Download 1 Intro / Review : Chemical Kinetics

NAME __________________________________ AP CHEMISTRY SCIENCE: PRACTICES & BIG IDEAS
Science Practices: The student must be able to:
Science Practice 1: use representations and models to communicate scientific phenomena and solve scientific
problems.
Science Practice 2: use mathematics appropriately.
Science Practice 3: engage in scientific questioning to extend thinking or to guide investigations within the
context of the AP course.
Science Practice 4: plan and implement data collection strategies in relation to a particular scientific
question. [Note: Data can be collected from many different sources, e.g., investigations, scientific
observations, the findings of others, historic reconstruction, and/or archived data.].
Science Practice 5: perform data analysis and evaluation of evidence.
Science Practice 6: work with scientific explanations and theories.
Science Practice 7: connect and relate knowledge across various scales, concepts, and representations in
and across domains.
***********************************
Big Idea 1: The chemical elements are fundamental building materials of matter, and all matter can be
understood in terms of arrangements of atoms. These atoms retain their identity in chemical
reactions. (e.g. Atomic Structure, Quantum Mechanics, Matter, Periodicity, Stoichiometry)
Big Idea 2: Chemical and physical properties of materials can be explained by the structure and the
arrangement of atoms, ions, or molecules and the forces between them. (e.g. Gas Laws, Bonding,
Inter-molecular Forces of Attraction, Calorimetry, Acid/Base Strength, Gas Laws)
Big Idea 3: Changes in matter involve the rearrangement and/or reorganization of atoms and/or the transfer of
electrons. (e.g. Stoichiometry, Balancing, Redox Reactions, Acid/Base Reactions, Exothermic and
Endothermic Reactions)
Big Idea 4: Rates of chemical reactions are determined by details of the molecular collisions. (Kinetics)
Big Idea 5: The laws of thermodynamics describe the essential role of energy and explain and predict the
direction of changes in matter. (e.g. Thermodynamics, Entropy, Enthalpy, Free Energy)
Big Idea 6: Any bond or intermolecular attraction that can be formed can be broken. These two processes are in
a dynamic competition, sensitive to initial conditions and external perturbations. (e.g. Equilibrium,
Le Chatlier’s Principle, Weak Acids/Bases)
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A Comparison of The Big Ideas of AP Chemistry & Mr. D’s Major Reasons from Honors Chemistry
1
AP Chemistry
The chemical elements are fundamental
building materials of matter, and all
matter can be understood in terms of
arrangements of atoms. These atoms
retain their identity in chemical
reactions.
Honors Chemistry
Law of the Conservation of Mass, Energy & Charge (1)
Chemical and physical properties of
materials can be explained by the
structure and the arrangement
of atoms, ions, or molecules and
the forces between them.
The Physical Characteristics of Compounds are influenced by
the Ionic Bond or IMF (5) and
Changes in matter involve the
rearrangement and/or reorganization
of atoms and/or the transfer of electrons.
The Concept of Charge (3) and
4
Rates of chemical reactions are
determined by details of the
molecular collisions.
Collision Theory (6) Any factor which alters / affects the
number of effective collisions between the reacting particles
will help increase (or inhibit) the rate of a reaction. A
reaction rate may be controlled, by controlling the number of
effective collisions between the species capable of making
new bonds. The reacting species must have sufficient energy
and proper orientation.
5
The laws of thermodynamics describe
the essential role of energy and explain
and predict the direction of changes
in matter.
Potential Energy (2) and
Any bond or intermolecular attraction
that can be formed can be broken.
These two processes are in a dynamic
competition, sensitive to initial conditions
and external perturbations.
Potential Energy (2) and
2
3
6
Substances can be described as chemically reactive or
unreactive* due to a combination of competing factors. (7)
The Siblings: Effective Nuclear Charge / Shielding Effect /
Electronegativity (4)
Substances can be described as chemically reactive or
unreactive* due to a combination of competing factors. (7)
• charge neutrality in that, #p = #e• a best-case proximity to a stable valence octet, which is
often achieved by making new bonds
• lowering energy (decreasing enthalpy) which is often
accomplished by making bonds, hybridizing orbitals,
achieving a best-case molecular geometry &/or
• increasing intermolecular forces of attraction
• increasing entropy
The Physical Characteristics of Compounds are influenced by
the Ionic Bond or IMF (5)
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NAME ______________________________________________ AP NOTES: UNIT 1 KINETICS (1)
Big Idea 4: Rates of chemical reactions are determined by details of the molecular collisions.
Enduring understanding 4.A: Reaction rates that depend on temperature and other environmental factors are determined by measuring changes in
concentrations of reactants or products over time.
Essential knowledge 4.A.2: The rate law shows how the rate depends on reactant concentrations.
Essential knowledge 4.A.1: The rate of a reaction is influenced by the concentration or pressure of reactants, the phase of the reactants and products,
and environmental factors such as temperature and solvent
Essential knowledge 4.A.3: The magnitude and temperature dependence of the rate of reaction is contained quantitatively in the rate constant.
Enduring understanding 4.B: Elementary reactions are mediated by collisions between molecules. Only collisions having sufficient energy and proper
relative orientation of reactants lead to products. Enduring understanding 4.C: Many reactions proceed via a series of elementary reactions.
Essential knowledge 4.B.1: Elementary reactions can be uni-molecular or involve collisions between two or more molecules.
Essential knowledge 4.B.2: Not all collisions are successful. To get over the activation energy barrier, the colliding species need sufficient energy.
Also, the orientations of the reactant molecules during the collision must allow for the rearrangement of reactant bonds to form product bonds.
Essential knowledge 4.B.3: A successful collision can be viewed as following a reaction path with an associated energy profile.
Enduring understanding 4.D: Reaction rates may be increased by the presence of a catalyst.
Essential knowledge 4.C.1: The mechanism of a multistep reaction consists of a series of elementary reactions that add up to the overall reaction.
Essential knowledge 4.C.2: In many reactions, the rate is set by the slowest elementary reaction, or rate-limiting step.
Essential knowledge 4.C.3: Reaction intermediates, which are formed during the reaction but not present in the overall reaction, play an important
role in multistep reactions.
Essential knowledge 4.D.1: Catalysts function by lowering the activation energy of an elementary step in a reaction mechanism, and by providing a
new and faster reaction mechanism.
Essential knowledge 4.D.2: Important classes in catalysis include acid-base catalysis, surface catalysis, and enzyme catalysis.
Questions on Kinetics appear on every AP exam. Questions may appear on the multiple choice
(2 to 3 of 60 questions) section as well as in both of the free-response sections (at least 1 of the
7). You should expect to analyze data, plan an experiment, and construct an explanation based
on evidence collected in an experiment.
So, let’s pick up the conversation, regarding KINETICS, by reviewing our work in Honors. From that
review we will build.
Mechanisms
Kinetics Definition
moving
on
to
Average Rate
Collision Theory
Instantaneous Rate
Factors Affecting
Rate Laws
Kinetics
P.E. Diagrams
Change in Enthalpy
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NAME _____________________________
AP NOTES: UNIT 1 KINETICS (1)
I) Kinetics (DEFINITION): studies of the rate of a reaction
the change over time of bond breaking &/or of bond making
A) Chemical Kinetics deals with the speed at which changes in the molar concentration of the
reactant(s) [due to consumption] and/or the speed at which changes in the molar concentration of
the product(s) occur.
1) The rate of consumption of reactants does not necessarily equal the rate of product formation
2) The speed or average rate, of a process may be defined as the change in a given quantity over
a specific period of time. Rates are generally discussed as positive values.
a) For a chemical reaction that is essentially the change in molar concentration of a
reactant (or product) per unit time.
b) So, assume a reaction:
R→P
Consider the average rate of consumption of reactant R.
Avg Rate = -Δ[R]
Δt
Or, in terms of the average rate of formation of product P, Avg Rate = Δ[P]
Δt
Anyone notice a difference
between the last 2 equations?
i) We’ll discuss this particular average rate equation, in more detail later
… but for now it is critical that you grasp that average rate is
*the change in the molar concentration of a species, per unit time
3) As a point of order … many students confuse the energy release (change in enthalpy) with the
“speed” of the reaction. Yet, thermodynamic does not really inform us of the rate of reaction.
Recall from your first year course, that some thermodynamically favored reactions can still be
excruciatingly slow to occur… e.g. conversion of diamond to graphite (sp3 to sp2)
decomposition of benzene to individual C and H atoms
Some thermodynamically favored reactions occur very quickly … such as the proton transfer
in a neutralization reaction.
4) Issues which affect the kinetics of a chemical reaction included, but are not necessarily
limited to:
a) the reaction’s stoichiometry
b) the thermodynamics (the relationship between “heat and work”)
c) the factors of concentration, temperature, surface area, and catalysis
d) the reaction mechanism
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5) With kinetics we must concern ourselves with the Reaction Mechanism
a) Study the rather simple looking reaction of cellular respiration in a eukaryotic cell:
C6H12O6 + 6 O2  6 CO2 + 6 H2O + ~ 38 ATP
but in reality it is a multi-step process:
Check Out: http://evillusion.files.wordpress.com/2008/01/krebs-citric-acid.jpg
b) A reaction mechanism is essentially a step-by-step description of what occurs
at a molecular level, as a chemical reaction proceeds.
i) Each step of a reaction mechanism is known as an elementary process &
describes production of relatively unstable but representable
intermediary products which


exhibit changes in molecular geometry or
is perturbed by the addition or omission of another interacting species.
ii) Hence, a reaction mechanism is often multiple elementary processes, which
express our best guess as to how a reaction proceeds from initiation to
completion. We’ll see this again…a bit later on.
http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Rate_Laws/Reaction_Mechanisms/Reaction_Mechanisms
6) Reactions with multiple elementary processes, may very well have what is termed a
a slow step and a fast step.
i) The slow step is the rate determining step, when all is said and done. That is, the
reaction cannot proceed at a faster rate than the slowest step of the mechanism.
ii) Think about the passage of water through a series of funnels….HUH?
Oh yeah, it’s time for a (simple) demo!
7) All of these issues (and many more) are affected by one of my major reasons, called
the Collision Theory.
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II) The Collision Theory:
A) The Collision Theory is a series of ideas which help explain or predict the kinetics of a reaction.
Any factor which increases (or decreases) the number of effective collisions between the reacting
particles will help speed up (or slow down) a reaction. WE CAN CONTROL A REACTION, BY
CONTROLLING THE NUMBER OF EFFECTIVE COLLISIONS.
(THIS IS THE VERY HEART OF BIG IDEA 4)
1) Effective Collision Any collision between two reactant species which results in a
* a new bond being produced (products being made)
2) There are 2 necessary conditions for an effective collision:
a) * sufficient kinetic energy
b) * proper orientation
3) Complete the statement: The greater the number of * effective collisions
the greater the rate of a chemical reaction.
4) Activation Energy: A compilation of ideas related to:
The minimum amount of energy required to break the bonds of the reacting species, in
preparation for new bonds to be made or * the extra energy needed to be added
The bond energy of the reacting species AND
The energy required to raise the energy levels of the species to a point of highest potential
energy (called the activated complex)…where those intermediate products exist.
B) Collision Theory assumes that most reactions occur via reaction mechanism, in which one or more
reactant particles collide, often producing unstable, intermediate products, which then re-organize
or continue to collide via elementary processes moving towards a more stable end product.
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C) We will deal with the issue of energy, in just a bit, when we attack the factor of temperature. But,
let’s take a closer look at the issue of orientation.
Consider the reaction between ethene and hydrogen bromide, to produce, bromoethane.
C2H4
(Rewritten using condensed formulae):
CH2 = CH2
+
HBr → C2H5Br
+ H-Br → CH3CH2Br
1) By analyzing the condensed formulae, what must happen in order to synthesize the
bromoethane?
* the double bond must be broken,
bonds must break, new bonds must be made…
2) Given that the *double bond of the ethene must be broken, how can HBr accomplish that?
* it collides, it must draw electrons away from the ethene, the electropositive H
must get close enough to the double bond portion of the molecule…
3) What might that look like?
remove
Citation: http://www.chemguide.co.uk/mechanisms/eladd/symhbr.html#top
Notice that the more * electropositive hydrogen, must attack the double bond (proper
orientation). When the collision is between the area of the double bond and the partially
negative bromide, the electrons of the double bond will not be drawn away from the
carbon’s, thus the bond will not break. Proper orientation is an important piece of an
effective collision.
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III) Fundamental factors which affect the ability to achieve the requirements of effective collisions
A)5 factors
1) Temperature:
Basic Idea: As a rule, an increase in temperature * increases the rate of both
endothermic and exothermic reactions.
However, endothermic reactions are increased a greater extent
Reason: An increase in temperature, indicates that more energy has been added to the
reacting species. This increase in energy is reflected in the faster movement of
the species.
In theory then, faster moving species have a statistically greater chance of
colliding with other reactive species with sufficient energy and orientation.
a) When we consider how ideas of Collision Theory & Kinetic Molecular Theory
overlap we may begin to understand the impact of temperature upon the kinetics of a
reaction, with a Maxwell-Boltzmann energy diagram
Increasing temperature (average kinetic energy) corresponds to faster and faster
motions of the species. … As a rule of thumb, every 10°C temperature increase,
doubles the rate of a chemical reaction.
This means that a greater number of energetic collisions occur at higher temperatures
A Generalized Maxwell-Boltzmann Energy Diagram: With a Constant Number of Gas Molecules
At lower temperature, only a few molecules under this curve have
enough energy to collide effectively to make bonds (shaded area in the
lower right.)
When temperature is increased, a greater number of molecules
possess sufficient average K.E. to collide effectively and to bond
(larger shaded area)
Note: Point A represents the necessary activation energy needed to
break the reactant species’ bonds and thus initiate a reaction.
Check Out: https://www.youtube.com/watch?v=66AxPceP5QY (up to 2:45 only)
Check out: For the shift in graph: https://www.youtube.com/watch?v=YnHIfqUZi48
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Please be sure that you grasp that as the absolute temperature of the system is increased,
the Maxwell-Boltzmann energy diagram, shows a “flattening of the curve and a shift to the
right.
Consider a closed system with a constant number of gas molecules. The area under the curve
represents that total number of species.
The curve(s) flatten(s) out, because the number of particles is fixed, yet more and more exhibit
an increased molecular speed.
http://ibchem.com/IB/ibnotes/full/sta_htm/Maxwell_Boltzmann.htm
Consider this: Go to Doc Brown’s Website: http://www.docbrown.info/BBCbasic/kpts.htm
and run/save the program. Look at #3 on the dialogue box which opens.
Try a temperature of 400 K and then try 700 K
TN:
1) Temperature (continued…)
b) The opposite of every exothermic reaction is a(n) *endothermic reaction
b) Endothermic reactions are affected MORE by changes in temperature than
exothermic reactions, because changes in temperature may change indicate changes
in the system’s energy. This energy is a necessary part of endothermic reactions.
c) Don’t forget, when you read: average kinetic energy you think of * temperature
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2) Concentration (note: for gases, changes in concentration = changes in pressure)
a) symbol for molar concentration is a set of brackets: [ ] Thus the when discussing the
molar concentration of dioxygen gas, a chemist may write the phrase as [O2]
b) M or M = moles/Liter of solution
c) How do changes in concentration affect collisions?
i) With a greater concentration, the likelihood of appropriate orientation
increases, statistically, as does the probable number of actual collisions. Each
leads to a statistically greater number of effective collisions.
d) Note: Increasing the pressure on a gas is synonymous with increasing its
concentration, due to an increase in the number of moles per unit volume
(Reference: Molarity: M = moles / Liter the same # of species, in a smaller volume,
effectively increases the molar concentration … and the density of the gas…
What will such an increase, do, to the likelihood of effective collisions?
3) surface area (particle size): Greater surface area allows for greater collisions and greater
probability for proper orientation by exposing more of the reactive species to the other
reactant(s)
Action Sought On Dust Explosions
The Chemical Safety & Hazard Investigation Board (CSB) recommended last week that federal agencies take
action to protect workers from industrial dust explosions and fires. The recommendation grew from a two-year
investigation of the hazards of combustible dust. In 2003, 14 workers were killed and another 81 were injured
in three industrial accidents that were fueled by combustible dust. Between 1980 and 2005, CSB adds that 119
workers were killed and 718 were injured in nearly 300 dust-related accidents. Combustible dust accumulates
in many chemically related manufacturing facilities, CSB notes, and once ignited, the dust leads to explosive
accidents. Chemical and Engineering News 13 Nov 2006 p. 30
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4) nature of the reactants (dissolved ionic compounds vs. molecular (covalent) compounds)
Basic Idea: Solutions of ionic compounds react * faster (form new bonds)
combination than solutions of * molecular (covalent)
upon
substances.
Recall: There are both molecular * inorganic and molecular organic compounds.
Reason: Often, and in the simplest of reactions, WATER can disrupt the ionic bond
and thus breaks the compound into ions (electrolytes), ready to make new bonds.
Water can NOT break the bonds of most covalent compounds. The consequence is
that ionic compounds (pre-dissolved in water) react faster, than dissolved covalent,
compounds because the ionic bonds are * broken already (dissociated))
Whereas, the bonds of dissolved covalent compounds must first * be broken
and then the species can * react
Think about it:
In the beakers =
Pb(NO3)2(aq) +
Pb2+(aq) + 2 NO31-(aq)
2 KI(aq)

PbI2(s)↓ + 2 KNO3(aq)
2 K1+(aq)+ I1-(aq) 
◘ Do you see how the ionic bonds of the reactants are broken already?
◘ Note: As a comparison to the Bosch-Haber Process, this ionic reaction occurs at room
temperature and standard pressure…
Think about it: Bosch-Haber Process:
N2(g) + 3H2(g) 2NH3(g)
While not in aqueous solution, the gaseous reactants are bonded strongly enough that this
reaction requires an iron-based catalyst, (mixed with a potassium hydroxide promoter), a
temperature of 400°C to 450°C and approximately 200. atmospheres of pressure.
5) Catalysts:
a) Catalysts work by allowing the reaction to proceed along an * alternate pathway
or mechanism that has a lower activation energy requirement (bond breaking thus
occurs “faster).
b) Catalysts are often written above the reaction arrow because * they are not consumed
during the reaction (they are regenerated). They are neither reactants nor products.
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c) enzymes are biological catalysts (remember the names of enzymes often end in –ase)
d) many metals (Pt, Zn and even Na) and compounds may be effective catalysts.
*A catalyst must be a substance
e) Note: Energy (okay, “heat”) IS NOT A CATALYST!!
f) The opposite of a catalyst is an inhibitor
g) The addition of a catalyst does NOT change a reaction’s H (change in enthalpy)
NOR the energy of the reactants & products. The catalyst enables more species to
reach a required activation energy so that the reaction proceeds faster. Often this
is done by some alternate mechanism.
h) We will expand this conversation when we work on Potential Energy Diagrams.
PRACTICE:
For questions 1 and 2 use your grasp of chemistry the given reaction equation and the data:
Zn(s) + 2HCl(aq)
Molarity of HCl
H(2)(g) + ZnCl2(aq)
Description of Zn
Temperature
A
2M
10 grams in small lumps
26 C
B
1M
10 grams of fine powder
30 C
C
1M
10 grams in small lumps
26 C
D
2M
10 grams of fine powder
30 C
Which of the listed conditions will result in the fastest generation of hydrogen gas? *D Defend your answer:
_____________________________________________________________________________________
___________________________________________________________________________________
Which set of conditions probably results in the slowest generation of hydrogen? *C
Why? ________
____________________________________________________________________________________
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For questions 3 - 5 use your grasp of chemistry and the following passage.
ISSUE: A kitchen grease fire can range (get it?...a “range???”) from a troublesome mishap to a home disaster.
Grease can reach its flash point, relatively quickly. The flash point is the temperature at which the
vapors spontaneously combust in the presence of atmospheric oxygen. The combustion of grease
produces combustion products, such as carbon dioxide and water, and some nasty smelly once such
as acrolein.
But the take home message is that grease will ignite when there is sufficient oxygen gas present
and when the grease is sufficiently vaporized at high temperature. The vaporization of the grease
is important, for you will recall that organic compounds combust, when in the gaseous phase.
Throwing baking soda (sodium hydrogen carbonate) onto the fire, often puts out the fire. Baking sod
is an absorbent, often finely divided inorganic compound and it decomposes under high heat to
produce sodium carbonate, water and carbon dioxide, according to the reaction equation:
683kJ + *2 NaHCO3(s)  Na2CO3(s) + H2O(𝓁 ) + CO2(g)
3) Write the reaction equation for the decomposition of sodium hydrogen carbonate.
4) Once the reaction equation is written, balance it by inspection, so that coefficients represent
the simplest whole number ratio.
5) Using the ideas of collision theory and the factors that help control the kinetics of a
reaction, please explain how the kinetics of the combustion reaction are controlled (diminished in
this case) by the addition of baking soda? (How does its absorbency and decomposition help reduce
the rate of combustion?)
* Baking soda can absorb liquid grease, preventing the vaporization process. This limits one of the
reactants. Baking soda decomposes under high energy and produces a denser carbon dioxide,
which is present just long enough to displace (push out) dioxygen gas (thus limiting another
reactant).
*2 6) A student is trying to find the activation energy of a reaction. What technique would be most effective?
1) Vary the concentrations of all reactants in several different experiments at a constant temperature
2) Choose one set of reactant concentrations and run the same reaction at different temperatures
3) Use a spectrophotometer to monitor the time required for a reaction to go to completion.
4) Add a catalyst to the reaction to see how much quicker the reaction goes.
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* 1 7) The mechanism below shows how ethanol (C2H5OH) is converted into ethane (C2H4), catalyzed by
the activity of an acid (HA).
H H
| |
H
H–C–C–O
| |
H H
H H
| |
H
H–C–C–O+
| |
H
H H
H–A
AH
|
H–C–C–O-H
| |
H H
H
H
C=C
H
H
H–A
1)
2)
3)
4)
HA
HA
HA
HA
reacts with
reacts with
reacts with
reacts with
CH3CH2OH to produce CH2CH2
CH3CH2OH to produce CH2CH2
CH3CH2OH to produce CH2CH2
CH3CH2OH to produce CH2CH2
+
+
+
+
H2O + HA
H3OA
HA
H2O
IV) Potential Energy Diagrams:
LEARNING GOALS : As a first year chemistry student you should be able to:




identify a potential energy diagram as representing either an exothermic or an endothermic reaction.
label any portion of a potential energy diagram.
judge, using a diagram, as to whether the reactants are more or less stable than the products.
sketch a rough potential energy diagram for any given chemical equation.
 The Potential Energy Diagram is a qualitative visual means of representing the potential energy changes
and the progress of a chemical reaction, with potential energy as a function of time.
 The Change in Enthalpy (new term) is the difference in the potential energy of the products and
reactants. Its symbol is ∆H or q
 Since the graph is concerned with the "time for a reaction", it has a direct link to the study of kinetics.
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 Each diagram is divided into:
 potential energy of the reactants
 activation energy (the energy gained by the reactants
needed to break the existing bonds)
 the potential energy of the activated complex
 the energy lost due to bond formation
 the potential energy of the products.
 the change in enthalpy also called the heat of reaction
(∆H, change in enthalpy)
A) Putting all the pieces together:
PE
Reaction Coordinate
35
PE
Reaction Coordinate
NOT:
PE
Reaction Coordinate
B) Change in Enthalpy and Generalized /Labeling Potential Energy Diagrams
1) Every chemical reaction is a combination of exothermic and endothermic processes.
The reaction is classified according to whichever process is greater. There is a
"competition of sorts" between the acts of bond breaking and bond formation.
2) The H or the "q" of a reaction is equal to the difference in energy possessed by the
products and reactants...it is the * change in enthalpy or heat of reaction
the net change in energy of a reaction
a) H= Hproducts - Hreactants
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Regents Table I: Heats of Reaction at 101.3 kPa and 298 K
∆H (kJ)*
Reaction
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(𝓁)
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(𝓁)
2 C8H18((𝓁) + 25 O2(g)  16 CO2(g) + 18 H2O(𝓁)
2 CH3OH (𝓁) + 3 O2(g)  2 CO2(g) + 4 H2O(𝓁)
C2H5OH(𝓁) + 3 O(g)  2 CO2(g) + 3 H2O(𝓁)
C6H12O6(s) + 6 O(g)  6 CO2(g) + 6 H2O(𝓁)
2 CO(g) + O2(g)  2 CO2(g)
C(s) + O2(g)  CO2(g)
4 Al(s) + 3 O2(g)  2 Al2O3(s)
N2(g) + O2(g)  2 NO(g)
N2(g) + 2 O2(g)  2 NO2(g)
2 H2(g) + O2(g)  2 H2O (g)
2 H2(g) + O2(g)  2 H2O (𝓁)
N2(g) + 3 H2(g)  2 NH3(g)
2 C(s) + 3 H2(g)  C2H6(g)
2 C(s) + 2 H2(g)  C2H4(g)
2 C(s) + H2(g)  C2H2(g)
H2(s) + I2(g)  2 HI(g)
𝑤𝑎𝑡𝑒𝑟
KNO3(s) →
𝑤𝑎𝑡𝑒𝑟
NaOH(s) →
𝑤𝑎𝑡𝑒𝑟
NH4Cl(s) →
K+(aq) + NO3- (aq)
-890.4
-2219.2
-10943
-1452
-1367
-2804
-566.0
-393.5
-3351
+182.6
+66.4
-483.6
-571.6
-91.8
-84.0
+52.4
+227.4
+53.0
+34.89
Na+(aq) + OH- (aq)
-44.51
NH4+(aq) + Cl- (aq)
+14.78
𝑤𝑎𝑡𝑒𝑟
NH4NO3(s) →
𝑤𝑎𝑡𝑒𝑟
NaCl(s) →
+25.69
NH4+(aq) + NO3- (aq)
+3.88
Na+(aq) + Cl- (aq)
𝑤𝑎𝑡𝑒𝑟
-48.83
LiBr(s) →
Li+(aq) + Br- (aq)
-55.8
H+(aq) + OH- (aq)  H2O(𝓁)
*The ∆H values are based on molar quantities represented in the equations. A minus
sign indicates an exothermic reaction.
3) Exothermic Reaction:
The products of an exothermic reaction are * more stable (less likely to react than the reactants
The products are * lower
in energy
The change in enthalpy of an exothermic reaction is written with a *negative
∆H
heat of reaction
e.g) 2 NO(g) + O2(g) → 2 NO2(g) + 112.86 kJ
sign.
∆H = * - 112.86 kJ
2 C7H5N3O6(s)  3 N2(g) + 12 CO(g) + 5 H2(g) + 2C(s) + 2,092kJ ∆H =* -2,092 kJ
37
4) ENDOthermic Reaction:
The products of an endothermic reaction are * less stable
The products are * higher (greater)
than the reactants
in energy
The change in enthalpy of an endothermic reaction is written with a * positive
∆H
heat of reaction
53 kJ + H2(g) + I2(g)  2 HI(g)
sign.
∆ H = * + 53 kJ
5) For synthesis reactions we can speak of “heat of formation”. This is the energy released or
absorbed for the production of ONLY 1 mole of compound from its elements.
This is the critical attribute of the definition
53 kJ + H2(g) + I2(g)  2 HI(g)
∆H = +53 kJ
but
∆Hf = *+26.5 kJ
N2(g) + 3H2(g)  2 NH3(g) + 91.8 kJ
∆H = -91.8 kJ
but
∆Hf = *-45.9 kJ
C(s) + O2(g)  CO2(g) + 393.5 kJ
∆H = -393.5 kJ and
∆Hf = *-393.5
66.4 kJ + N2(g) + 2 O2(g)  2 NO2(g)
∆H = +66.4 kJ
but
∆Hf = *+ 33.2
2 Al(s) + 3 O2(g)  2 Al2O3(s) + 3351 kJ
∆H = -3351 kJ
but
∆Hf = *-1,675.5 kJ
2 C(s) + 3 H2(g)  C2H6(g) + 84 kJ
∆H = -84 kJ
and
∆Hf =*-84 kJ
C) Labeling Potential Energy Diagrams (continued)
1) Activation Energy * the energy required to break the bonds of the reactants and raise their
potential energy to the activated complex (The endothermic portion of a reaction)
a) Activation Energy of the Reverse Reaction: * Represented by the line from the
products to the activated complex.
38
2) Activated Complex A point at which intermediate associations are made. The species are at
the highest potential energy ... in that bond breaking has been achieved and intermediate
species exist temporarily... just before, all out new bond formation occurs. This is a point of
transition.The species are not isolated easily and are often hypothesized.
See: http://en.wikipedia.org/wiki/Sulfur_dichloride for an application of the concept of the
“intermediate”
3) Energy Lost during Bond Formation (There is no "technical" name – but it is a useful
concept)
a) ALL bond formation * releases
regardless of the overall reaction.
energy (it is *exothermic
)
D) Effects of a catalyst and P.E. Diagrams
The Effect of a Catalyst on Reaction Rate*
1400
uncatalyzed
1300
1200
catalyzed
potential
energy 1100
(kJ)
1000
900
800
1.0
1.5
2.0
2.5
3.0 3.5 4.0
Time (minutes)
4.5
5.0 5.5
(should be reaction coordinate ... see *)
A catalyst works by allowing the reaction to occur via an alternate pathway, one which has * a lower
activation energy. The level of activation energy and the activated complex are lessened … but
H and the heat of the reactants and heat of the products remain unchanged.
*Please note that I have taken liberties with the interpretation of the reaction coordinate by equating it with time. I have done this only for the
illustrative purpose of demonstrating a lesser amount of time in order to initiate a reaction, with the use of a catalyst. The concept of the reaction
coordinate is a far more complex issue and as a rule should not be simplified as time. Again, I do so here only to emphasize the impact of the
catalyst on the rate of reaction .
39
E) Practice Labeling P.E. Diagrams
1) Given the following reaction equation and potential energy diagram answer the questions.
An answer may be used once, more than once or not at all.
4Al(s) + 3O2(g)  2 Al2O3(s) + 3,351 kJ
2
3
Potential Energy
4Al + 3O2
4
1
2 Al2O3
5
Reaction Coordinate
_____ a) Which arrow represents the energy required to initiate the reaction by breaking
the bonds of the reactants?
_____ b) Which arrow represents the heat of reaction (∆H or change in enthalpy)?
_____ c) Which arrow represents the chemical energy possessed by the reactants?
__&__ d) Which arrows would be most affected by the addition of a catalyst?
_________e) Which type of reaction is represented: exothermic or endothermic?
______ f) Which arrow represents the value: 3,351 kJ?
______ g) Which is / are more stable?
Al & O2 or Al2O3
___________h) According to the diagram, was more energy absorbed from the
surroundings or released into the surroundings, as the reaction occurred?
2) Given the following reaction equation , answer each question. 53.0 kJ + H2(g) + I2(g) 2 HI(g)
a)
b)
c)
d)
Sketch the potential energy diagram.
Label Hreactants and the Hproducts
Label the area which represents the 53.0 kJ
Identify the area representing the activation
energy
40
3)
a ) The reaction is
Exothermic / Endothermic
5
4
2 Elost by new bonding
b) Ereactants is represented by arrow
c) E products is represented by arrow
Ep
3
1
d) Eactivated complex is rep. by arrow
e) Eactivation is represented by arrow
6
f) Heat of reaction (H) is rep. by arrow
Reaction Coordinate
4)
2
a ) The reaction is
Exothermic / Endothermic
3
4 E lost by new bonding
b) Ereactants is represented by arrow
c) E products is represented by arrow
Ep
1
d) Eactivated complex is rep. by arrow
5
6
Reaction Coordinate
e) Eactivation is represented by arrow
f) Heat of reaction (H) is rep. by arrow
41
NAME _________________________________________
BASIICS OF KINETICS PRACTICE
DIRECTIONS: Study first and then attack the worksheet. All of the answers are given at the end.
For 1-4 one or more of the responses given is (are)
correct. Using your understanding of chemistry
decide which of the responses is (are) correct. Then
choose :
1)
2)
3)
4)
5)
when only I is correct
when only II is correct
when only I and II are correct
when only II and III are correct
when I, II, and III are correct
___1) Given the following reaction, which
statement(s) is (are) true ?
C(s) + O2(g)  CO2(g) + 393 kJ
I)
___4)Given: 80.0 kJ + X2 + 2Y + 8Z  2 XYZ4
According to the above reaction equation,
I) the ∆H of the forward reaction is + 80 kJ
II) the reaction is exothermic
III) products are more stable than reactants
For questions 5 - 18, only a single answer per
question is most correct.
___5) The "heat of reaction", ∆H, is equal to
1) Hproducts - Hreactants
2) Hproducts + Hreactants
The forward reaction is exothermic
3) Hproducts / Hreactants
II) The product is more stable that the
reactants
III) The heat of reaction (∆H) for the
forward reaction is +393 kJ
___2) Which of the following statements, regarding
the role of catalysts is (are) true?
I) A catalyst reduces the activation energy
of a reaction
II) A catalyst reduces the value for the heat
of reaction (∆H)
III) A catalyst gets used up and converted
into product
___3) Given:
4) (Hproducts) (Hreactants)
___6) Given the reaction:
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
The reaction occurs more slowly when a single
lump of zinc is used when compared to the rate of
the reaction when the same mass of powdered zinc
is used. This is best explained because the
powdered zinc:
1)
2)
3)
4)
is more concentrated
has a greater surface area
requires a lower activation energy
generates more heat
2 Pb(s) + O2(s)  2 PbO(s) + 438 kJ
According to the above thermochemical equation,
I)
the forward reaction is exothermic
II) the heat of reaction equals -438 kJ
III) products are more stable than reactants
42
___7) Given the reaction:
A + B  C + D
The reaction most likely will occur at the greatest
rate if A and B represent:
a) metal elements and nonmetal elements
in the solid phase
For questions 12-16 use the given reaction and potential
energy diagram.
N2(g) + O2(g) + 66 kJ  2 NO2(g)
B
C
b) ionic compounds in the solid phase
D
E
c) organic compounds in the solid phase
d) solutions of ionic compounds
___8) Given
A
A + B  C + 100 kJ
the potential energy of the products when
compared to the potential energy of the
reactants is:
a) less, so the reaction is exothermic
b) more, so the reaction is endothermic
c) less, so the reaction is endothermic
d) more, so the reaction is exothermic
___9) According to Table I which compound
forms via an endothermic reaction ?
a) carbon dioxide
b) aluminum oxide
c) nitrogen monoxide
d) ammonia
___10) The minimum energy required to break
bonds and initiate a chemical reaction is
___12) Which arrow(s) is/are the equivalent(s) of
the energy of activation for the forward
reaction?
a) A + E
b) B + A
___13) The heat of formation is the heat of reaction
when only 1 mol is produced. According
to the given equation, the heat of formation
of NO2 is equal to :
a) -33 kJ
b) -132 kJ
___11) Assume that the potential energy of the
products in a chemical reaction is 60. kJ.
The reaction which formed these products
would be exothermic if the potential
energy of the reactants were equal to:
a) 50. kJ
b) 20. kJ
c) 30. kJ
d) 80. kJ
c) +132kJ
d) +33 kJ
___14) According to the diagram, the heat of
reaction is best represented by arrow :
a) A
a) activated complex c) change in enthalpy
b) heat of reaction
d) activation energy
c) C + D
d) just arrow E
b) B
c) C
d) D
___15) The activation energy of the REVERSE
reaction is represented by arrow
a) A
b) B
c) C
d) D
___16) According to the diagram, the reaction is
______ and the products are _____ stable.
a)
b)
c)
d)
endothermic, more
exothermic , more
endothermic, less
exothermic, less
43
For question 17 use the following table and equation. The concentration and surface area of X remain constant,
but the concentration and surface area of Y may be changed in each trial.
X(l) + Y(s) + heat  Z(g)
Temp (◦C)
Trial Concentratio Surface
n of Y (M) area of Y
a
1
cube
b
1
powdered
c
2
powdered
d
2
cube
100
50
150
100
____17) Which Trial, most probably has the fastest rate of reaction?
____18) Using the provided information, the formation of which compound is best represented by the potential
energy diagram?
Compound
P.E
1
2
3
4
ethene
ethane
nitrogen dioxide
nitrogen monoxide
∆H kJ/mol
(at 101.3 kPa and 298 K)
+52.4
-84.0
+66.4
+182.6
Reaction Coordinate
Go on to the next page
44
19) Given: 4 Al(s) + 3 O2(g)  2 Al2O3(s) + 3,351 kJ
For the above reaction:
a) what is the change in enthalpy?
* -3,351
kJ
b) what is the heat of formation for aluminum oxide? * -1676
kJ
c) what would be the change in enthalpy were 8.0 moles of aluminum oxide produced? * -13408
kJ
d) what would be the heat of formation were 8.0 moles of aluminum oxide produced?
e) sketch the potential energy diagram for the production of aluminum oxide.
potential
energy
Reaction Coordinate
e) sketch the effect of a catalyst on the production of the aluminum oxide, using the above
potential energy diagram.
f) the reaction is
endothermic
/
exothermic
g) the products are less stable than the reactants
(select one)
/
more stable than the reactants (select one)
ANSWERS :
1) c the heat is to the right of the  , so it's an exothermic reaction. Since it is exothermic carbon dioxide would be more stable
than the reactants (if you draw a quick p.e. diagram you can see that the products of exothermic reactions are lower in energy
thus more thermodynamically favored)
However, the change in enthalpy is NOT a POSITIVE 393 kJ. The ∆H must be a NEGATIVE 393
2) a catalysts affect the activation energy and therefore the activated complex - but nothing else. See your Potential Energy notes
under catalysts. Catalysts are not "consumed" or used up.
3) e
4)
5)
6)
7)
The energy (kJ) is to the right of the -----> , so it's an exothermic reaction.
Since it is an exothermic reaction, the products are lower in energy (more “stable” or more thermodynamically favored)
a The total heat is +80 kJ, but that means it is an endothermic reaction
a definition : see your notes
2
4 the bonds are already broken and you have free moving ions. Reactions of solid organic compounds are just slower. Even
reactions of pre-dissolved organic compounds are slower than reactions of dissolved ionic compounds. Reactions
between ionic compounds in aqueous solution just tend to be faster, due to the already broken bonds.
45
8) 1
9) 3 it has a + value
10) 4 not a bad definition
11) 4 this is the only way to get an exothermic reaction. An exo. means a -∆H. Using ∆H = Hp -Hr , only 80 kJ gives you a
negative value.
12) 3 Combine the two arrows, their sum extends from the reactants of to the pinnacle of the graph. Yes, arrow B is the activation
energy of the forward, but I wanted you to analyze the relationships between the other energies, and so, I did not give it to you
as a choice.
13) 4 The reaction is endothermic, so the ∆H and the ∆Hƒ must be positive. Since 2 moles of product are produced the ∆H must be
divided by 2.Note how the heat of formation is defined in the question. Be prepared to flex your brain like that ... according to
the Regents rules ... it is perfectly acceptable to ask a question like this as long as the concept is defined first.
14)
15)
16)
17)
18)
4 it is the difference between the products and the reactants
3
3
3 the molarity is the greatest, the surface area is the greatest and the temperature is the greatest
2 the PE Diagram represents an exothermic reaction. According to Table I, C 2H6 is the only choice offered which is produced
exothermically
19) a) -3,351 kJ
b) -16.75.5 kJ/mol
c) -13408 kJ
d) -16.75.5 kJ/mol
e)
f) exothermic g) more stable
46