1. No category

AP CHEMISTRY
REVIEW
MR. DAVID GRAY
TEACHER OF AP CHEMISTRY
SCIENCE DEPARTMENT HEAD
NORTHEAST HIGH SCHOOL
[email protected]
AP CHEMISTRY REVIEW
WITH
MR. DAVID GRAY
Welcome to the AP Chemistry review. This will be a two hour review of the
main topics that are usually tested on the free response portion of the exam.
I must warn you that this review does not replace the amount of time that you
must put in reviewing, studying and practicing for this very intense exam.
We will be reviewing the following topics:
1. The periodic table and its trends
2. Stoichiometry
3. The tables included with the exam. You do not have to memorize
formulas but you do need to be able to use the formulas to solve the free
response questions.
4. Atomic structure - electron configurations, orbital diagrams, and
quantum numbers.
5. Reactions - basic types and redox, solubility rules, oxidation numbers.
6. Electrochemistry
7. Equilibrium
8. Kinetics
9. Thermodynamics
10.States of Matter
I have included a review package that you may take with you. The package
includes review information and practice questions that come directly from
the College Board.
If you go on the College Board AP Chemistry website you will find many sets of
free response questions that have been released. They are actual questions
from previous exams. They are a great way to practice for the free response
portion of the exam. You should also look for a practice multiple-choice exam.
You should take the exam yourself and time it. If you score between 50-75, you
are well on your way to a passing grade on the exam. I have included a scoring
sheet so you may see how various sections of the exam are scored.
~~
~~
::jo
0~
1QQ.~Qi~~i~~fl._i.~_Q~_-;Q_Q_~~--· ·1
PERIODIC TABLE OF THE ELEMENTS
I
H
He
1.0079
4.0026
3
Li
I
4
Be
'tl
llo
C)
!"
B
N
0
F
Ne
10.811 112.011 I 14.007 I 16.00
13
I
20
Ca
I
21
22
23
Sc
Ti
v
40.08 I 44.96 I 47.90 I 50.94
Fe
I
27
Co
I
28
Ni
Pd
91.22
92.91
95.94
72
73
74
102.91
77
106.42
78
Sr
Y
Zr
85.47
87.62
88.91
55
56
57
*La Hf
89
26
Rh
Rb
88
Mn
I
Mo
41
138.91
25
Nb
40
137.33
Cr
II
58.93 I 58.69
46
45
39
Ba
24
52.00 I 54.938 I 55.85
42 I 43
44
38
Ta
178.49 180.95
w
104
105
183.85
106
I Tc
I
Ru
(98) I 101.1
75
Re
1
·1•
76
Ir
Pt
108
195.08
llO
190.2
107
I 30 I
Zn
63.55 I 65.39
47 ! 48
Ag
l Cd
107.87 1112.41
79
l
80
Au
j Hg
Ra tAc
Rf
Db
Sg
Bh
Hs
Mt
Ds
Rg I §
226.02 227.03
(261)
(262)
(266)
(264)
(277)
(268)
(271)
(272)
*Lanthanide Series
58
59
Ce
Pr
140.12 140.91
t Actinide Series
60
61
62
Nd
Pm Sm
144.24
(145)
150.4
63
64
Eu
Gd
151.97 157.25
65
Tb
158.93
90
91
92
93
94
Th
Pa
u
Np
Pu
Am Cm Bk
(244)
(243)
232.04 231.04 238.03 237.05
95
31
Ga
96
(247)
97
I
14
Si
I
15
P
I
20.179
16
17
1 18
S
Cl
Ar
I 32 I 33 I 34 I 35 I 36
Ge
As
69.72 I 72.59 I 74.92
49
In
I
50
Sn
I
51
Sb
Se
Br
Kr
78.96 I 79.90
53
83.80
54
I
Xe
52
Te
I
114.82 1118.71 1121.75 1127.60 1126.91 1131.29
81
TI
196.97 . 200.59 204.38
Ill j 112
i (223)
I
19.00
26.98 I 28.09 130.974 I 32.06 I 35.453 I 39.948
j
192.2
109
Os
186.21
29
Cu
I Fr
z
m
z
m
~
lO
AI
132.91
; 87
-t
J:
9
Mg
Cs
a
8
Na
39.10
37
0
0
7
12
K
i
6
c
9.012
19
C)
5
6.941
II
22.99 I 24.30
N
2
82
83
84
85
86
Pb
Bi
Po
At
Rn
207.2
208.98
(209)
(210)
_Qlli
~z
Cll:z
:z~
~~
Cll6;
g~
oo:l
z~
~~
~~
~:z
:z~
~~
~E
~
~
!
"fl
VI
§Not yet named
(277)
o:l
trl
66
67
68
Dy
Ho
Er
69
Tm Yb
162.50 164.93 167.26 168.93
99
100
101
98
I Cf
(247J J _(251)
Es
Fm Md
{252]_ (257)
70
125_81
173.04
102
71
Lu
174.97
103
No
Lr
Q59)
(262)
c
!:z
~
;
STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT 25°C
E 0 (V)
Half-reaction
2FCo 2+
2.87
Au(s)
2Cl-
1.50
2 H 20(1)
2Br-
1.23
~
~
Hg22+
~
Hg(l)
0.92
0.85
0.80
F2 (g)+2e-
~
Co 3+ +
e3eCl 2 (g)+ 2e-
~
Au 3+ +
~
0 2 (g)+4H+ +4eBr2 (l)+ 2e2Hg 2+ + 2eHg 2+ + 2eAg+ + eHg 22+ + 2 eFe 3+ + e-
~
I 2 (s)+ 2ecu+ + eCu 2+ + 2eCu 2+ + eSn 4 + + 2 eS(s)+2H+ +2e2H++2ePb 2+ + 2 eSn 2++2eNi 2+ + 2 eCo 2++
2e-
Cd 2++2eCr 3+ + eFe 2+ + 2eCr 3+ + 3eZn 2++2e2 H 20(/) + 2 eMn 2+ + 2eAl 3+ + 3eBe 2+ + 2eMg 2+ + 2eNa+ + eCa 2+ + 2 eSr 2+ + 2eBa 2+ + 2 eRb+ +eK+ +ecs+ + eLi+ + e-
~
1.82
1.36
1.07
~
Ag(s)
~
0.79
~
2 Hg(l)
Fe 2+
~
21-
0.53
Cu(s)
Cu(s)
cu+
Sn 2+
0.52
0.34
0.15
0.15
~
H 2S(g)
0.14
~
~
H2(g)
Pb(s)
0.00
-0.13
~
Sn(s)
~
Ni(s)
-0.14
-0.25
~
Co(s)
Cd(s)
Cr 2+
-0.28
Fe(s)
Cr(s)
Zn(s)
H 2 (g) + 20H-
-0.44
-0.74
~
~
~
~
~
~
~
~
0.77
-0.40
-0.41
~
Mn(s)
-0.76
-0.83
-l.l8
~
Al(s)
-1 .66
~
Be(s)
-1.70
~
Mg(s)
~
Na(s)
Ca(s)
Sr(s)
Ba(s)
-2.37
-2.71
-2.87
-2.89
~
~
~
~
~
~
~
~
~
-2.90
Rb(s)
K(s)
Cs(s)
-2.92
Li(s)
-3 .05
-2.92
-2.92
GO ON TO THE NEXT PAGE.
3
ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS
ATOMIC STRUCTURE
E
= hv
c
h
A=-
p = mv
mv
En
= Av
= -2.178 X 102
n
18
. I
JOU e
E =energy
v = velocity
v = frequency
A = wavelength
p =momentum
n = principal quantum number
m =mass
Speed of light, c = 3.0 x 108 m s- 1
EQUILIBRIUM
X
w-34
Js
Boltzmann's constant, k = 1.38 X
w- 23
J K-l
Planck's constant, h = 6.63
[H+] [A-]
Ka = [HA]
Avogadro's number = 6.022 x 10 23 mol- 1
Kw
= [OH-] [H+]
= K(/
X
= 1.0 X
w- 14
@
Electron charge, e = -1.602
25°C
X
w- 19
coulomb
I electron volt per atom = 96.5 kJ mol- 1
Kb
pH = - log [H+ ], pOH = - log [OH-]
Equilibrium Constants
14 =pH+ pOH
Ka
Kb
Kw
KP
[A-]
pH = pKa + log [HA]
[HB+]
pOH = pKb + log ---rB)
pK0 = -log K0 , pKb = -log Kb
KP = Kc(RT)
an
(weak acid)
(weak base)
(water)
(gas pressure)
Kc (molar concentrations)
,
s·
where t::.n = moles product gas - moles reactant gas
= standard entropy
H• = standard enthalpy
THERMOCHEMISTRY /KINETICS
t::.c•
:L s· products - z. s· reactants
= L MIJ products - L MIJ reactants
= L t::.GJ products - L t::.GJ reactants
t::.c•
= Mi.
G 0 = standard free energy
Eo = standard reduction potential
t::.s· =
Mf•
T = temperature
n =moles
m
q
c
CP
- Tt::.s •
= -RT InK = -2.303 RT log K
= -n ?Ji E o
q = mct::.T
Mf
cp = t::.T
1
-
= heat
= specific heat capacity
= molar heat capacity at constant pressure
Ea = activation energy
k = rate constant
A = frequency factor
t::.G = t::.G• + RT in Q = t::.G • + 2.303 RT log Q
ln[A] 1
=mass
Faraday's constant, ?Ji = 96,500 coulombs per mole
of electrons
ln[A] 0 = -kt
Gas constant, R = 8.31 J mol-l K -l
1
= 0.0821 L atm mol- 1 K- 1
[A]t - [A)o = kt
= 8.31 volt coulomb mol- 1 K- 1
T 1)
ink= -E ( T +In A
GO ON TO THE NEXT PAGE.
4
GASES, LIQUIDS, AND SOLUTIONS
P = pressure
PV = nRT
V =volume
T = temperature
PA
P,otal
= P,0101
x X A. where X A =
n = number of moles
moles A
total moles
--=c:..::..:..::-=-----
D = density
m =mass
v = velocity
= PA + PB + Pc + ...
m
n =M
·c + 273
K =
urm .. =
KE =
r =
M =
1t =
i =
K1 =
~~
P2 V2
-=T1
T2
D='!!:.
v
urms
= pkT = tRT
m
! mv 2
KE per molecule =
2
KE per mole =
~
r2
=
M
root-mean-square speed
kinetic energy
rate of effusion
molar mass
osmotic pressure
van't Hoff factor
molal freezing-point depression constant
Kb = molal boiling-point elevation constant
~ RT
A = absorbance
2
a
rg_
= molar absorptivity
b =
c=
Q=
I =
path length
concentration
reaction quotient
current (amperes)
q = charge (coulombs)
t = time (seconds)
VM,
molarity, M = moles solute per liter solution
molality = moles solute per kilogram solvent
f:.T1 = iK1 x molality
f:.Tb = iKb x molality
1t = iMRT
A = abc
E o = standard reduction potential
K = equilibrium constant
Gas constant, R = 8.31 J mol - 1 K- 1
OXIDATION-REDUCTION; ELECTROCHEMISTRY
= 0.0821 L atm mol- 1 K- 1
Q=
[C] c [D]d
[A]a [B]b ,
= 8.31 volt coulomb mol-l K- 1
where a A + b B ~ c C + d D
Boltzmann's constant, k = 1.38
J K- l
Kb for H 2 0 = 0.512 K kg mol- 1
t
E~ell
n£
w-23
K 1 for H 2 0 = 1.86 K kg mol-l
1=!1..
£cell =
X
-
R! In Q = £~.11 -
n~
0 0592
·
log Q
n
@
25°C
1 atm = 760 mm Hg
= 760
torr
STP = O.Ooo•c and 1.000 atm
Faraday's constant, <JF = 96,500 coulombs per mole
of electrons
0
logK = - 0.0592
GO ON TO THE NEXT PAGE.
5
AP CHEMISTRY SCORING WORKSHEET
NAME_______________________
SECTION I-MULTIPLE-CHOICE
MULTIPLE-CHOICE SCORE - - - SECTION 2-FREE RESPONSE
QUESTION 1 _______
0-10 PTS
X 1.2667 = _____
QUESTION 2 OR 3 _ _ __
0-10 PTS
QUESTION 4 _____
0-15 PTS
QUESTION 5 - - - - 0-10 PTS
QUESTION 6 _____
0-10 PTS
QUESTION 7 OR 8 _____
0-8 PTS
X 1.2667 =
X .5700 =
X 1.0688 = _ _ _ __
X 1.0688 = _ _ _ __
X 1.0688 =
TOTAL FOR FREE RESPONSE
COMPOSITE SCORE .9600 X
- _ _ _ _ _ __
MULTIPLE- CHOICE
1.5440 X ______ - _ _ _ _ __
FREE RESPONSE
TOTAL
'1:t-..- .
L~
.'2i,l.
99-160
76-98
49-75
25-48
0-24
APGRADE
5
4
3
2
1
Atomic structure, theory, and periodicity
lesson 1
Even at our level of technology, we have yet to see an atom. It bas been one of our most
coveted secrets. What we do know about the atom comes from the way that the atom
behaves under certain circumstances. It is know that once an atom absorbs energy, it will
then release that energy. It is also known that the release of energy is predictable, and
never varies. Excited atoms of copper always bum green. They never bum red. Keeping
this in mind, it is obvious how the development of the atom as a rigid model of specific
"quantized" energy levels came about. Further investigation yielded the spectral series
(figure one) which provided information on the size of various energy levels, and the
distance from the nucleus. As such, an electron dropping from the third energy level to the
first energy level would always release a specific quantity (quanta) of energy. According to
figure one, this quantity is assigned the designation "Lymaa Series"
0
t
Ioni zatio n occurs
...··
I
J
Balmer series
transi tions
I
...
........ "
.-...... n •
I
.
Som~ ~n~rgy l~v~ls and transitions in th~
atom. Lines in the Balmer series arise from
transitions from upper levels (D > 2) to the a • 2 lcv~l.
3 In the Lyman series, the lower level is a • l.
6
n • 5
II'"
h ydrog~n
n=2
..
Lyman series
tra nsitio ns
-8
Ground state
n'" I
Figure one: the spectral series
The emissions of the spectral series are seen as specific wavelengths with the spectrum.
Using a spectrometer, the wavelengths may be seen as bright lines corresponding to these
specific wavelengths. A table of these spectrums is called the bright-line spectrum for each
element. The wavelength and frequency may be related as:
c wavelength x frequency
(cis the speed of light)
=
1
In this modet; the e"'ectrons in atoms can have oniy certain discrete energies, referred to as
energy states or energy levels. Normally, the electron is in the state of lowest energy,
called the ground state. By absorbing a certain definite amo'unts of energy, the electron can
move to a higher level, called an excited state. When electrons return to lower energy
levels, energy may be given off as light The difference in energy between levels can be
deduced from wavelength or frequency of the light:
•
E = hf or E = hc/w
1
where E is the energy of the emitted light, f is the frequency, w is the wavelength cis the
'
speed of light and h is Planes constant.
Bohr postulated that electrons move about the nucleus in circular orbits of fixed radius.
By absorbing energy, it moves to a higher orbit of larger radius. The energy is given off as
the electron returns to the ground state.
According to Bohr, the energy of any level could be calculated as:
E
=·1312 I
n2 (kVmol) where n
or E
= the
=·2.179 x 10 _,. I n
2
energy level
= 1,2,3•..
Jlpartlcle
In the ground state, n = 1, in excited states, n = 2,3 ... If an electron moves from the third
energy level to the first energy level, the energy of the emitted light may be calculated as E1
- E2, or Energy (high)- Energy (low). This is in accordance with observations of excited
electrons. In this case, the light emitted is that associated with the Balmer Series. .
The Bohr model did not agree with experiment for any atom with more than one electron
(quite a draw back!). The idea of elec~ns moving about the nucleus in circular orbits had
to be abandoned. However, the calculations of energies do still follow observed emissions.
A better explanation of electron locations could be expressed through the use of the
Quantum Mechanical model.
In the Quantum Meelumic&l atom, four quantum numbers are required to desCribe the
energy of the electrons completely. It is only hindered by its ability to refer only to the
probability of findings an electron in a region, but cannot specify the electrons path. The
four quantum numbers which designate an electrons probable location are n, 1, m and s.
Electrons in atom are quantized, which means that electrons in atoms have only certain
allowed energy states. The energy of an electron in an atom is determined mainly by the
· value of (n): the principal quantum number. The principal quantum nnmber (n)
defines a shell, at some average distance from the nucleus, capable of holding electrons
having about the same energy. It designates the energy level of the electron. It is the main
factor that determines the energy of an electron and its distance from the nucleus. The
value of the principal quantum number can be any positive integer. The lowest energy
electrons have n=l. Only values up to n=7 are necessary to describe the electrons in the
first 118 elements. Each energy level has a maximum electron capacity of:
2n2
The sublevels in each energy level are designated by l, where l = 0,1,2....(n-1).
Therefore, in the n=l energy level, there may be only one sublevel :0 (1-1=0). The 0
sublevel is oftentimes discussed as the "s" sublevel. They are one in the same. In the
second energy level, there may be two sublevels: 0,1 (2-1= 1). In this case, the 1 refers to
the "p" sublevel-.. Oftler designations for the sublevels of l are 1=3 ("d") and'l=4 ("f').
You may recall that each sublevel is divided into specific orbitals. The "p" sublevel for
example has 3 orbitals; often assigned x,y and z. Individual orbitals are assigned by the
third quantum number m. The values form are determined as -1 to +I. For example, the
"p" sublevel is I = l. Therefore, values form are -1,0 and + l. These are the three orbitals
discussed earlier as x,y and z. You may further recall that the 1=2 ("d") sublevel lias 5
orbitals given as -2,-1,0,+1 and +2, and that 1=3 ("f') has 7.
·
Each individual orbital may ho~d only 2 electrons, and those two electrons must have
different spin characteristics. This allow for repulsion to be overcome by angular
momentum. The difference in spin is assigned by the fourth quantum numbers, and may
2
have the values of+ 112 and -1/2. The assignment of the value is arbitrary, as long as no
two electrons within an orbital have the same value.
To illustrate this example of quantum number electron designation, the quantum numbers
of the last electron in carbon may be determined as:
Tlie Sixth electron in carbon known otherwise as the 2p2 electron:
n=2
l 1 ("p")
m 0 (in they orbital)
s -1/2 (arbitrary assignment)
=
=
=
The electronic configuration of any atom in its ground (most stable and lowest energy) state
can be determined by using the Aofbao (building up) procedure. Two electrons are added
per orbital (lowest energy orbitals first) until the number of added electrons equals the
number of protons in the atom (making the atoms electrically neutral).
Electrons placed in a subshell consisting of more than one orbital will go into an empty
orbital rather than pair tip with the first electron in a half-filled orbital (Hands's role).
And its spin direction will be the same as that of the first electron (parallel spin). Electron
configurations and the order of filling subshells can readily be obtained from a periodic
table.
·
The periodic table is _now able to reflect our knowledge of electronic structure of the atom.
All of the elements in a given group have the same outer electron configuration. As such,
elements in groups 1 and 2 fill s sublevels. Elements in gtoups 3-8 fill p sublevels.
Elements in the transition series fill d sublevels, and Lanthanides and Actinides fill the f
sublevels. With the placement of atoms in proper electronic order, various trends become
·
obvious on the table itself.
In general, the atomic radiu decreases going across a period from left to right and
increases going down a group:
Na
=.186nm
Mg
=.160nm
S
=.104nm
Cl = .099nm
K= .231nm
The explanation of such a trend is best seen by examining the increase of energy levels as
you move down a group wbich in~ases the size of each atom. The decrease as you move
across each period may be explained as an increase in nuclear attraction with increased
number of protons and electrons as you move from left to right. This attraction pulls the
atoms more tightly together.
'
-
The ionic radiu of each atom parallels those of the atomic radius for the same reasons.
Furthermore, the cations are smaller than the corresponding atoms due to the fact ·that an
energy level is emptied when the electrons are donated, and anions are larger since the
added electrons fill outer levels of the atom.
·
.
It should be obvious·that electrons closer to the nucleus would be harder to remove that
electrons farther away. This is indeed the case, as the ionization energy, the energy
required to remove the last electron, increases with the decrease of atomic size. It may also
be apparent that the successive ionization energies increase steadily as they become more
difficult to remove, and large jumps in ionization energies occur when electrons are
removed from inner levels.
3
For those atoms with the smallest radius, the attraction for electrons is greatest. This
attraction is called amnity, and increases as the size of the atom decreases. Auorine is the
most electronegative element located in the upper right hand comer of the periodic table. In
general, electron affinity is greatest for small highly nonmetallic atoms.
Trends in Metallic character may also be observed. Metallic character decreases left to
right in the periodic table, and increases going down. Metals are found below and to the
left of the diagonal stairway, and nonmetals above and to the right. Metals have high
electrical and thermal conductivity; ductility and luster. Elements along the stairway (B,
Si, Ge) are referred to as metalloids and are often semiconductors.
4
buffer solution is prepared by dissolving some solid NaOCl in a solution of HOCI at 298 K . _.Uioli~o.JU='l"eJ~
the u
· n is determined to be 6.48.
_..(.ii~ftd+cmieewwh ich
of HOCl(aq) or OCqaq) is present
Support your answer with a calculation.
aflT1~~lP.
2. A student was assigned the task of determining the molar mass of an unknown gas. The student measured the
mass of a sealed 843 mL rigid flask that contained dry air. The student then flushed the flask with the unknown
gas, resealed it, and measured the mass again. Both the air and the unknown gas were at 23.0°C and 750. torr.
The data for the experiment are shown in the table below.
Volume of sealed flask
843 mL
Mass of sealed flask and dry air
157.70 g
Mass of sealed flask and unknown gas
158.08 g
(a) Calculate the mass, in grams, of the dry air that was in the sealed flask. (The density of dry air is 1.18 g L-I
at 23.0°C and 750. torr.)
(b) Calculate the mass, in grams, of the sealed flask itself (i.e., if it had no air in it).
(c) Calculate the mass, in grams, of the unknown gas that was added to the sealed flask.
(d) Using the information above, calculate the value of the molar mass of the unknown gas.
After the experiment was completed, the instructor informed the student that the unknown gas was carbon
dioxide (44.0 g mol- 1).
(e) Calculate the percent error in the value of the molar mass calculated in part (d).
(f) For each of the following two possible occurrences, indicate whether it by itself could have been responsible
for the error in the student's experimental result. You need not include any calculations with your answer.
For each of the possible occurrences, justify your answer.
Occurrence I : The flask was incompletely flushed with C0 2 (g), resulting in some dry air remaining
in the flask.
Occurrence 2: The temperature of the air was 23.0°C, but the temperature of the C02(g) was lower than
the reported 23.0°C.
(g) Describe the steps of a laboratory method that the student could use to verify that the volume of the rigid
flask is 843 mL at 23.0°C. You need not include any calculations with your answer.
© 2009 The College Board. All rights reserved.
Visit the College Board on the Web: www.collegeboard.com.
-7-
GO ON TO THE NEXT PAGE.
AP® CHEMISTRY
2009 SCORING GUIDELINES
Question 2 (10 points)
A student was assigned the task of determining the molar mass of an unknown gas. The student measured the
mass of a sealed 843 mL rigid flask that contained dry air. The student then flushed the flask with the unknown
gas, resealed it, and measured the mass again. Both the air and the unknown gas were at 23.0°C and 750. torr. The
data for the experiment are shown in the table below.
Volume of sealed flask
843 mL
Mass of sealed flask and dry air
157.70 g
Mass of sealed flask and unknown gas
158.08 g
(a) Calculate the mass , in grams, of the dry air that was in the sealed flask. (The density of dry air is
1.18 g L- 1 at 23 .0°C and 750. torr.)
m
=DxV
= (1.18 g
V 1)(0.843 L)
=
One point is earned for the correct
setup and calculation of mass.
0.995 g
(b) Calculate the mass, in grams, of the sealed flask itself (i.e., if it had no air in it) .
157.70 g- 0.995 g
=
One point is earned for subtracting
the answer in part (a) from 157.70 g.
156.71 g
(c) Calculate the mass, in grams, of the unknown gas that was added to the sealed flask.
,,..,,.
One point is earned for subtracting
the answer in part (b) from 158.08 g.
158.08 g - 156.71 g = 1.37 g
(d) Using the information above, calculate the value of the molar mass of the unknown gas.
(~~~· atm )<o .843 L)
PV
n =- =
RT
molar mass
(0.0821 Latm mol -
=
1
K- 1)(296K)
= 0.0342mol
-1
1.37 g
0 .0342 mol = 40 · 1 g mol
OR
molar mass
One point is earned for the correct
setup and calculation of moles of gas.
One point is earned for the correct setup
and calculation of molar mass.
DRT
p
OR
(o\~~)gL)co.0821
1
1
Latmmol- K- )(296K)
=~----~----~~----~-----------
750 .
)
( 760 atm
= 40.0 g mol- 1
One point is earned for the conversion of
pressure (if necessary) and temperature
and the use of the appropriate R.
If calculation is done in a single step,
1 point is earned for the correct
P and T, 1 point is earned for the
correct density, and 1 point is earned
for the correct answer.
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2009 SCORING GUIDELINES
Question 2 (continued)
After the experiment was completed, the instructor informed the student that the unknown gas was carbon dioxide
(44.0 g mol- 1).
(e) Calculate the percent error in the value of the molar mass calculated in pa1t (d).
percent error
144.0 gmol-
1
- __40.I_gmol_-_i
44.0 gmol- 1
I
>; 100
One point is earned for the
correct setup and answer.
g_9%
(f) For each of the following two possible occurrences , indicate whether it by itself could have been
responsible for the error in the student's experimental result. You need not incl ude any calculations with
your answer. For each of the possible occurrences ,justify your answer.
Occurrence l: The flask was incompletely flushed with C02(g), resulting in some dry air remaining
in the flask .
This occurrence could have been responsible.
The dry air left in the flask is less dense (or has a lower
molar mass) than C02 gas at the given T and P. This
would result in a lower mass of gas in the flask and a lower
result for the molar mass of the unknown gas .
.. ....
One point is earned for the correct
reasoning and conclusion.
Occurrence 2: The temperature of the air was 23.0°C, but the temperature of the C0 2(g) was lower
than the reported 23 .0°C .
This occurrence could not have been responsible.
The density of C0 2 is greater at the lower temperature. A
larger mass of C0 2 would be in the fla sk than if the C02
One point is earned for the correct
reasoning and conclusion.
had been at 23.0°C, resulting in a higher calculated m0lar
mass for the unknown gas.
(g) Describe the steps of a laboratory method that the student could use to verify that the volume of the rigid
flask is 843 mL at 23.0°C. You need not include any calculations with your answer.
Valid methods include the following :
1. Find the mass of the empty flask. Fill the flask with a liquid of
known density (e.g., water at 23°C), and measure the mass ofthe
liquid-filled flask . Subtract to find the mass of the liquid. Using the
known den sity and mass, calculate the volume.
2. Measure 843 mL of a liquid (e.g. , water) in a 1,000 mL graduated
cylinder and transfer the liquid quantitatively into the flask to see if
the water fill s the flask completely.
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One point is earned
for a valid method.
AP® CHEMISTRY
2009 SCORING GUIDELINES
Question 2 (continued)
Note: Significant figures were checked in this problem: parts (a) and (d) were scored with ±1 significant figure
needed, and parts (b) and (c) were scored with the correct number of significant figures needed for the subtraction.
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REACTIONS
group one and group two metal reactions
lesson 3
With help from the periodic table, certain trends may be plotted with regards to reactions.
Luckily, reactions in one group are similar for most items within that group. For that
reason, the reactions of the group one and group two metals may be q,uantified.
-In general, these metals, except for Be, exist as cations in their compounds (+ 1 for group 1
and +2 for group two). Aluminum behaves in a similar manner ( +3). In all cases, the
general formulas for halides of these compounds is MX 1 (group 1), MX 2 (group 2) and
A1X3 • Their reactions with hydrogen gas all produce the metal hydride:
ca <•> + ·t\ --.=.>CaH
(a)
2
The group one and group two metals will also combine with water to form hydrogen gas
and the metal hydroxide:
Na + t\0
-~ ~
+ NaOH
The reactionS with oxygen are somewhat less obvious. Lithium, the group two metals and
aluminum all react with oxygen to form normal (0.2) oxides. Na and Ba react to form
peroxides (02-2), and potassium, rubidium, and cesium react to form superoxides (02-).
It is best to commit these trends to memory!
Net Ionic representations:
All reactions may be represented as a net-ionic equation. Here, only the reacting species
which undergo a change are represented. Those items which remain the same as both
reactants and products are called" spectators", and are eliminated. The net-ionic then
shows only the participating ions and those species which undergo change. For example,
the reaction of strong acid HCl with strong·base NaOH may be represent as:
HO + NaOH
--7- HOH + NaQ
The HQ, NaOH and NaO are all ionic. Water is molecular. Therefore, the above
equation may now be written as:
H+ +
ct·
+ Na• + QH·
------?> HOH
+ Na+
+ c1·
Notice that the chloJine and sodium remain the same as reactants and produ~ts. They did
not undergo a change. They did not participate in the reaction. While it is true that they are
present in the reaction solution, their presence does not alter the.product formed: HOH.
These ions are called spectators are may be removed from the equation:
H+
+
OH-
--------~
HOH
The equation is now simplified to represent only the participating species. As a rule, ions
which do not change are eliminated, molecular species and precipitates must remain. As in
the case ofweak: Acetic acid and strong sodium hydroxide:
·
HAc +
NaOH
-------~ HOH
+ NaAc
10
The HAc is a weak acid, and dissociates only slightly. To represent it in its ionic form
would be incorrect.. Therefore, the ionic equation may be written as:
HAc + Na • +
Ofr ------;7 HOH
+ Na• +
Ac·
and the net ionic, which eliminates the spectator (sodium):
HAc + OH·
-------~
HOH
·+ Ac·
Having a good working relationship with solubility rules and common strong acids and
bases will help greatly in detennining which species may be written ionically in a net ionic
equation.
Preeipitatioo Reactions:
In reactions which follow the double displacement pattern:
AB +CD ---? -AD + CB
a reaction will only occur is AD or CB is water or a precipitate. Water is easy enough to
recognize. The precipitates are, however, less obvious. For this reason, the solubility rules
of common elements must be committed to memory. There is no other way to predict these
types of reactions. For example, in the combination of Ba(N03)2 + Na2C03 two possible
products will form: Ba(C03 )2 and NaN03 • According to the solubility rules, BaC03 is
insoluble, and this reaction my be represented by the net-ionic:
Ba• 2 + C03· 2
-------7 BaC03 (s)
The solubility rules are included below, however, they will NOT be included on the AP
exam. It can not be emphasized enough: These rules must be committed to memory!
all nitrates are soluble
all chlorides, bromides and iodides are soluble except: silver, mercury and lead.
most sulfates are soluble except strontium, barium and lead.
all carbonates are insoluble except those of the group one elements an~ NH+4
all hydroxides are insoluble except those of the group one elements, strontium and barium.
(calcium is slightly soluble)
all sulfides except tAPse of the group one and two elements and NH+4 are insoluble.
Recall that if no precipitate or water forms, then no reaction will occur.
•
11
Acid base- reactions (overview)
Many of the common reactions may be cla8sified as acid base reactions. Applying basic
definitions, an acid forms H+ ions in water. Strong acids are completely dissociated in
water; Some common strong acids include:
·
HCI,
HBr, HI, HN07 HC/04 and H2S04
Each of these exist as nearly 100% dissociated ions in solution. There would be virtually
no undissociated (molecular) HO in solution. Weak acids, however, do exist as
undissociated molecular species. For·this reason, a solution of HF (weak acid) will exist
as H+ , F- and HF. The true concentration of each of these species is dependent upon the
dissociation constant. ... a topic for another day.
Bases form OH- ions in solution. They follow the same patterns as weak and strong acids.
Strong bases dissociate nearly 100%. The strong bases will include the group 1 and
heavier group 2 hydroxides. Weak bases partially dissociate.
Reactions between acids and bases will follow o~e of the following pathways:
l•
strong acid and strong base-
H• + OH-
2•
HF
3•
------7 HOH
stroq base and weak acid
+ OH-
------~
HOH
+ F
strong acid and weak base
H• + ~ -----7' NH~ ..,
4.
weak acid and weak base (to be discussed at a later date)
Salts may also be acids or bases. To determine the pH, we must consider the effect of the
cation and the anion separately. then combine these effects to give overall results for the
salt.
Neutral cations are derived from strong bases:
-
Li, Na, K, Ca, Sr, Ba
Acidic cations are all other cations including those of the transition metals.
Neutral anions are derived from strong acids:
0 , Br, I, N0 3 , Cl0 4 , S04
Acidic anions include HSO,; arid ~P04 -
12
Basic anions include all other anions. In general, they are derived from a weak acid and is
expected to be basic..
'
F + ~0 --~ FOH +H•
Redox:
In instances where the oxidation state of one or more of the reacting species changes, and
electron exchange has taken pla~e. These reactions are termed Redox: reduction and
oxidation. In these instances, a species which accepts electrons is reduced, and the
donating species is oxidize<L Determination of which species is oxidized and which is reduced may be enhanced by following set patterns in oxidation states: The oxidation
number of elements in elementary substances (F2 , 0~, Mg) is zero. The oxidation number
of elements in monatomic ion form is the charge of that ion (Fe•2 is 2), the oxidation
number of group one elements in their compounds is+ 1, group two is +2, and of the
halides (F) -1. The oxidation state of hydrogen is almost always +1, and of oxygen -2.
The sum of oxidation numbers of all atoms in a neutral molecule must= 0. In polyatomic
.
ions, the sum is the charge of the ion.
Using these rules, assignment of oxidation state may be done easily. For example:
HO + HN03 ---~ N02 + 0 2 + ~0
Since hydrogen is assigned a+ l, the reactant Cl is a -1 charge. The elemental diatomic
chlorine product is assigned a charge of 0. Apparently, the chlorine has undergone an
oxidation, loss of electrons, to go from -1 to 0. For every oxidation, there must be a
reduction. Inspection of the nitrogen in the nitric acid shows a reduction from +5 to +4.
The reaction above is not balanced. It could be balanced easily by inspection. It could also
be balanced by examining the number of electrons exchanged. Realistically, the number of
electrons lost must equal the number gained. Free electrons do not fly about freely, nor do
the appear magically. In the above reaction, nitrogen gains one electron, but two chlorine's
give an electron. .. for a total of two electrons donated. This violation of conservation of
charge and electrons must be accounted for. Luckily, the electron is a useful tool in
balancing these reactions.
Since the number of electrons lost and gained must be equal, balancing each half reaction (
the reduction and the oxidation) and then combining them will balance the equation.
Examine the following redox equation:
Cr2Clr "2 +
r
---~ cr•3 + 12 (s) (in acidic solution)
Using common rules for assigning charge, it can be determined quickly that the chromium
is being reduced and the iodine is being oxidized:
-
1·
Cr2Clr
-l
---~
I2 (oxidation)
---~ cr•3 . ( reduction)
Each of these half reactions may be balanced separately and then combined to remake the
complete.. and now balanced net ionic equation. The following rules may be applied:
1.
2.
3.
4.
Balance the main species in each half reaction first.
Balance the oxygen by adding water to either side (these are done in solution)
Balance the hydrogen by adding H+ to either side (the solution is acidic)
Determine TarAL charge of each side and add electrons where necessary to make
each side equal
·
5. · Do whatever (mathematically) is necessary to make the number of electrons lost= gain.
6. Combine the two halves and simplify where needed.
13
The previous steps will now be applied to the equation above:
step 1: balancing the main species:
zr
Cr2 0, -2
--~
------:)
12
zcr•3
step 2: balance the oxygen:
· 21"
---~I.,.._
Cr20,-2 ----~ 2Cr•3 + 7 HZO
step 3: balancing the hydrogen:
21"
14 H+. + Cr20,"2
----~
12
---~ 2Cr•3
+ 7H20
step 4: balancing the charge:
ZI" ••• ;:,. 12 + 2e6e· + 14 H+ + Cr20, -2 --~-=... 2Cr +J + 7 H20
step 5: making the electrons equal:
multiply the flrst reaction by~····
61- --:=;., 312
+ 6e-
6e- + 14 H+ + Cr20, -2 ----.>2 cr•3 + 7 H20
notice that the number of electrons in each half reactioa is now the same
step 6: combine the equations and simplify.
The number of electrons on the left and right are equal and may be canceled. The resulting
equation is now:
+ Cr 0, -2 ----=> 2Cr•3 + 31., + 7H,O
AND THE RFACilON IS BAlANCED IN ACIDIC SOLlJTION!
61- + 14H+
However, if the sollftion were basic, 3 more steps must now be added:
7. convert all H+ to H OH ( water) by adding OH- to each of them
8. for every OH- you add to the left, you must add to the right or visa versa
9. simplify
to continue in basic solution:
61- + 14H+
+ Cr2 0, -2 ----~ 2Cr•3 + 312 + ~0
step 7 and 8: converting H+ to water and balancing the hydroxide:
14
adding l40H- to the left means adding l40H- to the right:
61- + 14 HOH + Cr20, ·2 ~--~ 2 cr•3 + 3I2 + 7 ~0 + 14 ORstep 9: simplify.
since there are 14 waters on the left and 7 waters on the right, 7 may be eliminated from
each side. The final balanced equation in base is:
. 61- + 7HOH + <;~<?r ·2 ----~ 2 Cr +l + 3I~..:_ 14 OHNOW IT IS BALANCED IN BASIC SOLunON
Now, with a balanced equation the typical calculations of stoichiometry may be applied.
Remember that in stoichiometry, the task is to convert any given material into moles so that
you may use the ratio from your newly balanced equation to find your new information.
There will be some reinforcement in the assignment package.
-
15
2010 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
CHEMISTRY
PartB
Time-40 minutes
NO CALCULATORS MAY BE USED FOR PART B.
Answer Question 4 below. The Section IT score weighting for this question is I 0 percent.
4. For each of the following three reactions, write a balanced equation for the reaction in part (i) and answer the
question about the reaction in part (ii). In part (i), coefficients should be in terms of lowest whole numbers.
Assume that solutions are aqueous unless otherwise indicated. Represent substances in solutions as ions if the
substances are extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction.
You may use the empty space at the bottom of the next page for scratch work, but only equations that are written
in the answer boxes provided will be scored.
EXAMPLE:
A strip of magnesium metal is added to a solution of silver(!) nitrate.
(i) Balanced equation:
M~ -t ~A-t t" ~
Mg. Z"f"
1-
2. ~
(ii) Which substance is oxidized in the reaction?
M~ ja, ¢tJIA;JY0.
(a) A 0.2 M potassium hydroxide solution is titrated with a 0.1 M nitric acid solution.
I 0) Balanced equation'
(ii) What would be observed if the solution was titrated well past the equivalence point using bromthymol
blue as the indicator? (Bromthymol blue is yellow in acidic solution and blue in basic solution.)
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-9-
2010 APe CHEMISTRY FREE-RESPONSE QUESTIONS
(b) Propane is burned completely in excess oxygen gas.
I (i) Balanced equation:
(ii) When the products of the reaction are bubbled through distilled water, is the resulting solution neutral,
acidic, or basic? Explain.
(c) A solution of hydrogen peroxide is heated, and a gas is produced.
I (i) Balanced equation:
(ii) Identify the oxidation state of oxygen in hydrogen peroxide.
YOU MAY USE THE SPACE BELOW FOR SCRATCH WORK, BUT ONLY EQUATIONS
THAT ARE WRITTEN IN THE ANSWER BOXES PROVIDED WILL BE SCORED.
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-10-
AP® CHEMISTRY
2010 SCORING GUIDELINES
Ouestion4
(15 points)
For each of the following three reactions, write a balanced equation for the reaction in part (i) and answer the
question about the reaction in part (ii). In part (i), coefficients should be in tenns of lowest whole numbers.
Assume that solutions are aqueous unless otherwise indicated. Represent substances in solutions as ions if the
substances are extensively ionized. Omit fonnulas for any ions or molecules that are unchanged by the
reaction. You may use the empty space at the bottom of the next page for scratch work, but only equations
that are written in the answer boxes provided will be scored.
EXAMPLE:
A strip of magnesium metal is added to a solution of silver(l) nitrate.
(i) Balanced equation:
Mlj
-t ~A-t.,. ~
Mg. t.-r r z ~
(ii) Which substance is oxidized in the reaction?
M~ ktjitPr~.
(a) A 0 .2 M potassium hydroxide solution is titrated with a 0.1 M nitric acid solution.
(i) Balanced equation:
.
-
One point is earned for each correct reactant.
H 3o+ + OH- ---+ 2 H 20
OR
One point is earned for the correct product.
One point is earned for correctly balancing (mass
and charge) the equation .
(ii) What would be observed if the solution was titrated well past the equivalence point using bromthymol
blue as the indicator? (Bromthymol blue is yellow in acidic solution and blue in basic solution.)
The solution would appear yellow.
One point is earned for the correct description of the solution.
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APe CHEMISTRY
2010 SCORING GUIDELINES
Question 4 (continued)
(b) Propane is burned completely in excess oxygen gas.
One point is earned for both correct reactants.
(i) Balanced equation:
C 3H 8 + 5 0 2
~
3 C02 + 4 H20
Two points are earned for the correct products.
One point is earned for correctly balancing the equation.
(ii) When the products of the reaction are bubbled through distilled water, is the resulting solution neutral,
acidic, or basic? Explain.
The resulting solution would be acidic because C02
reacts with water as a weak acid.
One point is earned for the correct choice with
justification.
(c) A solution of hydrogen peroxide is heated, and a gas is produced.
One point is earned for the correct reactant.
(i) Balanced equation:
2 H 20 2
--+
2 H 20 + 0 2
Two points are earned for the correct products.
One point is earned for correctly balancing the equation.
(ii) Identify the oxidation state of oxygen in hydrogen peroxide.
The oxidation state of 0 in H20 2 is -1 .
One point is earned for the correct oxidation state.
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2006 AP~~' CHEMISTRY FREE-RESPONSE QUESTIONS
CHEMISTRY
PartB
Time-50 minutes
NO CALCULATORS MAY BE USED FOR PART B.
Answer Question 4 below. The Section II score weighting for this question is 15 percent.
4. Write the formulas to show the reactants and the products for any FIVE of the laboratory situations described
below. No more than five choices will be graded. In all cases, a reaction occurs. Assume that solutions are
aqueous unless otherwise indicated. Represent substances in solution as ions if the substances are extensively
ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. You need not balance the
equations.
Example: A strip of magnesium is added to a solution of silver nitrate.
(a) Solid potassium chlorate is strongly heated.
(b) Solid silver chloride is added to a solution of concentrated hydrochloric acid.
(c) A solution of ethanoic (acetic) acid is added to a solution of barium hydroxide.
(d) Ammonia gas is bubbled into a solution of hydrofluoric acid.
(e) Zinc metal is placed in a solution of copper(II) sulfate.
(f) Hydrogen phosphide (phosphine) gas is added to boron trichloride gas.
(g) A solution of nickel(II) bromide is added to a solution of potassium hydroxide.
(h) Hexane is combusted in air.
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9
AP® CHEMISTRY
2006 SCORING GUIDELINES
Question 4
4. Write the formulas to show the reactants and the products for any FIVE of the laboratory situations described
below. Answers to more than five choices will not be graded. In all cases, a reaction occurs. Assume that
solutions are aqueous unless otherwise indicated. Represent substances in solution as ions if the substances are
extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. You need not
balance the equations.
General Scoring: Three points are earned for each reaction: 1 point for correct reactant(s) and 2 points for
correct product(s). Designation of physical states is not required.
(a) Solid potassium chlorate is strongly heated.
(b) Solid silver chloride is added to a solution of concentrated hydrochloric acid.
AgCl + Cl-
~
[AgC1 2
r
(c) A solution of ethanoic (acetic) acid is added to a solution of barium hydroxide.
(d) Ammonia gas is bubbled into a solution of hydrofluoric acid.
(e) Zinc metal is placed in a solution of copper(II) sulfate.
Zn + Cu 2+ ~ Zn 2+ + Cu
(f) Hydrogen phosphide (phosphine) gas is added to boron trichloride gas.
Note: PH 3BC1 3 also acceptable as a product.
(g) A solution of nickel(II) bromide is added to a solution of potassium hydroxide.
(h) Hexane is combusted in air.
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8
Electrochemistry .
lesson 4
In an electrolytic cell electrical energy is used to bring about a nonspontaneous electrical
change. Electrolysis always results in a redox chemical reaction.
electrochemical cells, both primary (voltaic) and electrolytic (secondary), consist of an
electrolyte and electrodes. An electrolyte is a substance that exists as ions in either the
molten fused) state or in water solution. The electrodes consist of an anode and a cathode
which are generally metals or graphite, C, rods. The electrodes conduct electrons to an
external circuit which in electrolysis is a source of energy. Electrons always flow from the
anode to the cathode.
The anode is always where oxidation occurs. The oxidation reaction is the most
spontaneous reaction of the reducing agents (loss of electrons). The cathode is always
where reduction occurs. The reduction reaction involves the most active of the given
oxidizing agents.
Inert electrodes that will not take·part in tlie electrolysis reaction are often selected.
common examples are graphite,C, platinum, Pt, gold, Au, stainless steel or nichrome wire.
Anodes which are not inert will oxidize during electrolysis
0
0
Copper
Zinc
.
Figure 3: the .voltaic ceU
(
In a voltaic cell (figure 3), electrical energy is produced by a spontaneous oxidationreduction reaction., this spontaneous reaction will occur whenever two dissimilar metallic
electrodes are immersed in an electrolyte and connected to an external circuit.
·
The electrodes of a voltaic cell have charges which are opposite to those of an electrolytic
cell. After all, it is working in an opposite manner. The anode is the negative electrode and
the cathode is the positive electrode. The flow of electrons is from the anode to the
16
cathode. The figure above shows the zinc and copper electrodes. The electrolytes in each
half-cell include Zn+2 and Cu+2 ( in salt form)
Zinc is the anode,-and the oxidation of zinc to Zn+2 occurs at the electrode. The reduction
of copper ions. CU+2, to copper occurs at the·other electrode, which is the cathode. the
electrons move from the anode to the cathode in the external circuit of every voltaic cell.
A "salt bridge" containing a nonreactive electrolyte completes the circuit. The salt bridge
allows current to flow but prevents contact between zinc and copper, which would shortcircuit the cell. Potassium chloride solution is used in this system. The positive ions in the
salt bridge move from the anode to the cathode because reduction is depleting the
concentration of positive ions at the cathode. The negative ions move from the cathode to
the anode to balance the charge since oxidation is increasing the concentration of positive
ions. The above voltaic cell may be represented as:
Zn/Zn+2 II Cu+21Cu (oxidation /saltbridgelreduction)
Tables of standard reduction potentials are used to determine cell voltages and to predict
whether a redox reaction is spontaneous. The standard conditions for these reactions are:
1.0 molar concentration, 1atm and 2SOC
The more positive the standard reduction potential, the stronger is the oxidizing agent and
the weaker is its conjugate reducing agent. The "E'' for an oxidation half reaction is the
standard reduction potential with the opposite sign. The table of standard electrode
potentials is supplied with this packet, and will be supplied on the AP exam. All values are
compared to hydrogen, whose reduction potential has been assigned a value of O.OOV
Notice that on your table, the large (more positive) the value of E, the stronger the
oxidizing agent ( oxidizing agents like to be reduced).
The cell voltage is determined by adding the potentials of the oxidation and reduction halfreactions. The half-reaction potential does not change when the reaction is multiplied to
balance the electrons.
·
If the cell voltage is positive, the reaction will occur spontaneously. if the voltaic cell has a
negative potential, the reverse reaction will be spontaneous. When E is 0, the reaction will
be at equilibrium in the standard state.
Use the table to determine which will be oxidized by Ni+2: Ag, Br, Cu, F2, Fe. Your
answer will be Fe. It has an oxidation potential of +.45volts, high enough to give a
positive overall potential;
Fe --~ Fe -+-2 + 2e
E= +.45V
Ni+1 + 2e --~ Ni
E=- .26V
Fe- + Ni+z __,.. Fe +2 + Ni
....
E(total) = +.19volts
I
An electrochemical cell is a system in the process of achieving chemical equilibrium. The
highest voltage occurs when the :system is far from equilibritim. LeChatelier's Principle is
used to analyze the effect of stresses which are applied or removed. Cell potential is
independent of the size of the electrodes, the size of the cells and the volume of electrolyte
(since the constant concentration of solids and liquids does not affect the kinetics of the
equilibrium pr:ocess.. they are not included in the equation). concentration does affect the
voltage. higher reactant and lower product concentration will increase cell potential.
the Nemst equation is used to determine the voltage of a voltaic cell when the
concentrations are not 1.0 molar:
l7
E = E' - .0592/n log Q
Where E is the new voltage, E' is the voltage at standard conditions, n is the number of
exchanged electrons, and Q is the concentrations of the reactants and products:
Q = products I reac.tanta
Calculate the voltage of the copper/zinc battery in the figure above. You should arrive at
the answer: 1.10 volts at standard conditions. Calculate the voltage if zinc+ 2 is .OSmolar
and copper +2 is 1.0 molar:
E = 1.10- .0592/ 2 log .OSM I l.OM = 1.14 volts (2moles of electrons transferred)
In an electrolytic cell, energy is used to bring about nonspontaneous redox. reactions such
as:
2Na+ + 2Cl-
-----~
2Na (s) + C12 (gas)
E' =-4.474 V
This reaction has a negative potential, and must be forced in this direction.
Many electrolytic reactions are done in aqueous solution. Under These circumstances, all
reactions must be considered: including the oxidation and reduction of water! For example,
the electrolysis of sodium chloride in water. Will this solution produce sodium metal and
chlorine gas?
reduction possibilities:
Na+ +
1~ -----~
Na (s)
H 20 + 2e -------------~ ~
E' =- 2.714 V
E' =- .828V
The reduction of the water is more spontaneous (more positive, less negative) and will
occur. The reduction of the sodium will not occur.
· Oxicbition possibilities:
20· -------")1 Cl2 + 2e
E' = - 1.36 V
E'
= - 1.229 V
The oxidation of water is more spontaneous (more positive, less negative) and will occur.
The oxidation of chlori~e to form chlorine gas will not occur. In this case, the electrolysis
of sodium chloride is not possible in aqueous solution. The electrolysis of sodium chloride
must be done in the molten state.
The amount of product formed in an electrolytic cell may be done stoichiometrically as in
1
any other reaction qantity.
Na+ + le
:;:. Na (s)
knowing the moles of electrons put in to the reactions will ~ow you to calculate the moles
of sodium you will get out. In this case, the ratio is one to one. To help solve these types
of problems the following information is needed:
1. one mole of electrons= 96485 Coulombs (C)
2. number of coulombs = number of amps x. number of seconds
3. number of joules= number of coulombs x number of volts
18
Consider this stoichiometric problem:
How much-sliver Ia plated from a
2.60 AmpS- In one hour?
.
Ag+ + 1e
--~
AgNO~
solution by a current of
Ag(s)
=(2.60) (3600) =9360 C
mole of e =9360 C x 1mole e /96485 C =JH70 mole e
answer: coulombs
And according to the equation, the ratio Ia 1:1 therefore;
.CY70 mole e x 1mole Ag I 1mole e x 107.9g Ag I 1 mole Ag =10.5 grams
-
19
2010 A P® CHEMISTRY FREE-RESPONSE QUESTIONS
2 Al(s) + 3 Zn 2+(aq) ~ 2 Al 3+(aq) + 3 Zn(s)
6. Respond to the following statements and questions that relate to the species and the reaction represented above.
(a) Write the complete electron configuration (e.g., l s2 2s 2
... )
for Zn 2+.
(b) Which species, Zn or Zn 2+, has the greater ionization energy? Justify your answer.
(c) Identify the species that is oxidized in the reaction.
The diagram below shows a galvanic cell based on the reaction. Assume that the temperature is 25°C.
Voltmeter
j
l
Salt Bridge
K+
N0 3
f-.;
AI
1'-'
-
t--
No 3-
No 3-
Al3+
No 3-
~
1'-'
Zn2+
No 3-
Zn
No 3-
(d) The diagram includes a salt bridge that is filled with a saturated solution of KN0 3 . Describe what happens
in the salt bridge as the cell operates.
(e) Determine the value of the standard voltage, £0, for the cell.
(f) Indicate whether the value of the standard free-energy change, I:!.G 0 , for the cell reaction is positive,
negative, or zero. Justify your answer.
(g) If the concentration of Al(N0 3h in the Al(s)/At 3+(aq) half-cell is lowered from 1.0 M to 0.01 Mat 25°C,
does the cell voltage increase, decrease, or remain the same? Justify your answer.
STOP
END OF EXAM
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-12-
AP® CHEMISTRY
2010 SCORING GUIDELINES
Question 6
(9 points)
2 Al(s) + 3 Zn 2+(aq) ~ 2 AJ 3+(aq) + 3 Zn(s)
Respond to the following statements and questions that relate to the species and the reaction represented
above.
(a) Write the complete electron configuration (e.g., I s 2 2s2 ... ) for Zn 2+.
One point is earned for the correct configuration.
(b) Which species, Zn or Zn 2+, has the greater ionization energy? Justify your answer.
Zn 2+ has the greater ionization energy. The electron being
removed from Zn 2+ experiences a larger effective nuclear
charge than the electron being removed from Zn because
Zn 2+ has two fewer electrons shielding the nucleus.
One point is earned for identifying Zn 2+
with justification.
OR
It takes more energy to remove a negatively charged electron
from a positive ion than from a neutral atom.
(c) Identify the species that is oxidized in the reaction.
Al(s)
One point is earned for identifying AI.
··~ ··
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2010 SCORING GUIDELINES
Question 6 (continued)
The diagram below shows a galvanic cell based on the reaction . Assume that the temperature is 25°C.
Voltmeter
j
l
Salt Bridge
No3K+
~
N '-
AI
'·
-
'--'
-
·"~
·-..
- r._..
No3-
No3-
Af3+
Zn 2+
No 3- No3-
Zn
! ''~;;
'--
NO3
(d) The diagram incl udes a salt bridge that is filled with a saturated solution of KN0 3 . Describe what
happens in the salt bridge as the cell operates.
As the cell operates, N0 3- ions flow toward the AI half-cell
and K+ ions flow toward the Zn half-cell .
One point is earned for correctly
indicating the direction of ion flow.
(e) Determine the value of the standard voltage, E', for the cell.
E0
= (-0.76 V)- (-1.66 V) = 0.90 V
One point is earned for the correct Eo .
(f) Indicate whether the value of the standard free-energy change, ;').G 0 , for the cell reaction is positive,
negative, or zero. Justify your answer.
;').Go
is negative since E0 is positive and
;').G o
= -n '?f £0.
OR
f').G 0 must be negative because the reaction is spontaneous
under standard conditions.
One point is earned for indicating that
f').G 0 is negative.
One point is earned for a correct
justification.
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AP® CHEMISTRY
2010 SCORING GUIDELINES
Question 6 (continued)
(g) If the concentration of AI(N03 h in the Al(s)/AJ 3 +(aq) half-cell is lowered from I .0 M to 0.01 Mat
25°C, does the cell voltage increase, decrease, or remain the same? Justify your answer.
Lowering LAJ 3+1causes an increase in the cell voltage.
The value of Q will fall below 1.0 and the log term in the Nernst
equation will become negative. This causes the value of Eceu to
become more positive.
OR
A decrease in a product concentration will increase the spontaneity
One point is earned for indicating
that E cell increases.
One point is earned for the correct
justification.
of the reaction, increasing the value of E ceLL·
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.E quilibrium
less9n 5
The equilibrium constant, K. expresses arithmetically the extent to which a reaction will
proceed at a given temperature~ The equilibrium law is defined in terms of the
concentrations or, in the case of gases, the pressures that exist at equilibrium. The
equilibrium law expression takes two forms:
l. when concentrations are expressed as ·molarities:
Kc = molar concentrations of products I molar concentrations of
reactant
aA + bB ---~
...- cC + dD
=
Kc (A)a (B)b I (C)c (D)d
2. when concentrations are expressed as partial pressures:
Kp
=Atm
pressures products I atm pressures of reactants
aA + bB --.;;;:> cC + dD
Kp may be related to Kc via:
Kc
= Kp (1/RT) .v.
Where R is the gas constant: .00 ll..atm/nK, T is the temperature (K) and &1 is the change
in moles (products- reactants). In the equation:
·
2NH3
N2 + 3 H2 -----:>r
~
The An Is : 2 - 4
= -2
In an equilibrium system, if the concentration of one of the constituents changes at constant
temperature, the equilibrium will not be reestablished until all of the equilibrium
concentrations change and the ratio of products over reactants is equal to K. To keep tract
of the changes occUlTing as equilibrium is established or re-established, the concentrations
changes are foUowed by using a table that shows the start, the change and the equilibrium.
In many instances, the problems will seem to require that the complete soluti9n will involve
the use of the quadrAtic equation (thank heavens.for programmable calculators). Only in
cases where K is less than .01 or greater than 100 may simplifications be made.
For example: starting with .2M HI decomposing according to:
=
2lll ~ ~ -+ 12 : Kc .016
Establish a start, change, equilibrium table:
Start:
change
equilibrium
2Hl
-:F
.2M
-2x
.2-2x
H2
0
· +X
x
+
12
0
+x
x
20
The equilibrium expression:
Kc
=(x)(x) I (.2 - x~
simplification yields:
Kc
solving for x:
= x I .2 • 2x =.018
.020
Therefore, at equilibrium, the concentration of HI is .2- .02 =.160M, and the
concentration of the ~ and 12 are .02 M
X=
In general, if K is large. the equilibrium lies far to the right (products favored). If K is
very small, the equilibrium favors the reactants.
Equilibrium is affected as defined by LeChatelier:s Principle. The effect of changes in
pressure, concentration and temperature on the constituent species and on the equilibrium
constant can be predicted. LeChatlelier's Principle states that a system in equilibrium will
·
react to a stress in a way that relieves the stress. Specifically:
1. if the concentration of a chemical on the reactant side of the reaction is increased, the
reaction to products is favored with a higher rate. The concentration of the reactants is
reduced, relieving the stress. This will result in a shift toward increased concentration of
the products.... and visa versa. ...
2. If the pressure of a system containing gaseous molecules is inc~ then the number
of molecules must e reduced to relieve the stress. The reaction will shift to the side with the
smallest moles of gaseous_molecules. An increase in volume has an immediate effect to
. lower concentration of molecules. To counteract this, reactions occur which increase the
number of molecules in the gas phase.
3. The reaction to products is favored because of a higher forward reaction rate when heat
is added to an endothermic reaction(heat is on the reactant side of the reaction). The heat is
"used up" by the endothermic reaction, relieving the stress. .The reaction rates is equal after
a shift toward higher concentration of products and a lower concentration of the reactants.
The equilibrium constant, K, is larger when heat is added to an endothermic reaction.
When heat is added to an exothermic reaction (heat is on the product side of the reaction),
the reaction rate of the products is favored. The result is a decreased product concentration
and an increased reactant concentration. The equilibrium constant, K, is smaller.
4. Catalysts do not affect equilibrium concentrations. The addition of the catalyst speeds
up the rate of both the forward and reverse reactions. At equilibrium these reaction rates
are already equal-. 'Qle effect of the catalyst is to make these equal rates fast~r.
The determination of the direction of a reaction may be done using Q, which has the same
form of K, but is not at equilibriUIIL In comparing Q to K, the direction of the reaction
may be determined. if Q is less thanK, the systems moves to the right (products formed).
If Q is greater thanK, the system moves to the left (reactants formed).
21
Solubility and the Ksp constant
Equilibrium may also be established for the formation of an insoluble product. When a
slightly soluble of insoluble salt is mixed with water, a saturated solution results and
solubility equilibrium is established. The rate of dissociation of the ions from the solid
equals the rate of precipitation of the salt. The equilibrium rules apply to the dissolving of
slightly soluble salts to form saturated solutions:
salts ___.:- anions + cations
~
The unique equilibrium constan~ solubility product (Ksp) is used to describe this common
equilibrium condition. The equation for Ksp takes the form:
Ksp
= products I
reactants
·since solids, liquids and water are not included in the equilibrium equation, the Ksp
becomes:
Ksp. ::a products
=(anlons)(catlons)
For example:
start:
change
equilibrium
BaCr04(s) •._..~
Original X
- .0013
X-.0013
Cr04-2
0
+.0013
.0013
Ba+2 +
0
+.0013
.0013
Ksp expression ::a (Ba+2)(Cr04 -2)
= (.0013)(.0013) = .00000169
Common Ion effect:
it is possible to shift the solubility equilibrium to favor the reactants. this is accomplished
by adding a solution which contain an ion in common with the salt. The result will be a
saturated solution which has:
l. a lower solubility of the starting solid
2. more undissolved solid and a lower ion concentration.
for example:
What is the solubility of Mgfl? (Ksp= 6.4 X w-~
a. in pure water'
b. in a .1M solution of NaF
-
In a, there is no common ion effect. Therefore;
____
Mg+2 +
Mgfl
~
0.0
original
start:
-X
+X
change
X
orig-x
equilibrium
expression: Ksp
...
2F
0.0
+2X
2x
=(x)(2x)2 =6.4 x 10·
9
x = 1.2 x 10"3 M TinS IS THE MOLAR SOLUBIUTY IN PURE WATER
22
b. In NaF, there is the common ion: .10M F
start
change
equilibrium
___...,_
Mgfz
original
~
Mg+2 +
0.0
-X
+X
orig- x
X
2F
.10M
+2x
.10 + 2x
expression: Ksp = (x)(.10-2x)2 ...... worth investing in a programmable! However you
can approximate if you assume that the 2x is small compared to .1. In any case, the answer
for the molar solubility in Naf is:
6.4 x 10"7M .....• this is only .05% of the solubility in pure water. Quite a change!
The addition of a common ion is a common practice in removing more of the desired
species from solution. In the above case, the addition of NaF greatly increaSes the amount
of Mgf2 that may be recovered from this solution.
In many reactions, two reactants are combined in the hopes of producing an insoluble
product.. This may not always be the case. It should now be obvious that in order for a
precipitate to form, the concentrations of the ions in solution must be equal to, or less than
the solubility product constant Ksp. This is another application ofthe non-equilibrium
entity: Q. If the noa-equilibrium product coastant (Q) is greater thaD the
Ksp, the solution has exceeded saturation. Precipitation of the solute will
occur until the Ksp = Q. If the Q is less than Ksp, the solution is
onsaturatecL Precipitatioa will not occur until the Q is greater than Ksp.
examine the following problem:
A solution with a final volume of20ml is made by mixing 10 ml of .10M Pb(N03 ) 2 and 10
ml of .001M NazS04 • The Ksp of PbS04 is 1.06 x 1o-s. Will a precipitate form?
=
Concentration of Pb+2 .O.SOM
Concentration of S04-2 = .OOO.SOM
=
=
and Q (.050)(.00050) 2.5 x to·'
This value for Q is greater than the Ksp. A precipitate will form.
K as a dissociation constant:
So far, K has been used to see to what extent a reaction wilt' proceed and to determine an
objects solubility. Now, K may also be used to determine the extent of dissociation of an
acid or a base..
Remember that the extent of dissociation determines the strength of the acid or base.
Strong acids dissociate nearly 100%, as do stro~g bases. For this reason, discussion of
the dissociation of mese items is pointless:
f
STRONG ACIDS AND BASES DISSOCIATE STRONGLY
However, the dissociation of weak acids and bases is much different. Luckily, much like
solubility, the amount, or degree to which an acid or a base may dissociate can be
determined through K.
Water is the most common item to examine.. since it is both acid and base. It dissociates
according to the equation:
H 20 --~~ H+ + OH23
Its dissociation constan~ K. has the fonn of products over reactants (but water never getsincluded in the equation).
Kw
=(H+)(OH-)
And has the numerical value: 1 x 10" 14 = K w. This simply means, that in a water
.
solution, the concentration of the hydrogen is equal to the concentration of the hydroxide,
and the numerical value for each is 1 x 10"7M.
Log scales allow us to express the concentration of hydrogen or hydroxide another way:
the pH or pOH scale, where p means -log~ In the case above, the concentration of the
hydrogen is 1 x 10"7M . The p of this His -log 1xl0"7 = 7. Recall that 7 is neutral on the
pH scale.
Other weak acids and weak bases may be represented in the same manner; In the case of
acids, the dissociation is represented by Ka. In the case of bases: Kb.
Most acids and bases are weak. The acid or base reacts with water and is in equilibrium
with the resulting ions~ the most common definition of acids and bases used in AP
Chemistry is the Bronsted-Lowry theory. Ac~ordingly, an acid is a proton donor and a
base is a proton acceptor. Water is ampbiprotic, and can act u either an. acid or a
base and its. role is determiDed by the acid or a base added to it.
The extent of an acid base equilibrium reaction is measured by Ka and Kb. the ionization
constants are determined by measuring the pH of the solution or are given as data. Ka and
Kb are related by the equation:
·
Ka x Kb
=Kw
Follow this example closely;
A .05M solution of acetic acid has a pH of 3.03. What is the value of Ka?
solution:
The concentration of the equilibrium hydrogen may be determined using the pH. Taking
the inverse log of - 3.03, the concentration of Hydrogen is determined to be:
(H+) = 9.49 X 10 4 M
Now then:
HAc ---,~
start:
change
equilibriwn
.OSM
-9.49 X Jey-4.
.OS - 9.49 X 1<r
.Ksp
=(H)(Ael I Jlae
Ksp
= 1.8 s
H+
0.0
+
+ 9.49 X 104
+9.49 X 104
= (9.49 s lr}( 9.49
Ac-
0.0
+9.49 X 104
+9.49 X 104
~ 10~ I .05 • 9.49
,s
tcr
ur•·
The acetate ion is the conjugate base of acetic acid. Its Kb may be determined as:
Ka xKb= Kw
AccordiDRiy, Kb
= 5.6 s 10 ' 10
K is also useful in espressiDR the dissociation of POLYPRariC acids. Each
proton of a polyrotic acid comes off with more dimculty thaa the previous
24
one. LeChateUer•s Principle predicts the H+ ions from a first ionizatioa
wiD repren the second ionization. The second ionization can be ignored.
The equilibrium is always written for the loss of only one proton.
-
25
2005 AP8 CHEMISTRY FREE-RESPONSE QUESTIONS
CHEMISTRY
Section II
(Total time-90 minutes)
Part A
Time-40 minutes
YOU MAY USE YOUR CALCULATOR FOR PART A.
CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS.
It is to your advantage to do this, since you may obtain partial credit if you do and you will receive little or no credit if
you do not. Attention should be paid to significant figures.
Be sure to write all your answers to the questions on the lined pages following each question in the booklet with the
pink cover. Do NOT write your answers on the green insert.
Answer Question 1 below. The Section II score weighting for this question is 20 percent.
Ka
= 1.34x w- 5
I. Propanoic acid, HC 3H 50 2 , ionizes in water according to the equation above.
(a) Write the equilibrium-constant expression for the reaction.
(b) Calculate the pH of a 0.265 M solution of propanoic acid.
• ,~,
(c) A 0.496 g sample of sodium propanoate, NaC3H 50 2 , is added to a 50.0 mL sample of a 0.265 M solution
of propanoic acid. Assuming that no change in the volume of the solution occurs, calculate each of the
following.
(i) The concentration of the propanoate ion, C 3H 5o 2- (aq), in the solution
(ii) The concentration of the H+(aq) ion in the solution
The methanoate ion, HC0 2-(aq), reacts with water to form methanoic acid and hydroxide ion, as shown in the
.,.,., · following equation.
(d) Given that [OH-] is 4.18 x to- 6 Min a 0.309 M solution of sodium methanoate, calculate each of the
following.
(i) The value of Kb for the methanoate ion, HCo 2-(aq)
(ii) The value of Ka for methanoic acid, HC0 2H
(e) Which acid is stronger, propanoic acid or methanoic acid? Justify your answer.
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6
AP® CHEMISTRY
2005 SCORING GUIDELINES
Question 1
Propanoic acid, HC 3 H50 2, ionizes in water according to the equation above.
(a) Write the equilibrium-constant expression for the reaction.
K
=
a
[H+][C3Hs02-]
[HC3Hs02]
One point is earned for the correct equilibrium
expression.
Notes: Correct expression without K 0 earns 1 point.
Entering the value of K 0 is acceptable.
Charges must be correct to earn 1 point.
(b) Calculate the pH of a 0.265 M solution of propanoic acid.
HC3H50 2(aq) ~ C 3H 5o2-(aq) + H+(aq)
I
c
0.265
0
~o
-x
+x
+x
+x
+x
0.265 -x
E
[H+ ][C 3 H 50 2-]
[HC 3 H 5 0 2 ]
Ka
One point is earned for recognizing that [H+] and
[C3 H 50 2-] have the same value in the equilibrium
expression.
(x)(x)
(0.265- x)
Assume that 0.265- x ::::: 0.265 ,
then 1.34
( 1.34
3.55
X
X
}
x
x2
Io- 5
=
One point is earned for calculating [H+].
--
0.265
o- 5)(0.265) = X 2
x 10- 6
=
x
One point is earned for calculating the correct pH.
2
= [H+] = 1.88
X
10- 3 M
pH = -log [H+] = -log (1.88
X
I o-3)
pH = 2.725
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2
AP® CHEMISTRY
2005 SCORING GUIDELINES
Question 1 (continued)
(c) A 0.496 g sample of sodium propanoate, NaC 3H50 2 , is added to a 50.0 mL sample of a 0.265 Msolution
of propanoic acid . Assuming that no change in the volume ofthe solution occurs, calculate each of the
following.
(i) The concentration of the propanoate ion, C3H50 2- (aq) in the solution
mol NaC H 0 = 0.496 NaC H 0 x 1 mol NaC3H502
g
3 5 2 96.0gNaC3H502
3 5 2
mol NaC3H50 2 = 5.17
[C 3Hs0 2-]
10- 3 mol NaC 3H50 2
x
=
mol C3H50 2-
One point is earned for the
molarity ofthe solution.
mol C3H502.
vo Iume o f so Iutton
=
One point is earned for
calculating the number of
moles ofNaC3H 50 2.
(ii) The concentration of the H+(aq) ion in the solution
I
0.265
0.103
~o
C
-X
+x
E
0.265
+x
+x
-X
0.103 +X
= [H+][C3H50 2-] = (x)(0.103+x)
K
[HC 3H 50 2]
a
One point is earned for calculating the
value of [H+].
(0.265- x)
Assume that 0.1 03 + x ::::: 0.103 and 0.265 - x ::::: 0.265
(x)(0. 103 )
a
0.265
0 265
= 3.45 x 10- 5 M
x = [H+] = (1.34 x 10- 5) x ·
0.103
K
=
1.34
x
10- 5
=
The methanoate ion, HC0 2- (aq), reacts with water to form methanoic acid and hydroxide ion, as shown in the
following equation .
(d) Given that [OH-] is 4.18 x I o-6 Min a 0.309 M solution of sodium methanoate, calculate each of the
following.
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3
AP® CHEMISTRY
2005 SCORING GUIDELINES
Question 1 (continued)
(i) The value of K 6 for the methanoate ion, HC0 2- (aq)
HC0 2-(aq) + H20(/)
I
c
0
-0
-x
+x
+x
+x
+x
0.309 -x
X=
[OH-] = 4.18
X
=
=
10-6 M
X
[OH-][HC0 2 H]
[HC0 2-]
is Very small (4.18
Kb
HC0 2H + OH-(aq)
0.309
E
Kb
p
(4.18x!0-6)2
0.309
X
=
(x)(x)
(4.18x10- 6 ) 2
(0.309-x)
(0.309 -x)
1o- 6 M), therefore 0.309- X
=
5.65 x
:::::
One point is earned for substituting
4.18 x 10-6 for both [OH-] and
[HC02H], and for calculating'the
value of K 6 _
0.309
10-1 1
(ii) The value of Ka for methanoic acid, HC0 2H
·"·
....
~
= Ka x Kb
14
K = Kw = I.OOx!0a
Kb
5.65x!0- 11
Ka
= 1.77
X
One point is earned for calculating a value of Ka from
the value of K 6 determined in part (d)(i).
10- 4
(e) Which acid is stronger, propanoic acid or methanoic acid? Justify your answer.
Ka for propanoic acid is 1.34 x 10- 5, and Ka for methanoic acid is
I. 77 X 1o- . For acids, the larger the value of Ka' the greater the
,.. ... strength; therefore methanoic acid is the stronger acid because
1.77 X 10- 4 > 1.34 X 10- 5.
4
One point is earned for the correct
choice and explanation based on
the Ka calculated for methanoic
acid in part (d)(ii).
. ..
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4
Buffers
lesson 6
It may seem like overkill to dedicate a whole day on buffers. I would agree, in that the
material is simple, however, the calculations are involved, and should be given equal time.
I have never seen an exam without buffers.... and so it goes...
Special applications of K are apparent with buffers. a buffer is an aqueous solution
prepared to contain roughly equal amounts of a weak acid and its conjugate base. The
buffer resists a change of pH when hydrogen ions from a strong acid or hydroxide ions
from a strong base are added. The equilibrium buffer system works by removing the
added hydrogen or hydroxide ions by a reaction with an amount of the corresponding base
or acid in the system. The basic equation for buffered solutions which represents the
general equilibrium between the acid and its conjugate is:
-
HA ----------:::... H+ + Aand is based on the equilibrium expression (K). Therefore;
(H) = Ka (HA)/(A-) and pH = pKa + log (A-) I (HA)
The system is buffered because addition of an acid is neutralized by the A-,
addition of a base is neutralized by the HA. The pH of the system .
fluctuates only slightly as long as the buffer remain intact.
The (H+) of a buffer is close to the Ka if the amounts of the acid and conjugate base are
near to being equal, and the pH of the buffer i~ close to the pKa.
The following examples may help to remind you of the concept:
. 1.
A buffer system may be made from the combination of 1.0 M acetic acid and 1.0 M
sodium acetate. What is the pH of this buffer? (Ka = 1.8 x 10"5 ).
solution:
Concentration of the acid: 1.0M
Concentration of the salt of the acid (base): 1.0 M
Ka 1.8 x HY5
=
=
pH
4.14 + log ( 1.0)./( 1.0)
equal.. ..
=4.74 !
the pH is equal to the pKa when concentrations are
2.
What is the pH of the solution when .1 0 M HCl is added to 1.0 Liter of the buffer
prepared in the preftous problem? (the acid is going to react with the conjugate base)
We need to consider a start. change, equilibrium table for the buffer system of acid and salt
of the acid:.
Hac
start:
change .
equilibrium
1.0
+.1
1.1
---
---------==-
H+
+
original
change
o+change
Ac-
1.0 .
-. 1
.90
26
=
=
pH (-log or tu originpj + th• change)
4.74 + log (.9)1(1.1)
4.65Not a
very big change in pH••• but exactly what we would expect from a buffer/
A common way to prepare a buffer is as follows: add to a weak acid half as many moles of
strong base(like NaOH). The resulting buffer will contain equal moles of acid and
conjugate base. The buffer equation may then be used. To illustrate this point, imagine the
combination of 20 ml .2M NaOH to 50.ml of a .1M acetic acid. What will be the resulting
pH?
The reaction:
Hac + NaOH
----==- Na+ + Ac ~
+ HOH
=
moles of Hac .0714n
moles of NaOH = .0571n
moles of sodium acetate = 0.0
The reaction will reduce the amount of acid by .057lmoles and produce .0571 moles of the
sodium acetate. When the reaction is over, the resulting quantities will be:
moles of Hac= .0143n
moles of NaOH = 0 .0
moles of sodium acetate= .0571n
We now have a buffer which contains the acid, and the salt of the acid. The Buffer
equation may now be used:
pH= 4.74 +log
(.0~71)
/(.0143)
=5.35
The pH of this buffer system is 535. Notice that the use of molarity in the equation is not
necessary!
As seen in a start change equilibrium table:
· start
change
equilibrium
•
Hac +
.0714
-.0571
*.0143
NaOH
.0571
-.0571
0 .0
--~
(Na+ + Ac -)
0.0
+.0571
*.0571
+ HOH
0.0
+.0571
.0571
The acid, and the salt of the acid (conjugate base). This is now a buffer system.
This example illustrates the combination of a weak acid and a strong base. This is the same
process that takes place during a titration of a weak acid with a strong base. As long as the
moles of the base remains less that the moles of the weak acid, the buffer equation may be
used to determine the pH of the resulting solution. At the instant the moles of acid and base
are equal, the pH is determined as a function of the conjugate base, and its Kb. As soon as
the base exceeds th~moles of the weak acid, the pH will be based only on the amount of
the excess hydroxide in solution.
27
to strong ase
1. As long as the moles of the base remains less that the moles of the weak acid, the buffer
equation may be used to determine the pH of the resulting solution.
2. At the instant the moles of acid and base are equal, the pH.is determined as a function of
the conjugate base, and its Kb.
3. As soon as the base exceeds the moles of the weak add, the pH will be based only on
the amount of the excess hydroxide in solution.
This is not the case with a strong acid and a strong base. At the endpoint, where moles of
acid and base are equal, the pH is 7.
Many examples of this will be illustrated for your understanding, and there will be a few in
the problem set for you to attempt, achieve and remember. If you have any questions,
remember to e-mail them or phone them in live.
-
28
Reaction Kinetics
·tesson 7
The rate of a chemical reaction depends on the frequency and force of collisions between
molecules. changes in the concentration and the temperature of the system are observed
experimentally to eStablish a rate equation for the reaction. Conceptually speaking:
rate = change in concentration of a species/change in time
or
R
= Initial concentration-final concentration I final time - Initial time
For a hypothetical reaction:
2A + 48 ----~ 5C + 30
The rate may be expressed as a function of 112 the change in A over time, 114 the change in
B over time, 1/5 the change inC over time or 113 the change of Dover time. A great
opportunity for calculus buffs, as it may be"seen as 112 dA/dt! In our example, two A's are
reacting, and A is disappearing as time goes on. Because A is decreasing, a minus sing
will be used in the rate equation. Half of the rate of disappearance of A with time will give
the s~cific rate of the reaction. 5C's are begin made as a product, so 115 the rate of
appearance (plus sign) of C will give the same rate for the reaction. The rate will be
expressed as a positive quantity regardless of what is formed when.
In a simple one reactant equation, the rate is expressed as a change in_that reactant.
Mathematically;
Rate = k( cone. reactant)m
where k =rate constant and m is the order.
For our two reactant equation from above,
=
Rate
k(A)"'(B)n
The order of a reaction is determined experimentally and is dependent on reactant
concentration. It is shown as an exponent of the molarity, and is generally a simple whole
number. Sometimes it might be a square root.
The order may or may not be the same as the stoichiometric coefficients of the balaneed
equation. Generally they will not be the same. In the sample reaction:
m is the order with respect to A
n is the order with respect to B
m + n is the overall order.
(m-and n are positive integers ( 1,2,3 ... ) However, they can be 0 or a fraction such as 112)
.
.
The specific rate constant, k, is dependent on reaction temperature, and must be determined
experimentally.
29
For a simple single reactant problem, if the rates and concentrations at various times are
known, then the order of the reaction may be detennined as
Rate 2 I Rate 1 = (cone. 2 I cone 1)m and solving for
Once m is known, the k may be solved for using:
m.
Rate = k( concentration reactant)m
A simple approach for simple one reactant problems. The approach is more rigorous for
multiple reactant situations. when more than one reactant is involved in a reaction,k the
order is somewhat more difficult to detennine experimentally. One rather straight-forward
approach involves holding the initial concentration of one reactant constant while varying
that of the other reactant. From the rates measured under these conditions we can find the
order of the reaction with respect to the reactant whose initial concentration is changing.
Consider this example:
For the reaction Hz + 2NO
---~
N2 0 +
HzO
at SOOOC the following data is obtained:
initial cone. ~ nil
initial cone Ndnll
rate mils
exp 1
exp2
exp3
.10
.10
.12
.20
.1 0
.24
.20
.20
.96
Detennine the order with respect to both reactants.
Examine the data. The concentration of NO is constant at .10M for the first two
experiments, so we use that data to determine the order with respect to Hz
Rate 1 I Rate 2
.24/.12 = (.20/.lO)m
=(cone~ exp2 I
cone ~ exp1)
m
-·
2=2m
m= 1
The reaction is first or.:der with respect to hydrogen. To determine the order of NO, we
may use the second and third experiments where the cone. of Hz is constant at .2M, and
follow the same procedure as shown above:
.961.24 =(.2/ .1)0
4=2°
-
n=2
The reaction is second order with respect to NO. Choosing.any one of the .experiments,
substitution will yield the rate constant:
(arbitrary choice experiment 1)
.12 =k(.1)1 (.1) 2 = 120L2/n2 s(gottalovethoseunits!)
30
One of the greatest applications is detennining the concentration of various products and
reactants over time. If the reaction is first order, a Half life relationship may be used
(nuclear reactions are examples of first order reactions).
,
InN/No= kt
where No is the original amount; and N is the amount at timet.
The half-life is the amount of time required for half of a given amount of reactant to be
consumed and is independent of the initial concentration of the reactant:-
N =No /2
· therefore No = 2N and No/N = 2
our equation becomes In 2 = kt
solving fort:
t
=ln2/k =.693/k
This is the half-life expression and is most often seen as:
tl/2
=.693 I k
For a first order reaction, t 1/2 is independent of original concentrations. It takes as long for
the concentration to drop from one point as it does for another. It is also inversely related
to k. If tl/2 is small, k is large and vice versa.
what atTects rate?
l. Activation energy and catalysis!
The collision theory of reaction rates describes reactions in terms of collisions between
reacting molecules. To be effective, a collision must be both aligned properly and energetic
enough to form an intermediate species called the~' activated complex". the number of
effective collisions is a small fraction of the total number of collisions between reacting
molecules. Most of the molecules bounce off one another and remain unchanged. the "
activation energy", Ea, is the minimum energy required to cause a reaction between
molecules colliding with the proper geometry. In other words, In order for the reaction to
occur upon collision, the reactant molecules must possess a certain minimum energy.
Othetwise, the collision is elastic; nothing happens. anything that affects the activation
energy will affect the rate~
31
.!!
Product•
=
o Reactant•
~
~
Reaction Coordinate
Reaction Coordinate ·
The dia ram on the left for an exothermic reaction. Ri ht is endothermic.
A catalyst increases the reaction rate without being consumed in the reaction (trudl is it is
consumed in the fJISt step, but then regenerated in the second). This happens because a
catalyst furnishes an alternative path for a reaction with a lower activation energy. Two
additional points n~d to be stated about a catalyst:
1. The heats of reactionfor an uncatalyzed and catalyzed reactions are the same. The
energy of the reactants and products is the same for both reactions.
2. The step of a catalyzed reaction with the highest activation energy is the rate determining
(slowest) step. there are fewer molecules with the required energy.
2. Temperature:
In general, increase in temperature increases rate. Rate is approximately doubles when T
increases by 100C. The first quantitative relationship between the rate constant, k, and
temperature, T, may be shown via the Arrhenius Equation. An integrated version of the
Arrhenius equation is used to determine:
1. the value of the rate constant at different temperatures when the activation energy is
known.
··
·
2. the value of the activation energy when the rate constants at two temperatures are
known.
The equation has the form:
In k/k7 = - Ea/R (lff2 - lff 1)
The value of the gas constant, R, used to calculate the activation energy i~ usually 831
joulelmoleK
32
Reactioa Mechanisms:
Most reactions take place in a series of steps. A series of steps (usually 2-4) that add up to
the overall reaction is called the reaction mechanism. This ordinarily results in a rather
complex rate expression. To fmd the rate expression for a multi-step mechanism:
1. focus on the slow step: assume rate for that step is the overall rate
2. Coefficients in the rate determining step = order of the reaction
3. Eliminate any unstable intermediates from the rate expression
There will be many. examples shown, and a few in the packet to help you strengthen you
understanding of mechanis~s. i' , , ·
..
<
·:
33
_ - - -- - - - - _ __ _
_ .......... ,.... ..,.
"aw~a;.
~u-=-~
£ &V&,..,
3. Answer the following questions related to the kinetics of chemical reactions.
IO - (aq) + CI -(aq)
Iodide ion, I - , is oxidized to hypoiodite ion, Io-, by hypochlorite, Cio-, in basic solution according to the
equation above. Three initial-rate experiments were conducted; the results are shown in the following table.
,,,.
(
Experimeryt
'
'[I-]
tm~l
t-')
[CIO-]
(p1ol L- 1)
Initial Rate of
Formation of Io(mol L-1 s-')
I
0.017
0.015
0.156
2
0.052
0.015
0.476
3
0.016
0.061
0.596
(a) Determine the order of the reaction with respect to each reactant listed below . Show your work.
(i) I -(aq)
(ii) ClO -(aq)
(b) For the reaction,
(i) write the rate law that is consistent with the calculations in part (a);
(ii) calculate the value of the specific rate constant, k, and specify units.
The catalyzed decomposition of hydrogen peroxide, H 20 2(aq), is represented by the following equation.
catalyst
The kinetics of the decomposition reaction were studied and the analysis of the results show that it is a
first-order reaction. Some of the experimental data are shown in the table below.
Time . .,...'"
(minutes)
'·. lfl202]
(nl()l
' ::~~:;;::\
L- 1)
1.00
0.0
0.78
5.0
0.61
10.0
:
..
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8
2005 APe CHEMISTRY FREE-RESPONSE QUESTIONS
(c) During the analysis of the data, the graph below was produced.
Time (minutes)
(i) Label the vertical axis of the graph.
(ii) What are the units of the rate constant, k, for the decomposition of H 20 2(aq) ?
(iii) On the graph, draw the line that represents the plot of the uncatalyzed first-order decomposition
of 1.00 M H 20 2(aq) .
STOP
If you finish before time is called, you may check your work on this part only.
Do not turn to the other part of the test until you are told to do so.
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9
AP® CHEMISTRY
2005 SCORING GUIDELINES
Question 3
Answer the following questions related to the kinetics of chemical reactions.
ow
Iodide ion, 1-, is oxidized to hypoiodite ion, 10-, by hypochlorite, CIO-, in basic solution according to the
equation above. Three initial-rate experiments were conducted; the results are shown in the following table.
'i,
(mol L- 1 y~
[ClO-]
(mol L- 1)
lnitial Rate'of
Fonnation of 10(mol L -t s- 1)
I
0.017
0.015
0.156
2
0.052
0.015
0.476
3
0.016
0.061
0.596
~
'
Experime~t
..
[q
(a) Determine the order of the reaction with respect to each reactant listed below. Show your work.
(i) qaq)
From experiments I and 2:
rate
k[I-]2 [CIO- ]~
rate!
k[l- ]f [CIO- ]j
- -2 = --='----=0.476
. ,.
0.156
3.05
=
One point is earned for the correct order of the
reaction with respect to r, with justification.
k(0.052t (0.0 IS)Y
=--'----'---'----'-
k(O.OI7)x(O.OI5)Y
(0.052)x
=
3.1x, therefore x
=
1,
(0.017t
The reaction is first order with respect to
r.
From experiments 1 and 3:
rate3 - k[r]; [Clo- ]r
rate 1 k[r] ~ [ClO-]j
~-· .
0.596 k(O .O 16)x(0.06 1)Y
--=
0.156
k(O .O17)x (0.0 15)Y
k(0.016)' (0.06 J)Y
One point is earned for the correct order of the
k(O .O17)' (0.0 15)Y
reaction with respect to CIO-, with justification.
=---'--_...:..~-~
3.82 = (0.94) (0.0 61 )Y
(0.0 15)Y
J!Jr!·... ,
4.06 = 4.1 Y, SOy = 1,
The reaction is first order with respect to CIO-.
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7
AP® CHEMISTRY
2005 SCORING GUIDELINES
Question 3 (continued)
(b) For the reaction
(i) write the rate law that is consistent with the calculations in part (a);
One point is earned for the correct rate law based on
exponents as determined in part (a).
(ii) calculate the value of the specific rate constant, k, and specify units.
k
k=
rate
[1-] [CI0-] 1
1
One point is earned for the value of k.
0.156 mol L- 1 s- 1
One point is earned for the correct units
(consistent with orders found).
(0.017 mol L- 1)(0.015 mol L- 1)
The catalyzed decomposition of hydrogen peroxide, H20 2(aq), is represented by the following equation.
catalyst
2 H 20(/) + 0 2(g)
The kinetics of the decomposition reaction were studied and the analysis of the results show that it is a firstorder reaction. Some of the experimental data are shown in the table below.
[Hz02]
(mol L- 1)
Time
(minutes)
1.00
0.0
0.78
5.0
0.61
10.0
(c) During the analysis of the data, the graph below was produced.
Time (minutes)
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8
AP® CHEMISTRY
2005 SCORING GUIDELINES
Question 3 (continued)
(i) Label the vertical axis of the graph.
One point is earned for they-axis label.
(ii) What are the units of the rate constant, k, for the decomposition of H20 2(aq)?
One point is earned for the correct units for k.
(iii) On the graph, draw the line that represents the plot of the uncatalyzed first-order decomposition
of 1.00 MH 20 2 (aq).
Two points are earned for all three features (same origin,
straight line, smaller negative slope), or one point for
any two features .
The line should have the same origin, be a straight
line, and have a smaller negative slope.
.....~.~.I.J::I.?.S?..?.J .....
Time (minutes)
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9
Thermodynamics
lesson 8
The first law of thermodynamics states that energy can be converted from one form into
another but cannot be created or destroyed. The energy is either in the form of heat or
work. heat energy can be converted into work, and work can be converted into heat. The
system does not contain heat or work• .heat and work are ways in which energy is
transferred to the system.
The transfer of work involves the action of a force that causes a change of the volume of a
constant pressure system. This is, most importantly, the compression or expansion of a
system that contains gases.
Energy transfer as heat occurs when there is a difference in temperature between the system
and the surroundings. The heat transfer ~ways occurs from a region of high-temperature
to a region of low temperature.
The convention that is used is as follows:
Heat put in to a system: endothermic ( +)
Heat put out of the system: exothermic (-)
The thermodynamic ·state functions ( dE, MiAS and AG ) are independent of how a
chemical system comes to a 'state'. These "state functions", and their numerical values, are
found as data in chemical references. Thermodynamics is concerned with the chariges in
state functions during chemical and physical changes. thermodynamic data may be used to
make predictions about a specified chemical reaction. The predictions possible include:
1. The heat of the reaction (Ali)
2. Whether the reaction is spontaneous~AG)
3. The value of the equilibrium constant (K)
AE:
Internal energy change is a measure of the change of kinetic energy of the system by the
application of heat and/or work. Using the convention that heat or work done on the
system is absorbed, and heat or work done by the system is evolved:
AE = q+w
Where work (w)
=PAV =&lRT
-
All
Enthalpy is a state function, which implies that the route taken has no consequence on the
value of AH. Enthalpy change is a measure of the change of potential energy of a system.
It is defined in terms of AE.
Mi = AE + w
=m cAT
34
Suppose a gas absorbs 50 kJ of heat and does- 20 kJ of work In
expansion~ then:
AE = 50kJ - 20 kJ
(w is- because it has done work on the system)
AE=30kJ
Mi is directly proportional to the amount of reactants or products. For example, when one
=
mole of ice melts, 6kJ of heat is absorbed (Ml +6.0kJ). If 1/3 of a mole of ice melts,
then MI is 1/3 (+6.0 kJ). In any reaction, the Mi may be calculated as:
Mi reaction = (Ml products- Mi reactants)
It should be easy to see why an exothermic reaction is (-). The products have less heat
(energy) than the reactants. The energy is released to the surroundings.
·
Hess' Law states that the Mi for any reaction is a constant, which implies that the heats of
reaction are additive. The overall heat of reaction is the algebraic sum of the heats of
reaction ~or the reactions that add up to th~ overall reaction.
RFACfiON 1
RFACfiON2
Mi2
RfAOION3
AH3
Mil
~REACfiONS
Enthalpy changes slightly with temperature and pressure, but changes dramatically with
changes in state. In order for a series of reactions to be added, the states under which the
heats of reaction are ;measured must be known. The heat of reaction where all substances
are in their standard states (2SOC and 1.0 atm) are denoted with a superscript as AH0 • For
example, the standard state for water is the liquid state, not the gaseous or the solid state....
Oxygen is a gas, and iron is a solid.
·
Summary of Hess' Law:
a. The overall reaction is used as a guide to manipulate the steps so that they will add up to
the overall reaction. A substance on the product side of a equation should cancel equal
moles on the reactant side of the next equation.
b. Whatever is done to the reaction is reflected in AH0 • If the reaction is doubles, AH0 is
doubled. If the reaction is reversed, the sign of Mi0 ' s is changed.
c. When the steps add up to the overall reaction, the L\H0 's of the steps should equal the
overall AH0 of the reaction.
2(REACfiON 1)
5(RFACfiON 2)
-I CB£ACfiON 3)
~
2(Mil)
5(Mi2)
-HAH3)
REACfiONS
35
Many common reaction•, and their /Uis' may be found In reference
texts
AH 0 f
The standard molar heat of formation is the enthalpy change for the reaction where 1.0
mole of a compound (in its standard state) is made from its elements in their standard states
at 250C. Heats of formation are usually negative; heat is evolved when a compound is
formed. The heats of formation are additive in a reaction. where the overall heat of reaction
may still be calculated as:
Mf' = 1:AHof (products)- 1:Airf (reactants)
In a the reaction:
HO + NaOH
-------~
HOH
+ NaO
The heat of reaction'may be calculated as:
(Mif HOH + Mlf NaCl ) - ( Mlf HO + Mlf NaOH)
The heat of formation of all elements is zero, since the el~ment in such a process would be
made from itself!
An additional method of calculating heat of reaction would be from standard heat of
combustion data. The sum of the heats of combustion in reactions is additive, as is
predicted by Hess' law.
The values for the heats of formation may be found in tables within many reference texts.
Values for the heats of formation are provided on the exam.
AH 0 c
The standard molar heat of combustion is the enthalpy change for the reaction where 1.0
mole of compound is burned in oxygen to form carbon dioxide and water, with all
substances being in their standard states at 250C.
Entropy
.
There are two driving forces for chemical reactions•. One is enthalpy change. which
restricts spontaneous reactions to a state of minimum energy. An exothermic reaction, with
a AH less than: zero. is more likely to be spontaneous than an endothermic reaction.
The other driving terce is Entropy (S). Entropy is a measure of randomneS's or chaos. The
second law of thermodynamics restricts spontaneous reactions to the direction of maximum
chaos. Processes with an entropy change, AS, greater than zero have increased iil chaos or
randomness and are more likely to occur.
. The third law of thermodynamics defines minimum entropy and maximum order. The
absolute entropy is zero only for pure crystalline solids.at absolute zero (0 K). On the basis
of the third law absolute entropy values can be calculated.
Entropy changes can be calculated from absolute entropies using the procedure similar to
that used for enthalpy change:
ASo =!So (products) - 1:So (reactants)
36
as· may be found in the tables
Entropy:
solid to liquid; as +
liquid to gas; AS -
* AS is usually positive for a reaction in which the number of moles of gas increases:
2S03
---~
2S02 + 0 2
. *Reactions for which as is positive tend to be spontaneous reactions at high temperatures.
H 2 0 (s)
---~
~0 (1)
Free energy
The standard free energy change~ aG•, for a reaction at constant temperature and pressure
is defined as:
AGO = Mio- TASO (Gibbs-Helmholtz equation)
Note that AG, like AS, is dependent on P. Unlike Mi. and AS•, AG is strongly
temperature dependent.
The free energy change, or net driving force, has no physical significance. It is tlie
solution to an equation which contains the two diving forces for chemical reactions,
enthalpy and entropy change. it is possible to determine reaction spontaneity from the
addition of the two terms. The result depends on the value and the sign of Mi and T AS.
If aG• ~s negative, the reaction is spontaneous
If aG· = 0 , the reaction is at equilibrium.
If aG· is positive, the reaction is not spontaneous.
The standard free energy change of formation of a compound, aG·f, is the change in net
driving force for a reaction at 2S'C. The reactants are elements in their standard states.
One mole of compound is formed in its standard state. The fee energy of formation of
elements is zero, just as it was for enthalpy of formation. The free energy change of a
reaction is calculated from the free energy of formation of the compounds involved in a
reaction using the equation:
-
AG.
=IaG·f ( products)- IaG·f (reactants)
An important use of free energy change is to determine values of equilibrium constant (K).
The equation relating AG(nonstandard), AG. and K is:
AG = AG• + RT In K
At equilibrium, aG
=0, and the equation becomes:
AG• =- RTlnK
37
2011 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
3. Hydrogen gas burns in air according to the equation below.
2 H2(g) + 0 2(g)
~
2 H20(l)
(a) Calculate the standard enthalpy change, 11H2_98 , for the reaction represented by the equation above.
(The molar enthalpy of formation,
11HJ, for H20(l) is -285.8 kJ mol- 1 at 298 K.)
(b) Calculate the amount of heat, in kJ, that is released when 10.0 g of Hz(g) is burned in air.
(c) Given that the molar enthalpy of vaporization, 11H~ap, for H20(l) is 44.0 kJ mol- 1 at 298 K, what is the
standard enthalpy change, !1H2_98 , for the reaction 2 H 2(g) + 0 2(g)
~
2 H20(g)?
A fuel cell is an electrochemical cell that converts the chemical energy stored in a fuel into electrical energy.
A cell that uses H 2 as the fuel can be constructed based on the following half-reactions.
E o (298 K)
Half-reaction
2 H 20(l) + 0 2(g) + 4 e2 H 20(l) + 2 e-
~
~
4 OH-(aq)
Hz(g) + 2 OH-(aq)
0.40V
-0.83
v
(d) Write the equation for the overall cell reaction.
(e) Calculate the standard potential for the cell at 298 K.
(f) Assume that 0.93 mol of H2(g) is consumed as the cell operates for 600. seconds.
(i) Calculate the number of moles of electrons that pass through the cell.
(ii) Calculate the average current, in amperes, that passes through the cell.
(g) Some fuel cells use butane gas, C4H 10 , rather than hydrogen gas. The overall reaction that occurs in a butane
fuel cell is 2 C 4H 10(g) + 13 0 2(g) ~ 8 C02(g) + 10 H20(l). What is one environmental advantage of using
fuel cells that are based on hydrogen rather than on hydrocarbons such as butane?
STOP
If you finish before time is called, you may check your work on this part only.
Do not turn to the other part of the test until you are told to do so.
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-8-
AP\!l) CHEMISTRY
2011 SCORING GUIDELINES
Question 3
Hydrogen gas burns in air according to the equation below.
(a) Calculate the standard enthalpy change,
t:Uf;98 , for the reaction represented by the equation above.
(The molar enthalpy of formation, D.H(, for H20(l) is -285 .8 kJ mol- 1 at 298 K.)
t:U~298
= [2 (-285.8)]- [2(0) + 1(0)] = -571.6 kJ mol- 1
I point is earned for the correct answer.
(b) Calculate the amount of heat, in kJ, that is released when 10.0 g of H2(g) is burned in air.
1 mol H 2
285.8 kJ = 1_42
q = 10
g H 2 x 2.016 g H 2 x I mol H 2
x
103 kJ
1 point is earned for the correct setup.
1 point is earned for the correct answer.
(c) Given that the molar enthalpy of vaporization, !1H~ap, for H20(l) is 44.0 kJ mol- 1 at 298 K, what is the
standard enthalpy change, t:U~298 , for the reaction 2 H 2(g) + 0 2(g)
2 H2(g) + 0 2(g)
~
2 H20(/)
-571.6 kJ
2 H 20(!)
~
2 H20(g)
+2(44.0) kJ
~
2 H 20(g)?
1 point is earned for the correct answer.
-483.6 kJ
A fuel cell is an electrochemical cell that converts the chemical energy stored in a fuel into electrical energy.
A cell that uses H2 as the fuel can be constructed based on the following half-reactions.
E o (298 K)
Half-reaction
2 H20(l) + 0 2(g) + 4 e2 H20(l) + 2 e-
~
~
4 OH-(aq)
Hig) + 2 OH-(aq)
0.40V
-0.83
v
© 2011 The College Board.
Visit the College Board on the Web: www.collegeboard.org.
AP® CHEMISTRY
2011 SCORING GUIDELINES
Question 3 (continued)
(d) Write the equation for the overall cell reaction.
2 H 20(l) + 0 2(g) + 4 e-
~
4 OH- (aq)
2 H 2(g) + 4 OH-(aq)
~
4 H 20(!) + 4 e-
I point is earned for the correct equation.
(e) Calculate the standard potential for the cell at 298 K.
£0
= 0.40 V -
(-0.83 V)
= 1.23 V
I point is earned for the correct answer.
(f) Assume that 0.93 mol of H 2(g) is consumed as the cell operates for 600. seconds.
(i) Calculate the number of moles of electrons that pass through the cell.
0.93 mol H 2 x 2 mol e1 mol H 2
= 1.9 mol e-
1 point is earned for the correct answer.
(ii) Calculate the average current, in amperes, that passes through the cell.
I point is earned for calculation of the charge in
coulombs.
1.9 mole- x %,SOOC
I mol e-
/ = !1_ =
t
5
J.8 X J0 C
600. s
= 3.0 x
102 amps
I point is earned for calculation of the current in
amperes.
(g) Some fuel cells use butane gas, C4 H 10 , rather than hydrogen gas. The overall reaction that occurs in a
butane fuel cell is 2 C4 H 10(g) + 13 0 2(g) ~ 8 C02 (g) + 10 H 20(!). What is one environmental
advantage of using fuel cells that are based on hydrogen rather than on hydrocarbons such as butane?
Hydrogen fuel cells produce only water as a product,
unlike fuel cells that use hydrocarbons, which release
carbon dioxide. Carbon dioxide contributes to global
warming via the enhanced atmospheric greenhouse
effect.
1 point is earned for an
acceptable environmental advantage.
© 2011 The College Board.
Visit the College Board on the Web : www.collegeboard.org.