Download - Chapter 13 Outline
Chapter 13 Chemical Kine1cs
Kine%cs • kine%cs is the study of the factors that affect the _____ of a reac%on and the __________ by which a reac%on proceeds.
• experimentally it is shown that there are 4 factors that influence the speed of a reac%on: – nature of the reactants, – temperature, – catalysts, – concentra1on
2
Defining Rate
• _____ is how much a quan%ty changes in a given period of %me
• the speed you drive your car is a rate – the distance your car travels (miles) in a given period of %me (1 hour) – so the rate of your car has units of mi/hr
3
Defining Reac%on Rate
• the rate of a chemical reac%on is generally measured in terms of how much the concentra%on of a reactant decreases in a given period of %me – or product concentra%on increases
• for reactants, a nega%ve sign is placed in front of the defini%on
4
Reac%on Rate Changes Over Time
• as %me goes on, the rate of a reac%on generally __________ – because the concentra%on of the reactants decreases.
• at some %me the reac%on stops, either because the reactants run out or because the system has reached equilibrium.
5
6
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
Rate = !" A[ ]"t
= !A[ ] 2
! A[ ]1( )t2 ! t1( )
Rate =Rate = !
" X[ ]"t
= !X[ ] 2
! X[ ]1( )t2 ! t1( )
Rate =
7
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
Rate =! C[ ]!t
=C[ ] 2
" C[ ]1( )t2 " t1( )
Rate =
Rate =! Z[ ]!t
=Z[ ] 2
" Z[ ]1( )t2 " t1( )
Rate =
8
Rate = !" X[ ]"t
= !X[ ] 2
! X[ ]1( )t2 ! t1( )
Rate =
Rate = !" A[ ]"t
= !A[ ] 2
! A[ ]1( )t2 ! t1( )
Rate =
at t = 16 [A] = 4 [B] = 4 [C] = 4
at t = 16 [X] = 7 [Y] = 7 [Z] = 1
at t = 32 [A] = 2 [B] = 2 [C] = 6
at t = 32 [X] = 6 [Y] = 6 [Z] = 2
9
Rate =! C[ ]!t
=C[ ] 2
" C[ ]1( )t2 " t1( )
Rate =
at t = 16 [A] = 4 [B] = 4 [C] = 4
at t = 16 [X] = 7 [Y] = 7 [Z] = 1
at t = 32 [A] = 2 [B] = 2 [C] = 6
at t = 32 [X] = 6 [Y] = 6 [Z] = 2
Rate =! Z[ ]!t
=Z[ ] 2
" Z[ ]1( )t2 " t1( )
Rate =
10
Rate = !" X[ ]"t
= !X[ ] 2
! X[ ]1( )t2 ! t1( )
Rate =
Rate = !" A[ ]"t
= !A[ ] 2
! A[ ]1( )t2 ! t1( )
Rate =
at t = 32 [A] = 2 [B] = 2 [C] = 6
at t = 32 [X] = 6 [Y] = 6 [Z] = 2
at t = 48 [A] = 0 [B] = 0 [C] = 8
at t = 48 [X] = 5 [Y] = 5 [Z] = 3
11
at t = 32 [A] = 2 [B] = 2 [C] = 6
at t = 32 [X] = 6 [Y] = 6 [Z] = 2
at t = 48 [A] = 0 [B] = 0 [C] = 8
at t = 48 [X] = 5 [Y] = 5 [Z] = 3
Rate =! C[ ]!t
=C[ ] 2
" C[ ]1( )t2 " t1( )
Rate =
Rate =! Z[ ]!t
=Z[ ] 2
" Z[ ]1( )t2 " t1( )
Rate =
Hypothe%cal Reac%on Red → Blue
12
Time (sec)
Number Red
Number Blue
0 100 0 5 84 16
10 71 29 15 59 41 20 50 50 25 42 58 30 35 65 35 30 70 40 25 75 45 21 79 50 18 82
in this reaction, one molecule of Red turns into one molecule of Blue
the number of molecules will always total 100
the rate of the reaction can be measured as the speed of loss of Red molecules over time, or the speed of gain of Blue molecules over time
Hypothe%cal Reac%on Red → Blue
13
100
84
71
59
50
42
35 30
25 21
18
0
16
29
41
50
58
65 70
75 79
82
0
10
20
30
40
50
60
70
80
90
100
0 5 10 15 20 25 30 35 40 45 50
Nu
mb
er o
f M
olec
ule
s
Time (sec)
Concentration vs Time for Red -> Blue
Number Red
Number Blue
Hypothe%cal Reac%on Red → Blue
14
Rate of Reaction Red -> Blue
5, 3.2
10, 2.615, 2.4
20, 1.825, 1.6
30, 1.4
35, 1 40, 145, 0.8
50, 0.6
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 10 20 30 40 50
time, (sec)
Rat
e, Δ
[Blu
e]/Δ
t
Reac%on Rate and Stoichiometry • in most reac%ons, the coefficients of the balanced equa%on are not all the same
• for these reac%ons, the change in the number of molecules of one substance is a mul%ple of the change in the number of molecules of another – for the above reac%on, 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 • in order to be consistent, the change in the concentra%on of each substance is mul%plied by 1/coefficient
15
a) 0.0014 Ms–1
b) 0.00070 Ms–1
c) 0.0028 Ms–1
d) None of the above are correct
H2O2 can be used as a disinfectant; it decomposes as:
2 H2O2 → 2 H2O + O2
If the rate of appearance of O2 is 0.0014 Ms–1, what is the rate of disappearance of H2O2?
Copyright © 2011 Pearson Education, Inc.
a) 3.4 x 10–4 M/s
b) 5.9 x 10–4 M/s
c) 4.1 x 10–4 M/s
d) 1.6 x 10–4 M/s
e) 2.1 x 10–4 M/s
Determine the rate of disappearance of NO in the first 100 seconds.
[NO] time (s)
0.100 0 0.078 50 0.059 100 0.043 150 0.031 200
Copyright © 2011 Pearson Education, Inc.
Average Rate
• the average rate is the change in measured __________ in any par%cular %me period – linear approxima%on of a curve
• the larger the %me interval, the more the average rate deviates from the instantaneous rate
18
Hypothe%cal Reac%on Red → Blue
19
Avg. Rate Avg. Rate Avg. Rate
Time (sec)
Number Red
Number Blue
(5 sec intervals)
(10 sec intervals)
(25 sec intervals)
0 100 0 5 84 16 3.2
10 71 29 2.6 2.9 15 59 41 2.4 20 50 50 1.8 2.1 25 42 58 1.6 2.3
30 35 65 1.4 1.5 35 30 70 1 40 25 75 1 1 45 21 79 0.8 50 18 82 0.6 0.7 1
20
H2 I2 HI
Avg. Rate, M/s Avg. Rate, M/s
Time (s) [H2], M [HI], M -Δ[H2]/Δt 1/2 Δ[HI]/Δt 0.000 1.000
10.000 0.819
20.000 0.670
30.000 0.549
40.000 0.449
50.000 0.368
60.000 0.301
70.000 0.247
80.000 0.202
90.000 0.165
100.000 0.135
Avg. Rate, M/s Avg. Rate, M/s
Time (s) [H2], M [HI], M -Δ[H2]/Δt 1/2 Δ[HI]/Δt 0.000 1.000 0.000
10.000 0.819 0.362
20.000 0.670 0.660
30.000 0.549 0.902
40.000 0.449 1.102
50.000 0.368 1.264
60.000 0.301 1.398
70.000 0.247 1.506
80.000 0.202 1.596
90.000 0.165 1.670
100.000 0.135 1.730
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
21
H2 I2 HI
Avg. Rate, M/s Avg. Rate, M/s
Time (s) [H2], M [HI], M -Δ[H2]/Δt 1/2 Δ[HI]/Δt 0.000 1.000
10.000 0.819
20.000 0.670
30.000 0.549
40.000 0.449
50.000 0.368
60.000 0.301
70.000 0.247
80.000 0.202
90.000 0.165
100.000 0.135
Avg. Rate, M/s Avg. Rate, M/s
Time (s) [H2], M [HI], M -Δ[H2]/Δt 1/2 Δ[HI]/Δt 0.000 1.000 0.000
10.000 0.819 0.362
20.000 0.670 0.660
30.000 0.549 0.902
40.000 0.449 1.102
50.000 0.368 1.264
60.000 0.301 1.398
70.000 0.247 1.506
80.000 0.202 1.596
90.000 0.165 1.670
100.000 0.135 1.730
Avg. Rate, M/s
Time (s) [H2], M [HI], M -Δ[H2]/Δt 0.000 1.000 0.000
10.000 0.819 0.362 0.0181
20.000 0.670 0.660 0.0149
30.000 0.549 0.902 0.0121
40.000 0.449 1.102 0.0100
50.000 0.368 1.264 0.0081
60.000 0.301 1.398 0.0067
70.000 0.247 1.506 0.0054
80.000 0.202 1.596 0.0045
90.000 0.165 1.670 0.0037
100.000 0.135 1.730 0.0030
The average rate is the change in the concentration in a given time period.
22
H2 I2 HI
Avg. Rate, M/s Avg. Rate, M/s
Time (s) [H2], M [HI], M -Δ[H2]/Δt 1/2 Δ[HI]/Δt 0.000 1.000 0.000
10.000 0.819 0.362 0.0181 0.0181
20.000 0.670 0.660 0.0149 0.0149
30.000 0.549 0.902 0.0121 0.0121
40.000 0.449 1.102 0.0100 0.0100
50.000 0.368 1.264 0.0081 0.0081
60.000 0.301 1.398 0.0067 0.0067
70.000 0.247 1.506 0.0054 0.0054
80.000 0.202 1.596 0.0045 0.0045
90.000 0.165 1.670 0.0037 0.0037
100.000 0.135 1.730 0.0030 0.0030
23
Concentration vs. Time for H2 + I2 --> 2HI
0.000
0.200
0.400
0.600
0.800
1.000
1.200
1.400
1.600
1.800
2.000
0.000 10.000 20.000 30.000 40.000 50.000 60.000 70.000 80.000 90.000 100.000
time, (s)
conc
entr
atio
n, (M
)
[H2], M
[HI], M
average rate in a given time period = - slope of the line connecting the _____ points; and ½ +slope of the line for _____
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
Instantaneous Rate
• the instantaneous rate is the change in concentra%on at any one par%cular %me – slope at __________ of a curve
• determined by taking the slope of a line ______ to the curve at that par%cular point – _______________ of the func%on
24
H2 (g) + I2 (g) → 2 HI (g)
25
Using [H2], the instantaneous rate at 50 s is:
Using [HI], the instantaneous rate at 50 s is:
Example -‐ For the reac%on given, the [I-‐] changes from 2.000 M to 0.735 M in the first 25 s. Calculate the average rate in the
first 25 s and the Δ[H+]. H2O2 (aq) + 3 I-‐(aq) + 2 H+
(aq) → I3-‐(aq) + 2 H2O(l)
Solve the equation for the Rate (in terms of the change in concentration of the Given quantity)
Solve the equation of the Rate (in terms of the change in the concentration for the quantity to Find) for the unknown value
Rate = ! 13"
#$%
&'([I! ](t
=
Rate =
Rate = ![H+ ]!t
=
![H+ ]!t
=
Measuring Reac%on Rate • in order to measure the reac%on rate you need to be able to measure the concentra%on of at least one component in the mixture at many points in %me
• there are two ways of approaching this problem • (1) for reac%ons that are complete in less than 1 hour, it is best to use __________ monitoring of the concentra%on,
• (2) for reac%ons that happen over a very long %me, _________ of the mixture at various %mes can be used – when sampling is used, o`en the reac%on in the sample is stopped by a quenching technique
27
Factors Affec%ng Reac%on Rate Nature of the Reactants
• nature of the reactants means what kind of reactant molecules and what physical condi%on they are in. – small molecules tend to react _____ than large molecules; – gases tend to react _____ than liquids which react _____ than solids;
– powdered solids are _____ reac%ve than “blocks” • more surface area for contact with other reactants
– certain types of chemicals are more reac%ve than others • e.g., the ac%vity series of metals
– ions react faster than molecules • no bonds need to be broken
28
Factors Affec%ng Reac%on Rate Temperature
• increasing temperature _________ reac%on rate – chemist’s rule of thumb -‐ for each 10°C rise in temperature, the speed of the reac%on doubles
• for many reac%ons
• there is a mathema%cal rela%onship between the absolute temperature and the speed of a reac%on discovered by Svante Arrhenius which will be examined later
29
Factors Affec%ng Reac%on Rate Catalysts
• catalysts are substances which affect the speed of a reac%on _______________consumed. – homogeneous = present in _____ phase – heterogeneous = present in ______ phase
• how catalysts work will be examined later
30
Factors Affec%ng Reac%on Rate Reactant Concentra%on
• generally, the larger the concentra%on of reactant molecules, the _____ the reac%on – increases the __________ of reactant molecule contact
– concentra%on of gases depends on the par%al pressure of the gas • higher pressure = higher concentra%on
• concentra%on of solu%ons depends on the solute to solu%on ra%o (__________)
31
The Rate Law • the Rate Law of a reac%on is the mathema%cal rela%onship
between the rate of the reac%on and the concentra%ons of the reactants – and homogeneous catalysts as well
• the rate of a reac%on is _______________ to the concentra%on of each reactant raised to a power
• for the reac%on nA + mB → products the rate law would have the form given below – n and m are called the _____ for each reactant – k is called the __________
32
The Integrated Rate Law
• Rela%onship between the concentra%on of reactants and %me
• Obtained from integra%on of differen%al rate laws
• It has the form of an equa%on for a straight line y = mx + b
33
Reac%on Order • the exponent on each reactant in the rate law is called the _____ with respect to that __________
• the sum of the exponents on the reactants is called the _______________
• The rate law for the reac%on:
34
2 NO(g) + O2(g) → 2 NO2(g)
is Rate = The reaction is
second order with respect to [NO], first order with respect to [O2],
and third order overall
Reactant Concentra%on vs. Time A → Products
35
Half-‐Life • the half-‐life, ___, of a reac%on is the length of %me it takes for the concentra%on of the reactants to fall ____ its ini%al value
• the half-‐life of the reac%on depends on the _____ of the reac%on
36
Zero Order Reac%ons • Rate = =
– constant rate reac1ons • Integrated Rate Law: [A] = • graph of [A] vs. %me is straight line with slope = ___ and y-‐intercept = ____
• t ½ = • when Rate = M/sec, k = M/sec
37
[A]0
[A]
time
Half Life from Rate Law
38
First Order Reac%ons • Rate = • Integrated Rate Law: ln[A] = • graph ln[A] vs. %me gives straight line with slope = ___ and y-‐
intercept = ______ – used to determine the rate constant
• t½ = • the half-‐life of a first order reac1on is constant • the when Rate = M/sec, k = sec-‐1
39
ln[A]0
ln[A]
time
Second Order Reac%ons • Rate = • Integrated Rate Law: 1/[A] = • graph 1/[A] vs. %me gives straight line with slope = __ and
y-‐intercept = ____ – used to determine the rate constant
• t½ = • when Rate = M/sec, k = M-‐1·sec-‐1
40
l/[A]0
1/[A]
time
Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions
Order Rate Law Concentration-Time
Equation Half-Life
0
1
2
rate = k
rate = k [A]
rate = k [A]2
ln[A] = ln[A]0 - kt
1 [A]
= 1
[A]0 + kt
[A] = [A]0 - kt
t½ ln2 k
=
t½ = [A]0 2k
t½ = 1 k[A]0
13.3
Ex. 13.4 – The reac%on 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 condi%ons. Find the [SO2Cl2] at
865 s when [SO2Cl2]0 = 0.0225 M
the new concentration is less than the original, as expected
[SO2Cl2]0 = 0.0225 M, t = 865, k = 2.90 x 10-4 s-1
[SO2Cl2]
Check:
Solution:
Concept Plan:
Relationships:
Given:
Find:
[SO2Cl2] [SO2Cl2]0, t, k
for a 1st order process:
ln[SO2Cl2 ]=ln[SO2Cl2 ]=ln[SO2Cl2 ]=[SO2Cl2 ]=
What is the order of a reaction that has the following time and concentration data?
Time (s) [Concentration] 0 0.01000 50 0.00887 100 0.00797 150 0.00723 200 0.00662 250 0.00611
a) Zero order b) First order c) Second order d) None of the above
Copyright © 2011 Pearson Education, Inc.
a) Second order
b) Third order
c) Fourth order
d) Fifth order e) Sixth order
Determine the overall order for a reaction with the following rate law.
Rate = k[A]2[B][C]3
Copyright © 2011 Pearson Education, Inc.
Prac%ce -‐ Determine the Rate Equa%on for the Reac%on A → 2 Prod
45
[A], (M) [Prod], (M) Time (sec) ln([A]) 1/[A]
0.100 0 0 -2.3 10
0.067 0.066 50 -2.7 15
0.050 0.100 100 -3.0 20
0.040 0.120 150 -3.2 25
0.033 0.134 200 -3.4 30
0.029 0.142 250 -3.5 35
46
[A] vs. Time
0
0.02
0.04
0.06
0.08
0.1
0.12
0 50 100 150 200 250
time, (s)
conc
entr
atio
n, M
47
LN([A]) vs. Time
-3.8
-3.6
-3.4
-3.2
-3
-2.8
-2.6
-2.4
-2.2
-2
0 50 100 150 200 250
time, (s)
Ln(c
once
ntra
tion)
48
1/([A]) vs. Timey = 0.1x + 10
0
5
10
15
20
25
30
35
40
0 50 100 150 200 250
time, (s)
inve
rse
conc
entr
atio
n, M
-1
Ini%al Rate Method • another method for determining the order of a reactant is to see the effect on the ini%al rate of the reac%on when the ini%al concentra%on of that reactant is changed – for mul1ple reactants, keep ini1al concentra1on of all reactants constant except one
– zero order = changing the concentra%on has _____ on the rate
– first order = the rate changes by the _____ _____as the concentra%on
• doubling the ini%al concentra%on will double the rate – second order = the rate changes by the _____ of the factor the concentra%on changes
• doubling the ini%al concentra%on will quadruple the rate
49
Ex 13.2 – Determine the rate law and rate constant for the reac%on NO2(g) + CO(g) → NO(g) + CO2(g)
given the data below.
50
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
Write a general rate law including all reactants Examine the data and find two experiments in which the concentration of one reactant changes, but the other concentrations are the same
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
Comparing Expt #1 and Expt #2, the [NO2] changes but the [CO] does not
Rate =
Ex 13.2 – Determine the rate law and rate constant for the reac%on NO2(g) + CO(g) → NO(g) + CO2(g)
given the data below.
51
Determine by what factor the concentrations and rates change in these two experiments.
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
[NO2 ]expt 2
[NO2 ]expt 1
=Rateexpt 2
Rateexpt 1
=
Ex 13.2 – Determine the rate law and rate constant for the reac%on NO2(g) + CO(g) → NO(g) + CO2(g)
given the data below.
52
Determine to what power the concentration factor must be raised to equal the rate factor.
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
[NO2 ]expt 2
[NO2 ]expt 1
!
"##
$
%&&
n
=Rateexpt 2
Rateexpt 1
Ex 13.2 – Determine the rate law and rate constant for the reac%on NO2(g) + CO(g) → NO(g) + CO2(g)
given the data below.
53
Repeat for the other reactants
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
[CO]expt 3
[CO]expt 2
=Rateexpt 3
Rateexpt 2
=
[CO]expt 3
[CO]expt 2
!
"##
$
%&&
m
=Rateexpt 3
Rateexpt 2
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
Ex 13.2 – Determine the rate law and rate constant for the reac%on NO2(g) + CO(g) → NO(g) + CO2(g)
given the data below.
54
Substitute the exponents into the general rate law to get the rate law for the reaction
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
mnk [CO]][NO Rate 2=n = 2, m = 0
Rate = Rate =
Ex 13.2 – Determine the rate law and rate constant for the reac%on NO2(g) + CO(g) → NO(g) + CO2(g)
given the data below.
55
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
a) 1 b) 0 c) 2 d) 3 e) 4
What is the order of the reaction with respect to OH–? 2 ClO2 (aq) + 2 OH– (aq) → ClO3
– (aq) + ClO2– (aq) + H2O (l)
[ClO2] [OH–] Initial Rate (M/s)
0.060 0.030 0.0248 0.020 0.030 0.00276 0.020 0.090 0.00828
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The Effect of Temperature on Rate • changing the temperature changes the rate constant of the rate law
• Svante Arrhenius inves%gated this rela%onship and showed that:
57
R is the gas constant in energy units, _______________ where T is the temperature in _____
A is a factor called the frequency factor Ea is the _______________ , the extra energy needed to start the molecules reacting
58
The Arrhenius Equa%on: The Exponen%al Factor
• the exponen%al factor in the Arrhenius equa%on is a number between 0 and 1
• it represents the frac%on of reactant molecules with sufficient energy so they can make it over the energy barrier – the higher the energy barrier (larger ac%va%on energy), the fewer
molecules that have sufficient energy to overcome it • that extra energy comes from conver%ng the kine%c energy of
mo%on to poten%al energy in the molecule when the molecules collide – increasing the temperature increases the average kine%c energy of the
molecules – therefore, increasing the temperature will increase the number of
molecules with sufficient energy to overcome the energy barrier – therefore increasing the temperature will increase the reac%on rate
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Arrhenius Plots • the Arrhenius Equa%on can be algebraically solved to give the following form:
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this equation is in the form y = mx + b where y = and x = a graph of ln(k) vs. (1/T) is a straight line
(-8.314 J/mol·K)(slope of the line) = Ea, (in Joules)
ey-intercept = A, (unit is the same as k)
Ex. 13.7 Determine the ac%va%on energy and frequency factor for the reac%on O3(g) → O2(g) + O(g) given the following data:
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Temp, K k, M-1·s-1 Temp, K k, M-1·s-1 600 3.37 x 103 1300 7.83 x 107 700 4.83 x 104 1400 1.45 x 108
800 3.58 x 105 1500 2.46 x 108
900 1.70 x 106 1600 3.93 x 108
1000 5.90 x 106 1700 5.93 x 108
1100 1.63 x 107 1800 8.55 x 108
1200 3.81 x 107 1900 1.19 x 109
Ex. 13.7 Determine the ac%va%on energy and frequency factor for the reac%on O3(g) → O2(g) + O(g) given the following data:
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use a spreadsheet to graph ln(k) vs. (1/T)
Ex. 13.7 Determine the ac%va%on energy and frequency factor for the reac%on O3(g) → O2(g) + O(g) given the following data:
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Ea = m·(-R) solve for Ea
Ea =
Ea =
A = ey-intercept solve for A
A =A =
Arrhenius Equa%on: Two-‐Point Form
• if you only have two (T,k) data points, the following forms of the Arrhenius Equa%on can be used:
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⎟⎟⎠
⎞⎜⎜⎝
⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛
211
2T1
T1ln
RE
kk a
Ex. 13.8 – The reac%on 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 ac%va%on energy
in kJ/mol
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:
Concept Plan:
Relationships:
Given:
Find:
Ea T1, k1, T2, k2
⎟⎟⎠
⎞⎜⎜⎝
⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛
211
2T1
T1ln
RE
kk a
The chirping of tree crickets has sometimes been used to predict temperatures. At 25.0 oC the rate was 179 chirps/min. At 21.7 oC it was 142 chirps/min. What is the Ea of the process?
a) 316 J b) 51,300 J c) 0.749 kJ d) 6.16 kJ
Copyright © 2011 Pearson Education, Inc.
Collision Theory of Kine%cs
• for most reac%ons, in order for a reac%on to take place, the reac%ng molecules must collide into each other.
• once molecules collide they may react together or they may not, depending on two factors -‐
1. whether the collision has enough energy to "_______________holding reactant molecules together";
2. whether the reac%ng molecules collide in the _______________for new bonds to form.
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Effec%ve Collisions • ____________________: collisions in which these two condi%ons are met
• the higher the frequency of effec%ve collisions, the faster the reac%on rate
• when two molecules have an effec%ve collision, a temporary, high energy (unstable) chemical species is formed -‐ called an ac1vated complex or _______________
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Effec%ve Collisions Orienta%on Effect
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Collision Theory and the Arrhenius Equa%on
• A is the factor called the frequency factor and is the number of molecules that can approach overcoming the energy barrier
• there are two factors that make up the frequency factor – __________ factor (p) – ____________________factor (z)
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!
k = A e"EaRT
#
$ %
&
' ( = pze
"EaRT
Orienta%on Factor • the proper orienta%on 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 orienta%on – reac%ons between atoms generally have p = 1
• for most reac%ons, the orienta%on factor is less than 1 – for many, p << 1
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Reac%on Mechanisms • we generally describe chemical reac%ons with an equa%on lis%ng all the reactant molecules and product molecules
• but the probability of more than 3 molecules colliding at the same instant with the proper orienta%on and sufficient energy to overcome the energy barrier is negligible
• most reac%ons occur in a series of small reac%ons involving 1, 2, or at most 3 molecules
• describing the series of steps that occur to produce the overall observed reac%on is called a __________ __________
• knowing the rate law of the reac%on helps us understand the sequence of steps in the mechanism
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An Example of a Reac%on Mechanism • Overall reac%on:
H2(g) + 2 ICl(g) → 2 HCl(g) + I2(g) • Mechanism:
1) H2(g) + ICl(g) → HCl(g) + HI(g) 2) HI(g) + ICl(g) → HCl(g) + I2(g)
• the steps in this mechanism are __________ _____, 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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Elements of a Mechanism Intermediates
• no%ce that the HI is a product in Step 1, but then a reactant in Step 2
• since HI is made but then consumed, HI does not show up in the overall reac%on
• materials that are products in an early step, but then a reactant in a later step are called ______________
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H2(g) + 2 ICl(g) → 2 HCl(g) + I2(g) 1) H2(g) + ICl(g) → HCl(g) + HI(g) 2) HI(g) + ICl(g) → HCl(g) + I2(g)
Molecularity
• the number of reactant par%cles in an elementary step is called its _______________
• unimolecular step: involves __ reactant par%cle • bimolecular step: involves __ reactant par%cles
– though they may be the same kind of par%cle
• termolecular step: involves __ reactant par%cles – 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 lirle reac%on – with its own ac%va%on energy and own rate law
• the rate law for an overall reac%on must be determined experimentally
• but the rate law of an elementary step can be deduced from the equa%on of the step
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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]
Rate Determining Step • in most mechanisms, one step occurs slower than the other steps
• the result is that product produc%on cannot occur any faster than the _______________ – the step determines the rate of the overall reac%on
• we call the slowest step in the mechanism the _____ _______________step – the slowest step has the largest ac%va%on energy
• the rate law of the rate determining step determines the rate law of the overall reac%on
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Another Reac%on Mechanism
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NO2(g) + CO(g) → NO(g) + CO2(g) Rateobs = 1) NO2(g) + NO2(g) → NO3(g) + NO(g) Rate = .
2) 2) NO3(g) + CO(g) → NO2(g) + CO2(g) Rate = .
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.
Valida%ng a Mechanism
• in order to validate (not prove) a mechanism, two condi%ons must be met:
1. the elementary steps must _____ to the overall reac%on
2. the rate law predicted by the __________ must be consistent with the experimentally observed __________
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Catalysts • catalysts are substances that affect the rate of a reac%on without being consumed
• catalysts work by providing an _________________ for the reac%on – with a ____________________
• catalysts are consumed in an early mechanism step, then made in a later step
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mechanism without catalyst
O3(g) + O(g) → 2 O2(g) V. Slow
mechanism with catalyst
Cl(g) + O3(g) ⇔ O2(g) + ClO(g) Fast ClO(g) + O(g) → O2(g) + Cl(g) Slow
Catalysts • ____________________ catalysts are in the same phase as the reactant par%cles – Cl(g) in the destruc%on of O3(g)
• ____________________ catalysts are in a different phase than the reactant par%cles – solid cataly%c converter in a car’s exhaust system
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Types of Catalysts
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