NAME _____________________________
NOTES: UNIT 8: KINETICS / EQUILIBRIUM (1)
I) Kinetics (DEFINITION): studies of the rate of a
    the change over time of bond breaking &/or of bond making
 READ THIS!
Too often, the concept of kinetics is viewed as how we may speed up a chemical reaction, only .
But, chemists are interested in slowing down (inhibiting) chemical reactions as well as in speeding up reaction rates. Chemists work to control everything from the economically important to the biologically very important reactions.
For example, chemists wish to learn how to control (and slow down) the rate of oxidation of metal bridges. They wish to speed the refractory period (the recharging time) of a digital camera’s screen image, so we can take another photo, in less time.
Chemists wish to learn how to control the processes of aging. They want to know the kinetics as to how antioxidants (Vitamin A,C and E for instance) help to keep our cytoplasmic materials, DNA and proteins from damage, (if they do so at all). Scientists want to learn how to control the rates at which cancer cells reproduce and how to increase the rate of cancer cell contact inhibition. Scientists want to get a more predictable life span out of light bulbs, batteries, etc.
All of these issues and many more are connected to the specialized field of kinetics. Kinetics is explained best by one of the six to eight big ideas in Regents chemistry: The Collision Theory.
You want to know this theory – it comes in very handy...
II) The Collision Theory: (T
EXT
R
EADING
:
P
.541-547)
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. W E CAN CONTROL A REACTION , BY
CONTROLLING THE NUMBER OF EFFECTIVE COLLISIONS .
(M AJOR R EASON # 6)
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 reaction.
(called the activated complex)
554

III) There are 5 fundamental factors which affect the ability to achieve the requirements of effective collisions
A)  5 factors
1) nature of the reactants ( dissolved ionic compounds vs. molecular (covalent) compounds )
Basic Idea : Solutions of ionic compounds react * faster (form new bonds) upon
combination than solutions of * molecular (covalent) 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 : Pb(NO
3
)
2
(aq)
+ 2 KI
(aq)

PbI
2
(s)
+ 2 KNO
3
(aq)
In the beakers = Pb 2+
(aq)
+ 2 NO
3
1-
(aq)
2 K 1+
(aq)
+ I 1-
(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: N
2(g)
+ 3H
2(g)

2 NH
3(g)
While not in aqueous solution, the gaseous reactants are bonded strongly enough that this reaction requires
a catalyst of iron, (mixed with a potassium hydroxide promoter), a temperature of 400°C to 450°C and approximately
200. atmospheres of pressure.
Think about it: C
12
H
25
OH
(l)
+ H
2
SO
4(aq)

HC
12
H
25
SO
4
+ H
2
O
lauryl alcohol dodecyl hydrogen sulfate
Bonds are NOT yet broken
In the beakers: = C
12
H
25
Water disrupts the ionic bond, but the covalent bonds of the polyatomic ion are NOT broken
OH + 2 H
+1
(aq)
+ (SO
4
2-
)
(aq)

H H H H H H H H H H H H
| | | | | | | | | | | |
H  C  C  C  C  C  C  C  C  C  C  C  C  O  H
O
| | | | | | | | | | | |
H H H H H H H H H H H H O
1
│
O ═ S ─ O -1
555

2) 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 with other reactive species with sufficient energy and orientation. a) 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 systems thermal energy. This thermal energy is a necessary part of endothermic
reactions. c) Don’t forget, when you read: average kinetic energy you think of * temperature
3) Concentration (note: for gases, changes in concentration = changes in pressure for gases ) a) symbol for concentration is a set of brackets: [ ] Thus the when discussing the
concentration of dioxygen gas, a chemist may write the phrase as [O
2
] b) M , % mass/mass and ppm are three concentration equations you must know
Use Table T to complete each of the following equations
M =
% m/m = ppm =
556

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)
4) 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, 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
5) Catalysts: a) Catalysts work by allowing the reaction to proceed along an * alternate pathway
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. c) enzymes are biological catalysts (remember the names of enzymes often end in –ase
* eg. lactase vs. lactose, amylase vs amylose, acetylcholinesterase vs. acetylcholine
) d) many metals (Pt, Zn and even Na) and compounds may be effective catalysts.
*  A catalyst must be a substance e) Note:
Thermal energy (okay, “heat”)
IS NOT A CATALYST!! f) The opposite of a catalyst is an inhibitor g) Note: The addition of a catalyst does NOT change a reaction’s 
H
(change in enthalpy)
557

QUESTIONS: Given the reaction: Zn
(s)
+ 2HCl
(aq)
H
(2)(g)
+ ZnCl
2(aq)
Molarity of HCl Description of Zn Temperature
A
B
C
D
2 M
1 M
1 M
2 M
10 grams in small lumps
10 grams of fine powder
10 grams in small lumps
10 grams of fine powder
26

C
30

C
26

C
30

C
Which of the listed conditions will result in the fastest generation of hydrogen gas?______ Defend your answer:
_____________________________________________________________________________________
___________________________________________________________________________________
Which set of conditions probably results in the slowest generation of hydrogen? ________ Why? ________
____________________________________________________________________________________
ISSUE: A kitchen grease fire can range (get it?) from a troublesome mishap to a home disaster. Grease can
reach its “flash point”. 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 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.
Throwing baking soda (sodium hydrogen carbonate) onto the fire, often puts out the fire. Baking soda
is absorbent and it decomposes under high heat according to the reaction equation:
683kJ + 2 NaHCO
3 (s)
 Na
2
CO
3(s)
+ H
2
O
(l)
+ CO
2(g)
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? (That is… how does its absorbency and decomposition help?)
* 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).
558

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.

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)
559

A) Putting all the pieces together:
PE
Reaction Coordinate
PE
NOT:
PE
Reacton Coordinate
Reaction Coordinate
560

V) Potential Energy Diagrams
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
Table I: Heats of Reaction at 101.3 kPa and 298 K
Reaction ∆ H (kJ)*
CH
4(g)
+ 2 O
2(g)
 CO
2(g)
+ 2 H
2
O
( 𝓁 )
-890.4
C
3
H
8(g)
+ 5 O
2(g)
 3 CO
2(g)
+ 4 H
2
O
( 𝓁 )
2 C
8
H
18(( 𝓁 )
+ 25 O
2(g)
 16 CO
2(g)
+ 18 H
2
O
( 𝓁 )
-2219.2
-10943
2 CH
3
OH
( 𝓁 )
+ 3 O
2(g)
 2 CO
2(g)
+ 4 H
2
O
( 𝓁 )
-1452
-1367 C
2
H
5
OH
( 𝓁 )
+ 3 O
(g)
 2 CO
2(g)
+ 3 H
2
O
( 𝓁 )
C
6
H
12
O
6(s)
+ 6 O
(g)
 6 CO
2(g)
+ 6 H
2
O
( 𝓁 )
2 CO
(g)
+ O
2(g)
 2 CO
2(g)
C
(s)
+ O
2(g)
 CO
2(g)
4 Al
(s)
+ 3 O
2(g)
 2 Al
2
O
3(s)
N
2(g)
+ O
2(g)
 2 NO
(g)
N
2(g)
+ 2 O
2(g)
 2 NO
2(g)
2 H
2(g)
+ O
2(g)
 2 H
2
O
(g)
2 H
2(g)
+ O
2(g)
 2 H
2
O
( 𝓁 )
N
2(g)
+ 3 H
2(g)
 2 NH
3(g)
2 C
(s)
+ 3 H
2(g)
 C
2
H
6(g)
2 C
(s)
+ 2 H
2(g)
 C
2
H
4(g)
2 C
(s)
+ H
2(g)
 C
2
H
2(g)
H
2(s)
+ I
2(g)
 2 HI
(g)
KNO
3(s) 𝑤𝑎𝑡𝑒𝑟
→ K +
(aq)
+ NO
3
-
(aq)
NaOH
(s) 𝑤𝑎𝑡𝑒𝑟
→ Na +
(aq)
+ OH -
(aq)
NH
4
Cl
(s) 𝑤𝑎𝑡𝑒𝑟
→ NH
4
+
(aq)
+ Cl -
(aq)
NH
4
NO
3(s) 𝑤𝑎𝑡𝑒𝑟
→ NH
4
+
(aq)
+ NO
3
-
(aq)
-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
-44.51
+14.78
+25.69
NaCl
(s) 𝑤𝑎𝑡𝑒𝑟
→ Na +
(aq)
+ Cl -
(aq)
LiBr
(s) 𝑤𝑎𝑡𝑒𝑟
→ Li +
(aq)
+ Br -
(aq)
H +
(aq)
+ OH -
(aq)
 H
2
O
( 𝓁 )
+3.88
-48.83
-55.8
* The ∆ H values are based on molar quantities represented in the equations.
A minus sign indicates an exothermic reaction .
P
E
Reaction Coordinate
P
E
Reaction Coordinate
561

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 sign.
∆H
heat of reaction e.g) 2 NO
(g)
+ O
2(g)
 2 NO
2(g)
+ 112.86 kJ ∆ H = * - 112.86 kJ
2 C
7
H
5
N
3
O
6(s)

3 N
2(g)
+ 12 CO
(g)
+ 5 H
2(g)
+ 2C
(s)
+ 2,092kJ ∆ H =* -2,092 kJ
4 ) ENDOthermic Reaction :
The products of an endothermic reaction are * less stable than the reactants
The products are * higher (greater) in energy
The change in enthalpy of an endothermic reaction is written with a * positive sign.
∆H
heat of reaction
53 kJ + H
2(g)
+ I
2(g)
 2 HI
(g)
∆ H = * + 53 kJ
562

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 + H
2(g)
+ I
2(g)

2 HI
(g)
∆H = +53 kJ
but
∆H f
= *+26.5 kJ
N
2(g)
+ 3H
2(g)

2 NH
3(g)
+ 91.8 kJ ∆H = -91.8 kJ but
∆H f
= * -45.9 kJ
C
(s)
+ O
2(g)

CO
2(g)
+ 393.5 kJ ∆H = -393.5 kJ and
∆H f
= * -393.5
66.4 kJ + N
2(g)
+ 2 O
2(g)

2 NO
2(g)
∆H = +66.4 kJ but
∆H f
= * + 33.2
2 Al
(s)
+ 3 O
2(g)

2 Al
2
O
3(s)
+ 3351 kJ ∆H = -3351 kJ but
∆H f
= * -1,675.5 kJ
2 C
(s)
+ 3 H
2(g)

C
2
H
6(g)
+ 84 kJ ∆H = -84 kJ and
∆H f
=* -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.
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” (HONORS)
3) Energy Lost during Bond Formation (There is no "technical" name – but it is a useful concept) a) ALL bond formation * releases energy (
regardless of the overall reaction.
 it is *exothermic )
563

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 4.5 5.0 5.5
Time (minutes)
(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 (HONORS).
564

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)
+ 3O
2(g)
 2 Al
2
O
3(s)
+ 3,351 kJ
Potential Energy
2 3
4Al + 3O 2
4
1 2 Al
2
O
3
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 & O
2
or Al
2
O
3
___________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 + H
2(g)
+ I
2(g)

2 HI
(g) a) Sketch the potential energy diagram. b) Label H reactants
and the H products c) Label the area which represents the 53.0 kJ d) Identify the area representing the activation
energy
565

3)
5 4 2 E lost by new bonding
E p
3 1
6 a ) The reaction is
Exothermic / Endothermic b) E reactants is represented by arrow c) E products is represented by arrow d) E activated complex is rep. by arrow e) E activation is represented by arrow f) Heat of reaction (

H) is rep. by arrow
Reaction Coordinate
4)
2 3 4 E a ) The reaction is
Exothermic / Endothermic lost by new bonding b) E reactants is represented by arrow
E p
1 5
6
Reaction Coordinate c) E products is represented by arrow d) E activated complex is rep. by arrow e) E activation is represented by arrow f) Heat of reaction (

H) is rep. by arrow
566

NAME _________________________________________ KINETICS
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 :
___4)Given: 80.0 kJ + X2 + 2Y + 8Z

2 XYZ4 a) if only I is correct b) if only II is correct c) if only I and II are correct
According to the above reaction equation,
I) the ∆H of the forward reaction is + 80 kJ d) if only II and III are correct e) if I, II, and III are correct
II) the reaction is exothermic
III) products are more stable than reactants
___1) Given the following reaction, which
statement(s) is (are) true ?
******************
For questions 5 - 18, only a single answer per question is most correct.
C(s) + O2(g)

CO2(g) + 393 kJ
I) The forward reaction is exothermic
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?
___5) The "heat of reaction", ∆H, is equal to a) Hproducts - Hreactants b) Hproducts + Hreactants c) Hproducts / Hreactants d) (Hproducts) (Hreactants)
___6) According to Table I, which of these
compounds is "most stable" ?
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: a) NO
2(g)
c) CO
2(g) b) HI( g
) d) NH
3 (g)
___7) 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
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 powdered zinc: a) is more concentrated b) has a greater surface area c) requires a lower activation energy d) generates more heat
III) products are more stable than reactants
567

___8) 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 b) ionic compounds in the solid phase c) organic compounds in the solid phase d) solutions of ionic compounds
___9) Given 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
___10) According to Table I which compound
forms via an endothermic reaction ? a) carbon dioxide c) nitrogen monoxide b) aluminum oxide d) ammonia
___11) The minimum energy required to break
bonds and initiate a chemical reaction is a) activated complex c) change in enthalpy b) heat of reaction d) activation energy
___12) 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
For questions 13 -17 use the given reaction and potential energy diagram.
N2(g) + O2(g) + 66 kJ

2 NO2(g)
B C
D E
A
___13) Which arrow(s) is/are the equivalent(s) of
the energy of activation for the forward
reaction? c) C + D d) just arrow E a) A + E b) B + A
___14) 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 c) +132kJ d) +33 kJ
___15) According to the diagram, the heat of
reaction is best represented by arrow : a) A b) B c) C d) D
___16) The activation energy of the REVERSE
reaction is represented by arrow a) A b) B c) C d) D
568

17) According to the diagram, the reaction is
____________ and the products are _____ stable. a) endothermic, more b) exothermic , more c) endothermic, less d) exothermic, less
For question 18 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 Concentration of Y (M) a 1 b 1
Surface area of Y cube powdered
100
50 c d
2
2 powdered cube
150
100
___18) Which Trial, most probably has the fastest
rate of reaction ?
____19) The following potential energy diagram
represents the formation of which
compound?
P.E
Reaction Coordinate a) C
2
H
4(g) c) NO
2(g) b) C
2
H
6(g) d) NO
(g)
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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 stable)
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 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 (more stable) in energy
4) a The total heat is +80 kJ, but that means it is an endothermic reaction
5) a definition : see your notes
6) c check Table I. The compound with the MOST negative value is the most stable (This negative illustrates that the most energy was lost in the
formation of the compound, thus producing a compound that is lower in energy and hence, more stable, and thus more difficult to react, in its
own right.
7) b
8) d 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.
9) a
10) c it has a + value
11) d not a bad definition
12) d 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.
13) c 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.
14) d 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.
15) d it is the difference between the products and the reactants
16) c
17) c
18) c the molarity is the greatest, the surface area is the greatest and the temperature is the greatest
19) b the PE Diagram represents an exothermic reaction. According to Table I, C
2
H
6
is the only choice offered which is produced exothermically
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NAME ____________________________________ KINETICS: TRY THIS
DIRECTIONS: Use everything you know ... Go in groups, go into your notes, go big...
1) Given: 4 Al
(s)
+ 3 O
2(g)
 2 Al
2
O
3(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? * d) sketch the potential energy diagram for the production of aluminum oxide.
-13408 kJ
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 (select one) g) the products are less stable than the reactants / more stable than the reactants (select one)
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