1. H2S (g) + Zn2+ (aq) ⇆ ZnS (s) + 2H+ (aq)
CHEMICAL KINETICS
Chemical reactions can be viewed from different perspectives
H2S (g) + Zn2+ (aq) ⇆ ZnS (s) + 2H+ (aq)
STOICHIOMETRY
Describes relationships based on conservation of atoms – predicts reaction quantities
THERMODYNAMICS
Describes energy and entropy changes – predicts if a reaction will occur
KINETICS
Describes how a reaction occurs – predicts the speed of a reaction
Kinetics
Thermodynamics
4D-1 (of 21)

2. CHEMICAL KINETICS
1 - Describes the speed of a chemical reaction and the factors that affect it
2 - Determines the mechanism of a chemical reaction at the molecular level
COLLISION THEORY
1 - Molecules must collide to react
2 - Molecules must collide with sufficient energy to react (to break bonds)
3 - Molecules must collide in the proper orientation
4D-2 (of 21)

3. 4D-3 (of 21)

4. REACTION RATES
Measured by how fast a reactant reacts away, or how fast a product is produced
Rate =
- Δ[reactant]
_________________ Δt
= - d[reactant]
_________________ dt
Rate =
+ Δ[product]
_________________ Δt
= + d[product]
_________________ dt
4D-4 (of 21)

5. For the reaction:
2N2O5 (g) → 4NO2 (g) + O2 (g)
- d[N2O5] =
_____________ dt
0.30 M/min
Find the reaction rate with respect to NO2 and O2
+ d[NO2] =
____________ dt
0.30 M N2O5
_______________
min
x 4 M NO ____________
2 M N2O5
= M NO2/min
+ d[O2] =
____________ dt
0.30 M N2O5
_______________
min
x M O ____________
2 M N2O5
= M O2/min
4D-5 (of 21)

6. Concentration as a Function of Reaction Time
For the reaction:
2N2O5 (g) → 4NO2 (g) + O2 (g)
Concentration as a Function of Reaction Time
-0.30 M/min
Fastest Reaction Rates
0 Reaction Rate (equilibrium)
+0.60 M/min
+0.15 M/min
Reaction rates are the slope of the tangent line (at any particular point) of a concentration vs. reaction time graph
4D-6 (of 21)

7. Experimentally, rates of reactions are proportional to
Temperature
Concentration of reacting molecules
RATE LAW – An algebraic expression that relates the rate of a reaction (how fast a reactant disappears or how fast a product appears) to the concentrations of the reactants and the temperature
4D-7 (of 21)

8. 2N2O5 (g) → 4NO2 (g) + O2 (g)
Rate  T [N2O5 ]
Rate = k [N2O5 ]
SPECIFIC RATE CONSTANT (k) – The rate law constant that depends on temperature
FIRST-ORDER REACTION – One in which the rate is proportional to the concentration of reactants to the first power
4D-8 (of 21)

9. 2NO2 (g) → 2NO (g) + O2 (g)
Rate = k [NO2 ]2
This is a SECOND-ORDER REACTION
H2 (g) + I2 (g) → 2HI (g)
Rate = k [H2 ] [I2 ]
1st order in H2
1st order in I2
2nd order overall
Rate laws can only be determined experimentally
4D-9 (of 21)

10. Find the rate law given the following experimental data
Initial Rate (M/s)
0.040
0.010
0.005
[NO]
1.00
0.50
[Cl2]
1.00
0.50
R = k [NO]x [Cl2]y
Choose 2 trials where [Cl2] is constant
= k [1.00]x [1.00]y
________________________________
= k [0.50]x [1.00]y
4 = 2x
 = x
4D-10 (of 21)

11. Find the rate law given the following experimental data
Initial Rate (M/s)
0.040
0.010
0.005
[NO]
1.00
0.50
[Cl2]
1.00
0.50
R = k [NO]x [Cl2]y
Choose 2 trials where [NO] is constant
= k [0.50]2 [1.00]y
________________________________
= k [0.50]2 [0.50]y
2 = 2y
 = y
4D-11 (of 21)

12. Find the rate law given the following experimental data
Initial Rate (M/s)
0.040
0.010
0.005
[NO]
1.00
0.50
[Cl2]
1.00
0.50
R = k [NO]x [Cl2]y
R = k [NO]2 [Cl2]1
The reaction is 2nd order in NO, 1st order in Cl2, and 3rd order overall
4D-12 (of 21)

13. Find the rate law given the following experimental data
Initial Rate (M/s)
0.040
0.080
0.640
[HCl]
3.0
6.0
[NO2]
1.0
2.0
R = k [HCl]x [NO2]y
Choose 2 trials where [NO2] is constant
= k [6.0]x [1.0]y
____________________________
= k [3.0]x [1.0]y
2 = 2x
 = x
4D-13 (of 21)

14. Find the rate law given the following experimental data
Initial Rate (M/s)
0.040
0.080
0.640
[HCl]
3.0
6.0
[NO2]
1.0
2.0
R = k [HCl]x [NO2]y
Choose 2 trials where [HCl] is constant
= k [6.0]1 [2.0]y
____________________________
= k [6.0]1 [1.0]y
8 = 2y
 = y
4D-14 (of 21)

15. Find the rate law given the following experimental data
Initial Rate (M/s)
0.040
0.080
0.640
[HCl]
3.0
6.0
[NO2]
1.0
2.0
R = k [HCl]x [NO2]y
R = k [HCl]1 [NO2]3
The reaction is 1st order in HCl, 3rd order in NO2, and 4th order overall
4D-15 (of 21)

16. Find the rate law given the following experimental data
Initial Rate (M/s)
0.040
0.080
0.640
[HCl]
3.0
6.0
[NO2]
1.0
2.0
R = k [HCl]x [NO2]y
R = k [HCl]1 [NO2]3
Find the value of k, with its units
R = k
________________
[HCl]1 [NO2]3
= M/s
____________________
(3.0 M)1 (1.0 M)3
= M-3s-1
4D-16 (of 21)

17. Find the rate law given the following experimental data
Initial Rate (M/s)
0.040
0.080
0.640
[HCl]
3.0
6.0
[NO2]
1.0
2.0
R = k [HCl]x [NO2]y
R = k [HCl]1 [NO2]3
Calculate the rate of the reaction when [HCl] = 1.0 M and [NO2] = 3.0 M
R = k [HCl]1 [NO2]3
= ( M-3s-1) (1.0 M)1 (3.0 M)3
= Ms-1
4D-17 (of 21)

18. Find the rate law given the following experimental data
Initial Rate (M/min)
0.09
0.18
1.08
[F2]
0.15
0.30
0.60
[Cl2]
0.20
0.60
R = k [F2]x [Cl2]y
Choose 2 trials where [Cl2] is constant
= k [0.30]x [0.20]y
_____________________________
= k [0.15]x [0.20]y
2 = 2x
 = x
4D-18 (of 21)

19. Find the rate law given the following experimental data
Initial Rate (M/min)
0.09
0.18
1.08
[F2]
0.15
0.30
0.60
[Cl2]
0.20
0.60
R = k [F2]x [Cl2]y
Choose 2 trials where [F2] is constant
???????
1.08 = k [0.60]1 [0.60]y
_____________________________
0.18 = k [0.30]1 [0.20]y
6 = (2) 3y
3 = 3y
 = y
4D-19 (of 21)

20. Find the rate law given the following experimental data
Initial Rate (M/min)
0.09
0.18
1.08
[F2]
0.15
0.30
0.60
[Cl2]
0.20
0.60
R = k [F2]x [Cl2]y
R = k [F2]1 [Cl2]1
Find the value of k, with its units
R = k
_____________
[F2]1 [Cl2]1
= M/min
________________________
(0.30 M)1 (0.20 M)1
= M-1min-1
4D-20 (of 21)

21. Find the rate law given the following experimental data
Initial Rate (M/min)
0.09
0.18
1.08
[F2]
0.15
0.30
0.60
[Cl2]
0.20
0.60
R = k [F2]x [Cl2]y
R = k [F2]1 [Cl2]1
Calculate the rate of the reaction when [F2] = 0.20 M and [Cl2] = 0.40 M
R = k [F2]1 [Cl2]1
= (3.0 M-1min-1) (0.20 M)1 (0.40 M)1
= Mmin-1
4D-21 (of 21)

23. 2N2O (g) → 2N2 (g) + O2 (g)
The rate law for this reaction is:
Rate = k
Rate = k [N2O ]0
This is a ZERO-ORDER REACTION
, or 0º
– d[N2O] = k
___________ dt
4E-1 (of 15)

24. 2N2O (g) → 2N2 (g) + O2 (g)
– d[N2O] = k
___________ dt ∫ t ∫ t
d[N2O] = – k dt t t
[N2O] = – kt
[N2O]t – [N2O]0 = – kt – (– k0)
[N2O]t – [N2O]0 = – kt
[N2O]t = – kt + [N2O]0
4E-2 (of 15)

25. [X]t = -kt [X]0
y = mx + b
To plot a linear graph for zero-order kinetics:
y = [X]t
m = -k
x = t
b = [X]0
a plot of [reactant]t vs. t will yield a line
4E-3 (of 15)

26. N2O5 (g) → N2O (g) + 2O2 (g)
The rate law for this reaction is:
Rate = k[N2O5 ]1
This is a FIRST-ORDER REACTION
, or 1º:
-d[N2O5] = k[N2O5]
__________ dt
4E-4 (of 15)

27. ∫ N2O5 (g) → N2O (g) + 2O2 (g) -d[N2O5] = k[N2O5] __________ dt t ∫ tt ∫ t
d[N2O5] = – k dt
__________
[N2O5] t t
ln[N2O5] = – kt
ln[N2O5]t – ln[N2O5]0 = – kt – (– k0)
ln[N2O5]t – ln[N2O5]0 = – kt
ln[N2O5]t = – kt + ln[N2O5]0
4E-5 (of 15)

28. ln[X]t = -kt ln[X]o
To plot a linear graph for first-order kinetics:
y = ln[X]t
m = -k
x = t
b = ln[X]o
a plot of ln[reactant]t vs. t will yield a line
4E-6 (of 15)

29. ∫ ln[X]t = -kt + ln[X]o e [X]t = e-kt [X]o [X]t = [X]oe-kt
This is the same as radioactive decay
, which follows first order kinetics:
-d[X] = k[X]
_______ dt ∫ t ∫ t
d[X] = – k dt
______
[X]
4E-7 (of 15)

30. 2HI (g) → H2 (g) + I2 (g)
The rate law for this reaction is:
Rate = k[HI ]2
This is a SECOND-ORDER REACTION
, or 2º
-d[HI] = k[HI]2
_______ dt
4E-8 (of 15)

31. ∫ 2HI (g) → H2 (g) + I2 (g) -d[HI] = k[HI]2 _______ dt t ∫ tt ∫ t
d[HI] = – k dt
_______
[HI]2 t t
–1 = – kt
_____
[HI]
–1 – – = – kt
_____ _____
[HI]t [HI]0
–1 = – kt – 1
_____ _____
[HI]t [HI]o –
1 = kt
_____ _____
[HI]t [HI]o
4E-9 (of 15)

32. 1 = kt
____ _____
[X]t [X]o
To plot a linear graph for second-order kinetics:
y = 1/[X]t
m = k
x = t
b = 1/[X]o
a plot of 1/[reactant]t vs. t will yield a line
4E-10 (of 15)

33. Order 1 2
Rate Law
R = k
R = k [X]1
R = k [X]2
Equality
[X]t = -kt [X]o
ln [X]t = -kt ln [X]o
1 = kt
____ ____
[X]t [X]o
Linear Plot
[X] vs. t
ln [X] vs. t
1 vs. t
____
[X]
4E-11 (of 15)

34. Find the rate law for the reaction A → B given the following:
[A] (M) :
Time (s) :
0.2500
0.00
0.1250
5.00
0.0625
15.00
R = k [A]X
4E-12 (of 15)

35. Find the rate law for the reaction A → B given the following:
[A] (M) :
Time (s) :
0.2500
0.00
0.1250
5.00
0.0625
15.00
Test for 0º
Linear plot would be [A] vs. t
If linear, the slope calculated with any 2 points will be constant
M – M
___________________________
0.00 s – 5.00 s
= Ms-1
M – M
___________________________
5.00 s – s
= Ms-1
Slopes are not constant
 not 0º
4E-13 (of 15)

36. Find the rate law for the reaction A → B given the following:
[A] (M) :
Time (s) :
0.2500
0.00
0.1250
5.00
0.0625
15.00
Test for 1º
Linear plot would be ln[A] vs. t
ln ( M) – ln ( M)
____________________________________
0.00 s – 5.00 s
= s-1
ln ( M) – ln ( M)
____________________________________
5.00 s – s
= s-1
Slopes are not constant
 not 1º
4E-14 (of 15)

37. Find the rate law for the reaction A → B given the following:
[A] (M) :
Time (s) :
0.2500
0.00
0.1250
5.00
0.0625
15.00
Test for 2º
Linear plot would be 1/[A] vs. t
(1/ M) – (1/ M)
___________________________________
0.00 s – 5.00 s
= M-1s-1
(1/ M) – (1/ M)
___________________________________
5.00 s – s
= M-1s-1
Slopes are constant
 the reaction is 2º
R = k [A]2
4E-15 (of 15)

39. REACTION MECHANISMS
Chemical reactions occur as a specific series of collisions
and each collision is considered a STEP
ELEMENTARY REACTION – A reaction that occurs in only one step (or one collision)
Each step has a MOLECULARITY
a) If 1 molecule decomposes, the step is UNIMOLECULAR
If 2 molecules collide to react, the step is BIMOLECULAR
If 3 molecules collide to react, the step is TRIMOLECULAR (rare)
4F-1 (of 19)

40. REACTION MECHANISM – The series of steps that yield the balanced chemical reaction
RATE DETERMINING STEP (or RATE LIMITING STEP) – The slowest step in the reaction mechanism
The reactants in a reaction’s rate law are the reactants in the rate determining step
4F-2 (of 19)

41. 2NO (g) + 2H2 (g) → N2 (g) + 2H2O (g)
This reaction occurs via a 3 step mechanism: K1
(1) NO ⇆ N2O2 (fast equilibrium) k2
(2) N2O2 + H2 → N2O + H2O (slow) k3
(3) N2O + H2 → N2 + H2O (fast)
R = k2 [N2O2] [H2]
This is not a reactant in the reaction, it is a REACTION INTERMEDIATE
 [N2O2] must be substituted out
4F-3 (of 19)

42. 2NO (g) + 2H2 (g) → N2 (g) + 2H2O (g)
This reaction occurs via a 3 step mechanism: K1
(1) NO ⇆ N2O2 (fast equilibrium)
K1 = [N2O2]
________
[NO]2 k2
(2) N2O2 + H2 → N2O + H2O (slow) k3
(3) N2O + H2 → N2 + H2O (fast)
K1 [NO]2 = [N2O2]
R = k2 [N2O2] [H2]
R = k2 K1 [NO]2 [H2]
R = k [NO]2 [H2]
4F-4 (of 19)

43. CHCl3 (g) + Cl2 (g) → CCl4 (g) + HCl (g)
This reaction occurs via a 3 step mechanism: K1
(1) Cl2 ⇆ 2Cl (fast equilibrium)
K1 = [Cl]2
______
[Cl2] k2
(2) CHCl3 + Cl → CCl3 + HCl (slow) k3
(3) CCl3 + Cl → CCl4 (fast)
K1 [Cl2] = [Cl]2
R = k2 [CHCl3] [Cl]
K1½ [Cl2]½ = [Cl]
R = k2 [CHCl3] K1½ [Cl2]½
R = k [CHCl3] [Cl2]½
4F-5 (of 19)

44. This reaction occurs via a 3 step mechanism:
H2 (g) + l2 (g) → 2HI (g)
This reaction occurs via a 3 step mechanism: K1
(1) l2 ⇆ 2l (fast equilibrium)
K2 = [H2I]
________
[H2][I] K2
(2) H2 + l ⇆ H2l (fast equilibrium) k3
(3) H2I + l → 2HI (slow)
K2 [H2] [I] = [H2I]
R = k3 [H2I] [l]
R = k3 K2 [H2] [l] [I]
4F-6 (of 19)

45. This reaction occurs via a 3 step mechanism:
H2 (g) + l2 (g) → 2HI (g)
This reaction occurs via a 3 step mechanism: K1
(1) l2 ⇆ 2l (fast equilibrium) K2
(2) H2 + l ⇆ H2l (fast equilibrium)
K1 = [I]2
____
[I2] k3
(3) H2I + l → 2HI (slow)
R = k3 [H2I] [l]
K1 [I2] = [I]2
R = k3 K2 [H2] [l] [I]
R = k3 K2 [H2] K1 [I2]
R = k [H2] [I2]
4F-7 (of 19)

46. ACTIVATION ENERGY (Ea) – The minimum energy needed by the reacting molecules for an effective collision
Reactants
Products Ea ΔH
4F-8 (of 19)

47. CALCULATING THE ACTIVATION ENERGY
On the macroscopic level, rates of reactions depend on:
Concentrations of reactants
Temperature
On the microscopic level, rates of reactions depend on:
The collision frequency of the reacting molecules
The fraction of collisions that have the proper orientation
The fraction of collisions that have the activation energy
4F-9 (of 19)

48. SVANTE ARRHENIUS
Proposed that the specific rate constant, k, is the product of the collision frequency, the fraction of collisions with the proper orientation, and the fraction of collisions with the activation energy
z = collision frequency
p = fraction of collisions with the proper orientation
e-Ea/RT = fraction of the collisions with the activation energy
k = zpe-Ea/RT
k = A e-Ea/RT
A is the PRE-EXPONENTIAL FACTOR
Ea is the ARRHENIUS ACTIVATION ENERGY
4F-10 (of 19)

49. The activation energy can be determined graphically by knowing specific rate constants at different temperatures
k = Ae-Ea/RT
ln k = ln A – Ea
____ RT
ln k = -Ea ln A
_____ RT
4F-11 (of 19)

50. The activation energy can be determined graphically by knowing specific rate constants at different temperatures
ln k = -Ea ln A
_____ ___
R T
y = ln k
m = -Ea/R
x = 1/T
b = ln A
4F-12 (of 19)

51. The activation energy can be determined graphically by knowing specific rate constants at different temperatures
ln k = -Ea ln A
_____ ___
R T
-Ea = m
_____ R
Ea = -Rm
Ea = -(8.314 J/K)( K)
= 28,400 J
4F-13 (of 19)

52. Exothermic Negative ΔH Endothermic Positive ΔH
REACTION ENERGY PROFILES
Reactants
Products Ea Ea
Products ΔH ΔH
Reactants
Exothermic
Negative ΔH
Endothermic
Positive ΔH
4F-14 (of 19)

53. Forward Reaction Ea (Ea-for) Reverse Reaction Ea (Ea-rev)
Reactants
Products
Enthalpy Change (ΔH)
Forward Reaction Ea (Ea-for)
Reverse Reaction Ea (Ea-rev)
Ea-for – Ea-rev = ΔH
4F-15 (of 19)

54. Reactants
Products ǂ
At the highest point of the graph, the reactants go through a high energy state called the TRANSITION STATE
At the transition state, the colliding molecules form a single unit called the ACTIVATED COMPLEX (ǂ)
4F-16 (of 19)

55. The Activated Complex
A single unit in which old bonds are breaking and new bonds are forming
4F-17 (of 19)

56. CATALYSIS
Increases the rate of a reaction by allowing the reaction to take place via a different pathway with a lower activation energy Ea Ea
A catalyst brings a reaction to equilibrium faster, it does not change the equilibrium concentrations
4F-18 (of 19)

57. Metal surfaces can act as catalysts for gaseous reactions
N2 (g) + 3H2 (g) → 2NH3 (g)
4F-19 (of 19)

58. REVIEW FOR TEST 4
Coordination Complex
Ligand
Monodentate, Polydentate
Chelate
Coordinate Covalent Bond
Coordination Number
Nomenclature
Isomers
Structural
Geometrical
Optical

59. REVIEW FOR TEST 4
Formation Constants
Crystal Field Theory
d-Orbital Splitting for Crystal Fields
Octahedral
Square Planar
Tetrahedral
Linear
Low-Spin and High-Spin Complex
Paramagnetic and Diamagnetic
Weak-Field and Strong-Field Ligands
Calculation of Splitting Energy
Lewis Acid, Lewis Base

60. REVIEW FOR TEST 4
Kinetics
Collision Theory
Reaction Rates – disappearance of reactant or formation of product
Rate Law
Specific Rate Constant
Order
Rate Law from Concentrations and Initial Rates
Algebraic Equations and Linear Plots for 0º, 1º, 2º
Rate Law from Concentrations and Time
Reaction Mechanisms
Rate Law from Reaction Mechanism

61. REVIEW FOR TEST 4
Activation Energy
Energy Profile
Transition State, Activated Complex
Catalysis
Experiments 22, 23, 26
Reading