ch 1 chemical kinetics
TRANSCRIPT
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Chapter 1
Chemical Kinetics
Copyright 2011 Pearson Education, Inc.
Kinetics: Rates and Mechanisms of Chemical Reactions
1.1 Expressing Reaction Rate: Differential rate law
Graphical representation of reaction rate
1.2 The Rate Law : Order of reaction, Initial rate method
Integrated rate law, rate constant, half -life
1.3 Collision Theory: Activation energy, effective collisions
1.6 Arrhenius Equation: Determination of rate constant and
activation energy
1.7 Reaction Mechanism: Intermediates, molecularity,
rate determining step
1.4 Factors affecting reaction rate: Concentration, temperature, catalyst,
nature of reactants
1.5 Transition State Theory: Activated complex
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Chemical Kinetics
• The speed of a chemical reaction is called its
reaction rate
• The rate of a reaction is a measure of how fast
the reaction makes products, or how fast the
reactants are consumed.
• Chemical kinetics is the study of the factors
that affect the rates of chemical reactions
Such as temperature, reactant concentrations
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Defining Reaction Rate
• The rate of a chemical reaction is generally
measured in terms of how much the concentration
of a reactant decreases in a given period of time
or product concentration increases
• For reactants, a negative sign is placed in front of
the definition
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AA
t
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12
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at t = 0
[A] = 8
[B] = 8
[C] = 0
at t = 0
[X] = 8
[Y] = 8
[Z] = 0
at t = 16
[A] = 4
[B] = 4
[C] = 4
at t = 16
[X] = 7
[Y] = 7
[Z] = 1
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Reaction Rate Changes Over Time
• As time goes on, the rate of a reaction generally
slows down
because the concentration of the reactants
decreases
• At some time the reaction stops, either because
the reactants run out or because the system has
reached equilibrium
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Average Rate
• The average rate is the change in measured
concentrations in any particular time period
linear approximation of a curve
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Reaction Rate and Stoichiometry
• In most reactions, the coefficients of the balanced equation are not all the same
H2 (g) + I2 (g) 2 HI(g)
• For these reactions, the change in the number of molecules of one substance is a multiple of the change in the number of molecules of another for the above reaction, for every 1 mole of H2 used, 1
mole of I2 will also be used and 2 moles of HI made
therefore the rate of change will be different
• To be consistent, the change in the concentration of each substance is multiplied by 1/coefficient
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H2 I2 HI
Stoichiometry tells us that for
every 1 mole/L of H2 used,
2 moles/L of HI are made.
Assuming a 1 L container, at
10 s, we used 0.181 moles of
H2. Therefore the amount of
HI made is 2(0.181 moles) =
0.362 moles.
At 60 s, we used 0.699 moles
of H2. Therefore the amount
of HI made is 2(0.699 moles)
= 1.398 moles.
The average rate is
the change in the
concentration in a
given time period
In the first 10 s, the
[H2] is −0.181 M,
so the rate is
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average rate in a given
time period = slope of
the line connecting the
[H2] points; and ½ +slope
of the line for [HI]
the average rate for the
first 10 s is 0.0181 M/s
the average rate for the
first 40 s is 0.0150 M/s
the average rate for the
first 80 s is 0.0108 M/s
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Instantaneous Rate
• The instantaneous rate is the change in
concentration at any one particular time
Determined by taking the slope of a line tangent to the
“concentartion vs time” curve at any particular point.
• The initial rate is the reaction rate at the
beginning of a reaction
slope of a line tangent to the curve at time t = 0
No product formed yet, reversed rate = 0
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H2 (g) + I2 (g) 2 HI (g)
Using [H2], the
instantaneous
rate at 50 s is
Using [HI], the
instantaneous
rate at 50 s is
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Problem 1.1: For the reaction given, the [I] changes from1.000 M to 0.868 M
in the first 10 s. Calculate the average rate in the first 10 s, and [H+]/t.
H2O2 (aq) + 3 I(aq) + 2 H+(aq) I3
(aq) + 2 H2O(l)
13
solve the rate
equation for the rate
(in terms of the
change in
concentration of the
given quantity)
solve the rate
equation (in terms of
the change in the
concentration for the
quantity to find) for
the unknown value
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Practice :
If 2.4 x 102 g of NOBr (MM 109.91 g) decomposes in a 2.0 x
102 mL flask in 5.0 minutes, find the average rate of Br2
production in M/s.
2 NOBr(g) 2 NO(g) + Br2(l)
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2 mol NOBr:1 mol Br2, 1 mol = 109.91g, 1 mL=0.001 L
Solve:
Conceptual
Plan:
Relationships:
240.0 g NOBr, 200.0 mL, 5.0 min.
[Br2]/t, M/s
Given:
Find:
Practice – If 2.4 x 102 g of NOBr decomposes in a 2.0 x 102 mL
flask in 5.0 minutes, find the average rate of Br2 production
2 NOBr(g) 2 NO(g) + Br2(l)
15
M mole Br2
Rate min s
}
5.45 M Br2, 3.0 x 102 s
[Br2]/t, M/s
240.0 g NOBr, 0.2000 L, 5.0 min.
[Br2]/t, M/s
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Practice: Consider the reaction:
8A(g) + 5B(g) 8C(g) + 6D(g)
If [C] is increasing at the rate of 4.0 mol L¯1s¯1, at what rate is
[B] changing?
A. -0.40 mol L¯1s¯1
B. -2.5 mol L¯1s¯1
C. -4.0 mol L¯1s¯1
D. -6.4 mol L¯1s¯1
E. None of these choices is correct, since its rate of
change must be positive.
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In general, the differential rate law for the reaction
aA + bB cC + dD
rate = 1
a - = -
[A]
t
1
b
[B]
t
1
c
[C]
t = +
1
d
[D]
t = +
The numerical value of the rate depends upon the substance that
serves as the reference. The rest is relative to the balanced
chemical equation.
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SOLUTION:
Expressing Rate in Terms of Changes in Concentration with Time
PROBLEM 1.2: Because it has a nonpolluting product (water vapor), hydrogen
gas is used for fuel aboard the space shuttle and may be used
by earthbound engines in the near future.
2H2(g) + O2(g) 2H2O(g)
(a) Express the rate in terms of changes in [H2], [O2], and [H2O] with time.
(b) When [O2] is decreasing at 0.23 mol/L*s, at what rate is [H2O] increasing?
- 1
2
[H2]
t = -
[O2]
t
= + [H2O]
t
1
2
0.23 mol/L*s = + [H2O]
t
1
2
= 0.46 mol/L*s [H2O]
t
rate = (a)
[O2]
t - = - (b)
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The Rate Law
• The rate law of a reaction is the mathematical
relationship between the rate of the reaction and the
concentrations of the reactants
and homogeneous catalysts as well
• The rate law must be determined experimentally!!
• The rate of a reaction is directly proportional to the
concentration of each reactant raised to a power
• For the reaction aA + bB products the rate
law would have the form given below n and m are called the orders for each reactant
k is called the rate constant
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Reaction Order
• The exponent on each reactant in the rate law is called the order with respect to that reactant
• The sum of the exponents on the reactants is called the overall order of the reaction
• The rate law for the reaction given: 2 NO(g) + O2(g) 2 NO2(g)
is: Rate = k[NO]2[O2]
The reaction is:
second order with respect to [NO],
first order with respect to [O2],
and third order overall. 20 Tro: Chemistry: A Molecular Approach, 2/e
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SOLUTION:
Determining Reaction Order from Rate Laws
PROBLEM 1.3: For each of the following reactions, determine the
reaction order with respect to each reactant and the
overall order from the given rate law.
(a) 2NO(g) + O2(g) 2NO2(g) rate = k[NO]2[O2]
(b) H2O2(aq) + 3I-(aq) + 2H+(aq) I3-(aq) + 2H2O(l) rate = k[H2O2][I
-]
PLAN: Look at the rate law and not the coefficients of the chemical reaction.
(a) The reaction is 2nd order in NO, 1st order in O2, and 3rd order overall.
(b) The reaction is 1st order in H2O2, 1st order in I- and zero order in H+,
while being 2nd order overall.
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Rate = k[NO][O3]
Solve:
Conceptual
Plan:
Relationships:
[NO] = 1.00 x 10−6 M, [O3] = 3.00 x 10−6 M, Rate = 6.60 x 10−6 M/s
k, M−1s−1
Given:
Find:
Example: The rate equation for the reaction of NO with ozone is
Rate = k[NO][O3]. If the rate is 6.60 x 10−5 M/sec when [NO] = 1.00 x 10−6 M
and [O3] = 3.00 x 10−6 M, calculate the rate constant k
Rate, [NO], [O3] k
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Practice:
The rate law for the decomposition of acetaldehyde,
CH3CHO, is Rate = k[CH3CHO]2.
What is the rate of the reaction when the [CH3CHO]
= 1.75 x 10−3 M and the rate constant is 6.73 x 10−6
M−1•s−1?
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Rate = k[acetaldehyde]2
Solve:
Conceptual
Plan:
Relationships:
[CH3CHO] = 1.75 x 10−3 M, k= 6.73 x 10−6 M−1•s−1
Rate, M/s
Given:
Find:
Practice – What is the rate of the reaction when the
[CH3CHO] = 1.75 x 10−3 M and the rate constant is 6.73 x
10−6 M−1•s−1?
k, [acetaldehyde] Rate
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Finding the Rate Law:
the Initial Rate Method
• The rate law must be determined experimentally
• The rate law shows how the rate of a reaction
depends on the concentration of the reactants
• Changing the initial concentration of a reactant will
therefore affect the initial rate of the reaction
25
if for the reaction
A → Products
Rate = k[A]n
then doubling the
initial concentration
of A doubles the
initial reaction rate
then doubling the
initial concentration
of A does not
change the initial
reaction rate. Thus,
n = 0 (Zero Order)
Rate = k
then doubling the
initial concentration
of A quadruples the
initial reaction rate
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Proposed rate law: Rate = k[A]n
• If a reaction is Zero Order, the rate of the reaction
is always the same
doubling [A] will have no effect on the reaction rate
• If a reaction is First Order, the rate is directly
proportional to the reactant concentration
doubling [A] will double the rate of the reaction
• If a reaction is Second Order, the rate is directly
proportional to the square of the reactant
concentration
doubling [A] will quadruple the rate of the reaction
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Determining the Rate Law when there are Multiple
Reactants
• Changing each reactant will effect the overall rate
of the reaction
• By changing the initial concentration of one
reactant at a time, the effect of each reactant’s
concentration on the rate can be determined
• In examining results, we compare differences in
rate for reactions that only differ in the
concentration of one reactant
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Example 13.2: Determine the rate law and rate constant for the
reaction NO2(g) + CO(g) NO(g) + CO2(g)
given the data below
Comparing Expt #1 and Expt #2,
the [NO2] changes but the [CO]
does not 28 Tro: Chemistry: A Molecular Approach, 2/e
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Example 13.2: Determine the rate law and rate constant
for the reaction NO2(g) + CO(g) NO(g) + CO2(g)
given the data below
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Example 13.2: Determine the rate law and rate constant
for the reaction NO2(g) + CO(g) NO(g) + CO2(g)
given the data below
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Example 13.2: Determine the rate law and rate constant
for the reaction NO2(g) + CO(g) NO(g) + CO2(g)
given the data below
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Example 13.2: Determine the rate law and rate constant for the
reaction NO2(g) + CO(g) NO(g) + CO2(g)
given the data below
n = 2, m = 0
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Example 13.2: Determine the rate law and rate constant for the
reaction NO2(g) + CO(g) NO(g) + CO2(g)
given the data below
Substitute the
concentrations
and rate for
any experiment
into the rate
law and solve
for k
Expt.
Number
Initial
[NO2], (M)
Initial
[CO], (M)
Initial Rate
(M/s)
1. 0.10 0.10 0.0021
2. 0.20 0.10 0.0082
3. 0.20 0.20 0.0083
4. 0.40 0.10 0.033
Rate k[NO2]2
for expt 1
0.0021 Ms k 0.10 M
2
k 0.0021 M
s
0.01 M2 0.21 M1 s1
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Practice – Determine the rate law and rate constant for the
reaction NH4+ + NO2
− N2 + 2 H2O
given the data below
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Practice – Determine the rate law and rate constant for the
reaction NH4+ + NO2
− N2 + 2 H2O
given the data below
Rate = k[NH4+]n[NO2
]m
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Integrated Rate Laws
• For the reaction A Products, the rate law
depends on the concentration of A
• Applying calculus to integrate the rate law gives
another equation showing the relationship
between the concentration of A and the time of the
reaction – this is called the Integrated Rate Law
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First Order Reactions
• Rate = k[A]1 = k[A]
• ln[A] = −kt + ln[A]initial
• Graph ln[A] vs. time gives straight line with
slope = −k and y–intercept = ln[A]initial
used to determine the rate constant
• t½ = 0.693/k. The half-life of a first order
reaction is constant
• When Rate = M/sec, k = s–1
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ln[A]initial
ln[A]
time
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Half-Life
• The half-life, t1/2, of a reaction is the
time it takes for the concentration of the
reactant to fall to ½ its initial value
• The half-life of the reaction depends on
the order of the reaction
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The Half-Life of a First-Order Reaction Is Constant
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Rate Data for: C4H9Cl + H2O C4H9OH + HCl
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C4H9Cl + H2O C4H9OH + 2 HCl
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C4H9Cl + H2O C4H9OH + 2 HCl
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C4H9Cl + H2O C4H9OH + 2 HCl
slope =
−2.01 x 10−3
k =
2.01 x 10−3 s-1
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Zero Order Reactions
• Rate = k[A]0 = k
• Integrated rate law: [A] = −kt + [A]initial
• Graph of [A] vs. time is straight line with slope = −k and y–intercept = [A]initial
• t ½ = [Ainitial]/2k = [A0]/2k
• When Rate = M/sec, k = M/sec
[A]initial
[A]
time 45 Tro: Chemistry: A Molecular Approach, 2/e
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Second Order Reactions
• Rate = k[A]2
• 1/[A] = kt + 1/[A]initial
• Graph 1/[A] vs. time gives straight line with
slope = k and y–intercept = 1/[A]initial
used to determine the rate constant
• t½ = 1/(k[A0])
• When Rate = M/sec, k = M−1s−1
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1/[A]initial
1/[A]
time
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PLAN:
SOLUTION:
Determining the Half-Life of a First-Order Reaction
PROBLEM 1.3: Cyclopropane is the smallest cyclic hydrocarbon. It is
thermally unstable and rearranges to propene at 1000oC
via the following first-order reaction:
CH2
H2C CH2(g)
H3C CH CH2 (g)
The rate constant is 9.2 s-1, (a) What is the half-life of the reaction?
(b) How long does it take for the concentration of cyclopropane to
reach one-quarter of the initial value?
Use the half-life equation, t1/2 = 0.693
k
One-quarter of the initial value means two half-lives have passed.
t1/2 = 0.693/9.2 s-1 = 0.075 s (a) 2 t1/2 = 2(0.075 s) = 0.150 s (b)
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Table 1 Units of the Rate Constant k
Overall Reaction Order Units of k (t in seconds)
0 mol/L*s (or mol L-1 s-1)
1 1/s (or s-1)
2 L/mol*s (or L mol -1 s-1)
3 L2 / mol2 *s (or L2 mol-2 s-1)
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Example 13.4: The reaction SO2Cl2(g) SO2(g) + Cl2(g) is first order
with a rate constant of 2.90 x 10−4 s−1 at a given set of conditions.
Find the [SO2Cl2] at 865 s when [SO2Cl2]initial = 0.0225 M
51
the new concentration is less than the original,
as expected
[SO2Cl2]init = 0.0225 M, t = 865, k = 2.90 x 10-4 s−1
[SO2Cl2]
Check:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
[SO2Cl2] [SO2Cl2]init, t, k
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Practice
The reaction Q 2 R is second order in Q. If the
initial [Q] = 0.010 M and after 5.0 x 102 seconds the
[Q] = 0.0010 M, find the rate constant.
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Practice – The reaction Q 2 R is second order in Q. If the
initial [Q] = 0.010 M and after 5.0 x 102 seconds the [Q] =
0.0010 M, find the rate constant
53
[Q]init = 0.010 M, t = 5.0 x 102 s, [Q]t = 0.0010 M
k
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
k [Q]init, t, [Q]t
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Graphical Determination of
the Rate Law for A Product
• Plots of [A] vs. time, ln[A] vs. time, and 1/[A] vs.
time allow determination of whether a reaction is
zero, first, or second order
• Whichever plot gives a straight line determines the
order with respect to [A]
if linear is [A] vs. time, Rate = k[A]0
if linear is ln[A] vs. time, Rate = k[A]1
if linear is 1/[A] vs. time, Rate = k[A]2
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Practice – Complete the Table and Determine the
Rate Equation for the Reaction A 2 Product
55
What will the rate be when the [A] = 0.010 M?
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Practice – Complete the Table and Determine the
Rate Equation for the Reaction A 2 Product
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Conclusion for the determination of the rate
equation for the reaction A 2 Product
Because the graph 1/[A] vs. time is linear,
the reaction is second order,
60
Rate k[A]2
k slope of the line 0.10 M–1 s –1
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Relationship Between
Order and Half-Life
• For a zero order reaction, the half-life is directly proportional to the initial concentration. The lower the initial concentration of the reactants, the shorter the half-life t1/2
= [A]init/2k
• For a first order reaction, the half-life is independent of the concentration. (t1/2 = constant) t1/2 = ln(2)/k
• For a second order reaction, the half-life is inversely proportional to the initial concentration – increasing the initial concentration shortens the half-life t1/2 = 1/(k[A]init) 61 Tro: Chemistry: A Molecular Approach, 2/e
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Example 13.6: Molecular iodine dissociates at 625 K with a first-
order rate constant of 0.271 s−1. What is the half-life of this
reaction?
62
the new concentration is less than the original,
as expected
k = 0.271 s−1
t1/2
Check:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
t1/2 k
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Practice
The reaction Q 2 R is second order in Q. If the
initial [Q] = 0.010 M and the rate constant is 1.8
M−1•s−1 find the length of time for [Q] = ½[Q]init
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Practice : The reaction Q 2 R is second order in Q. If the
initial [Q] = 0.010 M and the rate constant is 1.8 M−1•s−1 find
the length of time for [Q] = ½[Q]init
64
[Q]init = 0.010 M, k = 1.8 M−1s−1
t1/2, s
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
t1/2 [Q]init, k
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Collision Theory
• For most reactions, for a reaction to take place, the
reacting molecules must collide with each other
on average about 109 collisions per second
• Once molecules collide they may react together or
they may not, depending on two factors –
1. whether the collision has enough energy to "break
the bonds holding reactant molecules together";
and
2. whether the reacting molecules collide in the
proper orientation for new bonds to form
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Activation Energy
• Molecules must possess a certain minimum kinetic
energy called the activation energy for a reaction
to occur
energy needed to break the bonds of the reacting
molecules
Reaction with low activation energy is easier to take
place, and is faster
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Effective Collisions
Kinetic Energy Factor
For a collision to be
effective, the reacting
molecules must have
sufficient kinetic
energy to overcome
the energy barrier.
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Orientation Factor
• The proper orientation results when the atoms are aligned in such a way that the old bonds can break and the new bonds can form
• The more complex the reactant molecules, the less frequently they will collide with the proper orientation
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Effective Collisions
Orientation Effect
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Collision Frequency
• The collision frequency is the number of collisions
that happen per unit time
• The more collisions per second there are, the
more collisions can be effective and lead to
product formation
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Effective Collisions
• Collisions in which these two conditions are met
(and therefore lead to reaction) are called
effective collisions
• The higher the frequency of effective collisions,
the faster the reaction rate
• Any factor that can increase the frequency of
effective collisions would increase the rate of a
reaction
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Factors Affecting Reaction Rate
Under specific conditions, every reaction has its own
characteristic rate, which may be controlled by four
factors:
1. Concentrations of reactants
2. Chemical nature of reactants
3. Temperature
4. The use of a catalyst
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Factors Affecting Reaction Rate:
Reactant Concentration
• Generally, the larger the concentration of
reactant molecules, the faster the reaction
increases the frequency of effective collisions
between reactant molecules
concentration of gases depends on the partial
pressure of the gas
higher pressure = higher concentration
• Concentrations of solutions depend on the
solute-to-solution ratio (molarity)
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Factors Affecting Rate – Temperature
• Increasing the temperature raises the average kinetic energy of the reactant molecules
• Increasing the temperature increases the number of molecules with sufficient kinetic energy to overcome the activation energy, leads to increase in frequancy of effective collisions. Hence, increasing the rate of reaction
• The distribution of kinetic energy is shown by the Boltzmann distribution curves
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Factors Affecting Rate – Catalyst
• Catalysts are substances that affect the rate of a
reaction without being consumed
• Catalysts generally speed up a reaction
• They give the reactant molecules a different path
to follow with a lower activation energy
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Catalysts
• Homogeneous catalysts are in the same phase
as the reactant particles
Cl(g) in the destruction of O3(g)
• Heterogeneous catalysts are in a different
phase than the reactant particles
solid catalytic converter in a car’s exhaust system
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Types of Catalysts
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Factors Affecting Reaction Rate:
Nature of the Reactants
• Nature of the reactants means what kind of reactant
molecules and what physical conditions they are in
small molecules tend to react faster than large molecules
gases tend to react faster than liquids, which react faster
than solids
powdered solids are more reactive than “blocks”
more surface area for contact with other reactants
certain types of chemicals are more reactive than others
e.g. potassium metal is more reactive than sodium
ions react faster than molecules
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The Transition State Theory
• There is an energy barrier to almost all reactions
• There exists an intermediate stage when reactant
molecules change into products.
• The activated complex or transition state is a
chemical species with partially broken and partially
formed bonds
an intermediate formed between reactant and
product molecules
has highest energy and extremely unstable
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Figure 2
Reaction energy diagrams and possible transition states.
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Energy Profile of a Catalyzed Reaction
polar stratospheric
clouds contain ice
crystals that catalyze
reactions that release
Cl from atmospheric
chemicals
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Reaction progress
Po
ten
tia
l E
ne
rgySOLUTION:
Drawing Reaction Energy Diagrams and Transition States
PROBLEM 1.5: A key reaction in the upper atmosphere is
O3(g) + O(g) 2O2(g)
The Ea(fwd) is 19 kJ, and the Hrxn for the reaction is -392 kJ. Draw a
reaction energy diagram for this reaction, postulate a transition state, and
calculate Ea(rev).
PLAN: Exothermic reaction. The reactants are at a higher energy level than the
products and the transition state is slightly higher than the reactants.
O3+O
2O2
Ea= 19 kJ
Hrxn = -392 kJ
Ea(rev)= (392 + 19) kJ =
411kJ
O O
O
O
breaking
bond
forming
bond
transition state
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• Changing the temperature changes the rate constant,
k, of the rate law
• The Arrhenius Equation shows the relationship
between absolute temperature, T, and rate constant,
k :
R is the gas constant in energy units, 8.314 J/(mol•K)
where T is the temperature in kelvins
A is called the Arrhenius constant (which includes the
frequency factor and the orientation factor)
Ea is the activation energy, the minimum energy
needed for the molecules to react
The Arrhenius Equation
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The Arrhenius Equation:
The Exponential Factor
• The exponential factor in the Arrhenius equation is a number between 0 and 1
• The larger the activation energy, Ea, the fewer molecules that have sufficient energy to overcome the energy barrier, the smaller the rate constant k.
• Increase in temperature, T, will gives a larger k value therefore increasing the temperature will increase the reaction rate
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Arrhenius Plots
• The Arrhenius Equation: The integrated form
this equation is in the linear form y = mx + b
where y = ln(k) and x = (1/T)
• A graph of ln(k) vs. (1/T) is a straight line
• (−R)(slope of the line) = Ea (in Joules), R = 8.314 J/(mol•K)
• ey–intercept = A (unit is the same as k)
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Example 13.7: Determine the activation energy and
frequency factor for the reaction O3(g) O2(g) + O(g)
given the following data:
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Example 13.7: Determine the activation energy and
frequency factor for the reaction O3(g) O2(g) + O(g)
given the following data:
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Example 13.7: Determine the Ea and the Arrhenius
constant, A, for the reaction O3(g) O2(g) + O(g)
given the following data:
90
slope, m = 1.12 x 104 K
y–intercept, b = 26.8
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Arrhenius Equation:
Two-Point Form
• If you only have two (T,k) data points, the following
forms of the Arrhenius Equation can be used:
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Example 13.8: The reaction NO2(g) + CO(g) CO2(g) + NO(g)
has a rate constant of 2.57 M−1∙s−1 at 701 K and 567 M−1∙s−1
at 895 K. Find the activation energy in kJ/mol
92
most activation energies are tens to hundreds
of kJ/mol – so the answer is reasonable
T1 = 701 K, k1 = 2.57 M−1∙s−1, T2 = 895 K, k2 = 567 M−1∙s−1
Ea, kJ/mol
Check:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
Ea T1, k1, T2, k2
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Practice: It is often said that the rate of a reaction doubles for
every 10 °C rise in temperature. Calculate the activation
energy for such a reaction.
93
Hint:
make T1 = 300 K, T2 = 310 K
and k2 = 2k1
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Practice – Find the activation energy in kJ/mol for
increasing the temperature 10 ºC doubling the rate
94
most activation energies are tens to hundreds
of kJ/mol – so the answer is reasonable
T1 = 300 K, T1 = 310 K, k2 = 2 k1
Ea, kJ/mol
Check:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
Ea T1, k1, T2, k2
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Reaction Mechanisms
• We generally describe chemical reactions with an equation listing all the reactant molecules and product molecules
• But the probability of more than 3 molecules colliding at the same instant with the proper orientation and sufficient energy to overcome the energy barrier is negligible
• Most reactions occur in a series of small reactions involving 1, 2, or at most 3 molecules
• Describing the series of reactions that occurs to produce the overall observed reaction is called a reaction mechanism
• Knowing the rate law of the reaction helps us understand the sequence of reactions in the mechanism
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An Example of a Reaction Mechanism
• Overall reaction:
H2(g) + 2 ICl(g) 2 HCl(g) + I2(g)
• Mechanism:
Step 1: H2(g) + ICl(g) HCl(g) + HI(g)
Step 2: HI(g) + ICl(g) HCl(g) + I2(g)
• The reactions (1) and (2) in this mechanism are called the elementary steps, meaning that they cannot be broken down into simpler steps, and that the molecules actually interact directly in this manner without any other steps
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Overall: H2(g) + 2 ICl(g) 2 HCl(g) + I2(g)
Step (1): H2(g) + ICl(g) HCl(g) + HI(g)
Step (2): HI(g) + ICl(g) HCl(g) + I2(g)
Intermediates
• Notice that the HI is a product in Step 1, but then a
reactant in Step 2
• Because HI is made but then consumed, HI does
not show up in the overall reaction
• Materials that are products in an early mechanism
step, but then a reactant in a later step, are called
intermediates
Adding equations (1) and (2), removing HI, will give
the overall equation.
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Molecularity
• The number of reactant particles in an elementary step
is called its molecularity
• A unimolecular step involves one particle
• A bimolecular step involves two particles
though they may be the same kind of particle
• A termolecular step involves three particles
though these are exceedingly rare in elementary
steps
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Rate Laws for Elementary Steps
• Each step in the mechanism is like its own little
reaction – with its own activation energy and own
rate law
• The rate law for an overall reaction must be
determined experimentally
• But the rate law of an elementary step can be
deduced directly from the equation of the step
H2(g) + 2 ICl(g) 2 HCl(g) + I2(g)
1) H2(g) + ICl(g) HCl(g) + HI(g) Rate = k1[H2][ICl]
2) HI(g) + ICl(g) HCl(g) + I2(g) Rate = k2[HI][ICl]
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Rate Laws of Elementary Steps
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Rate Determining Step
• In most mechanisms, one step occurs slower than the other steps
• The result is that product formation cannot occur any faster than the slowest step – the step determines the rate of the overall reaction
• We call the slowest step in the mechanism the rate determining step
the slowest step has the largest activation energy
• The rate law of the rate determining step determines the rate law of the overall reaction
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Another Reaction Mechanism
The first step in this
mechanism is the rate
determining step.
The first step is slower than the
second step because its
activation energy is larger.
The rate law of the first step is
the same as the rate law of the
overall reaction.
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NO2(g) + CO(g) NO(g) + CO2(g) Rateobs = k[NO2]2
1. NO2(g) + NO2(g) NO3(g) + NO(g) Rate = k1[NO2]2 Slow
2. NO3(g) + CO(g) NO2(g) + CO2(g) Rate = k2[NO3][CO] Fast
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Validating a Mechanism
• To validate (not prove) a mechanism, two
conditions must be met:
1. The elementary steps must sum to the overall
reaction
2. The rate law predicted by the mechanism must
be consistent with the experimentally observed
rate law
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Mechanisms with a Fast
Initial Step
• When a mechanism contains a fast initial step, the rate
limiting step may contain intermediates
• When a previous step is rapid and reaches equilibrium,
the forward and reverse reaction rates are equal – so
the concentrations of reactants and products of the
step are related
and the product is an intermediate
• Substituting into the rate law of the RDS will produce a
rate law in terms of just reactants
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An Example
1. 2 NO(g) N2O2(g) Fast
2. H2(g) + N2O2(g) H2O(g) + N2O(g) Slow Rate = k2[H2][N2O2]
3. H2(g) + N2O(g) H2O(g) + N2(g) Fast
k1
k−1
2 H2(g) + 2 NO(g) 2 H2O(g) + N2(g) Rateobs = k [H2][NO]2
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Example 13.9: Show that the proposed mechanism for the
reaction 2 O3(g) 3 O2(g) matches the observed rate law
Rate = k[O3]2[O2]
−1
1. O3(g) O2(g) + O(g) Fast
2. O3(g) + O(g) 2 O2(g) Slow Rate = k2[O3][O]
k1
k−1
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Practice – Mechanism
Determine the overall reaction, the rate determining
step, the rate law, and identify all
intermediates of the following mechanism
1. A + B2 AB + B Slow
2. A + B AB Fast
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Practice – Mechanism
Determine the overall reaction, the rate determining
step, the rate law, and identify all intermediates of
the following mechanism
1. A + B2 AB + B Slow
2. A + B AB Fast
reaction 2 A + B2 2 AB
B is an intermediate
The first step is the rate determining step
Rate = k[A][B2], same as RDS
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