Chemical Kinetics: Rates of Reaction

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Chemistry The Molecular Science Moore, Stanitski
and Jurs
Chapter 13 Chemical Kinetics Rates of Reactions
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Inquiring Minds Want to Know.

• Will a reaction happen? (Ch. 18)
• How fast or slow does a reaction happen? (Ch. 13)
• When will the reaction stop or reach
  equilibrium? (Ch. 14 and 17)
• Drugs in the body.
• Ozone in the atmosphere.
• Graphite to diamond.

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Chemical Kinetics

• Chemical kinetics studies the rate at which a
  chemical reaction occurs and the pathway taken.
  Rate is the change of something per unit time.
• Examples distance/time
• /time
• concentration/time
• C (graphite) C (diamond) excruciatingly
  slow rate -?C(graphite) / ?time
• CH4 (g) 2 O2 (g) CO2 (g) 2 H2O (g)
  rapid
• rate -?CH4(g) /?time

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Chemical Kinetics
Homogeneous - reactants products in one
phase. ex. molecules more likely to come together
in gas or liquid phase Heterogeneous -
species in multiple phases. ex. solid catalyst
and air (catalytic converter)
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Reaction Rate

• Factors affecting the speed of a reaction
• (Homogeneous)
• X, M or P
• Concentration of reactants
• k, units include s-1
• Properties (phase, structure)
• Temperature
• Catalysts

Rate -?Cv/?time k Cv n
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Reaction Rate

• Change in reactant (or product) per unit time.

Cresol violet (Cv a dye) decomposes in NaOH(aq)
Cv(aq) OH- (aq) ? CvOH(aq)
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Reaction Rates and Stoichiometry

• For any general reaction
• a A b B c C d D
• The overall rate of reaction is

Cv(aq) OH-(aq) ? CvOH(aq)
Loss of 1 Cv ? Gain of 1 CvOH Rate of Cv loss
Rate of CvOH gain Rate -?Cv/?time
?CvOH/?time
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Reaction Rates and Stoichiometry
H2 (g) I2 (g) ? 2 HI (g) The rate of loss of
I2 is 0.0040 mol L-1 s-1. What is the rate of
formation of HI ?
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Reaction Rate
The rate of the Cv reaction can be calculated

• Time, t Cv Average rate
• (s) (mol / L) (mol L-1 s-1)
• 0.0 5.000 x 10-5
• 10.0 3.680 x 10-5
• 20.0 2.710 x 10-5
• 30.0 1.990 x 10-5
• 40.0 1.460 x 10-5
• 50.0 1.078 x 10-5
• 60.0 0.793 x 10-5
• 80.0 0.429 x 10-5
• 100.0 0.232 x 10-5

13.2 x 10-7 9.70 x 10-7 7.20 x 10-7 5.30
x 10-7 3.82 x 10-7 2.85 x 10-7 1.82 x 10-7
0.99 x 10-7
trend? Rate depends on conc. Rate decrease with
conc. React.
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Average Rate and Instantaneous Rate
Graphical view of Cv reaction
5.0E-5 4.0E-5 3.0E-5 2.0E-5 1.0E-5 0
Rate -?Cv/?time
Cv (mol/L)
t (s)
0 20 40 60 80 100
Average rate slope of the blue or grey
triangle but the avg. rate depends on interval
chosen.
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Average Rate and Instantaneous Rate
5.0E-5 4.0E-5 3.0E-5 2.0E-5 1.0E-5 0
Cv (mol/L)
t (s)
0 20 40 60 80 100

• Instantaneous rate slope of a line tangent to
  the curve.
• t 5 s and t 75 s have different instantaneous
  rates
• Cannot predict Cv at a time too far away since
  the rate changes.

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The Rate Law

• Rate may change when reactant changes.

• Cv example shows this.
• For Cv the rate is proportional to concentration.

k
(10 second interval)
rate k Cvn
y m x
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The Rate Law
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Rate Law and Order of Reaction
A general reaction will usually have a rate
law rate k Am Bn . . .
where k rate constant (T, properties,
catalyst) m, n order for A B, respectively m
n overall order of the reaction

• The orders are usually integers (-2, -1, 0, 1,
  2), but may also be fractions (½, ?)
• The concentration of a reactant with a zero-order
  dependence has no effect on the rate of the
  reaction. (There just needs to be some reactant
  around.

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The Rate Law Reaction Order
The reaction is second order in NO first order
in H2 third order overall.
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The Rate Law Reaction Order

• 22.
• 2 NO (g) Br2 (g) 2 NOBr (g)
• Experiment shows that the reaction is first-order
  in Br2 and second-order in NO.
• Write the rate law for the reaction.
• Rate kBr2NO2
• If the concentration of Br2 is tripled, how will
  the reaction rate change?
• 3X
• What happens to the reaction rate when the
  concentration of NO is doubled?
• 4X

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Determining Rate Laws from Initial Rates
Rate laws must be measured experimentally. They
cannot be predicted from reaction stoichiometry.

• Initial Rate Method
• To find the order for a reactant
• Run the experiment with known reactant0.
• Measure the initial rate of reaction (slope at t
  0, lt2 consumed.)
• Change reactant0 of 1 reactant keep all others
  constant.
• Re-measure the initial rate.
• The ratio of the two rates gives the order for
  the chosen reactant.

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Determining Rate Laws from Initial Rates

• Data for the reaction of methyl acetate with base

change concentration of each reactant in turn
Initial concentration (M) Expt. CH3COOCH30 OH
-0 Initial rate (M/s) 1 0.040 0.040
2.2 x 10-4 2 0.040 0.080 4.5 x 10-4
3 0.080 0.080 9.0 x 10-4
gt x2
gt x2
Start with generic rate law rate k
CH3COOCH3m OH-n
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Determining Rate Laws from Initial Rates
Initial concentration (M) Expt. CH3COOC
H3 OH- Initial rate (M/s) 1
0.040 0.040 2.2 x 10-4 2 0.040
0.080 4.5 x 10-4 3 0.080 0.080
9.0 x 10-4

• Dividing the first two data sets (OH changing)
• 4.5 x 10-4 M/s k (0.040 M)m(0.080 M)n
• 2.2 x 10-4 M/s k (0.040 M)m(0.040 M)n

Thus 2.05 (2.00)n 2.05 (2.00)n and n
1
rate k CH3COOCH3m OH-1
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Determining Rate Laws from Initial Rates

• Use experiments 2 3 to find m (dependance on
  CH3COOCH3)

9.0 x 10-4 M/s k (0.080 M)m(0.080 M)n 4.5 x
10-4 M/s k (0.040 M)m(0.080 M)n
So 2.00 (2.00)m (1)n 2.00
(2.00)m and m 1 Also 1st order with respect
to CH3COOCH3.
rate k CH3COOCH31 OH-1
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Determining Rate Laws from Initial Rates

• The rate law is
• rate k CH3COOCH3OH-
• Overall order for the reaction is
• m n 1 1 2
• The reaction is 2nd order overall.

1st order in OH- 1st order in CH3COOCH3
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Determining Rate Laws from Initial Rates
If a rate law is known, k can be
determined rate k CH3COOCH3OH-
Using Exp. 1
k 0.1375 M-1 s-1 0.1375 L mol-1 s-1
Could repeat for each run, take an average But a
graphical method is better.
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Practice
Rate kNOnO2m

• 2NO(g) O2(g) ? 2NO2
• Initial rates were measured at 25 ºC starting
  with various concentrations of reactants.

What is the rate law? (m,n) Calculate the rate
constant (k).
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Practice What is n?
Rate kNOnO2m
4x
The reaction is second order with respect to NO.
Why cant we compare experiment 1 and 3 ???
n 2
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Practice What is m?
Rate kNO2O2m
2x
2x
The reaction is first order with respect to
oxygen.
m 1
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Practice What is k?
Rate kNO2O2
Using Exp. 1
Units?... rate always has units of M/s.
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Practice What is k?

• Iodide ion is oxidized in acidic solution to the
  triiodide ion, I3- , by hydrogen peroxide.

A series of four experiments were run at
different concentrations, and the initial rates
of I3- formation were determined. From the
following data, obtain the reaction orders with
respect to H2O2, I-, and H. Calculate the
numerical value of the rate constant.
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Practice What is m,n,p?
Rate kH2O2mI-nHp
Rate kH2O21I-2H0
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Because H 0 1, the rate law is
Rate kH2O2I-2
We can now calculate the rate constant by
substituting values from any of the experiments.
Using Experiment 1
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Change of Concentration with Time

• A rate law simply tells you how the rate of
  reaction changes as reactant concentrations
  change.

The integrated rate law shows how a reactant
concentration changes over a period of time.
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The Integrated Rate Law (Calculus)
Consider a 1st-order reaction
A products
A big change A little change (differential
equation)
Integrates to ln At -k t ln A0
y m x b (straight line)
If a reaction is 1st-order, a plot of ln A vs.
t will be linear.
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Graphing Kinetic Data

• In addition to the method of initial rates, rate
  laws and k can be deduced by graphical methods.

For first-order reaction
y mx b
This means if we plot lnAt versus t, we will
get a straight line for a first-order reaction.
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Concentration-Time Equations

• Second-Order Rate Law

Using calculus, you get the following equation.
Here At is the concentration of reactant A at
time t, and Ao is the initial concentration.
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Graphing Kinetic Data
For second-order reaction
y mx b

• This means if we plot 1/At versus time,
• we will get a straight line for a second-order
  reaction.

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Concentration-Time Equations
For a zero-order reaction we could write the rate
law in the form
Rearrange get the following equation.
Here At is the concentration of reactant A at
time t, and Ao is the initial concentration.
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Graphing Kinetic Data
For zero-order reaction
y mx b

• This means if we plot At versus time,
• we will get a straight line for a second-order
  reaction.

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The Integrated Rate Law
0 rate k At -kt A0 -k 1 rate kA
lnAt -kt lnA0 -k
y mx b
The most accurate k is obtained from the slope of
a plot.
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The Integrated Rate Law
Rate data for the decomposition of
cyclopentene C5H8(g) ? C5H6(g) H2(g) were
measured at 850C. Determine the order of the
reaction from the following plots of those data
NOT a line line! NOT a line

• The reaction is first order (the only linear
  plot)
• k -1 x (slope) of this plot.

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Example 1

• The decomposition of N2O5 to NO2 and O2 is first
  order with a rate constant of 4.800 x 10-4 s-1.
  If the initial concentration of N2O5 is 1.65 x
  10-2 mol/L, what is the concentration of N2O5
  after 825 seconds?

The first-order time-concentration equation for
this reaction would be
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Practice- using integrated rate laws

• The decomposition of N2O5 to NO2 and O2 is first
  order with a rate constant of 4.800 x 10-4s-1. If
  the initial concentration of N2O5 is 1.65x10-2
  mol/L, what is the concentration of N2O5 after
  825 seconds?

eln(x) x
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Example 2

• Consider the following first-order reaction
• CH3CH2Cl(g)? C2H4(g) HCl(g)
• The initial concentration of ethyl chloride was
  0.00100M. After heating at 500.0ºC for 155.00 s
  the concentration was reduced to 0.000670 M.
  What was the concentration of ethyl chloride
  after a total of 256.00 s?

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Example 1

• The initial concentration of ethyl chloride was
  0.00100M. After heating at 500.0ºCfor 155.00 s
  the concentration was reduced to 0.000670 M.
  What was the concentration of ethyl chloride
  after a total of 256.00 s?

C2H5Cl0 0.00100M C2H5Clt1 0.000670M t1
155.00s t2 256.00s, C2H5Clt2?
2.58 10-3 s-1
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Example 1
The initial concentration of ethyl chloride was
0.00100M. After heating at 500.0ºCfor 155.00 s
the concentration was reduced to 0.000670 M.
What was the concentration of ethyl chloride
after a total of 256.00 s?
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Half-Life

• t1/2 time for reactant to fall to
  ½reactant0.

• For 1st -order reactions
• t1/2 is independent of the starting
  concentration.
• only true for 1st order reactions (not 0th,
  2nd)
• t1/2 is constant for a given 1st -order reaction.

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Half-Life
For a 1st-order reaction lnAt -kt
lnA0
When t t1/2, At ½A0
Then ln(½A0) -kt1/2 lnA0
ln(½A0/A0) -kt1/2 note ln x ln
y ln(x/y) ln(½) -ln(2) -kt1/2 note
ln(1/y) ln y
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Half Life
t1/2 of a 1st-order reaction can be used to find
k.

• For cisplatin (a chemotherapy agent)
• 0.0100 M ? 0.0050 M, after 475 min
• (t½475minutes)

graphically
t½
t½
t½
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Calculating or t

• Use an integrated rate equation or half-life
  equation.

Example In a 1st-order reaction, reactant0
0.500 mol/L and t1/2 400.s. Calculate

• reactant,1600.s after initiation.
• t for reactant to drop to 1/16th of its initial
  value.
• t for reactant to drop to 0.0500 mol/L.

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Calculating or t

• In a 1st-order reaction, reactant0 0.500
  mol/L and t1/2 400.s
• (a) Calculate reactant ,1600.s after initiation.

1st order k ln 2/ t½ 0.6931/(400. s)
1.733x10-3 s-1 lnAt -(0.001733
s-1)(1600 s) ln(0.500) lnAt -2.773
-0.693 -3.466 At e-3.466 0.0312 mol/L
ln At -kt ln A0
OR1600 s 4 t1/2 so 0.500 ? 0.250 ? 0.125 ?
0.0625 ? 0.0313 M.
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Calculating or t

• In a 1st-order reaction, reactant0 0.500
  mol/L and t1/2 400.s
• (b) Calculate t for reactant to drop to 1/16th
  of its initial value.

4 (400 s) 1600 s
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Calculating or t
In a 1st-order reaction, reactant0 0.500
mol/L and t1/2 400.s (c) Calculate t for
reactant to drop to 0.0500 mol/L ?

• From part (a) k 1.733 x 10-3 s-1
• ln At -kt ln A0
• then ln (0.0500) -(0.001733 s-1) t
  ln(0.500)
• -2.996 -(0.001733 s-1) t 0.693
• t 1.33 x 103 s

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Elementary and Complex Reactions

• On the nanoscale level, a reaction may be
• unimolecular a single particle (atom, ion,
  molecule) rearranges into 1 or 2 different
  particles.
• (example some decomposition, geometric isomer)
• bimolecular two particles collide and rearrange
  
• (example composition)
• Observed reactions, on the macroscale level, may
  be
• elementary directly occur by one of these two
  processes, or
• complex occur as a series of elementary steps

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Elementary Reactions
Example Label these elementary reactions as
unimolecular or bimolecular
unimolecular
bimolecular
unimolecular
unimolecular
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Elementary Reactions - Unimolecular
2-butene isomerization is unimolecular

• cis-trans conversion twists the CC bond.
• This requires a lot of energy (Ea
  4.35x10-19J/molecule 262 kJ/mol)
• Even more (4.42x10-19J/molecule) to convert back.

cis-2-butene trans-2-butene
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Unimolecular Reactions
transition state or activated complex
500 400 300 200 100 0
Ea is the activation energy, the minimum E to go
over the barrier.
Potential energy (10-21 J)
Ea 435 x 10-21 J
?E -7 x 10-21 J
Initial state
Final state
-30 0 30 60 90
120 150 180 210
Exothermic overall
Reaction Progress (angle of twist)

• kinetics if lower Ea, increase the speed of the
  reaction
• thermodynamics if EproductltEreactant, the
  reaction is exothermic

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Transition State

• Occurs at the top of the activation barrier
• Exists for very short time (few fs, 1 fs 10-15
  s).
• Falls apart to form products or reactants.
• the molecules must have enough energy
  activation energy to form the transition state
  and convert it into products.

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Endothermic reaction
overall up hill EproductsgtEreactants
Ea
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Exothermic reaction
overall down hill EproductsgtEreactants
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Bimolecular Reactions
I- must collide with enough E and with the right
orientation to cause the inversion.
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Bimolecular Reactions
For the reaction to take place two molecules must
effectively collide. One out of four collisions
is successful. (backside of tetrahedron, opposite
Br). This geometric constraint on approach is
called a steric factor.
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Bimolecular Reaction
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Bimolecular Reaction
O3 NO O2 NO2
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Reaction Energy Diagrams
Forward reaction Ea 10kJ/mol Exothermic Backwa
rd reaction Ea? Exo or Endo?
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Factors Affecting Reaction Rates
Increasing T will speed up most reactions.
Higher T higher average Ek for the
reactants. larger fraction of the molecules
can overcome the activation barrier.
Many more molecules have enough E to react at
75C, so the reaction goes much faster.
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Temperature and Reaction Rate

• Rule of thumb approximately double the rate for
  every 10 K.

I- CH3Br ? Br - CH3I
Rate kI-mCH3Brn
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Temperature and Reaction Rate

• The Arrhenius equation shows how k varies with T

A Frequency factor. How often a collision
occurs with the correct orientation. (1/4 for I-
CH3Br) e-Ea/RT Fraction of the molecules
with enough Ea to cross the barrier.
T Temperature Must be in Kelvin. R Gas law
constant 8.314 J K-1 mol-1.
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Temperature and Reaction Rate
rate kI-mCH3Brn rate
(Ae-Ea/RT)I-mCH3Brn Can rearrange.
OWL 13.5a OWL link
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Factors Affecting Reaction Rates

• Presence of a catalyst.

A catalyst is a substance that increases the rate
of a reaction without being consumed in the
overall reaction. The catalyst generally does
not appear in the overall balanced chemical
equation (although its presence may be indicated
by writing its formula over the arrow).
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Catalysts

• A catalyst increases the rate of reaction by
  lowering the activation energy, Ea.
• To avoid being consumed, the catalyst must
  participate in at least one step of the reaction
  and then be regenerated in a later step.
• Ultimately a catalyzed reaction has a different
  reaction mechanism than the same reaction
  proceeding without the presence of catalyst.

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Catalyzed and Uncatalyzed Reactions
Without catalyst. With I- catalyst.
70
Catalyzed and Uncatalyzed Reactions
71
Reaction Mechanisms

• To explain how reactions happen on the molecular
  scale we need to look into individual steps that
  take place during the reaction.

The collection of individual steps that a
reaction goes through is referred to as the
reaction mechanism. The individual steps are
referred to as elementary reactions. The sum of
several elementary reactions is a complex
reaction.
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Elementary/Complex Reactions

• Consider the reaction of nitrogen dioxide with
  carbon monoxide.

reaction intermediate not in net equation
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Elementary/Complex Reactions
The overall chemical equation is obtained by
adding the two steps together and canceling any
species common to both sides.
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Rate Laws for Elementary Reactions

• Elementary reactions
• Occur as written and their rate laws are
  predictable.
• Unimolecular reactions are always 1st -order.
• Bimolecular reactions are always 2nd-order.

A Products rate kA A B Products
rate kAB A A Products rate
kA2
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Rate Laws for Elementary Reactions

• For an elementary reaction, the coefficient of
  each reactant becomes the power to which it is
  raised in the rate law for that reaction.

rate kNOO3 (determined experimentally)
76
Rate Laws for Complex Reactions

• Complex reactions
• Do not occur as written.
• They occur as a series of elementary steps.

rate kH2O2 (determined experimentally)
77
Rate Laws and Mechanisms

• Consider the reaction below.

This implies that the reaction above is not an
elementary reaction but rather the result of
multiple steps.
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Rate Laws and Mechanisms
2I- H2O2 2 H3O I2 4H2O Rate
I-H2O2
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Rate Laws and Mechanisms
What is the net reaction? What is the rate law
for this reaction?
(fast) (slow) (fast)
Rate N2O2H22