Transcript Document

Chemical Kinetics
http://www.chem1.com/acad/webtext/dynamics/
dynamics-3.html
http://ibchem.com/IB/ibnotes/brief/kin-sl.htm
http://www.docbrown.info/page03/3_31rates.ht
m#TheeffectofaCatalyst
http://www.practicalchemistry.org/experiments/
the-rate-of-reaction-of-magnesium-withhydrochloric-acid,100,EX.html
1.
What 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.
Reaction 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 = - Δ[A]/Δt = Δ[B]/Δt
mol dm-3 s-1
Chemical
Kinetics
How an Airbag Works: http://www.youtube.com/watch?v=dZfLOnXoVOQ
• 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)=> K2O (s) + 5 Na2 O(s) + 16N2 (g)
Chemical
Kinetics
A => B
Rate = - Δ[A]/Δt = Δ[B]/Δt until completion
Chemical
Kinetics
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
• The reaction slows
down with time
because the
concentration of the
reactant decreases.
Chemical
Kinetics
6.
Collision 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.
Chemical
Kinetics
• Activation Energy?
Minimum energy
required.
Maxwell Boltzmann? Read
CC
Chemical
Kinetics
Maxwell 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 10°C rise in temperature.
Chemical
Kinetics
Factors That Affect Reaction Rates
http://www.docbrown.info/page03/3_31rates.htm
http://www.blackgold.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
.
http://www.gcsescience.com/rc4-sodium-thiosulfate-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)Chemical
Kinetics
The 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 > X > original > Y > Z
See SG graphs
Chemical
Kinetics
Chemical
Kinetics
Measuring 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) => 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
Chemical
Kinetics
Measuring the volume of gas produced with time
Chemical
Kinetics
Mass Loss Method
Chemical
Kinetics
Graphs
•
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)
(ii)
Write a balanced equation for the reaction.
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.
Chemical
Kinetics
Graph mass x time
Chemical
Kinetics
SG page 36
Chemical
Kinetics
Example 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.
Rate =
[C4H9Cl]
t
20.
Reaction 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:
21.
Reaction 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.
22.
Reaction 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.
23.
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
• The reaction slows
down with time
because the
concentration of the
reactants decreases.
24.
Reaction Rates and Stoichiometry
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
• In this reaction, the
ratio of C4H9Cl to
C4H9OH is 1:1.
• Thus, the rate of
disappearance of
C4H9Cl is the same as
the rate of appearance
of C4H9OH.
Rate =
-[C4H9Cl]
=
t
[C4H9OH]
t
25.
Reaction Rates & Stoichiometry
Suppose that the mole ratio is not 1:1?
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
26.
Concentration 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 = k[A]x[B]y
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.
27.
Concentration and Rate
NH4+ (aq) + NO2- (aq)  N2 (g) + 2H2O (l)
Compare Experiments 1 and 2:
when [NH4+] doubles, the initial rate doubles.
28.
Concentration and Rate
NH4+ (aq) + NO2- (aq)  N2 (g) + 2H2O (l)
Likewise, compare Experiments 5 and 6:
when [NO2-] doubles, the initial rate doubles.
29.
Concentration and Rate
NH4+ (aq) + NO2- (aq)  N2 (g) + 2H2O (l)
This equation is called the rate law, and k is the
rate constant.
30.
The 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 herethe rate constant k is unique for each reaction
and the value of k changes with temperature
31.
The 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+ ]1[NO2- ]1
32.
Determining 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.
[NO] mol dm-3
0.40
0.40
0.20
[O2] mol dm-3
0.50
0.25
0.25
Rate mol dm-3 s-1
1.6 x 10-3
8.0 x 10-4
2.0 x 10-4
33.
Graphical Representation
• http://www.chem.purdue.edu/gchelp/howtos
olveit/Kinetics/IntegratedRateLaws.html
AB
Chemical
Kinetics
Zero Order
For a zero order reaction:
[A] versus t (linear for a zero order reaction)
rate = k
(k = - slope of line)
Chemical
Kinetics
First Order
For a 1st order reaction:
rate = k[A]
(k = - slope of line)
Chemical
Kinetics
First-Order Processes
Consider the process in which
methyl isonitrile is converted to
acetonitrile.
CH3NC
CH3CN
How do we know this is a
first order reaction?
37.
First-Order Processes
CH3NC
CH3CN
This data was collected
for this reaction at
198.9°C.
Does
rate=k[CH3NC]
for all time intervals?
38.
First-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.
39.
Half-Life of a Reaction
• Half-life is defined as the
time required for onehalf of a reactant to
react.
• Because [A] at t1/2 is
one-half of the original
[A],
[A]t = 0.5 [A]0.
40.
Half-Life of a First Order Reaction
For a first-order process, set [A]t=0.5 [A]0 in integrated rate
equation:
NOTE: For a first-order
process, the half-life
does not depend on the
initial concentration, [A]0.
41.
First Order Rate Calculation
Example 1: The decomposition of compound A is first
order. If the initial [A]0 = 0.80 mol dm-3. and the rate
constant is 0.010 s-1, what is the concentration of [A]
after 90 seconds?
42.
First Order Rate Calculation
Example 1: The decomposition of compound A is first
order. If the initial [A]0 = 0.80 mol dm-3. and the rate
constant is 0.010 s-1, what is the concentration of [A]
after 90 seconds?
Ln[A]t – Ln[A]o = -kt
Ln[A]t – Ln[0.80] = - (0.010 s-1 )(90 s)
Ln[A]t = - (0.010 s-1 )(90 s) + Ln[0.80]
Ln[A]t = -0.90 - 0.2231
Ln[A]t = -1.1231
[A]t =
0.325 mol dm-3
43.
First 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.
44.
First 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
Ln[A] – 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
45.
First 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?
46.
First 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
Ln[1] – Ln(100) = - (0.02406 yr-1)t = - 4.065
t = - 4.062
.
- 0.02406 yr-1
t = 168.8 years
47.
• [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)
Chemical
Kinetics
Second Order
For a 2nd order reaction:
rate = k[A]2
(k = slope of line)
Chemical
Kinetics
Determining Reaction Order
Distinguishing Between 1st and 2nd Order
The decomposition of NO2 at 300°C is described by the
equation:
NO2 (g)
NO (g) + 1/2 O2 (g)
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
50.
Determining Reaction Order
Distinguishing 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
51.
Second-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.
52.
Half-Life for 2nd Order Reactions
For a second-order process, set
[A]t=0.5 [A]0 in 2nd order equation.
In this case the half-life
depends on the initial
concentration of the
reactant A.
53.
Sample 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
.
.
= 0.334 mol-1dm3s-1 t + 1
.00150
.00750
666.7 = 0.334 t + 133.33
0.334 t = 533.4
t = 1600 seconds
54.
Sample 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
[A]t
1
[A]t
. = 0.334 mol-1dm3s-1 (1200s) +
1
.
0.00200 mol dm-3
. = 0.334 mol-1dm3 s-1 (1200s) + 500 mol-1dm3
= 900.8 mol-1dm3
[A]t
=
1 _____. = 0.00111 mol dm-3
900.8 mol-1dm3
55.
Summary of Kinetics Equations
First order
Second order
Second order
Rate
Laws
Integrat
ed Rate
Laws
complicated
Halflife
complicated
56.
Temperature 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
57.
The 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.
58.
The Collision Model
Furthermore, molecules must collide with
the correct orientation and with enough
energy to cause bonds to break and new
bonds to form
59.
Activation 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.
60.
Reaction 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.
61.
Reaction 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.
62.
Maxwell–Boltzmann 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.
63.
Maxwell–Boltzmann Distributions
• As the temperature
increases, the curve
flattens and broadens.
• Thus at higher
temperatures, a larger
population of molecules
has higher energy.
64.
Maxwell–Boltzmann 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.
65.
Maxwell–Boltzmann Distributions
This fraction of molecules can be found through the expression:
where R is the gas constant and T is the temperature in Kelvin .
66.
• http://www.science.uwaterloo.ca/~cchieh/cac
t/c123/eactivat.html
Chemical
Kinetics
Arrhenius 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
68.
• The effect of temperature on rate constants is given
quantitatively by the Arrhenius equation:
•
• k = Ae-Ea/RT
•
• A is the “frequency factor” (although I like to call it the
“orientation factor”)
• Ea is the activation energy.
• R is the ideal gas constant
• T is the absolute temperature
• e-Ea/RT is the fraction of molecules whose kinetic energy is
equal to or greater than Ea.
Chemical
Kinetics
Arrhenius Plot
Taking the natural
logarithm of both
sides, the equation
becomes
1
RT
y = mx + b
Gradient :
m = -Ea / R
When k is determined experimentally at several
temperatures, Ea can be calculated from the slope of a plot of
ln k vs. 1/T.
70.
Chemical
Kinetics
Chemical
Kinetics
slope = -6.12/(1,230 x 10-3) = -4.98 x 103 Kelvin
Ea = -R x slope = -8.31 J/molK x -4.98 x 103 K =
41.3 kJ/mol or 41,300 J/mol
Chemical
Kinetics
Arrhenius 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.
Arrhenius 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)
[
1
1
------ -- -----700K 650K
]
Ea = 180,000 J mol-1 = 180 kJ mol-1
75.
Overview of Kinetics Equations
First order
Second order
Second order
Rate
Laws
Integrated
Rate Laws
Half-life
complicated
complicated
Rate and
Temp (T)
76.
16.2. Reaction Mechanisms
The sequence of events that
describes the actual process by
which reactants become products
is called the reaction mechanism.
• Reactions may occur all at once or
through several discrete steps.
• Each of these processes is known
as an elementary process.
78.
• Sometimes the molecularity of a reaction
can be used to guess if an elementary
reaction is a rate–determining step.
• Molecularity is the count of reacting
particles in an elementary reaction.
Chemical
Kinetics
•
The molecularity of a process tells how many
molecules are involved in the process.
80.
Multistep 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.
• The slow step is the one that determines the
rate!!!!!!!!!!!!!!!!!!!!!
81.
I. Slow Initial Step
NO2 (g) + CO (g)  NO (g) + CO2 (g)
• The rate law for this reaction is found experimentally
to be
Rate = k [NO2]2
82.
• A proposed mechanism for this reaction is
Step 1: NO2 + NO2  NO3 + NO (slow)
Step 2: NO3 + CO  NO2 + CO2 (fast)
_______________________________________
NO2 + CO
=> CO2
+ NO
• 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.
83.
2. Fast 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.
84.
A proposed mechanism is
85.
Fast 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]?
• It comes from step 1
86.
Fast 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.
87.
Chemical
Kinetics
Chemical
Kinetics
Catalysts
• 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
90.
Catalysts
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
91.
Catalysts
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
92.
Catalysts
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+ + AlCl4Cl+ + C6H6  C6H5Cl + H+
H+ + AlCl4-  HCl + AlCl3
Overall reaction
C6H6 + Cl2  C6H5Cl + HCl
93.
Catalysts & 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.
94.
Catalysts & 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
95.
Enzymes
• Enzymes are catalysts in
biological systems.
• The substrate fits into
the active site of the
enzyme much like a key
fits into a lock.
96.
97.