regenerated to repeat the Process. The result is
that each chlorine free radical can Repeat this
process many many times. The result is
that Ozone is destroyed faster than it is formed,
causing its level to drop
94.
95Enzymes
- Enzymes are catalysts in biological systems.
- The substrate fits into the active site of the
enzyme much like a key fits into a lock.
95.
9696.