Module 5
Chemical Kinetics
Lesson 3. Rate LAw and its components
The Rate Law
● The rate constant (k) conveys important information about
the kinetics of a chemical reaction.
● If the rate constant is small, the reaction is likely to
proceed slowly.
● If the rate constant is large, the reaction is likely to
proceed quickly.
● The value of the rate constant, k, depends on the
temperature and describes temperature dependence of the
reaction rate.
Lesson 3. Rate LAw and its components
The Rate Law
●
The units for the rate constant, k, depends on the overall
order of the reaction and must be chosen to balance the units
in the rate law.
❖
❖
❖
❖
The
The
The
The
rate has units of mol L-1 s-1.
concentration has units of mol L-1.
unit of k for a first order reaction is s-1.
unit of k for a second order reaction is L mol-1 s-1.
Lesson 3. Rate LAw and its components
The Rate Law
Lesson 3. Rate LAw and its components
Determination of the Rate Law
The rate law can be determined two ways:
1) Measuring the initial rate of the reaction while adjusting
the concentrations of the various reactants.
1) Using a series of graphs to compare data to various
possible rate laws.
Lesson 3. Rate LAw and its components
Determination of the Rate Law
●
For a reaction with only one reactant, A, the rate of the
reaction is rate = k[A]n.
❖ The common possible orders with respect to A are 0, 1, 2.
●
If the concentration of A is doubled experimentally, the
rate of the reaction will change in a simple and predictable
way:
❖ If n = 0, doubling [A] does not change the reaction rate.
❖ If n = 1, doubling [A] doubles the reaction rate.
❖ If n = 2, doubling [A] quadruples the reaction rate.
Lesson 3. Rate LAw and its components
Determination of the Rate Law: Rate = k[A]n
Zero Order: Rate=k[A]0 = k
● The rate does not depend on the concentration. Whatever you do to the
concentration, the rate will not change.
First Order: Rate=k[A]1=k[A]
● The rate is directly proportional to the concentration.
● If you double the concentration, you double the rate.
● If you triple the concentration, you triple the rate.
● If you halve the concentration, you halve the rate, and so on.
Second Order: Rate=k[A]2
● The rate is proportional to the square of the concentration.
● If you double the concentration, you multiply the rate by four.
● If you triple the concentration, you multiply the rate by nine.
● If you halve the concentration, you divide the rate by four, and so
on.
Lesson 3. Rate LAw and its components
Sample Problem 5:
Consider the following data for the kinetics of this reaction:
2 N2O5(g) → 4 NO2(g) + O2(g)
Determine the rate law and rate constant for this reaction at the
temperature of these experiments.
Lesson 3. Rate LAw and its components
Sample Problem 5:
Lesson 3. Rate LAw and its components
Sample Problem 5:
Lesson 3. Rate LAw and its components
Determination of the Rate Law:
● For a reaction with two reactants, A and B, the rate of
the reaction is rate = k[A]m[B]n.
● To separate the influence of one reactant concentration
from the other, one reactant concentration is held
constant while changing the other to determine its effect
on the rate.
● To determine the order with respect to A and B, at least
three experiments must be carried out.
Lesson 3. Rate LAw and its components
Sample Problem 6:
The study of the kinetics of real systems can be complicated. For example,
there are several ways by which O3 can be converted to O2. One such reaction
is NO2 + O3 → NO3 + O2. Three experiments were run, and the following data
obtained.
Determine the rate law and rate constant for this reaction.
Lesson 3. Rate LAw and its components
Sample Problem 6:
Lesson 3. Rate LAw and its components
Sample Problem 6:
Lesson 3. Rate LAw and its components
Sample Problem 6:
Lesson 3. Rate LAw and its components
Sample Problem 7:
Ozone in the upper atmosphere is depleted when it reacts with nitrogen
oxides. The rates of the reactions of nitrogen oxides with ozone are
important factors in deciding how significant these reactions are in the
formation of the ozone hole over Antarctica. One such reaction is the
combination of nitric oxide, NO, with ozone, O3: NO(g)+O3(g)⟶NO2(g)+O2(g)
This reaction has been studied in the laboratory, and the following rate
data were determined at 25°C.
Determine the
rate law and the
rate constant for
the reaction at
25 °C.
Lesson 3. Rate LAw and its components
Integrated Rate Law
● Because the concentrations of reactants change over
time, the rate law does not let us easily predict the
concentrations or rate of a reaction at some later
time.
● The integrated rate law, derived from the rate law
itself, explicitly determines concentration as a
function of time.
● The form of the integrated rate law depends on the
order of the reaction.
Lesson 3. Rate LAw and its components
Integrated Rate Law
For a zero-order reaction, the rate law is:
The zero-order integrated rate law,
If a plot of [A] versus time is linear, the overall order is zero order and
the slope equals –k.
Lesson 3. Rate LAw and its components
Integrated Rate Law
• A reaction is zero-order if a
concentration versus time is linear.
• The slope of the plot is –k.
plot
of
reactant
Lesson 3. Rate LAw and its components
Integrated Rate Law
Lesson 3. Rate LAw and its components
Integrated Rate Law
Lesson 3. Rate LAw and its components
Integrated Rate Law
Lesson 3. Rate LAw and its components
Integrated Rate Law (Zero- Order Reaction)
Lesson 3. Rate LAw and its components
Integrated Rate Law (First- Order Reaction)
Lesson 3. Rate LAw and its components
Integrated Rate Law (Second- Order Reaction)
Lesson 3. Rate LAw and its components
Sample Problem 8:
The photodissociation of ozone by ultraviolet light in the upper atmosphere is a
first order reaction with a rate constant of 1.0 x 10-5 s-1 at 10 km above the
planet’s surface.
Consider a laboratory experiment in which a vessel of ozone is27
exposed to
UV radiation at an intensity chosen to mimic the conditions at that altitude. If
the initial O3 concentration is 5.0 mM, what will the concentration be after 1.0
day?
27
Lesson 3. Rate LAw and its components
Sample Problem 8:
28
28
Lesson 3. Rate LAw and its components
Sample Problem 9:
Lesson 3. Rate LAw and its components
Sample Problem 9:
Lesson 3. Rate LAw and its components
Sample Problem 9:
Lesson 3. Rate LAw and its components
Sample Problem 9:
Lesson 3. Rate LAw and its components
Sample Problem 9:
Lesson 3. Rate LAw and its components
Lesson 4. Half- Life
●
The half-life (t1/2) of a reactant is the time it takes for
its concentration to fall to one-half its original value.
●
When a reaction has proceeded for one half-life,
concentration of the reactant must be [A]t = 0.5[A]o.
the
Lesson 4. Half- Life
Sample Problem 10:
36
Lesson 4. Half- Life
Sample Problem 11:
37
Lesson 4. Half- Life
Sample Problem 12:
38
Lesson 5. Collision theory
Collision theory is a model used to explain the factors that
affect the rate of chemical reactions.
According to collision theory, chemical reactions occur when
reactant molecules collide with enough energy and proper
orientation.
The energy required for a reaction to occur is called activation
energy.
If reactant molecules do not have enough energy, they will not
be able to overcome the activation energy barrier and the
reaction will not proceed.
Lesson 5. Collision theory
● Increasing the concentration of reactants or decreasing
the size of reactant particles can increase the frequency
of collisions and therefore increase the rate of
reaction.
● Increasing temperature increases the kinetic energy of
molecules and the likelihood of successful collisions.
● Collision theory provides a framework for understanding
how different factors affect the rate of chemical
reactions and can be used to optimize reaction conditions
in the laboratory or industrial settings.
Lesson 5. Collision theory
Factors Affecting Collision Theory
1) Temperature: As the temperature increases, the kinetic energy of the particles
also increases, leading to more frequent collisions and higher collision rates.
1) Concentration: Higher concentration of reactants increases the chances of their
collision and hence, the collision rate.
1) Surface Area: Smaller particle size or increased surface area increases the
number of collisions per unit time.
1) Catalysts: Catalysts increase the collision rate by lowering the activation
energy required for a reaction to occur.
Lesson 5. Collision theory
Applications:
Combustion: In combustion reactions, oxygen molecules collide
with hydrocarbon molecules and have enough energy to break their
bonds and form carbon dioxide and water vapor.
Cooking: When food is cooked, the heat increases the kinetic
energy of the food particles, leading to more frequent
collisions and the formation of new compounds.
Corrosion: Corrosion occurs when metal surfaces are exposed to
oxygen and moisture, leading to the formation of new compounds.
Lesson 6. catalysis
Catalysis is the process by which a catalyst speeds up a
chemical reaction without being consumed in the process.
Catalysis is a fundamental concept in chemistry that is
involved
in
numerous
industrial,
biological,
and
environmental processes.
Lesson 6. catalysis
Types of Catalysis:
Homogeneous Catalysis: Catalyst and reactants are in the same
phase (e.g. gas or liquid).
Heterogeneous Catalysis: Catalyst and reactants are in
different phases (e.g. solid catalyst and gas or liquid
reactants).
Enzymatic Catalysis: Catalysis performed by enzymes, which
are biological catalysts.
Lesson 6. catalysis
Examples on the Types of Catalysis:
Homogeneous Catalysis: Include acid-catalyzed reactions, such as the
reaction between ethanol and acetic acid to form ethyl acetate.
Heterogeneous Catalysis: Include the use of platinum catalysts in
catalytic converters to reduce harmful emissions in vehicles, and the
use of zeolites in petroleum refining to break down complex
hydrocarbons.
Enzymatic
Catalysis:
Include
photosynthesis, and respiration.
enzymes
involved
in
digestion,
Lesson 6. catalysis
How Catalysis Work:
●
Catalysts lower the activation energy of a chemical reaction,
making it easier for the reaction to occur.
●
Catalysts provide an alternative reaction pathway with a lower
activation energy, allowing the reaction to proceed at a faster
rate.
●
Catalysts are not consumed in the reaction
repeatedly, making them highly efficient.
and
can
be
used
Lesson 6. catalysis
Examples of Catalysts:
●
●
●
●
●
●
Enzymatic detergents: Enzymes such as protease and amylase are used as catalysts in
laundry detergents to break down protein and starch stains.
Baking soda: Baking soda (sodium bicarbonate) can act as a catalyst in the process
of baking by releasing carbon dioxide gas, which helps the dough to rise.
Hydrogen peroxide: Hydrogen peroxide can act as a catalyst in the reaction with
yeast to produce oxygen gas, which is used in the process of baking.
Enzymatic cleaners: Enzymes such as protease, amylase, and lipase are used as
catalysts in household cleaners to break down stains and odors from carpets,
upholstery, and clothing.
Oxygen bleach: Oxygen bleach (sodium percarbonate) can act as a catalyst in the
process of stain removal by releasing oxygen gas, which helps to break down the
stain.
Vinegar: Vinegar (acetic acid) can act as a catalyst in the process of cleaning by
reacting with baking soda to produce carbon dioxide gas, which helps to lift dirt
and grime.
Resources
●
●
●
Purdue
University
Department
of
Chemistry.
(n.d.).
Integrated rate laws.
LibreTexts. (2021, August 24). 2.5: Reaction Rate. In
Physical
and
Theoretical
Chemistry
Textbook
Maps:
Supplemental Modules (Physical and Theoretical Chemistry):
Kinetics: Reaction Rates. LibreTexts.
Commission on Higher Education. (2019). General Chemistry
2 Teaching Guide for Senior High School. CHED.