Transcript CHAPTER 16

CHAPTER 15
Chemical
Kinetics
KINETICS
— the study of REACTION RATES and
their relation to the way the reaction
proceeds, i.e., its MECHANISM.
Reaction rate = change in concentration of
a reactant or product with time.
Three “types” of rates
initial rate
average rate
instantaneous rate
The Rate of Reaction
• Consider the hypothetical reaction,
A(g) + B(g)  C(g) + D(g)
• equimolar amounts of reactants, A and B, will be consumed while
products, C and D, will be formed as indicated in this graph:
Concentrations of
Reactants & Products
1.2
1
0.8
[A] & [B]
[C] & [D]
0.6
0.4
0.2
0
0
50
0
10
0
15
0
20
Time
0
25
30
0
0
35
The Rate of Reaction
• Mathematically, the rate of a reaction can be written as:
aA(g) + bB(g)  cC(g) + dD(g)
- A - B + C + D
Rate =

or

a t
b t
c t
d t
The Rate of Reaction
• The rate of a simple one-step reaction is directly proportional
to the concentration of the reacting substance.
A (g)  B(g) + C(g)
Rate  A or Rate = kA
• [A] is the concentration of A in molarity or moles/L.
• k is the specific rate constant.
– k is an important quantity in this chapter.
The Rate of Reaction
•
•
Important terminology for kinetics.
The order of a reaction can be expressed in terms of either:
1 each reactant in the reaction or
2 the overall reaction.
 Order for the overall reaction is the sum of the orders
for each reactant in the reaction.
•
For example:
2 N 2 O 5g   4 NO2 g  + O 2 g 
Rate = kN 2 O 5 
This reaction is first order in N 2 O 5
and first order overall.
In general, for
a A + b B --> x X with a catalyst C
Rate = k [A]m[B]n[C]p
The exponents m, n, and p
• are the reaction order
• can be 0, 1, 2 or fractions
• must be determined by experiment!
The Rate of Reaction
• Given the following one step reaction and its rate-law
expression.
– Remember, the rate expression would have to be
experimentally determined.
2 A  g   B g   C  g 
Rate = kA
2
• Because it is a second order rate-law expression:
– If the [A] is doubled the rate of the reaction will increase
by a factor of 4. 22 = 4
– If the [A] is halved the rate of the reaction will decrease by
a factor of 4. (1/2)2 = 1/4
Factors That Affect Reaction Rates
•
1.
2.
3.
4.
•
There are several factors that can influence the rate of a
reaction:
The nature of the reactants.
The concentration of the reactants.
The temperature of the reaction.
The presence of a catalyst.
We will look at each factor individually.
Nature of Reactants
• This is a very broad category that encompasses the different reacting
properties of substances.
• For example sodium reacts with water explosively at room temperature to
liberate hydrogen and form sodium hydroxide.
2 Na s   2 H 2 O    2 NaOHaq   H 2g 
This is a violent and rapid reaction.
The H 2 ignites and burns.
Nature of Reactants
• Calcium reacts with water only slowly at room temperature to
liberate hydrogen and form calcium hydroxide.
Ca s   2 H 2 O    Ca OH2 aq   H 2g 
This is a rather slow reaction.
Nature of Reactants
• The reaction of magnesium with water at room temperature is
so slow that that the evolution of hydrogen is not perceptible to
the human eye.
Mg s   H 2O   No reaction
Nature of Reactants
• However, Mg reacts with steam rapidly to liberate H2
and form magnesium oxide.
Mg s   H 2O   MgO s   H 2g 
100o C
• The differences in the rate of these three reactions can be
attributed to the changing “nature of the reactants”.
34
Rb
85
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Concentrations of Reactants: The Rate-Law
Expression
• This is a simplified representation of the effect of different
numbers of molecules in the same volume.
– The increase in the molecule numbers is indicative of an
increase in concentration.
A(g) + B (g)  Products
A
B
A B
4 different possible
A-B collisions
A B
B
A B
6 different possible
A-B collisions
A B
A B
A B
9 different possible
A-B collisions
Concentrations of Reactants: The Rate-Law
Expression
• Example 16-1: The following rate data were obtained at 25oC
for the following reaction. What are the rate-law expression
and the specific rate-constant for this reaction?
2 A(g) +
B(g)  3 C(g)
Experiment
Number
Initial [A]
(M)
Initial [B]
(M)
Initial rate of
formation of C
(M/s)
1
0.10
0.10
2.0 x 10-4
2
0.20
0.30
4.0 x 10-4
3
0.10
0.20
2.0 x 10-4
Concentrations of Reactants: The Rate-Law
Expression
Concentrations of Reactants: The Rate-Law Expression
• The following data were obtained for the following reaction at
25oC. What are the rate-law expression and the specific rate
constant for the reaction?
2 A(g) + B(g) + 2 C(g)  3 D(g) + 2 E(g)
Experiment
Initial [A]
(M)
Initial [B]
(M)
Initial [C]
(M)
Initial rate of
formation of D
(M/s)
1
0.20
0.10
0.10
2.0 x 10-4
2
0.20
0.30
0.20
6.0 x 10-4
3
0.20
0.10
0.30
2.0 x 10-4
4
0.60
0.30
0.40
1.8 x 10-3
Concentrations of Reactants: The Rate-Law
Expression
Concentrations of Reactants:
The Rate-Law Expression
• Consider a chemical reaction between compounds A and B that is
first order with respect to A, first order with respect to B, and
second order overall. From the information given below, fill in the
blanks.
Experiment
Initial Rate
(M/s)
Initial [A]
(M)
Initial [B]
(M)
1
4.0 x 10-3
0.20
0.050
2
1.6 x 10-2
?
0.050
3
3.2 x 10-2
0.40
?
Concentrations of Reactants: The Rate-Law
Expression
Concentration vs. Time: The Integrated Rate
Equation
• The integrated rate equation relates time and concentration
for chemical and nuclear reactions.
– From the integrated rate equation we can predict the
amount of product that is produced in a given amount of
time.
• Initially we will look at the integrated rate equation for
first order reactions.
These reactions are 1st order in the reactant and 1st order
overall.
Concentration vs. Time: The Integrated Rate
Equation
• An example of a reaction that is 1st order in the reactant
and 1st order overall is:
a A  products
This is a common reaction type for many chemical
reactions and all simple radioactive decays.
• Two examples of this type are:
2 N2O5(g)  2 N2O4(g) + O2(g)
238U  234Th + 4He
Concentration vs. Time: The Integrated Rate
Equation
• The integrated rate equation for first order reactions is:

A0
ln
akt
A
where:
[A]0= mol/L of A at time t=0.
[A] = mol/L of A at time t.
k = specific rate constant.
t = time elapsed since
beginning
of reaction.
a = stoichiometric coefficient of A in balanced overall
equation.
Concentration vs. Time: The Integrated Rate
Equation
• Solve the first order integrated rate equation for t.
1  A0 
t  ln 

a k  A 
• Define the half-life, t1/2, of a reactant as the time
required for half of the reactant to be consumed, or the
time at which [A]=1/2[A]0.
Concentration vs. Time: The Integrated Rate
Equation
• At time t = t1/2, the expression becomes:
Concentration vs. Time: The Integrated Rate
Equation
• Cyclopropane, an anesthetic, decomposes to propene
according to the following equation.
CH2
H2C
CH2
(g)
H3C
CH
CH2
(g)
The reaction is first order in cyclopropane with k = 9.2 s-1
at 10000C. Calculate the half life of cyclopropane at
10000C.
Concentration vs. Time: The Integrated Rate
Equation
• Refer to Previous Example: How much of a 3.0 g sample of
cyclopropane remains after 0.50 seconds?
– The integrated rate laws can be used for any unit that
represents moles or concentration.
– In this example we will use grams rather than mol/L.
Concentration vs. Time: The Integrated Rate Equation
• The half-life for the following first order reaction is 688 hours at
10000C. Calculate the specific rate constant, k, at 10000C and the
amount of a 3.0 g sample of CS2 that remains after 48 hours.
CS2(g)  CS(g) + S(g)
Concentration vs. Time: The Integrated Rate Equation
• For reactions that are second order with respect to a particular
reactant and second order overall, the rate equation is:
1
1

ak t
A A0
• Where:
[A]0= mol/L of A at time t=0. [A] = mol/L of A at time t.
k = specific rate constant.
t = time elapsed
since
beginning of reaction.
a = stoichiometric coefficient of A in balanced
Temperature:
The Arrhenius Equation
• Svante Arrhenius developed this relationship among (1) the
temperature (T), (2) the activation energy (Ea), and (3) the
specific rate constant (k).
E
- a
k = Ae
or
RT
Ea
ln k = ln A RT
• If the Arrhenius equation is written for two temperatures, T2
and T1 with T2 >T1.
Ea
ln k1  ln A RT1
and
Ea
ln k 2  ln A RT2
Temperature:
The Arrhenius Equation
1. Subtract one equation from the other.
Ea  Ea 
ln k 2  ln k1  ln A - ln A   
RT2  RT1 
Ea Ea
ln k 2  ln k1 
RT1 RT2
2.
Rearrange and solve for ln k2/k1.
k 2 Ea  1
1
ln

  
k1
R  T1 T2 
or
k 2 E a  T2 - T1 
ln



k1
R  T2 T1 
Temperature:
The Arrhenius Equation
• Consider the rate of a reaction for which Ea=50 kJ/mol, at
20oC (293 K) and at 30oC (303 K).
– How much do the two rates differ?
Temperature:
The Arrhenius Equation
• For reactions that have an Ea50 kJ/mol, the rate
approximately doubles for a 100C rise in temperature, near
room temperature.
• Consider:
2 ICl(g) + H2(g)  I2(g) + 2 HCl(g)
• The rate-law expression is known to be R=k[ICl][H2].
At 2300 C, k = 0.163 M -1s-1
At 2400 C, k = 0.348 M -1s-1
k approximately doubles
Catalysts
• Catalysts change reaction rates by providing an alternative reaction
pathway with a different activation energy.
• Homogeneous catalysts exist in same phase as the reactants.
• Heterogeneous catalysts exist in different phases than the
reactants.
– Catalysts are often solids.
Kinetics
Initial rate
Instantaneous rate
Average rate
Rate of reaction
Order of reaction
Overall order of reaction
Catalyst
Four factors that affect the
rate of reaction
nature of reactant
concentration
temperature
presence of a catalyst
[A] vs t
[A]
General rate expression
Rate =
Differential rate law (with respect to concentration)
Rate = k [A]m[B]n[C]p
If zero order Rate = k[A]0 = k
first order Rate = k[A]1 = k[A]
second order Rate = k[A]2
Integrated rate law (with respect to time)
If zero order [A]0 - [A] = ak t
first order A
ln
 0  a k t
A
second order 1
1
ak t
A A0

ln [A] vs t
ln [A]
t
- A - B + C + D

or

a t
b t
c t
d t
1/[A] vs t
1/[A]
t
t
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55. A reaction has the following experimental rate equation: Rate = k[A]2[B]. If the
concentration of A is doubled and the concentration of B is halved what happens
to the rate?
87. The following statements relate to the reaction with the following rate law:
Rate = k[H2][I2]
H2(g) + I2(g) → 2 HI(g)
Determine which of the following statements are true. If a statement is
false, indicate why it is incorrect.
a)
b)
c)
d)
e)
f)
The reaction must occur in a single step
This is a second-order reaction overall
Raising the temperature will cause the value of k to decrease
Raising the temperature lowers the activation energy for this reaction
If the concentration of both reactants are doubled, the rate will double
Adding a catalyst in the reaction will cause the initial rate to decrease
89. Describe each of the following statements as true or false. If false, rewrite the
sentence to make it correct.
a) The rate determining elementary step in a reaction is the slowest step in a
mechanism
b) It is possible to change the rate constant by changing the temperature
c) As a reaction proceeds at constant temperature, the rate remains constant
d) A reaction that is third order overall must involve more than one step
e) Reactions are faster at a higher temperature because activation energies
are lower
f) Rate increase with increasing concentration of reactants because there are
more collisions between reactant molecules
g) At higher temperatures a larger fraction of molecules have enough energy
to get over the activation energy barrier
h) Catalyzed and uncatalyzed reactions have identical mechanisms