1. Chapter 5 Rates of Chemical Reaction
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2. 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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3. 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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4. 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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5. A+3B → 2D (5-1)
(5-2) or
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6. △[A] 1 △[B] 1 △[D] Rate = - = － = + △t 3 △t 2 △t
The units of the rate are usually mol·L-1·s-1 ; mol·L-1·min-1 ; mol·L-1·h-1
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7. 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, Δc refers to change in concentration and Δt refers to change in time.
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8. Let us look at a specific example: N2 + 3H2 = 2NH3
c( mol/L, t=0)
c( mol/L, t=2s)
vNH3= (0.4-0)/2= 0.2 (mol·L-1·s-1) or
vH2= -( )/2= 0.3 (mol·L-1·s-1)
vN2= -( )/2= 0.1 (mol·L-1·s-1)
vN2 ：vH2 ：vNH3= 1 ：3 ： 2
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9. the average rate over an arbitrary short
Instantaneous rate:
the average rate over an arbitrary short
period of time
2NO2 = 2NO + O2
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10. Determination of instantaneous rate

11. 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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12. Types of Elementary Reactions
Overall reaction: A reaction was completed
through several elementary reactions.
Rate
controlling
step
For example
H2(g) + I2(g) HI(g)
First step I2(g) I(g) (fast)
Second step H2(g) + 2I(g) HI(g) (slow)
Types of Elementary Reactions
SO2 + Cl2
unimolecular SO2Cl2
NO + CO2
bimolecular NO2 + CO
termolecular
2I + H2
2HI
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13. 5-2 Theories of Reaction Rate
5-2.1 Collision Theory and Activation Energy
● Contents of Collision Theory
⑴ reacting molecules must come so close that they collide.
⑵ 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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14. 2NOCl ─→ 2NO + Cl2 ⑶ enough energy; proper orientation
effective collision
ineffective collision
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15. (a) ineffective collision
(b) effective collision

16. 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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17. 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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18. Activation molecule
is the molecule have enough energy and can produce effective collision E1 Ee Ea
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19. ● activation energy (Ea) : The minimum energy of a collision that leads to a reaction.
It has the symbol Ea and is expressed in kilojoules.
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20. In general: ＜ 40 kJ/mol very fast ＞120 kJ/mol slow Ea : 40～400kJ/mol
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:
＜ 40 kJ/mol very fast ＞120 kJ/mol slow
Ea : 40～400kJ/mol
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21. 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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22. Example 1: HI + HI → I•••H •••H ••• I → H2 + I2 activated complex
Give off
energy
Absorb
energy
Activated
process
Activation energy
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23. 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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24. 5-3.1 The Rate Law
The rate of a reaction is proportional to the product of the concentrations of the reactants raised to some power.
Consider the reaction:
a A+b B → c C+d D
v∝[A]m[B]n
v = k[A]m[B]n
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25. ① when [A]=[B]=1mol·L-1, v=k ② the greater the k , the faster the rate
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:
① when [A]=[B]=1mol·L-1, v=k
② the greater the k , the faster the rate
③ 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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26. 5-3.2 Order of A Reaction
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.
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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27. For equation v = k[A]m[B]n
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.
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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28. v = k[N2O5] For the thermal decomposition of N2O5
2N2O5(g) →4NO2(g) + O2(g)
the rate law is
v = k[N2O5]
and not v = k[N2O5]2, as we might have inferred from the balanced equation
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29. For the follow reaction C2H6(g) → 2CH3(g)
The rate expression has the form
v = k [C2H6]2
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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30. Example 5-1: The reaction of nitric oxide with hydrogen at 1280℃ is
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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31. Experiment [NO] [H2] Initial Rate (mol/L s)
× × ×10-5
× × ×10-5
× × ×10-5
Reasoning and Solution:
We assume that the rate law takes the form
v = k[NO]m[H2]n
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32. 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 = k[NO]2[H2]
which shows that it is a (1+2) 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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33. v k=
[NO]2[H2]
data from experiment 2 gives us
5×10-5 k=
(10 ×10-3)2(2 ×10-3)
=2.5 ×102/(mol/L)2·s
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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34. 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)
×10-4
×10-3
×10-2
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35. 2.5 = (2.5)n Solution: 30.3 = (5.5)m v2/v1 = ([OH-]2/[OH-]1)n
v3/v1= ([ClO2]3 / [ClO2]1)m
1.82×10-2/6.00×10-4 = (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
1.50×10-3/6.00×10-4 = (0.075/0.030)n
2.5 = (2.5)n
By inspection, n = 1
The overall rate expression is therefore
v = k[ClO2]2[OH-]
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36. 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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37. Thus Also, from the rate law we know that
Integrate the left side from c = c0 to c and the right from t = 0 to t.
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40. 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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41. 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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43. 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 4.03×104 seconds? (b) Under these conditions, what percent of the N2O molecules will not have reacted after one day?
Solution:
(a)
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44. (b)
Putting in the value for k and substituting t = 8.64×104 seconds (one day has 86,400 seconds) gives
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45. Hence
Therefore, 22.6% of the N2O5 molecules will not have reacted after one day at 25℃.
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46. Example 5-4 SO2Cl2 decomposes to sulfur dioxide and chlorine gas
Example 5-4 SO2Cl2 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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47. k= h-1
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48. 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.
v=k[A][B]
v=k[A]2
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50. 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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51. 2. The rate constant, k, has units of [c]-1[t]-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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52. Example 5-5 Butadiene dimerizes to form C8H12
Example 5-5 Butadiene 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 mol/L sample of butadiene to dimerize until the butadiene concentration is mol /L?
Solution:
t = 3.6 (min)
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53. 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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54. The characteristics of zero-order reactions：
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 t1/2=0.5c0/k.
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55. Example 5-6 The 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] = ×5
= 0.250(mol/L)
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56. Summary of First-order, Second-order and zero-order reactions
Order Rate Law Concentration-Time Equation Half-Life
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57. Exercises 2N2O5→4NO2 + O2. When the reaction temperature is T, the rate constant for the reaction is 1.68×10-2 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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58. 5-4 Effect of Temperature on Reaction Rates
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59. 2018/11/25

60. 5-4.1 Rule of Thumb The rate of a chemical reaction will
( Van ’ t Hoff Law)
The rate of a chemical reaction will
double for each 10℃ increase in the
temperature.
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61. Temperature coefficient
a A + b B = c C + d D
T : vT = kT [A]a [B]b
(T + 10) : v (T+10) = k（T+10） [A]a [B]b
v (T+10) k (T+10)
γ= = = 2～4
v T k T
Temperature coefficient
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63. 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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64. 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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66. 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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68. Example 5-6
The rate constant of a first-order reaction is 3.46×10-2 /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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69. 2018/11/25

70. 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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71. 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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72. kc＞k, Ea’＜Ea k ∝ 1/Ea Ea Ea’ Action mechanics of catalysis
A can be treated as a constant for a given reacting system, so,
k ∝ 1/Ea Ea
For
Ea’
For
kc＞k, Ea’＜Ea
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74. Catalysis can be classified into
homogeneous catalysis
heterogeneous catalysis
enzyme catalysis
three types:
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75. 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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76. 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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77. However, the reaction can be catalyzed by acids or bases.
v =k[CH3COOC2H5]
However, the reaction can be catalyzed by acids or bases.
In the presence of hydrochloric acid the rate is given by
v =k[CH3COOC2H5][H+]
2nd order
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78. 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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81. Enzyme Catalysis Three features Gentle High efficient
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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85. 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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86. 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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87. Find the rate law and rate constant for this reaction.
3. The initial rate of the reaction, 2A+2B = 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)
×10-4
×10-3
×10-4
×10-3
Find the rate law and rate constant
for this reaction.
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88. 4. The growth of pseudomonas bacteria is
modeled as a first-order process with
k= min-1 at 37 ℃. The initial
pseudomonas population density is
1.0×103 cells·L-1.
(a) What is the population density after 2 h?
(b) What is the time required for the
population to go from 1.0×103 cells·L-1
to 5.0×102 cells·L-1 ?
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89. 5. The activation energy of an enzyme-catalyzed reaction in human body (37℃) is 50.0 KJ·mol-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).
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