Chapter 5 Rates of Chemical Reaction

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Chapter 5 Rates of Chemical Reaction
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• 5-1 Rates and Mechanisms of Chemical
• Reactions
• 5-2 Theories of Reaction Rate
• 5-3 Reaction Rates and Concentrations
• 5-4 Effect of Temperature on Reaction
• Rates
• 5-5 Effect of Catalyst on Reaction Rates

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5-1 Rates and Mechanisms of Chemical
Reactions

• 5-1.1 The Rate of Chemical Reaction
• reaction rates
• the change in the concentration of a
• reactant or a product with time.
• The rate is defined to be a positive number.

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The rate of a chemical reaction is measured by
the decrease in concentration of a reactant or
the increase in concentration of a product in a
unit of time.
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A3B ? 2D (5-1)
(5-2)
or
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?A 1 ?B 1
?D Rate - -
?t 3 ?t
2 ?t
The units of the rate are usually
molL-1s-1 molL-1min-1 molL-1h-1
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• average reaction rate is obtained by dividing
• the change in concentration of a reactant
  or
• product by the time interval over which the
• change occurs.

v refers to average reaction rate, ?c refers to
change in concentration and ?t refers to change
in time.
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• Let us look at a specific example
• N2 3H2 2NH3
• c( mol/L, t0) 1.0 3.0 0
• c( mol/L, t2s) 0.8 2.4 0.4
• vNH3 (0.4-0)/2 0.2 (molL-1s-1)
• or
• vH2 -(2.4-3.0)/2 0.3 (molL-1s-1)
• vN2 -(0.8-1.0)/2 0.1 (molL-1s-1)
• vN2 vH2 vNH3 1 3 2
• or

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• Instantaneous rate
• the average rate over an arbitrary short
• period of time

2NO2 2NO O2
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Determination of instantaneous rate
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5-1.2 The Mechanisms of Chemical
Reactions

• Reaction Mechanisms is a description of the path
• that
  a reaction takes.
• Elementary reaction A reaction can complete
  directly
• by only
  one step or reactants
• can
  convert into products.
• 2NO2 2NO O2
• 2I H2 2HI

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termolecular

• Overall reaction A reaction was completed
• through several elementary
  reactions.

Rate controlling step
For example H2(g) I2(g)
2HI(g) First step I2(g)
2I(g) (fast) Second step
H2(g) 2I(g) 2HI(g) (slow)

• Types of Elementary Reactions

SO2 Cl2
unimolecular SO2Cl2
NO CO2
bimolecular NO2 CO
2HI
2I H2
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5-2 Theories of Reaction Rate

• 5-2.1 Collision Theory and Activation Energy
• ? Contents of Collision Theory
• ? reacting molecules must come so close that they
  collide.
• ? not every collision between molecules creates
  products,
• only few collisions between reactant
  molecules
• will react.

effective collision a collision that leads to a
reaction
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2NOCl -? 2NO Cl2
? enough energy proper orientation
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(a) ineffective collision (b) effective
collision
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For a collision to result in reaction, the
molecules must be properly oriented. For
the reaction CO(g) NO2(g) ? CO2(g)
NO(g) the carbon atom of the CO molecule must
strike an oxygen atom of the NO2 molecule,
forming CO2 as one product, NO as the other.
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• Collisions must occur with enough energy to break
  the bonds in the reactants so that new bonds can
  form in the products.

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• Activation molecule

is the molecule have enough energy and can
produce effective collision
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? activation energy (Ea) The minimum
energy of a collision that leads to a
reaction.
It has the symbol Ea and is expressed
inkilojoules.
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Figure As the activation energy of a reaction
decreases, the number of molecules with at least
this much energy increases, as shown by the
yellow shaded areas.
In general
lt 40 kJ/mol very fast gt120 kJ/mol slow
Ea 40400kJ/mol
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5-2.2 The Transition StateTheory

• Transition state theory (TST)
• is also called activated complex theory.
• reactants pass through high-energy transition
• states before forming products, they are
  associated in an unstable entity called an
  activated complex,
• then change into products.

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Example 1 HI HI ? IH H I ? H2
I2
activated complex
Give off energy
Absorb energy
Activated process
Activation energy
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5-3 Reaction Rates and Concentrations

• Chemical reactions are faster when the
  concentrations of the reactants are increased
  because more molecules will exist in a given
  volume. More collisions will occur and the
• rate of a reaction will increase.

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The rate of a reaction is proportional to the
product of the concentrations of the reactants
raised to some power.
5-3.1 The Rate Law
Consider the reaction
a Ab B ? c Cd D
v?AmBn
v kAmBn
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where k is the rate constant A, B are
the concentration of A and B m and n are
themselves constants for a given reaction

• Notice
• ? when AB1molL-1, vk
• ? the greater the k , the faster the rate
• ? m and n must be determined experimentally,
• in general, m and n are not equal to the
• stoichiometric coefficients a and b

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The order of a reaction with respect to one
of the reactants is equal to the power to which
the concentration of that reactant is raised in
the rate equation.
5-3.2 Order of A Reaction
The sum of the powers to which all reactant
concentrations appearing in the rate law are
raised is called the overall reaction order.

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m is the order of the reaction with respect to A,
n is the order of the reaction with respect to B.
The overall order of the reaction is the sum of m
and n.
For equation v kAmBn
the exponents m and n are not necessarily related
to the stoichiometric coefficients in the
balanced equation, that is, in general it is not
true that for a A b B ? c C d D, a m
and b n
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For the thermal decomposition of N2O5
2N2O5(g) ?4NO2(g) O2(g)

• the rate law is
• v kN2O5
• and not v kN2O52, as we might have inferred
  from the balanced equation

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For the follow reaction C2H6(g) ? 2CH3(g) The
rate expression has the form v k C2H62

• so that n 2 even though the coefficient of
  C2H6 in the chemical equation is 1. Thus, the
  decomposition of N2O5 is first order, whereas
  that of C2H6 is second order.

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The follow example illustrates the procedure for
determining the rate law of a reaction.
Example 5-1 The reaction of nitric oxide
with hydrogen at 1280? is

• 2NO(g) 2H2(g)

N2(g) 2H2O(g)
From the following data collected at this
temperature, determine the rate law and calculate
the rate constant.
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Experiment NO H2 Initial
Rate (mol/L s)

• 1 5.010-3 2.010-3 1.310-5

2 10.010-3 2.010-3 5.010-5
3 10.010-3 4.010-3 10.010-5
Reasoning and Solution

We assume that the rate law takes the form
v kNOmH2n
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Experiments 1 and 2 shows that when we double the
concentration of NO at constant concentration of
H2, the rate quadruples. Thus the reaction is
second order in NO.
Experiments 2 and 3 indicate that doubling H2
at constant NO doubles the rate the reaction
is first order in H2. The rate law is given by
v kNO2H2
which shows that it is a (12) or third-order
reaction overall.
The rate constant k can be calculated using the
data from any one of the experiments. Since
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v k-------------- NO2H2
data from experiment 2 gives us
510-5 k------------------------
---- (10 10-3)2(2 10-3)
2.5 102/(mol/L)2s
Comment Note that the reaction is first order in
H2, whereas the stoichiometric coefficient for H2
in the balanced equation is 2
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• Example 5-2
• Given the following data, what is the rate
  expression for the reaction between hydroxide ion
  and chlorine dioxide?
• 2ClO2(a q) 2OH-(a q) ?ClO3-(a q) ClO2-(a q)
  H2O
• ClO2 (mol l-1) OH- (mol l-1) Rate
  (mol L-1 s-1)
• 0.010 0.030
  6.0010-4
• 0.010 0.075
  1.5010-3
• 0.055 0.030
  1.8210-2

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• Solution
• v3/v1 (ClO23 / ClO21)m
• 1.8210-2/6.0010-4 (0.055/0.010)m
• 30.3
  (5.5)m
• By inspection, m 2. The reaction is 2nd order
  in ClO2
• v2/v1
  (OH-2/OH-1)n
• 1.5010-3/6.0010-4 (0.075/0.030)n
• 2.5 (2.5)n
• By inspection, n 1
• The overall rate expression is therefore
• v kClO22OH-

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First-order reactions

• A first-order reaction is a reaction whose rate
  depends on the reactant concentration raised to
  the first power.

A ? product
the rate is
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Also, from the rate law we know that
Thus

• Integrate the left side from c c0 to c and the
  right from t 0 to t.

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The characteristics of first-order
reactions

• 1. A plot of logc versus t (time) gives a
  straight line with a slope of -k/2.303.

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2. The rate constant, k, has units of time-1.

• 3. half-life (t1/2)
• is the time it takes for the concentration of
  a reactant A to fall to one half of its original
  value.

By definition, when t t1/2, c c0/2, so
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Example 5-3(a) What is the rate constant k for
the first-order decomposition of N2O5(g) at
25? if the half-life of N2O5(g) at that
temperature is 4.03104 seconds?(b) Under these
conditions, what percent of the N2O5
molecules will not have reacted after one day?

• Solution
• (a)

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• (b)

Putting in the value for k and substituting t
8.64104 seconds (one day has 86,400 seconds)
gives
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• Hence

Therefore, 22.6 of the N2O5 molecules will not
have reacted after one day at 25?.
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Example 5-4SO2Cl2 decomposes to sulfur dioxide
and chlorine gas. The reaction is first order
If it takes 13.7 hours for a 0.250mol/L solution
of SO2Cl2 to decompose into a 0.117mol/L
solution, what is the rate constant for the
reaction and what is the half-life of SO2Cl2
decomposition?

• Solution

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k0.0554 h-1
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Second - order reactions

• A second-order reaction is a reaction whose rate
• depends on reactant concentration raised to the
• second power or on the concentrations of two
• different reactants, each raised to the first
  power.

vkAB
vkA2
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The characteristics of second-
order reactions

• 1. A graph of 1/c against time is a straight line
  ,
• the slope of which gives the rate constant
  for the reaction

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2. The rate constant, k, has units of c-1t-1

• 3. The half-life of 2th-order reactions
  

Note that the half-life of a second-order
reaction is not independent of the initial
concentration, as in the case of a first-order
reaction. This is one way to distinguish a
first-order reaction from a second-order reaction.
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Example 5-5Butadiene dimerizes to form C8H12.
This reaction is 2nd order in butadiene. If the
rate constant for the reaction is 0.84 L
mol-1min-1, how long will it take for a 0.500
mol/L sample of butadiene to dimerize until the
butadiene concentration is 0.200 mol /L?

• Solution

t 3.6 (min)
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Zero - order reactions

• A zero-order reaction is one where the rate does
  not
• depend on the concentration of the species.

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c - k t c0
The characteristics of zero-order
reactions

• 1. A graph of c against t is a straight line

2. The rate constant, k, has units of c
t-1
3. The half-life of a zero-order reaction is
t1/20.5c0/k.
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Example 5-6The decomposition of HI into hydrogen
and iodine on a gold surface is 0th order in HI.
The rate constant for the reaction is 0.050mol/L
s. If you begin with a 0.500mol/L concentration
of HI, what is the concentration of HI after 5
seconds?

• Solution

HI 0.500-0.0505 0.250(mol/L)
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Summary of First-order, Second-order
and zero-order reactions

• Order Rate Law Concentration-Time Equation
  Half-Life

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Exercises2N2O5?4NO2 O2. When the reaction
temperature is T, the rate constant for the
reaction is 1.6810-2 s -1 If we add 2.5mol N2O5
to the container (V 5L). what is the
amount-of-substance of N2O5 and O2 after 1
minutes ?
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5-4 Effect of Temperature on Reaction Rates
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• 5-4.1 Rule of Thumb
• ( Van t Hoff Law)

The rate of a chemical reaction will double for
each 10? increase in the temperature.
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a A b B c C d D
T vT kT Aa Bb (T 10) v (T10)
k(T10) Aa Bb
v (T10) k (T10) ?
---------- ----------- 24 v T
k T
Temperature coefficient
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5-4.2 The Arrhenius Equation
Where Ea is the activation energy of the reaction
(in kJ/mol), R is the gas constant (8.314
JK-1mol-1), T is the absolute temperature, and e
is the base of the natural logarithm scale. The
quantity A represents the collision frequency,
and is called the frequency factor.
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Thus, a plot of log k versus 1/T gives a
straight line whose slope is equal to -E a
/2.303R and whose intercept with the ordinate is
log A.
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5-4.3 Application of Arrhenius Equation
According to this equation, we can calculate Ea
and k

• T1 ? k1

T2 ? k2
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Example 5-6

• The rate constant of a first-order reaction is
  3.4610-2 /s
• at 298 K. What is the rate constant at 350 K if
  the
• activation energy for the reaction is 50.2
  KJ/mol?

Answer
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5-5 Reaction Rates and Catalyst

• Catalyst is a substance that increase the rate
  of a chemical reaction without itself being
  consumed (changed).

negative catalyst slows the rate of a reaction
MnO2
inhibitor
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• The features of catalysts

• No change of mass and composition
• Selective
• Small amount can have big action
• Not only speed forward reaction but also speed
  reverse reaction ( effect velocity , not effect
  equilibrium constant )

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• Action mechanics of catalysis

• A can be treated as a constant for a given
  reacting system, so,
• k ? 1/Ea

Ea
For
Ea
For
kcgtk, EaltEa
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Catalysis can be classified into
homogeneous catalysis heterogeneous
catalysis enzyme catalysis
three types
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• Homogeneous catalysis

• In homogeneous catalysis, the catalyst is present
  in the same phase as the reactants, as when a
  gas-phase catalyst speeds up a gas-phase
  reaction, or a species dissolved in solution
  speeds up a reaction in solution

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Tl(aq) 2Ce4(aq) ? Tl3(aq) 2Ce3(aq)
The rate is very
slow,
but it can be catalyzed by Ag or Mn2
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• v kCH3COOC2H5
• However, the reaction can be catalyzed by acids
  or bases.
• In the presence of hydrochloric acid the rate is
  given by
• v kCH3COOC2H5H

2nd order
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• Heterogeneous catalysis

• In heterogeneous catalysis, the catalyst is
  present as a distinct phase. The most important
  case is the catalytic action of certain solid
  surfaces on gas-phase and solution-phase
  reactions.

C2H4 (g) H2(g) ? C2H6(g)
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• Enzyme Catalysis

• Enzymes are biological catalysts.
• An average living cell may contain some 3000
  different enzymes

Gentle High efficient High selective (special)
Three features
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Problems
1. A slightly bruised apple will rot extensively
in about 4 days at room temperature (20?). If it
is kept in the refrigerator at 0?, the same
extent of rotting takes about 16 days. Suppose
the rate of rotting is proportional to the time,
what is the activation energy for the rotting
reaction?
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2. Sucrose decomposes in acid solution into
glucose and fructose according to a first-order
rate law, with a half-life of 3.33 h at 25?. What
fraction of a sample of sucrose remains after
9.00 h?
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3. The initial rate of the reaction, 2A2B C
D, is determined for different initial
conditions, with the results listed in the
following table
Run number c(A)0(mol/L ) c(B)0(mol/L) Initial
rate (mol/L s)
1 0.185 0.133
3.3510-4
2 0.185 0.226
1.3410-3

3 0.370 0.133
6.7010-4
4 0.370 0.226
2.7010-3
Find the rate law and rate constant for this
reaction.
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4. The growth of pseudomonas bacteria is
modeled as a first-order process with k
0.035 min-1 at 37 ?. The initial
pseudomonas population density is 1.0103
cellsL-1. (a) What is the population density
after 2 h? (b) What is the time required for
the population to go from 1.0103
cellsL-1 to 5.0102 cellsL-1 ?
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5. The activation energy of an enzyme-catalyzed
reaction in human body (37?) is 50.0 KJmol-1.
How many times will the rate of the reaction be
increased if a patient has a fever up to 40 ?
(supposed that temperature has no effect on
enzyme activity).