1. Chemical Kinetics CHAPTER 14 Part B
Chemistry: The Molecular Nature of Matter, 6th edition
By Jesperson, Brady, & Hyslop

2. CHAPTER 14 Chemical Kinetics
Learning Objectives:
Factors Affecting Reaction Rate:
Concentration
State
Surface Area
Temperature
Catalyst
Collision Theory of Reactions and Effective Collisions
Determining Reaction Order and Rate Law from Data
Integrated Rate Laws
Rate Law  Concentration vs Rate
Integrated Rate Law  Concentration vs Time
Units of Rate Constant and Overall Reaction Order
Half Life vs Rate Constant (1st Order)
Arrhenius Equation
Mechanisms and Rate Laws
Catalysts
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

3. CHAPTER 14 Chemical Kinetics
Lecture Road Map:
Factors that affect reaction rates
Measuring rates of reactions
Rate Laws
Collision Theory
Transition State Theory & Activation Energies
Mechanisms
Catalysts
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

4. CHAPTER 14 Chemical Kinetics
Integrated Rate Laws
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

5. Rate law tells us how speed of reaction varies with concentrations.
Integrated Rate Laws
Concentration & Time
Rate law tells us how speed of reaction varies with concentrations.
Sometimes want to know
Concentrations of reactants and products at given time during reaction
How long for the concentration of reactants to drop below some minimum optimal value
 Need dependence of rate on time
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

6. First Order Integrated Rate Law
Integrated Rate Laws
First Order Integrated Rate Law
Corresponding to reactions
A  products
Integrating we get
Rearranging gives
Equation of line y = mx + b
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

7. Slope = –k Integrated Rate Laws First Order Integrated Rate Law
Yields straight line
Indicative of first order kinetics
Slope = –k
Intercept = ln [A]0
If we don't know already
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

8. 2nd Order Integrated Rate Law
Integrated Rate Laws
2nd Order Integrated Rate Law
Corresponding to special second order reaction
2B  products
Integrating we get
Rearranging gives
Equation of line y = mx + b
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

9. Slope = +k Integrated Rate Laws 2nd Order Integrated Rate Law
Yields straight line
Indicative of 2nd order kinetics
Slope = +k
Intercept = 1/[B]0
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

10. Graphically determining Order
Integrated Rate Laws
Graphically determining Order
Make two plots:
ln [A] vs. time
1/[A] vs. time
If ln [A] is linear and 1/[A] is curved, then reaction is 1st order in [A]
If 1/[A] plot is linear and ln [A] is curved, then reaction is 2nd order in [A]
If both plots give horizontal lines, then 0th order in [A]
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

11. Graphically determining Order
Integrated Rate Laws
Graphically determining Order
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

12. Integrated Rate Laws Example: SO2Cl2  SO2 + Cl2 Time, min
[SO2Cl2], M
ln[SO2Cl2]
1/[SO2Cl2] (L/mol)
0.1000
10.000
100
0.0876
11.416
200
0.0768
13.021
300
0.0673
14.859
400
0.0590
16.949
500
0.0517
19.342
600
0.0453
22.075
700
0.0397
25.189
800
0.0348
28.736
900
0.0305
32.787
1000
0.0267
37.453
1100
0.0234
42.735
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

13. Reaction is 1st order in SO2Cl2
Integrated Rate Laws
Example: SO2Cl2  SO2 + Cl2
Reaction is 1st order in SO2Cl2
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

14. Time (s) [HI] (mol/L) ln[HI] 1/[HI] (L/mol)
Integrated Rate Laws
Example: HI(g)  H2(g) + I2(g)
Time
(s)
[HI]
(mol/L)
ln[HI]
1/[HI]
(L/mol)
0.1000
10.000 50
0.0716
100
0.0558
150
0.0457
200
0.0387
25.840
250
0.0336
300
0.0296
350
0.0265
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

15. Reaction is second order in HI.
Integrated Rate Laws
Example: HI(g)  H2(g) + I2(g)
Reaction is second order in HI.
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

16. Group
Problem
A plot for a zeroth order reaction is shown. What is the proper label for the y-axis in the plot ?
A. Concentration
B. ln of Concentration
C. 1/Concentration
D. 1/ ln Concentration
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

17. Half Life (t1/2) for first order reactions
Integrated Rate Laws
Half Life (t1/2) for first order reactions
Half-life = t½ We often use the half life to describe how fast a reaction takes place
First Order Reactions
Set
Substituting into
Gives
Canceling gives ln 2 = kt½
Rearranging gives
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

18. Half Life (t1/2) for First Order Reactions
Integrated Rate Laws
Half Life (t1/2) for First Order Reactions
Observe:
t½ is independent of [A]o
For given reaction (and T)
Takes same time for concentration to fall from
2 M to 1 M as from
5.0  10–3 M to 2.5  10–3 M
k1 has units (time)–1, so t½ has units (time)
t½ called half-life
Time for ½ of sample to decay
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

19. Does this mean that all of sample is gone in two half-lives (2 × t½)?
Integrated Rate Laws
Half Life (t1/2)
Does this mean that all of sample is gone in two half-lives (2 × t½)?
No!
In 1st t½, it goes to ½[A]o
In 2nd t½, it goes to ½(½[A]o) = ¼[A]o
In 3rd t½, it goes to ½(¼[A]o) = ⅛[A]o
In nth t½, it goes to [A]o/2n
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

20. Integrated Rate Laws Half Life (t1/2)
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

21. Half Life (t1/2): First Order Example
Integrated Rate Laws
Half Life (t1/2): First Order Example
131I is used as a metabolic tracer in hospitals. It has a half-life, t½ = 8.07 days. How long before the activity falls to 1% of the initial value?
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

22. Group
Problem
The radioactive decay of a new atom occurs so that after 21 days, the original amount is reduced to 33%. What is the rate constant for the reaction in s–1?
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

23. Group
Problem
The half-life of I-132 is h. What percentage remains after 24 hours?
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

24. Half Life (t1/2): Carbon-14 Dating
Integrated Rate Laws
Half Life (t1/2): Carbon-14 Dating
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

25. Half Life (t1/2): Second Order Reactions
Integrated Rate Laws
Half Life (t1/2): Second Order Reactions
How long before [A] = ½[A]o?
t½, depends on [A]o
t½, not useful quantity for a second order reaction
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

26. Group
Problem
The rate constant for the second order reaction 2A → B is 5.3 × 10–5 M–1 s–1. What is the original amount present if, after 2 hours, there is 0.35 M available?
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

27. CHAPTER 14 Chemical Kinetics
Collision Theory
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

28. As the concentration of reactants increase
Collision Theory
Reaction Rates
Collision Theory
As the concentration of reactants increase
The number of collisions increases
Reaction rate increases
As temperature increases
Molecular speed increases
Higher proportion of collisions with enough force (energy)
There are more collisions per second
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

29. One that gives rise to product e.g. At room temperature and pressure
Collision Theory
Reaction Rates
Rate of reaction proportional to number of effective collisions/sec among reactant molecules
Effective collision
One that gives rise to product
e.g. At room temperature and pressure
H2 and I2 molecules undergoing 1010 collisions/sec
Yet reaction takes a long time
Not all collisions lead to reaction
Only very small percentage of all collisions lead to net change
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

30. Molecular Orientation
Collision Theory
Molecular Orientation
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

31. So more molecules undergo reaction
Collision Theory
Temperature
As T increases
More molecules have Ea
So more molecules undergo reaction
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

32. Activation Energy (Ea)
Collision Theory
Activation Energy (Ea)
Molecules must possess certain amount of kinetic energy (KE) in order to react
Activation Energy, Ea = Minimum KE needed for reaction to occur
Get energy from collision with other molecules
If molecules move too slowly, too little KE, they just bounce off each other
Without this minimum amount, reaction will not occur even when correctly oriented
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

33. CHAPTER 14 Chemical Kinetics Transition State Theory
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

34. Molecular Basis of Transition State TheoryKE KE
KE decreasing as PE increases
Is the combined KE of both molecules enough to overcome Activation Energy PE KE KE
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

35. Hreaction = Hproducts – Hreactants
Transition State
Molecular Basis of Transition State Theory
Activation energy (Ea)
= hill or barrier
between reactants
and products
Potential Energy
Heat of reaction (H) = difference in PE between
products and reactants
Hreaction = Hproducts – Hreactants
Products
Reaction Coordinate (progress of reaction)
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

36. Products lower PE than reactants
Transition State
Exothermic Reactions
Exothermic reaction
Products lower PE than reactants
Potential Energy
Exothermic
Reaction
H = –
Products
Reaction Coordinate (progress of reaction)
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

37. Hreaction < 0 (negative) Decrease in PE of system
Transition State
Exothermic Reactions
Hreaction < 0 (negative)
Decrease in PE of system
Appears as increase in KE
So the temperature of the system increases
Reaction gives off heat
Can’t say anything about Ea from size of H
Ea could be high and reaction slow even if Hrxn large and negative
Ea could be low and reaction rapid
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

38. Hreaction = Hproducts – Hreactants
Transition State
Endothermic Reactions
Endothermic
Reaction
H = +
Hreaction = Hproducts – Hreactants
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

39. Endothermic Reactions
Transition State
Endothermic Reactions
Hreaction > 0 (positive)
Increase in PE
Appears as decrease in KE
So temperature of the system decreases
Have to add E to get reaction to go
Ea  Hrxn as Ea includes Hrxn
If Hrxn large and positive
Ea must be high
Reaction very slow
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

40. Arrangement of atoms at top of activation barrier
Transition State
Activated Complex
Arrangement of atoms at top of activation barrier
Brief moment during successful collision when
bond to be broken is partially broken and
bond to be formed is partially formed
Example
Transition State
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

41. CH3CH2O- + H3O+  CH3CH2OH + H2O
Group
Problem
Draw the transition state complex, or the activated complex for the following reaction:
CH3CH2O- + H3O+  CH3CH2OH + H2O
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

42. CHAPTER 14 Chemical Kinetics
Activation Energies
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

43. Equation expressing temperature-dependence of kEa
Arrhenius Equation
The rate constant is dependent on Temperature, which allows us to calculate Activation Energy, Ea
Arrhenius Equation:
Equation expressing temperature-dependence of k
A = Frequency factor has same units as k
R = gas constant in energy units
= J mol–1 K–1
Ea = Activation Energy—has units of J/mol
T = Temperature in K
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

44. Calculating Activation EnergyEa
Calculating Activation Energy
Method 1. Graphically
Take natural logarithm of both sides
Rearranging
Equation for a line
y = b + mx
Arrhenius Plot
Plot ln k (y axis) vs. 1/T (x axis)  yield a straight line
Slope = -Ea/R
Intercept = A
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

45. Ea
Arrhenius Equation: Graphing Example
k (M/s)
T, ˚C
T, K 25
298 50
348
100
398
150
448 ? 75
Given the following data, predict k at 75 ˚C using the graphical approach
ln (k) = –36.025/T – 6.908
ln (k) = –36.025/(348) – = – 7.011
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

46. Arrhenius Equation: Graphing ExampleEa
Arrhenius Equation: Graphing Example
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

47. Sometimes a graph is not needed Only have two k s at two Ts Ea
Arrhenius Equation
Sometimes a graph is not needed
Only have two k s at two Ts
Here use van't Hoff Equation derived from Arrhenius equation:
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

48. Arrhenius Equation: Ex Vant Hoff EquationEa
Arrhenius Equation: Ex Vant Hoff Equation
CH S2  CS H2S
k (L/mol s)
T (˚C)
T (K)
1.1 = k1
550
823 = T1
6.4 = k2
625
898 = T2
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

49. Group
Problem
Given that k at 25 ˚C is 4.61 × 10–1 M/s and that at 50 ˚C it is × 10–1 M/s, what is the activation energy for the reaction?
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

50. A. Rate increases approximately 1.5 times
Group
Problem
A reaction has an activation energy of 40 kJ/mol. What happens to the rate if you increase the temperature from 70˚C to 80 ˚C?
A. Rate increases approximately 1.5 times
B. Rate increases approximately 5000 times
C. Rate does not increase
D. Rate increases approximately 3 times
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E