Transcript AP Kinetics Powerpoint part II

Two Types of Rate Laws
1. Differential- Data table contains
RATE AND CONCENTRATION DATA.
Uses “table logic” or algebra to find the
order of reaction and rate law
2. Integrated- Data table contains TIME
AND CONCENTRATION DATA. Uses
graphical methods to determine the
order of the given reactant. K=slope of
best fit line found through linear
Chemical
Kinetics
regressions
Integrated Rate Law
• Can be used when we want to know
how long a reaction has to proceed to
reach a predetermined concentration of
of some reagent
Chemical
Kinetics
Graphing Integrated Rate Law
• Time is always on x axis
• Plot concentration on y axis of 1st graph
• Plot ln [A] on the y axis of the second
graph
• Plot 1/[A] on the y axis of third graph
• Your are in search of a linear graph
Chemical
Kinetics
Results of linear graph
• Zero order: time vs concentration= line
y= mx+ b
[A]= -kt + [A0 ]
A- reactant A,
A0 - initial concentration of A at t=0
l slope l= k, since k cannot be negative,
and k will have a negative slope
Rate law will be rate=k[A]0
Chemical
Kinetics
Results of linear graph
• First order: time vs ln [ ]= line
y= mx+ b
ln [A]= -kt + ln [A0 ]
A- reactant A,
A0 - initial concentration of A at t=0
l slope l= k, since k cannot be negative,
and k will have a negative slope
Rate law will be rate=k[A]1
Chemical
Kinetics
Results of linear graph
• second order: time vs 1/ [ ]= line
y= mx+ b
1/[A]= kt + 1/ [A0 ]
A- reactant A,
A0 - initial concentration of A at t=0
k=slope
Rate law will be rate=k[A]1
Chemical
Kinetics
First-Order Processes
Consider the process in
which methyl isonitrile is
converted to acetonitrile.
CH3NC
CH3CN
Chemical
Kinetics
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First-Order Processes
CH3NC
CH3CN
This data were
collected for this
reaction at 198.9 C.
Chemical
Kinetics
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First-Order Processes
• When ln P is plotted as a function of time, a
straight line results.
• Therefore,
– The process is first-order.
– k is the negative of the slope: 5.1  105 s1.
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Chemical
Kinetics
Second-Order Processes
The decomposition of NO2 at 300 °C is described by
the equation
1
NO2(g)
NO(g) + 2 O2(g)
and yields data comparable to this table:
Time (s)
[NO2], M
0.0
0.01000
50.0
0.00787
100.0
0.00649
200.0
0.00481
300.0
0.00380
Chemical
Kinetics
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Using the graphing calculator
• L1=time
• L2=concentration-- if straight line, zero
order
• L3=ln concentration-- if straight line- 1st
order
• L4= 1/concentration– if straight line—2nd
order
• Perform 3 linear regressions
Chemical
Kinetics
Second-Order Processes
• The plot is a straight line,
so the process is secondorder 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
Chemical
Kinetics
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Second-Order Processes
1
[NOvs.
2]
• Graphing
t, however, gives
this plot
Fig. 14.9(b).
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
• Because this is a
straight line, the
process is secondorder in [A].
Chemical
Kinetics
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• Determine the rate law and calculate k
• What is the concentration of N2 O5 at
600s?
• At what time is the concentration equal
to 0.00150 M?
Chemical
Kinetics
The decomposition of N2 O5 was studied at constant temp
2 N2 O5 (g)  4 NO2(g) + O2(g)
[N2 O5 ]
Time (s)
0.1000
0
0.0707
50
0.0500
100
0.0250
200
0.0125
300
0.00625
400
Chemical
Kinetics
Half-Life
• 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],
[A]t = 0.5 [A]0.
Chemical
Kinetics
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Half-Life
For a first-order process, this becomes
0.5 [A]0
ln
=
kt
1/2
[A]0
ln 0.5 = kt1/2
0.693 = kt1/2
0.693
= t1/2
k
Note: For a first-order process, then, the half-life DOES
NOT DEPEND ON CONCENTRATION!!!!!!!!
Chemical
Kinetics
Half-Life
First order decay is what is seen in radioactive decay
0.693
= t1/2
k
This is the equation used to calculate the half-life of a
radioactive isotope
Chemical
Kinetics
Problem
• Exercise
• A certain first-order reaction has a halflife of 20.0 minutes.
• a. Calculate the rate constant for this
reaction.
• b. How much time is required for this
reaction to be 75% complete?
• 3.47 × 10−2 min−1;
40 minutes Chemical
Kinetics
Half-Life
For a second-order process,
1
1
= kt1/2 +
0.5 [A]0
[A]0
2
1
= kt1/2 +
[A]0
[A]0
2  1 = 1 = kt
1/2
[A]
[A]0
0
1
= t1/2
k[A]0
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Chemical
Kinetics
Half life zero order
Chemical
Kinetics
Half life first order
Chemical
Kinetics
Half life second order
Chemical
Kinetics
Summary table
order
zero
First
second
Rate Law
Rate=k
rate-=k[ A]
rate= k [A]2
Integrated rate
law in form of
y=mx+b
[A]t
Rate law in data
packet on AP
exam
Does not appear
ln[A]t
slope
Slope = -k
Slope= -k
Half-life
[A0 ]/2k
0.693/k
= -kt + [Ao
]
ln[A]t = -kt + ln[Ao ]
-ln[Ao ] = -kt
1/[A]t -1/[Ao ] = kt
slope=k
1/kA0
Chemical
Kinetics
Reaction Mechanisms
The molecularity of a process tells how
many molecules are involved in the process.
Chemical
Kinetics
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Reaction Mechanisms
• Chemical reactions proceed via a sequence of
distinct stages.
• The sequence is known as the mechanism and each
part of the mechanism is known as a “step”.
• The rate of the reaction is only dependant upon the
slowest step, also known as the rate determining
step or RDS.
• only reactants that appear in the rate determining
step appear in the rate equation and vice-versa
Chemical
Kinetics
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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.
Chemical
Kinetics
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• Example #1
The reaction below
W+YZ
Has the following mechanism
W  R slow
R + Y Q fast
QZ
fast
Here the rate only depends on the concentration of _______________
and therefore the rate equation only contains this reactant
The rate equation is, ______________ in the rate determining step is 1.
Note: R and Q are not reactants or products, but are rather they are called
________________________, produced in one step, but then are
used up in a subsequent step.
Chemical
Kinetics
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Note:
1. In all valid mechanisms the sum of the individual
fast and slow steps must be the same as the
overall chemical equation.
2.The stoichiometric coeffiecient of a substance that
appears in the slow step is the power that the
concentration of that substance is raised to in the
rate equation.
3. If a substance is present at the beginning of a
reaction AND present in the same form at the
end of the reaction, it can be identified as a catalyst
Chemical
Kinetics
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Slow 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.
Chemical
Kinetics
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The reaction below
A + B C + D
Has the following mechanism
AQ
fast equilibrium
Q+BC+D
slow
Here the slow step contains Q and B, and _____is an intermediate
Can intermediates be featured the rate equation? _____________
Since the formation of Q is dependent on A, Q can be replaced by
A in the rate equation. Therefore the rate equation is given
as_________________ The orders w.r.t A and B
are________________________________
since the stoichiometric coefficient of B in the rate determining step
is 1, and the stoichiometric coefficient of A (which Q depends
upon) is also 1.
Chemical
Kinetics
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Temperature and Rate
• Generally, as temperature
increases, so does the
reaction rate.
• This is because k is
temperature-dependent.
Chemical
Kinetics
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Reaction Coordinate Diagrams
• The diagram 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.
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Chemical
Kinetics
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.
Chemical
Kinetics
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Maxwell–Boltzmann Distributions
• As the temperature
increases, the
curve flattens and
broadens.
• Thus, at higher
temperatures, a
larger population of
molecules has
higher energy.
Chemical
Kinetics
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Maxwell–Boltzmann Distributions
• If the dotted line represents the activation
energy, then as the temperature increases,
so does the fraction of molecules that can
overcome the activation-energy barrier.
• As a result, the
reaction rate
increases.
Chemical
Kinetics
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Maxwell–Boltzmann Distributions
This fraction of molecules can be found through the expression
−Ea/RT
f=e
where R is the gas constant and T is the Kelvin temperature.
Chemical
Kinetics
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Arrhenius Equation
Svante Arrhenius developed a mathematical
relationship between k and Ea:
−Ea/RT
k = Ae
where A is the frequency factor, a number
that represents the likelihood that collisions
would occur with the proper orientation for
reaction.
Chemical
Kinetics
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Arrhenius Equation
Taking the natural
logarithm of both
sides, the equation
becomes
Ea
1
ln k = 
( T ) + ln A
R
y = mx + b
Therefore, if k is determined experimentally at
several temperatures, Ea can be calculated
1
from the slope of a plot of ln k vs. T .
© 2012 Pearson Education, Inc.
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.
Chemical
Kinetics
© 2012 Pearson Education, Inc.