Chemical
Kinetics
SCH4U: Grade 12 Chemistry
Unit Mind Map
Chemical Kinetics
• Chemical kinetics?
• Why do we study this?
• Who makes use of chemical kinetics?
Chemical Kinetics
• Chemical kinetics  studying how we can
make reactions go faster or slower
• Why do we study this?
• Who makes use of chemical kinetics?
Chemical Kinetics
• Chemical kinetics  studying how we can
make reactions go faster or slower
• Why should we study this?
• Economics effects Materials are expensive
• Developing pharmaceutical drugs
• Who makes use of chemical kinetics?
Chemical Kinetics
• Chemical kinetics  studying how we can make
reactions go faster or slower
• Why should we study this?
• Economics effects Materials are expensive
• Developing pharmaceutical drugs
• Who makes use of chemical kinetics?
• Biologists  metabolic rxns, food digestion, bone
regeneration
• Automobile engineers rate of rusting, decrease
pollutants
• Agriculture  slow down food ripening
Reaction Rates
• What is a chemical reaction rate?
• Possible formula for measuring rate?
Reaction Rates
• What is a chemical reaction rate?
• measure of how fast reactants are used up or how
fast products are produced
• Possible formula for measuring rate?
Reaction Rates
• What is the reaction rate?
• measure of how fast reactants are used up or how
fast products are produced
• Possible formula for measuring rate?
• Rate = Δ Concentration/Δ Time
Sample problem
N2 (g) + 3H2 (g)  2NH3(g)
• What is the average rate of production of
ammonia for the system if the concentration is
3.5mol/L after 1.0min and 6.2mol/L after 4.0 min?
Sample problem
N2 (g) + 3H2 (g)  2NH3(g)
• What is the average rate of production of
ammonia for the system if the concentration is
3.5mol/L after 1.0min and 6.2mol/L after 4.0 min?
• ANSWER: 0.9mol/(L*min)
Measuring reaction rates
Graphically
• Average rate of reaction  take the slope of the
secant of the line
• Instantaneous rate of reaction  take the slope
of the tangent of the line
Measuring reaction rates
Graphically
• Average rate of reaction  take the slope of the
secant of the line
1.
Draw a
secant line
between
two points
2. Calculate the
slope
Slope = Rise/Run
=
Δconcentration/
Δ time
Measuring reaction rates
Graphically
• Instantaneous rate of reaction  take the slope
of the tangent of the line
1.
Draw a
tangent line
to the graph
2. Calculate the
slope of the
tangent line
Slope = Rise/Run
=
Δconcentration/
Δ time
Measuring reaction rates
• What are some factors that we can measure
experimentally to determine the rate of reaction
?
Measuring reaction rates
• Production of a gas
• Production of ions
• Changes in colour
Reaction rates and
stoichiometry
CO(g) + NO2(g)  CO2(g) + NO(g)
In this reaction, the ratio of CO to NO is 1:1
Therefore, the disappearance of CO is the same
as the production of NO
Rate = - Δ [CO]/Δt = + Δ[NO]/Δt
Reaction rates and
stoichiometry
Suppose the ratio is NOT 1:1?
Example, H2 (g) + I2 (g)  2HI (g)
2 mols of HI are produced for every 1 mol H2 used
Rate =
-
D[H 2 ] D[HI]
=
´ molarratio
Dt
Dt
D[H 2 ] D[HI] 1molH 2
=
´
Dt
Dt
2molHI
The rate at which H2 is used up is only half of which
HI is produced
Sample Problem
• IO3- (aq) + 5I- (aq) + 6H+ (aq)  3I2 (aq) + 3H2O (l)
The rate of consumption of iodate ions (IO3-) is
determined experimentally to be 3.0 x 10-5 mol/(L*s).
What are the rates of reaction or the other reactants
and products in this reaction?
• Complete p.364 #3, 4
• Homework: practice problem package
Collision Theory
• Why do reactions occur the way that they do?
VS
Collision Theory
• Collision theory  reactions can only occur if:
1. There is a collision between molecules
Collision Theory
• Collision theory 
reactions can only occur if:
1. There is a collision between
molecules
2. the molecules are oriented
in the correct way
Collision Theory
• Collision theory  reactions
can only occur if:
1. There is a collision between
molecules
2. the molecules are oriented in
the correct way
3. enough energy is provided to
break the chemical bonds that
hold molecules together
(Activation energy)
Activation Energy
• Activation Energy  minimum potential energy
the system needs to overcome for the molecules to
react
Factors That Affect Rate
of Reaction
• What are some of the factors that may speed up or
slow down a chemical reaction? P.392
Factors That Affect Rate
of Reaction
• What are some of the factors that may speed up or
slow down a chemical reaction?
•
•
•
•
•
Chemical nature of the reactant
Concentration of reactants
Surface area of reactants
Temperature
Catalysts
• Get into groups, brainstorm and then explain these
factors to the rest of class in a creative way, making
use of collision theory
Lesson 2
• Review collision theory
• Rate Laws
• Group Practice problems
• Individual practice problems
• homework
Unit Mind Map
Collision theory
Collision Theory
• Collision theory  reactions
can only occur if:
1. There is a collision between
molecules
2. the molecules are oriented in
the correct way
3. enough energy is provided to
break the chemical bonds that
hold molecules together
(Activation energy)
Factors That Affect Rate
of Reaction
Factors That Affect Rate
of Reaction
•Chemical nature of the reactant
•Concentration of reactants
•Surface area of reactants
•Temperature
•Catalysts
The Rate Law
• Mathematical relationship between reaction rate and
factors that affect it
• Determined empirically (experimentally)
• Rate = k[X]m[Y]n
• e.g. 2NO2 + F2  2NO2F
The above reaction is 1st order with respect to NO2 and
2nd order with respect to F2. What is the rate law
equation?
The Rate Law
Answer:
• e.g. 2NO2 + F2  2NO2F
rate = k[NO2]1[F2]2
Steps to solve rate law
problems
eg. 2BrO3- (aq) + 5HSO3- (aq)  Br2 (g) + 5SO42- (aq) + H2O (l) + 3H+ (aq)
Rate = k [BrO3-]m [HSO3-]n
Steps to solve rate law
problems
eg. 2BrO3- (aq) + 5HSO3- (aq)  Br2 (g) + 5SO42- (aq) + H2O (l) + 3H+ (aq)
Write out rate equation: rate = k [BrO3-]m [HSO3-]n
pick a trial where [BrO3-] changes but [HSO3-] stays constant= trial 1
and trial 2
3. Using a ratio, determine the relationship between change in
concentration and change in rate = (trial 1/trial2) = (4.0/2.0) = (1.6/0.8)
=2=2
4. Using the following chart, determine the rate order of [BrO3-]
1.
2.
Steps to solve rate law
problems
5. If the concentration is doubled (2.0 to 4.0), the rate also doubles (0.80 to 1.60),
therefore the rate order is 1
m=1
Steps to solve rate law
problems
6. To find the rate order of the other reactant, follow the same steps but
pick a trial where [HSO3-] changes but [BrO3-] stays constant= trial 2 and
trial 3
= (6.0/3.0) = (0.8/0.2)
=2=4
7. Use the following table to determine the rate order
Steps to solve rate law
problems
8. As the concentration doubles the rate was multiplied by 4, therefore
rate order of HSO3- = 2 (n=2)
9. Plug values into rate equation rate = k [BrO3-]m [HSO3-]n and solve for k.
Example 1
Trial #
1
2
3
[A] Mol/L
5
10
10
[B] [Mol/L]
10
10
20
Rate [Mol/L*S)
6
6
12
From the data collected above, determine the rate law of the
following equation:
aA + bB  products
• Rate = k[A]m[B]n
Example 1
Trial #
1
2
3
[A] Mol/L
5
10
10
• aA + bB  products
• Rate = k[A]m[B]n
• m = 0; n = 1; k = 0.6s-1
• Rate = 0.6s-1 [B]1
[B] [Mol/L]
10
10
20
Rate [Mol/L*S)
6
6
12
Example 2
Trial #
1
2
3
[NO2] Mol/L
0.100
0.100
0.200
[F2] [Mol/L]
0.100
0.200
0.100
Rate [Mol/L*S)
4.0 * 10-5
4.0 * 10-5
16.0 * 10-5
From the data collected above, determine the rate law of the
following equation:
• 2NO2 + F2  2NO2F
• Rate = k[A]m[B]n
Example 2
Trial #
1
2
3
[NO2] Mol/L
0.100
0.100
0.200
• 2NO2 + F2  2NO2F
• Rate = k[A]m[B]n
• m = 2, n = 0, k = 4 x 10-3M-1s-1
• Rate =4 x 10-3M-1s-1 [NO2]2
[F2] [Mol/L]
0.100
0.200
0.100
Rate [Mol/L*S)
4.0 * 10-5
4.0 * 10-5
16.0 * 10-5
Team problem
• Get into groups of 3 and obtain problem clues
a) Determine the order with respect to each reactant
b) Determine the overall order of reaction
c) Write the rate expression for the reaction.
d) Find the value of the rate constant, k.
Practice Problem
Trial #
1
2
3
4
[IO3-] mol/L
0.10
0.20
0.10
0.10
[I-] mol/L
0.10
0.10
0.30
0.30
[H+] mol/L
0.10
0.10
0.10
0.20
Rate [Mol/L*S)
5.0 x 10-4
1.0 x 10-3
1.5 x 10-3
6.0 x 10-3
• From the data collected above, determine the rate
law for the following equation:
IO3- (aq) + 5I- (aq) + 6H+ (aq)  3I2 (aq) + 3H2O (l)
Practice Problem
Trial #
1
2
3
4
[IO3-] mol/L
0.10
0.20
0.10
0.10
[I-] mol/L
0.10
0.10
0.30
0.30
[H+] mol/L
0.10
0.10
0.10
0.20
Rate [Mol/L*S)
5.0 x 10-4
1.0 x 10-3
1.5 x 10-3
6.0 x 10-3
• From the data collected above, determine the rate
law for the following equation:
IO3- (aq) + 5I- (aq) + 6H+ (aq)  3I2 (aq) + 3H2O (l)
• m = 1; n = 1; o = 2; k = 5M-3s-1
• Rate = 5M-3s-1 [IO3-]1 [I-]1 [H+]2
Homework
• P. 377 practice # 1, 2, 3, 6,
• P. 415 # 15 a-e
Lesson 3
• Reaction mechanisms:
• Ping pong activity
• Assembly line
• Elephant toothpaste demonstration
• Practice problems
Unit Mind Map
Reaction mechanisms
• Rate Laws  determined experimentally
• Equation: Reactants
products
• This actually occurs in a series of steps called
elementary steps
• Analogy: Cooking takes place in several steps
Reaction mechanisms
• What are the chances of the following reaction
occurring in one step?
Hint: collision theory
• 4HBr (g) + O2 (g)  2H2O (g) + 2Br2 (g) actually occurs in 3
separate steps
Reaction mechanisms
4HBr (g) + O2 (g)  2H2O (g) + 2Br2 (g) actually occurs in 3
separate steps:
HBr (g) + O2 (g)  HOOBr (g)
(slow)
HOOBr (g) + HBr (g)  2HOBr (g)
(fast)
2HOBr (g) + 2HBr (g)  2H2O (g) + 2Br2 (g)
(fast)
4HBr (g) + O2 (g)  2H2O (g) + 2Br2
Reaction mechanisms
4HBr (g) + O2 (g)  2H2O (g) + 2Br2 (g) actually occurs in 3
separate steps:
HBr (g) + O2 (g)  HOOBr (g)
(slow)
HOOBr (g) + HBr (g)  2HOBr (g)
(fast)
2HOBr (g) + 2HBr (g)  2H2O (g) + 2Br2 (g)
(fast)
4HBr (g) + O2 (g)  2H2O (g) + 2Br2
Rate law: rate = k[HBr2][O2]
Reaction mechanisms:
working backwards
Overall equation: 4HBr (g) + O2 (g)  2H2O (g) + 2Br2 (g)
Rate law: rate = k[HBr]1[O2]1
Devise a proposed mechanism for this reaction
Reaction mechanisms:
working backwards
Overall equation: 4HBr (g) + O2 (g)  2H2O (g) + 2Br2 (g)
Rate law: rate = k[HBr][O2]
Devise a proposed mechanisms for this reaction
Step 1. using the rate law, write out the ratedetermining (slow) step
Step 2. use the overall equation to determine what
needs to be added to achieve the overall equation
Step 3. cross out intermediates and add up what is left
to produce the overall equation
Reaction mechanisms:
working backwards
4HBr (g) + O2 (g)  2H2O (g) + 2Br2 (g) actually occurs in 3
separate steps:
HBr (g) + O2 (g)  HOOBr (g)
(slow)
HOOBr (g) + HBr (g)  2HOBr (g)
(fast)
2HOBr (g) + 2HBr (g)  2H2O (g) + 2Br2 (g)
(fast)
4HBr (g) + O2 (g)  2H2O (g) + 2Br2
Potential Energy Diagram
Elephant toothpaste
demonstration
Elephant toothpaste
demonstration
Elephant toothpaste demonstration reaction:
2H2O2
Rate = k[H2O2]1[I-]1
O2 + 2H2O
Elephant toothpaste
demonstration
Elephant toothpaste demonstration reaction:
2H2O2
O2 + 2H2O
Rate = k[H2O2]1[I-]1
Proposed Reaction Mechanism:
H2O2 + I-  IO- + H2O
(Slow)
H2O2 + IO-  I- + H2O + O2
(Fast)
2H2O2  O2 + 2H2O
Elephant toothpaste
demonstration
Possible energy potential diagram?
Practice Problem
Propose a possible mechanism for the following
reaction:
2N2O5 (g)  2N2O4 (g) + O2 (g)
r = k[N2O5]1
Practice Problem
Propose a possible mechanism for the following
reaction:
2N2O5 (g)  2N2O4 (g) + O2 (g)
r = k[N2O5]1
Possible mechanism:
N2O5  N2O4 + O
O + N2O5  N2O4 + O2
2N2O5  2N2O4 + O2
(Slow)
(Fast)
Practice Problem
Propose a possible mechanism for the following
reaction:
2NO2 + F2  2NO2F
r = k[NO2]1[F2]1
Practice Problem
Propose a possible mechanism for the following
reaction:
2NO2 + F2  2NO2F
r = k[NO2]1[F2]1
Possible mechanism:
NO2 + F2  NO2F + F
(Slow)
F + NO2  NO2F
(Fast)
2NO2 + F2  2NO2F
Homework
p. 390 #2
p. 391 # 1, 2, 3
Also, complete worksheet.