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Course
203
Subject
Chemistry
Date
Jan 9, 2024
Type
Pages
93
Uploaded by SargentSeaLionMaster1039
General
Chemistry
11
material
begins
here
Sample
datasheet
and
periodic
table
provided
as
reference
on
the
next
page.
111
Periodic
Table and
sample
datasheet
AEC.S
Exams
INSTANTS
&
ABBREVIATIONS
AND
SYMBOLS
I
amount
of
substance
|
gas
constant
R
|
molar
mass
M
o
S
atmosphere
atm
|
gram
g
|mole
mol
R=8314Jmol"K
atomic
mass
unit
u
|
hour
h
|
Planck’s
constant
/|
|
R=0.0821
L-atm-mol-K™"
atomic
molar
mass
A
|
joule
J
|
pressure
2
Na
=
6.022
x
105
mol™!
Avogadro
constant
N
|
kelvin
K
|
second
s
h=6.626x
103
I's
Celsius
temperature
°C
|
kilo—
prefix
k
|
speed
of
light
c
2
;
centi—
prefix
c
|
liter
L
|temperature,
K
T
€=2.998
x
10°m-s”
energy
of
activation
E,
|
measure
of
pressure
mmHg
|
time
t
0°C=273.15K
efir;fll::};
1~\II
::)Ill;
prefix
1\‘2
volume
v
1
atm
=
760
mmHg
s
1
atm
=
760
torr
EQUATIONS
Equations
may
be
included
in
this
position
of
your
exam.
These
vary
depending
on
the
type
of
exam
you
are
taking.
For
example,
if
you
are
expected
to
do
a
calculation
involving
an
integrated
rate law,
you
may
be
provided
with
a
series
of
equations
in
this
position.
PERIODIC
TABLE
OF
THE
ELEMENTS
1
1
H
1,008
2
137145715116+
17+
sio0s.
3
4
5
6
7
8
9
Li
|
Be
B
Cil
N
.|
O
K,
6941
|
9.012
1081
|
1201
|
1401
|
1600
|
1900
|
2018
11
12
131415
)16
|
17
64
|
65
|
66
|
67
|
68
|
69
[
70
|
71
Ce
|
Pr
(Nd
|Pm
[Sm
|
Eu
|
Gd
|
Tb
|
Dy
[
Ho
|
Er
[
Tm
|
Yb
|
Lu
140.1
|
1409
|
1442
1504
|
1520
|
1573
|
1589
|
1625
|
1649
|
1673
|
1689
|
1730
|
1750
90
|
91
|
92|
93|94
(959
|97
|98
|99
[100]101]102]103
Th
|
Pa
|
U
|
Np
|
Pu|Am
(Cm
|
Bk
|
Cf
|
Es
|
Fm
[
Md
|
No
|
Lr
2320
|
2310
|
2380
Please
note
that
the
periodic
table
changes
to
keep
current
with
recent
discoveries.
The
periodic
table
you
use
may
vary
from
the
table
shown
here.
112
Chapter
9
—
Solutions
and
Aqueous
Reactions,
Part
2
Chapter
Summary:
This
chapter
will
focus
on
the
additional
aspects
of
solutions,
usually
found
in
the
second
semester
of
a
standard
general
chemistry
course
(whereas
the
first
part of
solutions
is
usually
found
in
the
first
semester).
Concepts
discussed
here
and
explained
in
example
items
go
into
more
detail
related
to
intermolecular
forces
associated
with
solution
formation
and
properties
of
solutions.
Reactions
of
solutions
are
discussed
separately
in
chapters
related
to
equilibrium.
Specific
topics
covered
in
this
chapter
are:
Solution
formation
including
intermolecular
forces
in
solutions
Solubility
as
it
relates
to
intermolecular
forces
Concentration
units
and
conversions
between
Solubility
of
solids
and
gases,
particularly
in
aqueous
solution
Colligative
properties
including
freezing
point
depression,
boiling
point
elevation
and
osmotic
pressure
Previous
material
that
is
relevant
to
your
understanding
of
questions
in
this
chapter
include:
The
mole
and
formula
calculations
(Chapter
3)
Stoichiometry
and
balancing equations
(Chapter
4)
Molar
concentrations
and
solution
stoichiometry
(Chapter
5)
Lewis
dot
structures
and
polarity
(Chapter
7)
Intermolecular
forces
(Chapter
8)
Common
representations
used
in
questions
related
to
this
material:
Name
Example
Used
in
questions
related
to
Structures
Intermolecular
forces
in
solutions
Temperature
dependence
of
Graphs
solubility
of
solids
0
10
20
30
40
50
60
70
80
90
Temperature,
°C
Where
to
find
this
in
your
textbook:
The
material
in
this
chapter
typically
aligns
to
a
solutions
chapter
(could
be
labeled
as
“Physical
Properties
of
Solutions™)
in
your
textbook.
The
name
of
your
chapter
may
vary.
Practice
exam:
There
are
practice
exam
questions
aligned
to
the
material
in
this
chapter.
Because
there
are
a
limited
number
of
questions
on
the
practice
exam,
a
review
of
the
breadth
of
the
material
in
this
chapter
is
advised
in
preparation
for
your
exam.
How
this
fits
into
the
big
picture:
The
material
in
this
chapter
aligns
to
Intermolecular
Interactions
(4)
as
listed
on
page
12
of
this
study
guide.
13
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Solutions
and
Aqueous
Reactions,
Part
2
cACS
w
Exams
Study
Questions
(SQ)
SQ-1.
Where
is
the
hydrophilic
(attracted
to
water)
region
Region
I
Region
2
Region
3
of
the
molecule?
HH
H
HeR]
(A)
Region
|
\/
I
”
(B)
Region2
N
R
(C)
Region3
/
|
/
(D)
The
three
regions
are
equally
hydrophilic
n
i
o
Knowledge
Required:
(1)
Structure
of
water.
(2)
Intermolecular
forces
(hydrogen
bonding).
Thinking
it
Through:
As
you
read
this
question.
you
realize
the
first
step
will
require
that
you
consider
the
structure
of
water
in
relation
to
the
structure
of
the
molecule
because
you
are
asked
to
consider
how
water
and
the
shown
molecule
will
interact.
Continuing
along
these
lines,
you
then
compare
the
molecules
and
consider
how
these
will
interact
within
the
context
of
hydrogen
bonding.
You
know
that
in
order
to
have
two
molecules
form
a
hydrogen
bond,
you
will
need
hydrogen
bonded
to
N,
O,
or
F
and
a
lone
pair
of
electrons
on
N,
O,
or
F.
In
a
pure
substance,
these
will
need
to
be
present
in
one
molecule
(as
all
molecules
are the
same
in
a
pure
substance).
Ina
solution,
molecules
could
have
only
part
of
these
requirements
with
the
other
molecules
having
the
other
components.
Therefore,
you
draw
water
molecules
in
relation
to
the
molecule,
including
the
relative
partial
charges
(to
help
make
the
point
of
attraction):
&
hydmgenbonding%
H/ \H
Therefore,
water
is
attracted
to
region
3
(Choice
(C))
and
not
region
1
(Choice
(A))
or
region
2
(Choice
(B))
or
all
three
equally
(Choice
(D)).
NOTE:
Drawing
your
own
representations
is
often
an
effective
problem-solving
strategy
in
chemistry.
Practice
Questions
Related
to
This:
PQ-1
and
PQ-2
SQ-2.
‘Which
molecule
is
most
soluble
in
water?
o~
o
o
@
!
®
g
PN
CH,
CH,
HiC
CHy
[o]
©
l
o
A
e
O
P
HeC
CHS
CHY
HaC
OH
114
.>ACS
Solutions
and
Aqueous
Reactions,
Part
2
W
Exams
Knowledge
Required:
(1)
Structure
of
water.
(2)
Intermolecular
forces
(hydrogen
bonding).
Thinking
it
Through:
Similar
to
the
previous
study
question,
to
solve
this
you
are
going
to
consider
each
structure
interacting
with
the
structure
of
water,
paying
close
attention
to
components
that
would
support
greater
attraction
due
to
hydrogen
bonding.
As
a
reminder
again,
in
a
solution
these
are:
1.
Hydrogen
bonded
to
N,
O,
or
F
2.
Alone
pair
of
electrons
on N,
O,
or
F
Molecules
interacting
with
water
can
have
one
or
more
locations
for
this
attraction
to
occur.
Considering
each
molecule
with
water,
including
relevant
partial
charges
and
hydrogen
bonding:
.
0.
W
hydrogen
&
bonding
<33
hydrogen
&
—eeeeT
bonding@
HsC
e}
CHy
N
N
cnz/
CH,/
no
hydrogen
bonding
&
CH,
CH;
CH;
O
A
Z\CH/
A
H/"\H
2
B
From
these
diagrams,
you
see that the
greater
the
number
of
attractions,
the
more
soluble
the
molecule
will be
in
water.
Adding
to
this
decision,
you
consider
the
relative
sizes
of
the
molecules
(as
solubility
can
be
conceptualized
as
a
“substitution”
process
of the
solute
molecules
into the
water
molecule’s
location,
thus
needing
the
same
relative
attractions
and
size).
You
see that the
molecules
are
reasonably
the
same
size.
Therefore,
you
can
see
that
one
molecule
will
not
hydrogen
bond
with
water
at
all
(Choice
(D)),
therefore
this
molecule
will be the least
soluble
in
water.
You
also
see
that
molecules
in
Choices
(A)
and
(B)
have
limited
attractions
to
water
making
these
incorrect
as
well.
Choice
(C)
is
ideal
for
substitution,
as
it
has
multiple
opportunities
for
hydrogen
bonding,
making
(C)
or
acetic
acid,
the
correct choice.
Practice
Questions
Related
to
This:
PQ-3. PQ-4,
and
PQ-5
SQ-3.
A
solution
of
NaCl
in
water
has
a
concentration
of
20.5%
by
mass.
Molar
mass
/
g'mol™
‘What
is
the
molal
concentration
of
the
solution?
NaCl
5844
(A)
0.205m
(B)
0258m
©)
35im
D)
441m
Knowledge
Required:
(1)
Concentration
unit
of
molality.
(2)
Concentration
unit
of
percent
by
mass.
Thinking
it
Through:
You
see
this
question
starts
by
giving
you percent
by
mass
and
asks
you
to
determine
molal
concentration.
So,
to
solve
this,
you
first
define
both
concentration
units.
First
for
percent
by
mass,
you
define
this
with
a
general
definition,
then
for
a
solution
and
finally
further
define
the
mass
of
the
solution
as
the
sum
of
the
masses
of
the
solvent and
solute:
mass
of
one
component
©100%
=
mass
solute
©100%
=
mass
solute
%100%
Percent
by
mass
=
"
total
mass
mass
solution
mass
solute
+
mass
solvent
mol
solute
Now
you
define
molal
concentration:
‘molal
concentration
=
——————
mass
solvent
(kg)
115
Solutions
and
Aqueous
Reactions,
Part
2
.
ACS
Exams
Before
you
set
up
your
solution
to
the
problem,
you
first
map
out
wfiat;ofi
will
do:
Set
your
Use
the
percent
by
Use
molar
Use
the
mass
of
Caleulate
sample
size
_y
|
mass
of
thesolute
to
|
_,
mass
to
_y
|
solute/mass
of
solution
|
_
the
molal
©0100g
give
the
mass
of
the
determine
1o
determine
the
mass
of
conc
solution
solute
solute
solvent
Now
setup and solve
the
problem:
NaCl
(20.5¢
NaCl)[
i‘il—ac—)
100.0
g
solution
—20.5
g
NaCl
=
79.5
g
water
=
w
441m
e
€
o€
0.0795
kg
water
J
.
‘Which
is
choice
(D).
Choice
(C)
uses
the
mass
of the
solution
(in
kg)
rather
than
the
mass
of
the
solvent
(water).
Choice
(B)
omits calculating
moles
and
use
the
mass
of
water
in
grams
instead
of
kilograms.
Choice
(A)
also
omits
calculating
moles
and
use
the
mass
of
the
solution (and
not
the
solvent
or
water)
in
grams
instead
of
kilograms.
Practice
Questions
Related
to
This:
PQ-6,
PQ-7,
PQ-8, PQ-9,
and
PQ-10
SQ-4.
‘What
is
the
mole
fraction
of
water
in
200.
g
of
89%
(by
mass)
ethanol,
Molar
mass
/
g'mol™!
CoHsOH?
C>HsOH
46
@A)
o011
(B)
024
©
032
D)
076
Knowledge
Required:
(1)
Concentration
unit of
mole
fraction.
(2)
Concentration
unit
of
percent
by
mass.
Thinking
it
Through:
Similar
to
the
previous
study
question,
you
see
this
question
starts
by
giving
you
percent
by
mass
and
asks
you
to
determine mole
fraction.
So,
again
to
solve
this,
you
first
define
both
concentration
units:
‘mass
solute
Percent
by
mass
=
——————————x100%
mole
fraction
(X,
)=
mol
solute
'mass
solute
+
mass
solvent
ol
mol
solute
+
mol
solvent
Before
you
set
up
your
solution
to
the
problem,
you
first
map
out
what
you
will
do:
Set
your
sample
size
to
100
g
solution,
Use
molar mass
Use
molar
mass
Calculate
making:
mass.owe
=%
by
massand
| =
|
(solute)to
|
=
|
(solventyto
|
—>
|
the
mole
masSaiven
=
100
—
%
by
mass
determine
nsoue
determine
riven
fraction
Now
setup
and
solve
the
problem
(norice
the
200.
g
sample
size
is
not
needed):
100
g
solution
89
g
C,H,OH
=11
g
water
(89
¢
CZH,OH)[M]
=1.9
mol
C,H,0H
46
¢
C,H,0H
mol
H,0
0.61
mol
H,0
1
g0
MO
ot
notto
x,,
——OSLmOIHO
o,
(11eH,
)(mgHZo,‘
o
9~
1.9
mol
C,H,0H
+
0.61
mol
H,0
Which
is
choice
(B).
Choice
(D)
is
the
mole
fraction
of
ethanol.
Choice
(A)
is
the
percent
by
mass
of
water
(as
a
fraction).
Choice
(C)
uses
the
number
of
moles
of
solvent
(ethanol)
only
and
not
the
total
number
of
moles.
Practice
Questions
Related
to
This:
PQ-6,
PQ-7,
PQ-8, PQ-9,
and
PQ-10
116
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“ACS
Solutions
and
Aqueous
Reactions,
Part
2
¥
Exams
SQ-5.
A
mixture
of
100
g
of
K2Cr:07
and
200
g
of
M
water
is
stirred
at
60
°C
until
no
more
of
the
salt
dissolves.
The
resulting
solution
is
poured
off
into
a
separate
beaker, leaving
the
undissolved
solid
behind.
The
solution
is
now
cooled
to
20
°C.
What
mass
of
K>Cr,0;
crystallizes
from
the
solution
during
the
cooling?
A)
9¢
(B)
18¢g
01002030
40
50
60
70
80
90
©
3lg
Temperature,
°C
Solubilify
KCr0./
100
¢
(
D)
62g
Knowledge
Required:
(1)
Interpretation
of
a
solubility
graph.
(2)
Definition
of
solubility.
Thinking
it
Through:
The
problems
provides
you
with both
quantitative
data
and
an
associated
process.
You
think
it
would
be
easiest
to
map
out
what
is
described
before
coordinating
this
with
the
graph:
60°C
60°C
20°C
You
start
with:
—
Solution
is
poured
off
(saturated
|
_,|
Cooled:
some
mass
of
K2Cr207
100
g
K2Cr207
in
200
g
water
solution
of
K2Cr207
in
200
g
water’
crystallizes
out
|
N
leaving
undissolved
solid
Now
you
coordinate
this
with
the
graph,
marking
the
solubility
of
K»Cr:0;
at
60
°C
and
20
°C:
You
see
that:
at
60
°C,
the
solubility
is
40
g/100
g
water
/
50
the
60
°C
solution
above
(once
poured
off)
contained
80
g
of
K>Cr:0-
in
200
g
of
water.
at
20
°C,
the
solubility
is
9
g/100
g
water
so
the
20
°
C
solution
above
(once
cooled)
contained
18
g
of
K2Cry07
in
200
g
of
water
in
solution.
Meaning,
the
cooled
solution
had
80
g
—
18
gor
62
g
of
K»Cr,0;
as
a
solid
or
crystallized
from
the
water,
0
10
20
30
40
50
60
70
BO
90
which
is
choice
(D).
Temperature,
"C
Choice
(B)
is
incorrect
because
this
is
the
mass
that
remained
in
solution.
Choice
(C)
is
incorrect
because
this
is
the
mass
that
crystallized
out
in
100
g
of
water,
rather
than
200
g
with
Choice
(A)
as
the
incorrect
choice
when
considering
100
g
of
water
and
the
mass
that
remained
in
solution.
Practice
Questions
Related
to
This:
PQ-11
and
PQ-12
SQ-6.
Carbonated
beverages
have
a
fizz
because
of
gas
dissolved
in
the
solution.
What
increases
the
a‘fi,\d
concentration
of
gas
in
a
solution?
(A)
cooling
the
solution
(B)
heating
the
solution
(C)
increasing
the
volume
of
solution
(D)
decreasing
the
volume
of
solution
Thinking
it
Through:
Y
ou
recall
that
solubility
of
gases
can
be
expressed
quantitatively
using
Henry’s
law
which
expresses
relationship
of the
concentration
of the
gas
(in
the
solution,
¢),
the
pressure
of
the
gas
above
the
solution
(P),
and
a
constant
for
the
gas
(k):
¢
=
kP
(also
written
as
¢
=
Ku/P).
From
this,
you
see that
as
pressure
of
the
gas
117
Solutions
and
Aqueous
Reactions,
Part
2
."ACS
W
Exams
increases,
the
concentration
of
the
gas
increases.
You
also
suspect
that
solubility
of
gases
vary
based
on
their
structure,
and
this
would
be
expressed
in
the
Henry’s
law
constant,
k&.
However,
from
this
you
also
see that the
volume
of the
solution
will
not
change
the
concentration
of the
gas
in
the
solution,
so
neither
choice
(C)
or
(D)
are
correct.
You
do
know
that
the
Henry’s
law
constant
is
a
constant
for
a
gas
at
a
given
temperature
and
as
temperature
increases,
solubility
of
the
gas
decreases
or
k
decreases.
Therefore,
heating
the
beverage
will
decrease
the
concentration
of
the
gas
(Choice
(B)),
but
cooling
the
beverage
will
increase
the
concentration
or
fizziness,
meaning
Choice
(A)
is
correct.
Practice
Questions
Related
to
This:
PQ-13
and
PQ-14
SQ-7.
When
100.
g
of an
unknown
compound
was dissolved
in
1.00
kg of
water,
the
freezing
point
was
lowered
by
6.36
°C.
What
is
the
identity
of
this
unknown
compound?
(K
for
water
=
1.86°C-m™')
(A)
CsCl
(B)
KCI
(©)
LiF
(D)
NaCl
Knowledge
Required:
(1)
Colligative
properties,
freezing
point
depression.
(2)
Ionic
éorfiponnds
and
van’t
Hoff
coefficients.
(2)
Molal
concentration
and
molar
mass.
Thinking
it
Through:
As
you
read
the
question,
you
approach
this
problem
by
first
evaluating
the
equation
for
freezing
point
depression
(which
you
know
is
the
appropriate
colligative
property
because
of the
cue
to
this
in
the
question):
AT
=
K,.m.
You
further
examine
the
responses
and
see
that
all
of
the
substances
are
ionic
with
the
same
1:1
ratio
of
cation
to
anion,
meaning
that
2
ions
form
for
each
formula
unit
of
the
compound.
Therefore,
you
refine
the
equation
for
freezing
point
depression
for the
van’t
Hoff
coefficient
of
2
(i
=
2):
AT
=iK;m=2K.m:
_AT
6.36°C
K,
2(1.86°C-m”)
From
here,
you
consider
molal
concentration
and
the
other
information
provided
in
the
problem
to
determine
first
the
number
of
moles
of
solute
and
then
(using
the
mass
of
solute),
the
molar
mass
of the
solute:
AT,
=iK;m
m
=17lm
molal
concentration
=~
SolUte_
L71mol
|1
00
kg
water)
=171
mol
solute
mass
solvent
kg
solvent
100.
g
solut
A
(&)
=58.5g-mol
'
which
is
NaCl
or
response
(D).
171 mol
solute
Choice
(A)
would
likely
be
selected
if
the
calculation
of
molal
concentration
was
inverted
(calculated
molar
mass
of
170
g'mol™').
Choice
(B)
may
be
selected
if
number
of
moles
are
calculated
incorrectly
by
using
the
mass
of
solution
rather
than
mass
of
solvent.
Choice
(C)
may
be
selected
if
the
van’t
Hoff
coefficient
was
omitted
(calculated
molar
mass
of
29
g'mol
™).
Practice
Questions
Related
to
This:
PQ-15,
PQ-16,
and
PQ-17
SQ-8.
Which
0.1
molal
aqueous
solution
will
have
the
lowest
freezing
point?
M‘
(A)
AI(NO3);
(B)
CaCl,
(C)
CHsOH
(D)
NaCl
Knowledge
Required:
(i
)
Collig’ativ’e
fimfieniés;
ffeéiing
}S{)int
dépression.
(i)
Tonic
conipourids“and
van’t
Hoff
|
coefficients.
Thinking
it
Through:
Y
ou
know
that
colligative
properties
depend
on
the
number
of
particles
in
solution
(the
molal
concentration)
but
not
the
identity
of
the
solute.
You
also
know
this
is
reflected
in
the
equation
for
freezing
point
depression
as
noted
in
the
previous
study
question.
Because
the
concentrations
are
all
the
same
(0.1
m)
and
the
solvent
is
the
same
(water
with
the
same
freezing
point
depression
constant,
K7),
the
only
component
that will
vary
will be the
van’t
Hoff
coefficient,
i.
So,
you
start
by
evaluating
the
number
of
particles
in
solution
for
each
solute:
118
*ACS
Solutions
and
Aqueous
Reactions,
Part
2
V
Exams
hydratic
hydration
equation
deale
AI(NOs)s(s)
=
Al**(aq)
+
3NOs™(aq)
CaCly(s)
—
Ca**(aq)
+
2CI(aq)
3
C2Hs0H(l)
=
C,HsOH(aq)
NaCl(s)
-
Na*(aq)
+
Cl-(aq)
2
Note:
icsic
is
used
here
because
you
don’t
have
the
measured
value
(imess);
in
calculations,
imcss
Would
be
better than
icalc.
You
are
asked
for
the
lowest
freezing
point
or
largest
ATr.
Therefore,
choice
(A)
is
correct
with
a
caleulated
van’t
Hoff
coefficient
of
4.
The
other
choices
are
incorrect
because
the
coefficients
are
all
smaller
than
the
value
for
AI(NOs)s.
Practice
Questions
Related
to
This:
PQ-18,
PQ-19,
and
PQ-20
Practice
Questions
(PQ)
M}
PQ-1.
Which
pair
of
compounds
will
form
hydrogen
bonds
with
one
another?
(A)
CHsand
H.O
(B)
CH:and
NH;
(C)
HF
and
CHs4
(D)
H.0
and
NH;
w‘"’”’\
PQ-2.
Which
pair
of
compounds
will
form
hydrogen
bonds
with one
another?
cl
H
HC,
CH;
@)
:
\\0th
\2
and
H—Cl
l
and
l
H,C™™
2™~CH,
//
H//
\
H
[
.
T
|
©)
C.
and
C.
D)
and
C.
w7
Son
w7
//
"™
H
H
H
PQ-3.
What
would
be
expected
to
be
the
most
soluble
in
ethanol
(CH;:CH,OH)?
(A)
CH;CH.CH3
(B)
CO:
(C)
NH;
(D)
SFe
M
PQ-4.
Which
substance
is
most
soluble
in
water?
(A)
ethane,
CH;CH3
(B)
ethanol,
CH;CH,OH
(C)
n-butane,
CH;CH,CH>CH;
(D)
1-butanol,
CH;CH,CH>CH,OH
PQ-5.
Consider
the
structure
of
Vitamin
A;
CHs
CH;
CHs
(C20H300).
Vitamin
A,
should
be
|
"
|
H
;
AN
NN
NN
(A)
insoluble
in
water
and
in
oil.
e
SN
N
e
N
N
N
(B)
less
soluble
in
water
than
in
oil.
H,l
l
N,
\\cnX
(C)
equally
soluble
in
water
and
in
oil.
i
Chy
(D)
more
soluble
in
water
than
in
oil.
PQ-6.
The
mass
percent
of
acetic
acid
(CH;COOH)
in
a
bottle
of
vinegar
is
Molar
mass
/
g'mol™!
5.45%
in
water.
What
is
the
molar
concentration
of
acetic
acid
in
CH;COOH
60.06
vinegar,
assuming
the
density
of
vinegar
is
1.005
g-mL™'?
(A)
0.545M
(B)
0.903M
(C)
0.908
M
(D)
0912M
119
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Solutions
and
Aqueous
Reactions,
Part
2
.
ACS
&
Exams
PQ-7.
A
250-mL
bottle
of
a
sports
drink
solution
contains
4.50%
by
mass
of
Molar
mass
/
g
sodium
chloride.
What
is
the
molal
concentration
of
sodium
chloride
in
NaCl
.
this
bottle
of
sports
drink?
(A)
0.806m
(B)
0.770m
(©)
0.180m
(D)
0.0113m
PQ-8.
A
1400
g
sample
of
stream
water contains
12.2
ppm
of
mercury.
What
is
the
mass
of
Hg
in
the
sample?
(A)
0.11mg
(B)
2.4mg
(C)
8.7mg
(D)
17
mg
PQ-9.
What
mass
of
water
is
needed
to
dissolve
292.5
g
of
NaCl
to
produce
a
Molar
mass
/
g-mol™!
0.25
m
aqueous
solution?
NaCl
58.44
(A)
20kg
(B)
5.0kg
(C)
0.80kg
(D)
0.050
kg
PQ-10.
The
density
of
a
3.539
M
HNO;
aqueous
solution
is
1.150
g-mL
"
at
Molar
mass
/
20°C.
What
is
the
molal
concentration?
HNO;3
(A)
3.077m
(B)
3818m
(C)
3.946m
(D)
5252m
PQ-11.
The
solubility
of
a
substance
is
60
g
per
100
mL
of
water
at
15
°C.
A
solution
of
this
substance
is
prepared
by
dissolving
75
g
in
100
mL
of
water
at
75 °C.
The
solution
is
then
cooled
slowly
to
15
°C
without
any
solid
separating.
The
solution
is.
(A)
supersaturated
at
75°C.
(B)
supersaturated
at
15°C.
(C)
unsaturated
at
15°C.
(D)
saturated
at
15°C.
PQ-12.
A
student
mixes
20.0
g
of
a
saltin
100.0
g
of
Solubility
at
20
°C
in
100.0
g
of
water
water
at
20
°C
and
obtains
a
homogeneous
KCl
100¢g
solution.
Which
salt
could
this
be
and
why?
KNO;
300g
(A)
KCl because
it
is
saturated.
(B)
KClI
because
it
is
unsaturated.
(€)
KNO;
because
it
is
saturated.
(D)
KNO:;
because
it
is
unsaturated.
PQ-13.
The
solubility
of
carbon
dioxide
in
water
is
very
low
in
air
(1.05x10°5
M
at
25
°C)
because
the
partial
pressure
of
carbon
dioxide
in
air
is
only
0.00030
atm.
What
partial
pressure
of
carbon
dioxide
is
needed
to
dissolve
100.0
mg
of
carbon
dioxide
in
1.00
L
of
water?
(A)
0.0649
atm
(B)
2.86atm
(C)
28.6atm
(D)
64.9
atm
Up"'vp\
PQ-14.
How
will
you
increase
the
solubility
of
oxygen
in
water?
The
partial
pressure
of
oxygen
(
B,
)is
0.21
atm
in
air
at
|
atm
(Pex).
(A)
increase
P,
but
keep
Pey
constant
(B)
decrease
£,
but
keep
Py
constant
(C)
increase
Pey
but
keep
B,
constant
(D)
decrease
Pe
but
keep
£,
constant
PQ-15.
A
2.50
g
sample
of
naphthalene,
CioHs,
was dissolved
in
100
g
of
Molar
mass
/
g'mol™!
benzene.
What
is
the
freezing
point
of the
benzene
solution?
The
freezing
CioHs
128.17
point
of
pure
benzene
is
5.45
°C;
Kr=5.07
°C-m™".
(A)
~0.989°C
(B)
0.989°C
(C)
446°C
(D)
6.44°C
120
g
cACS
Solutions
and
Aqueous
Reactions,
Part
2
Exams
PQ-16.
A
technician
has
the
measured
osmotic
pressure
of
a
solution
to
I
Temperature
determine
the
molar
mass
of
a
covalent
solute.
Which
other
1L
Volume
of
solution
information
would
need
to
be
measured
in
order
to
determine
the
1L
Mass
of
solute
molar
mass?
(A)
OnlyI
(B)
Only
I
(C)
OnlyIand
IT
(D)
LI,
and
I
PQ-17.
A
solution
was
made
by
adding
800
g
of
ethanol,
C;HsOH,
to
8.0x10°
g
Ki/°C-m!
of
water.
How
much
would
this
lower
the
freezing
point
H.0
1.86
(A)
32°C
(B)
4.1°C
(©)
82°C
(D)
16°C
M
PQ-18.
Which
aqueous
solution
will
have
the
lowest
osmotic
pressure?
(A)
0.04
M
MgCl,
(B)
0.05M
KBr
(©)
0.1
MLiCI
(D)
0.2
M
Na:SOs
PQ-19.
Consider
pure
water
separated
from
an
aqueous
sugar
solution
by
a
semipermeable
membrane.
After
some
time
has
passed,
what
(if
anything)
will
happen
to
the
concentration
of
the
sugar
solution?
(A)
It
will
decrease.
(B)
It
will
increase.
(C)
Ttwill
remain
the
same.
(D)
It
cannot
be
determined.
PQ-20.
Which
solution
has
the
|
Solution:
Molar
mass
/
g-mol™'
higher
boiling
point?
|
(1)
200.0
g
glucose dissolved
in
1.00
kg of
water
glucose,
180
(2)
200.0
g
sucrose
dissolved
in
1.00
kg
of
water
sucrose,
342
(A)
solution
1
(B)
solution
2
(C)
Both
would
boil
at
the
same
temperature
as
pure
water.
(D)
Both
would
boil
at
the
same
temperature,
but
above
that of
pure
water.
121
Solutions
and
Aqueous
Reactions,
Part
2
=
ACS
Y
Exams
Answers
to
Study
Questions
1.
C
5.
D
2.
C
6.
A
3
D
7.
D
4.
B
8
A
Answers
to
Practice
Questions
1.
D
11.
B
22
D
12.
D
3.
C
13.
A
4.
B
4.
A
5.
B
15.
C
6.
D
16.
D
7.
A
17.
B
8
D
18.
B
9.
A
19.
A
10.
B
20.
A
122
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Chapter
10
—
Kinetics
Chapter
Summary:
This
chapter
will
focus
on
chemical
kinetics.
This
includes
the
topics
of
reaction
rate
and
rate
laws.
Methods
of
determining
the
rate
laws
will be
presented
and
calculations
with
integrated
rate
laws
will be
reviewed.
The
collision
theory
of
chemical
kinetics
and reaction
mechanisms
will be
reviewed.
Specific
topics
covered
in
this
chapter
are:
Determining
a
rate
law
from
plots
of
data
Method
of
initial
rates
Solving
problems
using
integrated
rate
laws
Half-life
problems
Collision
theory
and
reaction
energy
diagrams
Calculation
of
activation
energy
for
a
reaction
Reaction
mechanisms
Relationship
of
kinetics
to
equilibrium
Previous
material
that
is
relevant
to
your
understanding
of
questions
in
this
chapter
include:
e«
Significant
figures
(Toolbox)
*
Scientific
notation
(Toolbox)
e«
Equilibrium
(Chapter
11)
«
Enthalpy
(Chapter
6)
Common
representations
used
in
questions
related
to
this
material:
Name
Example
Used
in
questions
related
to
Compound
units
M-s';
L-mol™"s
!
rates
and
rate
constants
Energy
diagrams
activation
energy
Where
to
find
this
in
your
textbook:
The
material
in
this
chapter
typically
aligns
to
“Kinetics”
or
“Chemical
Kinetics”
in
your
textbook.
The
name
of
your
chapter
may
vary.
Practice
exam:
There
are
practice
exam
questions
aligned
to
the
material
in
this
chapter.
Because
there
are
a
limited
number
of
questions
on
the
practice
exam,
a
review
of
the
breadth
of
the
material
in
this
chapter
is
advised
in
preparation
for
your
exam.
How
this
fits
into the
big
picture:
The
material
in
this
chapter
aligns
to
the
Big
Idea
of
Kinetics
(7)
as
listed
on
page
12
of
this
study
guide.
123
Kinetics
=
ACS
—¥
Exams
Study
Questions
(SQ)
SQ-1.
What
is
the
rate
law
of
this
reaction?
Trial#
|
[H2lo/M
|
[NOlo/M
|
Initial
rate
/
M-s~
1
LOx107
|
1.0%10°
2.0x10°
2Ha(g)
+
2NO(2)
—
Na(2)
—
2H:0(g
2
2.0x10°
|
2.0x10°
1.60%10*
“©
@~
©
3
[
20x107
[
1ox107
4.0410°
(A)
rate
=
[H2[NO]
(B)
rate
=
K[Hy][NOJ*
(€)
rate
=
A[H.2[NOJ®
(D)
rate
=
&{Ny)[H0}%/[H:]*[NOJ
|
Knowledge
Required:
(1)
Definition
of
rate
law.
(2)
How
to
use
the
method
of
initial
rates
to
determine
the
rate
law.
Thinking
it
Through:
You
are
given
data
with
varying
starting
amounts
(initial
concentration
noted
as
the
“0”
subscript,
i.e.
[Hz]o)
of
each reactant and
the
reaction
rate
that
was
measured.
You
write
the
generic
rate
law
for
the
reaction
as:
rate
=
A[H;NOJ
To
find
the
value
of
each
exponent,
you
select
a
pair
of
trials
that
holds
one
concentration
constant
and
allows
you
to
see
the
effect
on
the
rate
by
changing
the
other
concentration.
You
select
Trials
1
and
3
to
determine
the
value
of
x.
Taking
the ratio of the
trials
you
find:
(NOTE:
The
&
will
cancel
out.)
raie,
K{H,J'[NOJ
rate,
k[H,J'[NOJ
4.0x10"
_
k(2.0x107)"
(L0x107)"
9
(2.0x107)'
(1.0x107)’
20x10°
k(LOX10
)
(LOx107)"
(1.0x107) (1.0x107)
x=1
The
order
of
the
reaction with
respect
Hy
is
1.
Repeating
this
process
for
Trials
2
and
3
to
find
the
value
of
y.
rate,
_
k[H,T'[NOJ"
rate;
k[H,]'[NOJ"
1.6x10°
k(20x107)'(2.0x10°)'
.
(2.6x107)
(20107
4.0x10°
k(20x107)y
(1.0x10°)"
(20x107)(1.0x107)
y=2
The
order
of the
reaction with respect
NO
is
2.
The
rate
law
for
this
reaction
is
rate
=
[H2)[NOJ.
This
is
choice
(B).
Choice
(A)
is
incorrect
because
it
reverses
the
exponents.
Choice
(C)
is
incorrect
because
it
uses
the
coefficients
from
the
balanced
equation.
Choice
(D)
is
incorrect
because
it
is
the
equilibrium
constant
expression
for the
reaction.
Practice
Questions
Related
to
This:
PQ-1,
PQ-2,
and
PQ-3
124
*ACS
Kinetics
SQ-2.
The
rate
law
for
the
reaction
w.&“‘
A+B—C+D
is
first
order
in
[A]
and
second
order
in
[B].
If
[A]
is
halved
and
[B]
is
doubled,
the
rate
of
the
reaction
will
(A)
remain
the
same.
(B)
be
increased
by
a
factor
of
2.
(©)
be
increased
by
a
factor
of
4.
(D)
be
increased
by
a
factor
of
8.
Knowledge
Require&:'-(‘lj
The
meanings
of
the
terms
rate
law,
first
order,
and
second
order.
(2)
The
ability
to
use
the
rate
law
expression.
Thinking
it
Through:
You
are told the
order
of
the
reaction
with
respect
to
each
reactant.
You
recognize
that
this
is
telling
you
the
exponents
in
the rate
expression.
The
exponent
for
[A]
is
1
and
the
exponent
for
[B]is
2.
You
write
the
rate
expression
for
the
reaction:
rate
=
K[A](B]?
You
realize
the
problem
is
asking
you
to
determine
the
effect
of
changing
the
initial
concentrations
of
both
reactants
on
the
rate
of the
reaction.
This
is
a
similar
problem
to
SQ-1,
where
you
were
given
the
initial
amounts
and
the
rates
but
were
asked
to
find
the
exponents.
Here
you
know
the
exponents
and
the
amounts
and
are
being
asked
to
find
the ratio
of
the
rates.
As
before
you
set
this
up
as
a
ratio
of the
two
rates
(for
two
reaction
trials
noted
by
“17and
“2”
subscripts):
Where
rate:,
[A]:,
and
[B];
are
the
rate,
concentration
of A,
concentration
of
B
for
trial
#1
and
rate;,
[A],
and
[B],
are the
rate,
concentration
of A,
concentration
of
B
for
trial
#2.
The
ratio of the
rates
is
what
you
need
to
find.
You
also
know
the
relationship
between
[A];
and
[A]
and
[B];
and
[B],
from
the
information
in
the
question:
[A]2
=%
[A]:
(because
“[A]
is
halved™)
and
[B]2
=
2[B];
(because
“[B]
is
doubled™)
Substituting
these
into the
expression
you
get:
e,
_kALEE
_A(J)CB1)
k(Ja)4m)’
rate,
k[A)[B]
K[A),[BY
K[A][B]}
e,
(1)
The
correct
answer
is
choice
(B).
Choice
(A)
is
incorrect
because
the
2
was
not
squared
in
the
expression
(2[B])2.
Choice
(C)
is
incorrect
because
it
omitted
the
%2
term.
Choice
(D)
is
incorrect
because
the
%
term
was
mistakenly
put
in
the
denominator
rather
than
the
numerator.
Practice
Questions
Related
to
This:
PQ-4
and
PQ-5
SQ-3.
The
half-life
for the
first
order
conversion
of
cyclobutene
to
ethylene,
C:Hs(g)
—2C2Ha(g)
is
22.7
s
at
a
particular
temperature.
How
many
seconds
are
needed
for
the
partial
pressure
of
cyclobutane
to
decrease
from
100
mmHg
to
10
mmHg?
A)
0.101s
®)
5205
©
755s
D)
52335
125
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Kinetics
.
ACS
v
Exams
Knowledge
Required:
(1)
The
relationship
between
half-life
and
the
rate
constant
for
first-order
reaction.
(2)
The
ability
to
use
the
half-life
equation
and
the
integrated
first-law
equation.
Thinking
it
Through:
You
are
given
a
reaction
and
told
that
it
is
first-order.
You
need
to
select
and
use
the
first-
order
integrated
rate
law
to
find
the
amount
of
time
needed
for
the
pressure
to
fall
from
100
mmHg
to
10
mmHg.
You
realize
that
you
are
not
given
a
value
for
the
rate
constant,
.
You
recognize
that
because
you
know
the
half-
life,
you
can
find
the
value
of
the
rate
constant.
Because
the
reaction
is
first-order
you
use
the
half-life
expression
for
a
first-order
reaction:
_In2
k
Rearranging
and
substituting
the
given
half-life
you
calculate
the
value
of
k.
In2
In2
B
k=—=
=0.0305s
22.7s
You
then
use
this
value
of
k
in
the
first-order
intégrated
rate
law.
In[C,H,
]
=~k
+In[C
H,]
ICHI-IICHL
_
I0(10)~In(100)
-
which
is
choice
(C).
k
0.03055s"
’
Choice
(A)
is
incorrect
because
it
used
the
value
of
the
half-life
as
the rate
constant.
Choice
(B)
is
incorrect
because
it
used
1/half-life
as
the
rate
constant,
instead
of
In2/half-life.
Choice
(D)
is
incorrect
because
it
used
the
expression
for
the
second
order
half-life
to
find
the
value
of
k.
Practice
Questions
Related
to
This:
PQ-6
and
PQ-7
SQ-4.
The
half-life
for the
first-order
radioactive
decay
of
2P
is
14.2
days.
How
many
days
would
be
required
for
a
sample
of
a
radiopharmaceutical
containing
**P
to
decrease
to
20.0%
of
its
initial
activity?
(A)
33.0d
(B)
49.2d
©)
71.0d
(D)
286d
Knowledge
Required:
(1)
The
knowledge
that
radioactive
decay
is
first-order.
(2)
The
ability
to
use
the
integrated
form
of
the
first-order
rate
law.
(3)
Relationship
between
half-life
and
rate
constant.
Thinking
it
Through:
You
are
being
asked
to
find
the
time
for
a
process
to
occur.
You
know
that
you
have
to
use
the
integrated
rate
law.
You
also
know
that
nuclear
decay
is
a
first-order
process.
To
use
the
first-order
integrated
rate
law
you
need
a
value
for the
rate
constant.
You
use
the
relationship
for
the
first-order
half-life:
2
k
_
2
2
"
Solving
for
k
you
get:
k
=
—
=0.0488
d
4
v
14.2d
The
question
is
asking
you
for
the
time
required
for
the
sample
to
decrease
to
20.0%
of
its
original
activity.
This
is
different
from
other
problems
that
give
you
specific
amounts
of
initial
and
final
concentrations.
To
solve
the
problem,
you
realize
that
you
can
pick
any
initial
amount
and
the
final
amount
will
be
20.0%
of
that
initial
value.
To
keep
the
math
simple,
you
assume
you
started
with
100.0%.
You
then
need
to
find
the
time
required
for the
amount
to
decrease
to
20.0%
of
100%,
or
0.200%1.000
=
0.200.
You
now
know:
[Pl
=1.00;
[**P]=0.20;
k=0.0488d
¢
You
then
use
the
integrated
first-order
rate
law
to
find
the
time
needed.
126
-
ACS
Kinetics
W
Exams
In[“P]=~kr
+In{*P),
In[=P]—
]"[“P]u
~
In(0.200)~
In(1.00)
_
3304
which
is
choice
(A).
k
004884
'
Choice
(B)
is
incorrect
because
it
represents
5
half-lives
multiplied
by
In(2).
Choice
(C)
is
incorrect
because
it
assumes
it
took
5
half-lives
to
get
to
20.0%
left.
Choice
(D)
is
incorrect
because
it
is
20
half-lives.
Practice
Questions
Related
to
This:
PQ-8
and
PQ-9
SQ-5.
Plots
are
shown
for
the
reaction
NOx(g)
—
NO(g)
+
2
Oa(g).
What
is
the
rate
law
for
the
reaction?
=
5
g
time
time
time
k
B)
te
=k
!
Al
te
=
rate
=k
x
(A)
rate
(B)
NO.)
(€)
rate=k[NO,]
(D)
rate
=
k[NO,
|’
Knowledge
Required:
(1)
Definition
of
rate
law.
(2)
Ability
to
interpret
integrated
rate
law
equations.
Thinking
it
Through:
You
are
given
three
plots of
the
concentration
versus
time
data
for
a
reaction.
Each
plot
differs
in
how
it
plots
the
concentration
of
NO,.
You
recall
the
three
common
integrated
rate
law
expressions
for
reactions
with
a
single
reactant:
X
—
P
Order
Rate
Law
Integrated
Rate
Law
Linear
Plot
zero
rate
=
k
[X]e=—kt
+
[X]o
[X]
vs
time
first
rate
=
k[X]'
In[X],
=
—¢
+
In[X],
In[X]
vs
time
second
rate
=
K[XJ*
V[X]=
ke +
1/[X]o
1/[X]
vs
time
Each
of the
integrated
rate
laws
has
the
form
of
a y
=
mx
+
b
equation.
When
you
examine
the
plots
provided,
you
notice
that
the
plot
of
1/[NO;]
vs
time
is
linear
and
the
others
are
nonlinear.
You
decide
that this
means
the
rate
law
is
a
second
order
rate
law:
rate
=
k[NO,
]
which
is
choice
(D).
Choice
(A)
is
incorrect
because
this
is
the
zero-order
rate
law
and
the plot of
[NO]
is
not
linear.
Choice
(B)
is
incorrect
because
this
is
taken
directly
from
the
plot
with
1/[NO:]
on
the
y-axis.
Choice
(C)
is
incorrect
because
this
is
a
first-order
rate
law
and
the
plot
of
In[NO-]
is
not
linear.
Practice
Questions
Related
to
This:
PQ-10
and
PQ-11
127
Kinetics
@
ACS
SQ-6.
‘What
changes
when
a
catalyst
is
added
to
the
reaction
described
by
this
energy
diagram?
(A)
Tand
II
(B)
Land
11
©)
Tonly
(D)
I
only
Knowledge
Required:
(1)
Interpretation
of
reaction
energy
diagrams.
(2)
Meaning
of
catalyst
and
activation
energy.
Thinking
it
Through:
Y
ou
are
being asked
to
interpret
a
reaction
energy
diagram.
Several
parts of
the
diagram
have been
labeled.
Examining
the
diagram,
you
recall
what
each
labeled portion
represents.
1
s the
activation
energy
for
the
reaction
in
the
forward
direction.
11
is
the
activation
energy
for
the
reaction
in
the
reverse
direction.
II1
is
the
difference
in
energy
of
the
reactants and
products,
which
is
the
AH
for
the
reaction.
The
question
asks
you
to
determine
which
of
these
is
affected
by
the
addition
of
the
catalyst.
You
remember
that
a
catalyst
provides
an
alternative
pathway
for
the
reaction.
This alternative
pathway
has
a
lower
activation
energy.
The
presence
of
a
catalyst
does
not
affect
the
value
of
AH
for
the
reaction.
Because
arrows
I
and
I
represent
activation
energies,
these
will be
affected
by
the
addition
of
a
catalyst.
The
correct
answer
is
choice
(A).
Choice
(B)
and
(D)
are
incorrect
because
a
catalyst
does
not
affect
the
value
of
the
enthalpy
change,
IIL
Choice
(C)
is
incorrect
because
a
catalyst
will
affect
the
activation
energy
for
both
the
forward
and
reverse
reactions.
Practice
Questions
Related
to
This:
PQ-12,
PQ-13,
and
PQ-14
SQ-7.
Which
statement
regarding
chemical
reactions
is
true
according
to
collision
theory?
(A)
All
molecular
collisions
result
in
chemical
reactions.
(B)
Catalysts
make
individual
collisions
more
effective.
increasing
reaction
rates.
(©)
Proper
orientation
of
molecules
is
required
for
collision
to
result
in
chemical
reactions.
D)
Increasing
the
temperature
of
a
reaction
decreases
the
kinetic
energy
of
molecules,
making
collisions
more
effective.
|
Knowledge
Required:
(1)
Understanding
of
the
collision
theory
of
chemical
reactions.
(2)
Knowledge
of
what
factors
influence
the
rate
of
chemical
reactions.
Thinking
it
Through:
The
question
asks
you
to
evaluate
a
series
of
statements
about
chemical
reactions
and
collision
theory.
You
decide
to
evaluate
each
option
individually.
You
know
that
for
a
collision
to
lead
to
a
chemical
reaction
it
must
satisfy
two
criteria.
The
collision
must
have
sufficient
energy
for
the
reaction
to
occur.
Also,
the
molecules
must
be
in
the
proper
orientation
for the
collision
to
be
effective.
Therefore,
not
all
collisions
result
in
a
chemical
reaction.
Therefore
choice
(A)
is
incorrect.
The
presence
of
a
catalyst
does
increase
the
rate
of
a
reaction.
However,
the
way
a
catalyst
increases
the
rate
of
a
reaction
is
not
by
making
the
collisions
more
effective.
Instead
a
catalyst
increases
the
rate
of
a
reaction
by
providing
an
alternative
route
from
reactants
to
products
that
has
a
smaller
activation
energy.
Processes
with
smaller
activation
energies
result
in
a
faster
reaction
rate.
Therefore
choice
(B)
is
incorrect.
128
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cACS
Kinetics
—9Y
Exams
You
know
that
one
of the
two
requirements
for
a
collision
to
lead
to
a
chemical
reaction
is
that the
reactants
must
collide
with
the
proper
orientation.
Therefore
choice
(C)
is
correct.
Increasing
the
temperature
often
leads
to
an
increase
in
the
rate
of
a
reaction.
However,
the
reason
is
because
as
the
temperature
increases
the
average
kinetic
energy
of the
reactants
increases.
This
can
increase
the
rate
of
a
reaction
in
two
ways.
First,
there
are
more
collisions
that
have
the
minimum
energy
needed
for
a
reaction
to
occur.
Second,
the
higher
kinetic
energy
increases
the
number
of
collisions,
which
means
the
odds
of
a
collision
having
the
proper
orientation
increases.
This
increases
the
rate
of
a
reaction.
Therefore
choice
(D)
is
incorrect.
Practice
Questions
Related
to
This:
PQ-15,
PQ-16,
PQ-17,
PQ-18, PQ-19, PQ-20,
PQ-21, PQ-22,
PQ-23,
and
PQ-24
SQ-8.
The
activation
energy
for
a
particular
reaction
is
83.1
kJ-mol™'.
By
what
factor
will the
rate
constant
increase
when
the
temperature
is
increased
from
50.0
°C
to
60.0
°C?
(A)
253
(B)
1.00
©
0.927
(D)
0.395
Knowledge
Required:
(1)
How
to
use
the
Arrhenius
equation.
(2)
The
relationship
between
activation
energy
and
the
value
of the rate
constant.
Thinking
it
Through:
You
are
being
asked
to
determine
the
relative
change
in
the
value
of
the rate
constant
when
the
temperature
changes.
You
recall
the
Arrhenius
equation
which
relates
the
value
of
the
activation
energy,
the
temperature,
and
the
rate
constant.
You
also
remember
that
all
the
temperature
values
must
by
in
kelvin:
k=
A
BT
.
.
-E,
1
An
alternative
form
of
the
equation
is:
Ink=
(—)(—]
+In(4)
R
T
This
form
of
the
equation
is
useful
if
you
have
measured
values
of
&
at
different
temperatures.
You
recognize
that
this
equation
is
in
the
form:
y
=
mx
+
5.
If
you
plot
In{k)
versus
1/7
the
slope
of
this
plot
is
equal
to
(~<E./R).
Another
form
of
this
equation,
often
called
the
two-point
form,
is
used
when
you
only
have
two
values
of
k
or
temperature
values.
You
get
this
form
of
the
equation
by
subtracting
the
equation
at
the
first
temperature,
T,
from
that
at
the
second
temperature,
75:
Mz_.nk;(-_lg"-)[%}_(j-][%},nw_lnu,
o)
G
You
recognize
that
the ratio
of
k2/k)
is
what
the
problem
is
asking
for
and
you
substitute
the
given
values
into the
equation.
Note:
you
are
careful
to
make
sure
the
units of
R
and
E,
match
and
the
temperature
is
in
kelvin:
wl
~83.141
-
mol”
(10001)(
1
1
)
k
83143
mol
K"
U
13
JA33315K
323.15K
K,
In|
==
|=0.927
k\
The
correct
answer
is
choice
(A).
Choice
(B)
is
not
correct
because
it
is
the
answer
if
the
units
of
£,
are
not
converted
to
J.
Choice
(C)
is
not
correct
because
it
is
the
natural
log
(In)
of
the ratio
of
the
rate
constants.
Choice
(D)
is
not
correct
because
it
is
the ratio
ki/k.
"
=253
Practice
Questions
Related
to
This:
PQ-25
and
PQ-26
129
Kinetics
.o
ACS
/WY
Exams
SQ-9.
Consider
the
reaction
2NOs(g)
+
Falg)
=
2NO:F(g)
A
proposed
mechanism
for
the
reaction
is
NO;
+
Fa=
NO:F
-
F
(slow)
NO:
+
F=
NO;F
(fast)
What
is
the
rate
law
for
this
mechanism?
A
etk
®)
KINO_TE]
Tate
=k
—————
rate
=
LTI
‘
NOTIE]
¢
I
(©)
rate=kNO,][F]
(D)
rate=A[NO,J(F]
Knowledge
Required:
(1)
The
fact
that the
slowest
step
in
the
reaction
mechanism
determines
the
rate
law
for
the
|
reaction.
(2)
Being
able
to
write
a
rate
law
expression
from
an
elementary
step.
Thinking
it
Through:
You
are
given
a
mechanism
and
are
asked
to
write
the
rate
law
for
the
proposed
mechanism.
The
slowest
step
in
a
reaction
mechanism
is
the
step
with
the
highest
activation
energy
and
the
species
involved
in
this
step
determine
the rate
law
for
the
reaction.
You
note
that
you
are told the
first
step
is
the
slowest.
You
know
that
you
can
write
the
rate
law
for
this
elementary
step
using
the
coefficients
from
the
elementary
step
as
the
exponents
in
the
rate
law.
You
write
the
rate
law
for the
slow
step
as:
rate
=
([NO,
][F,
]
which
is
choice
(C).
Choice
(A)
is
not
correct
because
it
is
the
cquilibrium
expression
for
the
reaction,
not
a
rate
law.
Choice
(B)
is
not
correct
because
it
uses
the
stoichiometry
of the
overali
reaction.
Choice
(D)
is
not
correct
because
it
is
not
the
derived
from
the
slowest
step.
Practice
Questions
Related
to
This:
PQ-27, PQ-28,
and
PQ-29
SQ-10.
_
Consider
this
equilibrium:
280s(g)
+
Ox(g)
=
280s(g)
The forward
reaction
is
proceeding
at
a
certain
rate
at
some
temperature
and
pressure.
When
the
pressure
is
increased.
what
might
we
expect
for
the
forward
reaction?
(A)
agreater
rate
of
reaction
and
a
greater
yield
of
SO:
at
equilibrium
(B)
a
greater
rate
of
reaction
and
a
same
yield
of
SO;
at
equilibrium
(©)
a
lesser
rate
of
reaction
and
a
smaller
yield
of
SO;
at
equilibrium
(D)
alesser
rate
of
reaction and
a
greater
yield
of
SO;
at
equilibrium
Knowledge
Required:
(1)
How
to
use
LeChatlier’s
principle.
(2)
Qualitative
understanding
of
rate
laws.
Thinking
it
Through:
Y
ou
need
to
predict
what
will
happen
to
an
equilibrium
system
when
it
is
subjected
to
a
change.
You
know
that
a
change
in
pressure
may
shift
the
equilibrium
and
depends
on
the
moles
of
gaseous
substances
as
reactants and
products.
An
increase
in
pressure
will
shift
the
equilibrium
in
the
direction
that
minimizes
the
increase
in
pressure.
This
is
the
side
that
has
the
fewest
moles
of
gas.
For
this
reaction,
the
side
with
fewer
moles
of
gas
is
the
product
side.
You
realize
that
this
means
that
an
increase
in
pressure
will
cause
the
rate
of the
reaction
producing
SOs
to
increase.
The
correct
answer
is
choice
(A).
Choices
(B)
and
(C)
is
incorrect
because
SO;
will
increase.
Choice
(D)
is
incorrect
because
the rate of
the
reaction
will
increase.
Practice
Questions
Related
to
This:
PQ-30
130
cACS
Kinetics
Practice
Questions
(PQ)
w‘f\“\
PQ-1.
When
the
reaction:
"Exp
[CH:Cllo/M
[H:000/M
Initial
Rate
/
M-
CH;Cl(g)
+
H:0(g)
—
CH:OH(g)
+
HCI(g)
[
0.100
0.100
0.182
was
studied,
the
tabulated
data
were
2
0.200
0.200
1.45
obtained.
Based
on
these
data,
what
are
the
3
0.200
0.400
5.81
reaction
orders?
(A)
CH;Cl:
first
order
H,O:
first
order
(B)
CH:Cl:
first
order
H-0:
second
order
(C)
CH;CI:
second
order
H»O:
first
order
(D)
CH:Cl:
second
order
H>0:
second
order
PQ-2.
The
reaction
between
acetone
and
bromine
Ex
[CH3COCHso
[Br2Jo/
[H3O7o/
Initial
Rate
in
acidic
solution
is
represented
by the
P
/M
M
M
/M-s™!
equation:
1
0.30
0.050
0.050
5.8%107
CH;COCH;(aq)
+
Bra(aq)
+
H:O"(aq)
—
2
0.30
0.100
0.050
5.8x107
products
3
0.30
0.050
0.100
1.2x104
The
tabulated
kinetic
data
were
gathered.
4
0.40
0.050
0.200
3.2x107
Based
on
these
data,
the
experimental
rate
law
is
(A)
rate
=
{{CH;COCH;](Br2][H;0"]
(B)
rate
=
k[CH;COCH;][H;:O']
(C)
rate
=
k[H;0°]*
(D)
rate
=
K[CH;COCH;][Br]
PQ-3.
Initial
rate
data
for
the
reaction
Exp
[Hz]o/
M
[ClzJo/M
_Initial
Rate
/
M-s™!
2Hx(g)
+
Cl(g)
—
2HCI(g)
1
0.0020
0.0050
2.5%107%
are
given
in
the
table.
What
is
the
rate
law
|
2
0.0020
0.0025
1.3x107
for the
reaction?
3
0.0015
0.0025
1.3x107
4
0.0050
0.0010
0.5%107
(A)
rate
=
K[CLJ
(B)
rate
=[Cly]
(C)
rate
=
k[Ha)
(D)
rate=k{H:)[Cl>]
molecules.
The
rate
law
is
rate
=k[A2][B2]
1f
the
concentrations
of
both
A
and
B>
are
doubled,
the
reaction
rate
will
change
by
a
factor
of
(A)
%
®)
2.
©
2.
D)
4.
W’,&
PQ-4.
The
gas
phase
reaction,
A
+
B,
—2AB,
proceeds
by
bimolecular
collisions
between
Az
and
B>
PQ-5.
For
the
reaction
of
chorine
and
nitric
oxide,
2NO(g)
+
Clx(g)
—
2NOCl(g)
doubling
the
concentration
of
chlorine
doubles
the
rate
of
reaction.
Doubling
the
concentration
of
both
reactants
increases
the
rate
of
reaction
by
a
factor
of
eight.
The
reaction
is
(A)
first
order
in
both
NO
and
Clo.
(B)
first
order
on
NO
and
second
order
in
Cl,.
(C)
second
order
in
NO
and
first
order
in
Cl,.
(D)
second
order
in
both
NO
and
Cla.
131
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Kinetics
=
ACS
W
Exams
PQ-6.
The
decomposition
of
NOCI
is
second
order.
2NOCI(g)
—
2NO(g)
+
Clx(g)
The
initial
concentration
of
the
reactant
is
0.050
M.
The
rate
constant
equals
0.020
L-mol'-s™,
what
will
the
concentration
be after
45
min?
(A)
0.048M
(B)
0.029M
(©)
0.014M
(D)
0.0054M
PQ-7.
Ifthe
half-life
of
a
reaction
is
independent
of
concentration,
the
reaction
can
be
[
Ifirstorder
|
IIsecond
order
|
I
zeroorder
|
(A)
TandIIonly
(B)
IIand
Il
only
(C)
Tonly
(D)
Ionly
PQ-8.
The
activity
of
a
radioisotope
is
3000
counts
per
minute
at
one time and
2736
counts
per
minute
48
hours
later.
What
is
the
half-life,
in
hours,
of
the
radioisotope?
(A)
831hr
(B)
521hr
(©)
361hr
(D)
144hr
PQ-9.
After
55
years,
what mass
(in
g)
remains
of
a
200.0
g
sample
of
a
radioactive
isotope
with
a
half-life
of
10.0
years?
A)
022g
(B)
44g
©)
sig
D)
170¢g
PQ-10.
If
the
reaction
Os(g)
—
Ox(g)
+
O(g)
is
first
order
in
O3,
which
plot will be
linear?
(A)
[Os]
vs.
time
(B)
1/[0;]
vs.
time
(€)
[0:]
vs.
time
(D)
In[O5]
vs.
time
PQ-11.
The
experimental
data
from
a
certain
reaction
gives
these
three
graphs.
What
is
the
most
likely
order
for
this
reaction?
g
z
5
time
time
time
(A)
zero
(B)
first
(C)
second
(D)
third
PQ-12.
Consider
the
reaction
potential
energy
diagram.
‘What
describes
the
catalyzed
forward
reaction
pathway?
Ea
AH
(A)
10kJ-mol™
~15
kJ'mol™
(B)
10kJ-mol”
15
kJ-mol™!
(€)
25kJ'mol”!
—15
kJ-mol™!
(D)
25kJ-mol”’
15
kJ-mol"!
132
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cACS
Kinetics
P
Exams
PQ-13.
Which
line
segment
represents
the
activation
energy
for
the
reaction
between
C and
D
to
form
A
and
B?
T
£
£
5
=
5]
Reaction
Coordinate
—
A)
1,2,&3
®B)
2
©)
2&3
D)
3
PQ-14.
A
certain
reaction
has
AH
=
—
75
ki-mol™'
and
an
activation
energy
of 40
kJ'mol"'.
A
catalyst
is
found
that
lowers
the
activation
energy
of
the
forward
reaction
by
15
kJ-mol”.
What
is
the
activation
energy
of
the
reverse
reaction
in
the
presence
of
the
same
catalyst?
(A)
25
kJ'mol”!
(B)
60
kJ'mol"!
(©)
90
kl'mol
(D)
100
ki-mol™!
PQ-15.
Which
statement
best
explains
why
the
activation
energy
is
changed
by
adding
a
catalyst?
(A)
The
catalyst
changes
the
reaction
mechanism.
(B)
The
catalyst
changes
the free
energy
of
reaction.
(C)
The
catalyst
increases
the
kinetic
energy
of
the
reactants.
(D)
The
catalyst
lowers
the
reaction
volume,
increasing
the
concentration
of
reactants.
M‘d
PQ-16.
Which
reaction
rate
is
more
affected
by
a
change
in
temperature?
Reaction
E./kJ-mol!
[
Ha(g)
+
L(g)
—
2H(g)
173
I
CH;CHO(g)
—
CHa(g)
+
CO(g)
356
(A)
reaction
I
because
the
activation
energy
is
lower
(B)
reaction
I
because
the
number
of
moles
of
gas stays
the
same
(C)
reaction
II
because
number
of
moles
of
gases
increases
as
reaction
goes
forward
(D)
reaction
II
because
the
activation
energy
is
higher
PQ-17.
What
will
increase
the
value
of the
rate
constant
for
an
elementary
step?
(A)
adding
a
catalyst
(B)
raising
the
temperature
(C)
increasing
the
concentration
of
products
(D)
increasing
the
concentration
of
reactants
133
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Kinetics
we
ACS
M‘
PQ-18.
(A)
(B)
©)
()
MPQ—IQ.
(A)
(B)
©)
D)
‘,«’vpd
PQ-20.
(A)
(B)
©
(D)
“P‘JPQ-ZI.
[CY]
(B)
©)
(D)
M
PQ-22.
“)
®)
©)
(D)
w“’"’d
PQ-23.
“)
(B)
©)
(D)
Exams
The
Arrhenius
equation
describes
the
relationship
between
the
rate
constant,
&,
and
the
energy
of
activation,
E,
k=
de”
BT
In
this
equation,
4
is
an
empirical
constant,
R
is
the
ideal-gas
constant,
e
is
the
base
of
natural
logarithms,
and
7
is
the
absolute
temperature.
According
to
the
Arrhenius
equation,
at
constant
temperature,
reactions
with
lower
activation
energies
proceed
more
rapidly.
at
constant
temperature,
reactions
with
lower
activation
energies
proceed
less
rapidly.
at
constant
energy
of
activation,
reactions
at
lower
temperatures
proceed
more
rapidly.
at
constant
energy
of
activation,
reactions
with
smaller
values
of 4
proceed
more
rapidly.
‘What
will be the
effect
of
increasing
the
temperature
of
reactants
that are
known
to
undergo
an
endothermic
reaction?
Both
the
rate
of
reaction
and
the
value
of
the
equilibrium
constant
increases.
The
rate
of
reaction
increases
and
the
value
of the
equilibrium
constant
decreases.
The
rate
of
reaction
decreases
and
the
value
of
the
equilibrium
constant
increases.
The
rate
of
reaction
increases
and
the
value
of
the
equilibrium
constant
is
unchanged.
The
change
in
temperature
from
10
°C
to
20
°C
is
found
to
double
the
rate of
a
particular
chemical
reaction.
How
did
the
change
in
temperature
affect
the
reaction?
The
number
of
molecules
with
sufficient
energy
to
react
increased.
The
number
of
molecules
with
sufficient
energy
to
react
decreased.
The
activation
energy
increased.
The
activation
energy
decreased.
The
value
for
the
rate
constant
of
a
reaction
can
generally
be
expected
to
decrease
with
increasing
temperature.
increase
with
increasing
temperature.
decrease
with
increasing
temperature
only
when
the
reaction
is
exothermic.
increase
with
increasing
temperature
only
when
the
reaction
is
exothermic.
Two
reactions
with
different
activation
energies
have
the
same
rate
at
room
temperature.
Which
statement
correctly
describes
the
rates
of
these
two
reactions
at
the
same,
higher
temperature?
The
reaction with
the
larger
activation
energy
will
be
faster.
The
reaction
with
the
smaller
activation
energy
will be
faster.
The
two
reactions
will
continue
to
occur
at
the
same
rates.
A
prediction
cannot
be
made
without
additional
information.
A
catalyst
increases
the
rate
of
a
reaction
by
changing
the
mechanism
of
the
reaction.
increasing
the
activation
energy
of
the
reaction.
increasing
the
concentration
of
one
or
more
reactants.
decreasing
the
difference
in
relative
energy
of
the
reactants
and
products.
134
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Kinetics
.
ACS
¥
Exams
M
PQ-24.
Which
statement
most
accurately
describes
the
behavior
of
a
catalyst?
(A)
A
catalyst
increases
AG
of
a
reaction
and
hence
the
forward
rate.
(B)
A
catalyst
reduces
the
AH
of
a
reaction
and
hence
the
temperature
needed
to
produce
products.
(C)
A
catalyst
reduces
the
activation
energy
for
a
reaction
and
increases
the
rate
of
a
reaction.
(D)
A
catalyst
increases
the
equilibrium
constant
and
final
product
concentrations.
PQ-25.
The
rate
constants
for
a
specific
reaction
at
two
different
temperatures
are
given
in
the
table.
‘What
is
the
activation
energy
for
the
reaction?
(A)
-88.1
kJ'mol”’
(B)
“127Kmol!
(C)
127
Temperature
_Rate
constant
127
°C
3.0x10%s!
227°C
6.0x102s™"
KJmol”!
(D)
88.1kJmol”!
PQ-26.
Activation
energy
can
be
experimentally
determined
from
the
slope
of
the
plot
of
(A)
kversus
I/T.
(B)
kversus
In7.
(C)
Inkversus
T.
(D)
Ink
versus
1/T.
M&
PQ-27.
Which
energy
diagram
best
matches
the
proposed
mechanism
for
this
exothermic
reaction?
E+S=ES
(fast)
ES—E+P
(slow)
(A)
T
(B)
T
8
Reaction
Progress
—»
©
(D)
Fnergy
~—>
A
Reaction
Progress
—»
PQ-28.
Consider
the
reaction
Reaction
Progress
—»
—
L
Reaction
Progress
—»
Clx(g)
+
HaS(aq)
—
S(s)
+
2H*(aq)
+
2Cl(aq)
The
rate
equation
for
this
reaction
is
rate
=
k{CLy)[H.S}
Which
mechanisms
is
(or
are)
consistent
with
this
rate
equation
(if
any)?
1
Cly+
Hy8
—
H'
+CI
+
CI*
+
HS®
(slow)
CI'+HS"
—
H'+
Cl'+§
(fast)
i
HaS
—
H™+
HS™
(fast)
Cly+HS"—2CI+
H*
+8
(slow)
(A)
Tonly
(B)
Monly
(C)
BothIand
It
(D)
Neither
Lor
IT
135
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Kinetics
o
ACS
W
Exams
PQ-29.
The
decomposition
of
hydrogen
peroxide
in
the
presence
of
iodide
ion
is
believed
to
occur
via
this
mechanism.
H:0:(aq)
-
I
(aq)
—
H:0(1)
+
10(aq)
H:05(aq)
+
10-(ag)
—
H:0(1)
~
Ox(g)
+
I(aq)
In
this
mechanism,
[
(aq)
is
(A)
acatalyst.
(B)
areactant
in
the
overall
reaction.
(C)
the
activated
complex.
(D)
aproduct
of
the
overall
reaction.
PQ-30.
What
is
the
relationship
between
the
equilibrium
constant
(K.)
of
a
reaction and
the
rate
constants
for
the
forward
(k)
and
reverse
(,)
reactions
in
a
single
step
reaction?
(A)
K=k
B)
Ko
=klks
©
Ko
=1liksh,)
®
K=k-k
136
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o
ACS
Kinetics
Eak
ol
2l
o
OFE
Pownanpwnrs
CwanOATREE®
-
flyEmms
Answers
to
Study
Questions
RN
wA»T
-
Ze
>0
Answers
to
Practice
Questions
11.B
21.B
12.A
22.A
13.D
23.A
14.D
24.C
15.A
25.D
16.D
26.D
17.B
27.C
18.A
28.A
19.A
29.A
20.A
30.B
137
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Kinetics
138
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Chapter
11
—
Equilibrium
Chapter
Summary:
This
chapter
will
focus
on
equilibrium
including
equilibrium
constants
and
determining
the
amount
of
reactants and
products
at
equilibrium.
Specific
topics
covered
in
this
chapter
are:
Definition
of
equilibrium
Law
of
Mass
Action
K,
and
K.
Equilibrium
calculations
Reaction
quotient
(Q)
of
a
reaction
e
Le
Chatlier’s
Principle
Previous
material
that
is
relevant
to
your
understanding
of
questions
in
this
chapter
include:
*
Stoichiometry
(Chapter
4)
e
Solutions
and
Aqueous
Reactions,
Part
1
(Chapter
5)
*
Solutions
and
Aqueous
Reactions,
Part
2
(Chapter
9)
Where
to
find
this
in
your
textbook:
The
material
in
this
chapter
typically
aligns
to
“Fundamentals
of
Equilibrium
Concepts™
in
your textbcok.
The
name
of
your
chapter(s)
may
vary.
Practice
exam:
There
are
practice
exam
questions
aligned
to
the
material
in
this
chapter.
Because
there
are
a
limited
number
of
questions
on
the
practice
exam,
a
review
of
the
breadth
of
the
material
in
this
chapter
is
advised
in
preparation
for
your
exam.
How
this
fits
into
the
big
picture.
The
material
in
this
chapter
aligns
to
the
Big
Idea
of
Equilibrium
(8)
as
listed
on
page
12
of
this
study
guide.
Study
Questions
(SQ)
SQ-1.
What
is
the
equilibrium
expression
for
this
reaction?
Ni(CO),(g)
=
Ni(s)
+
4CO(g)
4
.
“
Ni(CO),
Ni(CO),
w
g
O8Nl
o
[Nico)]
<
~[NiCo).]
<
[Ni(co),]
[co]'
®
2
[Ni][co]*
Knowledge
Required:
(1)
How
to
determine
an
equilibrium
expression
from
a
balanced
equation.
Thinking
it
Through:
The
question
asks
you
determine
the
equilibrium
expression
for
a
given
reaction.
The
answer
options
are
in
Kc;
therefore,
you
will
consider
the
concentrations
of the
materials
when
constructing
the
equilibrium
expression.
An
equilibrium
constant
is
defined
as
the
concentrations
of
products
raised
to
a
value
equal
to
their
stoichiometric
coefficient
in
a
balanced chemical
equation
divided
by
the
concentrations
of
reactants
raised
to
a
value
equal
to
their
stoichiometric
coefficients
in
a
balanced
chemical
equation.
Because
the
concentrations
of
solids
and
liquids
do
not
change
measurably,
both
are
not
included
in
the
equilibrium
expression.
Therefore,
when
the
concentration
of
gaseous
or
solution
products
are
raised
to
their
power
you
have
a
numerator
of
[COJ*;
likewise
for
reactants,
when
the
concentration
of
gaseous
or
solution
reactants
are
raised
to
their
power
you
have
a
denominator
of
[Ni(CO)s].
Combined,
the
numerator
and
denominator
are
Choice
(A).
Choice
(B)
is
not
correct
because
it
includes
a
solid
in
the
expression.
Choice
(C)
is
not
correct
because
it
has
the
reactants
in
the
numerator
and
the
products
in
the
denominator,
which
is
reversed.
Choice
(D)
is
not
correct
because
it
includes
a
solid
in
the
expression
and
is
the
inverse
of
the
balanced
139
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Equilibrium
_*ACS
¥
Exams
chemical
equation
as
written.
Practice
Questions
Related
to
This:
PQ-1,
PQ-2, PQ-3, PQ-4, PQ-5,
PQ-6,
PQ-7, PQ-8,
PQ-9,
PQ-10,
and
PQ-11
SQ-2.
Given
the
reaction
and
equilibrium
constant:
2S0,(2)
=
280,(g)
+
0,(g)
Kc=23x107
What
is
the
equilibrium
constant
for
this
reaction
at
the
same
temperature?
SO,(2)
=280,(2)+
$0,(2)
Kc=?
(A)
12¢107
(B)
4.6x107
(©)
4.8x10*
(D)
43x10°
Knowledge
Required:
(1)
The
relationship
between
a
balanced
equation
and
an
equilibrium
constant.
Thinking
it
Through:
The
question
asks
you
to
determine
an
equilibrium
constant
for
a
given
balanced
chemical
equation
given
a
constant
for
a
related
balanced
chemical
equation.
You
note
that
the
chemical
equation
for
which
you
are
to
determine
an
equilibrium
constant
is
halved
compared
to
the
chemical
equation
for
which
the
equilibrium
constant
is
known.
You
can write
the
equilibrium
expressions
then
for
both
equations
and
compare:
[50.]'[0,]
250,(g)
<=
250,(g)
+
O,
g=l
]
1(g
1
(8)
+0,(8)
[so.f
-
1
_[so0."
v
50,(2)
=50,(2)
+
7
0(®)
K'=
[50.]
=Kk
=JK
‘When
the
reaction
is
halved,
the
equilibrium
constant
is
raised
to
the
0.5
power
(or
the
square
root).
Thus,
the
equilibrium
constant
needed
is
(2.3x10°7)**
or
4.8x10°%,
Choice
(C).
Choice
(A)
is
not
correct
because
it
is
the
known
equilibrium
constant
divided
by
2.
Choice
(B)
is
not
correct
because
it
is
the
known
equilibrium
constant
multiplied
by
2.
Choice
(D)
is
not
correct
because
it
is
the
inverse
of
the
known
equilibrium
constant.
Practice
Questions
Related
to
This:
PQ-12
$Q-3.
Consider
the
equilibrium
reactions:
250,(g)
=
0,(2)
=
280,(2)
Ki
2CO(g)
+
0,(g)
=2C0,(g)
K>
What
is
the
equilibrium
constant,
K,
for
this
reaction?
250,
(g)
+
2CO,
(g)
=
280;(g)
+
2CO(g)
-(
&
p.
-
=K+
@
K_(KZJ
®
K
(K]
©
K=KxK,
®
K=K+
Knowledge
Required:
(1)
How
to
combine
multiple
chemical
reactions
and
their
respective
equilibrium
constants
to
form
a
new
reaction
and
constant.
Thinking
it
Through:
The
question
asks
you
to
determine
the
equilibrium
constant
for
a
chemical
reaction
that
results
from
the
combination
of
two
reactions
with
given
equilibrium
constants.
First,
you
note
that
in
order
to
combine
the
given
reactions,
the
second
reaction
must
be
reversed;
therefore,
the
value
of
the
equilibrium
constant
for
the
reverse
reaction
is
the
inverse
of
the
given
K
or
K5™'.
250,(g)
+
0,(g)
=
280,(g)
K
2C0,(g)
=
2CO(g)
+
0,
(8)
K
280,
(g)
+
2CO,
(g)
=
280;(g)
+
2CO(g)
Kix
K™
The
desired
reaction
is
now
the
sum
of
the
first
reaction and
the
inverse
of the
second
reaction;
when
reactions
are
summed,
the
new
equilibrium
constant
is
the
product
of
the
two
known
equilibrium
constants
or
K,*
K;™!
for
the
140
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s
ACS
Equilibrium
reaction
given,
Choice
(A).
Choice
(B)
is
not
correct
because
it
represents
the
inverse
of
the
desired
chemical
reaction.
Choice
(C)
is
not
correct
because
it
does
not
account
for
the
reverse
of
the
second
chemical
reaction.
Choice
(D)
is
not
correct
because
it
is
the
sum
of
the
two
constants;
however,
when
reactions
are
summed,
the
respective
equilibrium
constants
are
multiplied.
Practice
Questions
Related
to
This:
PQ-13
and
PQ-14
SQ-4.
Consider
the
reaction:
X,
(g)
+2Y(g)
=
2Z(g)
.
12.00
moles
of
Z
are
placed
in
an
evacuated
2.00-
liter
flask.
After
the
reactants
and
products
reach
equilibrium,
the
flask
contains
6.00
moles
of
Y.
What
is
the
equilibrium
constant,
K,
for the
reaction?
(A)
0.333
(B)
0.667
(©)
150
(D)
3.00
Knowledge
Required:
(1)
How
to
determine
an
equilibrium
expression
from
a
balanced
equation.
(2)
How
to
calculate
equilibrium
amounts
from
an
equilibrium
expression.
Thinking
it
Through:
The
question
asks
you
to
determine
the
equilibrium
constant given
a
chemical
reaction,
the
amount
of
the
single
species present
before
equilibrium,
and
the
amount
of
one
species
after
equilibrium
is
reached.
First,
you
are told that
there
are
12.0
moles
of
Z
initially,
and
6.00
moles
of
Y
at
equilibrium.
You
then
also
consider
that
you
know
how
the
amounts
will
change
because
you
begin
with
only
product
(so the
change
for
Z
will
be
negative
and
the
change
for
X;
or
Y
will
be
positive)
and
the
amounts
of
these
changes
from
the
stoichiometry
of the
balanced
equation
(an
increase
of 2x
because
the
coefficient
for
Y
is
2,
etc.).
This
is
best
summarized
in
a
table:
X2
+
2y
=
2z
Initial
0
0
12.00
Change
+x
+2x
2x
Equilibrium
+x
6.00
12.0
-
2x
Now
you
can
see that
you
have
an
equivalency
of 2x
=
6.00
or
x
=
3.00.
Using
this
value
for
x,
you
can
complete
all
values
in
the
table:
X2
+
2Y
=
2Z
Tnitial
0
0
12.00
Change
+3.00
+2(3.00)
—2(3.00)
Equilibrium
3.00
6.00
6.00
You now
have
all
of
the
molar
amounts
of
the
substances
at
equilibrium.
To
obtain
the
molar
concentrations,
you
divide
each
value
by
2.00
L
(given
in
the
problem):
[Xa]=1.50M
[Y]=3.00M
(Z]=3.00M
To
determine
the
value
of
K,
you
write
an
equilibrium
expression
for
this
reaction using
the
concentration
of
the
product
raised
to
its
coefficient
divided
by
the
concentrations
of the
reactants
raised
to
their
coefficients;
you
can
then
substitute
the
equilibrium
concentrations
and calculate
K:
oz
oMy
1
DCIYE
(1,50
M)(3.00
MY
1.50
Choice
(A)
s
not
correct
because
it
is
determined
using
moles
and
not
molar
concentrations.
Choice
(C)
is
not
correct
because
it
is
determined
using
the
inverse
of
K.
Choice
(D)
is
not
correct
because
it
is
determined
using
moles
and
the
inverse
of K.
=0.667
Choice
(B)
Practice
Questions
Related
to
This:
PQ-15
and
PQ-16
SQ-5.
Phosgene
decomposes
into
carbon
monoxide
and
elemental
chlorine.
If
the
initial
concentration
of
COCly(g)
is
0.50 M,
what
is
the
equilibrium
concentration
of
CO(g)?
COCl,(g)
=
CO(g)
+Cly(g)
Ke=6.6x10*
(A)_1.8x10%M
(B)
9.1x10°M
(©
17x10%M
(D)
6.6x10°M
141
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Equilibrium
w.
ACS
Exams
Knowledge
Required:
(1)
How
to
write
an
equilibrium
expression
from
a
balanced
equation.
(2)
How
to
use
the
equilibrium
expression
to
find
equilibrium
amounts.
Thinking
it
Through:
The
question
asks
you
to
determine
the
equilibrium
concentration
of
a
product
given
that
only
the
reactant
is
present
before
it
is
allowed
to
reach
equilibrium.
You
can
assume
that the
reaction
is
done
at
constant
volume
and
therefore
concentrations
are
proportional
to
amount.
Begin
by
setting
up
a
table
of
initial-change-equilibrium
concentrations.
Initial
concentrations
are
provided
in
the
question.
Mole
ratios
are
used
to
determine
the
changes
in
concentration.
Equilibrium
values
are
determined
by
the
addition/subtraction
of
the
initial
and
change
values:
coc,
=
co
Cly
Tnitial
0.50
M
0
0
Change
X
+x
X
Equilibrium
0.50
—x
x
x
The
equilibrium
expression
for
this
reaction
is
the
concentration
of
products
raised
to
their
respective
coefficients
divided
by
the
concentration
of
the
starting
material
raised
to
its
respective
coefficient:
«
-
1C0ICL]
<
[cocL]
You
then
substitute
the
values
obtained
into
this
expression:
6.6x10™
=
%)—
.30
-x
2
And
simplified
further
to:
6.6x10™
=—*
0.50-x
At
this
point,
you
may
be
inclined
to
continue
the
algebra
to
set
up
the
equation
to
use
the
quadratic
equation.
However,
you
also
know
you
may
be
able
to
simplify
the
process.
The
value
of
the
equilibrium
constant
is
very
small.
You
know
this
means
that
most
of
the
species
at
equilibrium
are
reactants
with only
a
very
small
quantity
of
products
if
you
began
with
only
reactants
(which
you
did).
Therefore,
you
then note
that
any
change
to
the
initial
concentration
of
COCI;
will
be
very
small,
leaving
the
initial
concentration
(0.50
M)
essentially
unchanged
or:
COCl,
=
co
Ch
Tnitial
0.50
M
0
0
Change
—x
+x
+x
Equilibrium
0.50
-
x~0.50
x
x
‘Which
then
makes
the
equilibrium
expression:
6.6x10°
=
w
~
M
0.50-x
050
Which
can
now
be
solved
without
the
quadratic
equation:
6.6x10"
=
(L
050
0.50
Of
course,
you
also
know
there
is
nothing
wrong
with
continuing
from
before
and
setting
up
the
equation
and
using
the
quadric
equation:
6.6x10°(0.50~x)
=
x*
*+6.6x107
(x)-3.3x10"
The
quadratic
equation
can
then
be
used
to
determine
possible
values
of
x:
£1.8x107.
concentrations
of
the
products
cannot
be
negative,
x
must
be
equal
to
+1.8x 10~
and
thus
the
final
concentration
of
CO
is
1.8x10
M,
choice
(A),
which
was
the
same
answer
you
obtained
using
the
approximation.
Note:
You
can
check
your
approximation
by
using
the
value
of
x
you
calculated
and
using
this
in
the
equilibrium
expression
to
calculate
K.
—
the
value
will be
the
same
as
given
if
the
approximation
can
be
used.
You
can
also
approximate
in
the
reverse
if
the
value
of
K.
is
very
large
and
you
only
begin
with
product
(so
the
change
to
the
product
concentration
will be
small).
Choice
(B)
is
not
correct
because
it
would
be
the
answer
if
*“(2x)*”
was
the
numerator
of
K.
Choice
(C)
is
not
correct
because
it
would
be
the
answer
if
“2x”
was
the
numerator
of
K.
Choice
(D)
is
not
correct
because
the
final
concentration
of
CO
is
not
the
value
of
K.
Practice
Questions
Related
to
This:
PQ-17,
PQ-18,
and
PQ-19
x=1.8x10"
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ACS
Equilibrium
Exams
SQ-6.
BrCl(g)
is
in
equilibrium
with Brx(g) and
Cla(g)
at
25.0
°C:
2BrCl(g)
=
Br,(g)
+
Cl,(8)
K,
=0.130
Initially,
in
a
closed
container
at
25.0
°C,
BrCl(g)
has
a
partial
pressure
of
0.400
atm
and Brx(g) and
Cly(g)
each
have
partial
pressures
of
0.800
atm.
What
is
the
partial
pressure
of
BrC(g)
once
the
system
reaches
equilibrium?
(A)
0419
atm
(B)
0.781
atm
(C)
1.16atm
(D)
121 atm
Knowledge
Required:
(1)
How
to
determine
K;,
from
a
balanced
equation.
(2)
How
to
calculate
equilibrium
partial
pressures
from
a
K;,
expression.
Thinking
it
Through:
The
question
asks
you
to
determine
the
partial
pressure
of
a
species
given
initial
partial
pressures
of the
starting
material
and
products.
Because
this
is
a
closed
system,
partial
pressures
are
proportional
to
amount
and
therefore
changes
in
partial
pressures
are
proportional
to
changes
in
moles
of
a
given
species.
To
decide
which
way
the
reaction
proceeds,
you
calculate
a
value
of
Q
using
the
initial
pressures
given
in
the
question:
P,
0=
B
dolF
)y
Y
o
—(0'800)(0'2300)
=16
(K,
=0.130)
Bl
(0.400)
Because
Q
is
greater
than
K,
the
reaction
will
go
to
the
left.
You
then
continue
by
setting
up
a
table of
initial-change-equilibrium
(ICE)
partial
pressures.
Initial
partial
pressures
are
provided
in
the
question.
Your
decision
earlier
that the
reaction
will
proceed
to
the
left
is
indicated
in
the
ICE
table
by
the
negative
(-)
signs
in
the
change
line for
the
products
and
the
positive
(+)
sign
for
the
reactant.
Mole
ratios
are
used
to
determine
the
changes
in
partial
pressures
(so
+2x
for
2BrCl
and
—1x
for
1Brs,
etc.).
Equilibrium
values
are
determined
by
the
addition/subtraction
of
the
initial
and
change
values.
2BrCl(g)
=
Bn(g)
Ch(g)
Initial
0.400 atm
0.800atm
0.800
atm
Change
+2x
-x
-x
Equilibrium
0.400
+2x
0.800—x
0.800
-
x
The
Kj
expression
for
this
reaction
is
the
partial
pressures
of
products
raised
to
their
respective
coefficients
divided
by
the partial
pressure
of
the
reactant
raised
to
its
respective
coefficient:
You
then
substitute
the
values
obtained
into
this
expression:
.800—x)(0.800—
0.130
=
(800-3)(0800-x)
(0.400+
2x)
The
expression
is
then
further
simplified:
(The
perfect
square
is
used
in
the
simplification.)
0.800-x)"
—x
Y
0.130=
A0800=3)"
m;(,:[%]
(0.400+2x)
0.400+2x
0.800
V0130
=
|
—
0.361(0.400
+2x)
=
0.800
—
(0A400+2x
(
)
*
0.144+0.722x
=
0.800-x
1.722x
=
0.656
x=0381
Replacing
this
value
of
“x”
into the
table,
we
determine
the
partial
pressures
at
equilibrium;
2BrCl(g)
=
Bra(2)
Cly(g)
Initial
0.400
atm
0.800
atm
0.800 atm
Change
+2(0.381
atm)
—
(0.381atm)
(0381
atm)
Equilibrium
1.16
atm
0.419
atm
0.419
atm
Thus,
the
equilibrium
partial
pressure
of
BrCI(g)
is
1.16
atm,
Choice
(C).
Choice
(A)
is
not
correct
because
it
is
the
partial
pressure
of
either
product.
143
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oo
Equilibrium
.
ACS
%
Exams
Choice
(B)
is
not
correct
because
the
answer
comes
from
adding
one-half
of
x
to
the
initial
partial
pressure
of
BrCl(g)
instead
of
a
full
x;
this
is
likely
confused
with
subtracting
one-half
of
x
from
either
product
partial
pressure
to
determine
their
equilibrium
partial
pressures.
Choice
(D)
is
not
correct
because
K,
did not
include
the
“2”
exponent
for
the
partial
pressure
of
the
reactant.
Practice
Questions
Related
to
This:
PQ-20
and
PQ-21
SQ-7.
‘Which
will
drive
the
equilibrium
to
form
more
Cu(s)?
Cu,O(s)
+
CO(g)
=2Cu(s)
+
CO,
(2)
(A)
remove
CO(g)
(B)
remove
CO(g)
(C)
add
a
catalyst
(D)
increase
the
volume
of
the
container
Knowledge
Required:
(1)
Le
Chatelier’s
principle.
Thinking
it
Through:
The
question
asks
you
to
determine
what
would
influence
the
equilibrium
such
that
more
product
is
formed.
Le
Chatelier’s
principle
would
suggest
that
removing
a
reactant
would
lead
to
the
formation
of
more
reactants;
therefore,
choice
(A)
is
incorrect.
Adding
a
catalyst,
choice
(C),
would
result
in
the
rate
of
the
reaction
changing
without
impacting
the
position
of
equilibrium.
Le
Chatelier’s
principle
would
suggest
that
increasing
the
volume
would
favor
the
side
of
the
reaction
with
more
moles
of
gas;
however,
there
is
the
same
number
of
moles
of
gas
for
both
the
reactants and
products
of
this
reaction,
and
choice
(D),
therefore,
is
incorrect.
Finally,
Le
Chatelier’s
principle
would
suggest
that
removal
of
a
product
would
lead
to
the
formation
of
more
product;
therefore,
choice
(B)
is
correct,
i.e.,
“remove
COx(g)”
would
lead
to
the
formation
of
more
Cu(s).
Practice
Questions
Related
to
This:
PQ-22,
PQ-23,
PQ-24,
PQ-25,
PQ-26,
PQ-27,
PQ-28,
PQ-29,
and
PQ-30
Practice
Questions
(PQ)
a
PQ-1.
When
the
reversible
reaction
Na(g)
+
Ox(g)
==
2NO(g)
has
reached
a
state
of
equilibrium,
(A)
no
further
reactions
occurs.
(B)
the
total
moles
of
products
must
equal
the
remaining
moles
of
reactant.
(C)
the
addition
of
a
catalyst
will
cause
formation
of
more
NO.
(D)
the
concentration
of
each
substance
in
the
system
will
be
constant.
PQ-2.
Xenon
tetrafluoride,
XeFs,
can
be
prepared
by
heating
Xe
and
F
together:
Xe(g)
+
2,
(g)
=
XeF,
()
What
is
the
equilibrium
constant
expression
for
this
reaction?
A)
g
-
XE]
®
g
-
XE]
(o
g
o
[XE]
()
o
_[Xe]E]
©
[XellR,]
©2AXe]lR,]
©
[Xe]RT
¢
[XeR]
M\
PQ-3.
What
is
a
proper
description
of
chemical
equilibrium?
(A)
The
reaction
has
stopped.
(B)
The
concentrations
of
products
and
reactants
are
equal.
(C)
The
rates
of
the
forward
and
reverse
reactions
are
zero.
(D)
The
rates
of
the
forward
and reverse
reactions
are the
same.
144
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-
ACS
Equilibrium
¥
Exams
PQ-4.
What
is
the
K.
for
this
reaction?
P,(s)
+
6Cl,(g)
=
4PCl,(g)
@
g
oL
®
g
-
PN
o
g
_[PCL
®
g
-[RJCL]
©CL)
<RI
CeLy
<
e
PQ-5.
What
is
the
K,
for
this
reaction?
280,
(g)
=
250,(g)
+
O,(g)
W
_Rah
®
©
KD=(2PM,)
A,
TR,
o,
(2R,
PQ-6.
What
is
the
equilibrium
expression
for
this
reaction?
2C(s)
+
0,(g)
=
2CO(g)
@
g
=19
®)
g
_1COF
(¢
g
-_[2€0)
®)
g
[Cor
©[Q0,]
©[CT[0,]
©
[20)0,]
©
0]
M
PQ-7.
The
value
of an
equilibrium
constant
can
be
used
to
predict
each
of
these
except
the
(A)
extent
of
a
reaction.
(B)
direction
of
a
reaction.
(C)
time
required
to
reach
equilibrium.
(D)
quantity
of
reactant(s)
remaining
at
equilibrium.
PQ-8.
Chemical
equilibrium
is
the
result
of
(A)
astoppage
of
further
reaction.
(B)
the
unavailability
of
one
of the
reactants.
(C)
opposing
reactions
attaining equal
speeds.
(D)
formation
of
products
equal
in
mass
to
the
mass
of
the
reactants.
PQ-9.
The
photosynthetic
conversion
of
CO>
and
O,
can
be
represented
by
the
reaction
shown.
What
is
the
equilibrium
expression
for
this
reaction?
6CO,
(g)
+
6H,0(1)
=
C,H,,0,(s)
+
60,
()
@)
g
-[CO:TICH.O0]
®
g
_[CO.1
©
[0,T[H,0F
oy
©
g
.
L0:JH.0F
)
g
10T
©
[CO,JICH,,04)
©co,r”
PQ-10.
Which
statement
best
describes
general
equilibrium?
(A)
Equilibrium
is
reached
when
the
reaction
stops.
(B)
There
is
only one
set
of
equilibrium
concentrations
that
equals
the
K.
value.
(C)
At
equilibrium,
the
rate
of the
forward
reaction
is
the
same
as
the
rate
of
the
reverse
reaction.
(D)
At
equilibrium,
the
total
concentration
of
products
equals
the
total
concentration
of
reactants.
145
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Equilibrium
2
ACS
?
Exams
PQ-11.
For
this
chemical
reaction
at
25
°C,
the
concentration
of the
product
is
the
concentration
of
f
reactants
at
equilibrium.
C,H,(g)
+H,(&)
=
C,H(2)
Kc=13x10"
(A)
less
than
(B)
equal
to
(C)
greater
than
(D)
is
not
able
to
be
compared
to
PQ-12.
In
which
case
does
K.
=
K,?
(A)
2A(g)
+
B(s)=2C(s)
+
D(g)
(B)
2A(g)
+
B(s)
=
C(s)
+2D(g)
(C)
3A(g)
+B(s)
=22C(s)
+2D(g)
(D)
K.
and
K,
are
equivalent
in
more
than one
of
the
above
PQ-13.
Given
the
equilibrium
constants
for
these
reactions:
a,#“d
4Cu(s)
+
0,(8)
=
2Cu,06)
K,
2Cu0(s)
=
Cu,0(5)
+
30,(8)
K,
What
is
the
value
of
K.
for
this
reaction?
2Cu(s)
+
O,
(g)
=
2CuO(s)
K.
K,
@
K
®
©
&
<.,
®
g
ok,
K,
VK
PQ-14.
Given
the
equilibrium
constants
for
these
reactions:
2CH,
(2)
=
C,H,(g)
+
H,(2)
CH,(g)
+
H,0(g)
=
CH,;OH(g)
+
H,(g)
K.
What
is
the
value
of
K.
for
this
reaction?
2CH,OH(g)
+
H,
(g)
=
C,H,
()
+
2H,0(g)
(A)
9.5x10"7
(B)
2.9x107°
(C)
3.4x10°
(D)
1.2x10*
9.5x107"3
2.8x102
PQ-15.
Carbon
monoxide
gas
reacts
with
hydrogen
gas
at
elevated
temperatures
to
form
methanol.
CO(g)
+
2H,
(g)
==
CH,OH(g)
‘When
0.40
mol
of
CO
and
0.30
mol
of H;
are
allowed
to
reach
equilibrium
in
a
1.0 L
container,
0.060
mol
of
CH;OH
are
formed.
What
is
the
value
of
K.?
(A)
0.50
(B)
098
©
17
D)
54
PQ-16.
At
equilibrium
in
a
10
L
vessel,
there
are
7.60x10%
moles
of
SO,
8.60x10~
moles
of
O,
and
8.20x107
moles
of
SO3.
What
is
the
equilibrium
constant
K.
under
these
conditions?
250,
(g)
+
O,(2)
=
280,(2)
(A)
125
®)
135
(©
125
)
135
146
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>
ACS
Equilibrium
Exams
PQ-17.
A
mixture
of 2.0
mol
of
CO(g)
and
2.0
mol
of
HO(g)
was allowed
to
come
to
equilibrium
in
a
10.0-L.
flask
at
a
high
temperature.
If
K.
=
4.0,
what
is
the
molar
concentration
of
Hx(g)
in
the
equilibrium
mixture?
CO(g)
+
H,0(g)
=
CO,(g)
+
H,(g)
(A)
0.67M
(B)
0.40M
(©)
020M
(D)
0.13M
PQ-18.
1fa
1.0
L
flask
is
filled
with
0.22
mol
of
N; and
0.22
mol
of
O,
at
2000°C,
what
is
[NO]
after
the
reaction
establishes
equilibrium?
(K.
=
0.10
at
2000°C)
N,
()
+
0,()
=
2NO(g)
(A)
0.034M
(B)
0.060
M
(©)
0.079M
(D)
0.12M
PQ-19.
For
the
reaction,
A(g)
=B(g)
+
C(g),
5
moles
of
A
are
allowed
to
come
to
equilibrium
in
a
closed
rigid
container.
At
equilibrium,
how
much
of
A
and
B are
present
if
2
moles
of
C
are
formed?
(A)
0
moles
of
A
and
3
moles
of
B
(B)
1
mole
of
A
and
2
moles
of
B
(C)
2
moles
of
A
and
2
moles
of
B
(D)
3
moles
of
A
and
2
moles
of B
M
PQ-20.
The
equilibrium
constant,
K.
=
3.8x10°*
for
the
reaction,
L(g)=2(g)
What
is
the state
of
the
system
if
[[»(g)]
=
1.0
M
and
[I(g)}
=
1.0x10~*
M?
(A)
The
system
is
at
equilibrium.
(B)
The
system
is
shifting
towards
products.
(C)
The
system
is
shifting
towards
reactants.
(D)
There
is
insufficient
information
to
describe
the
status
of
the
system.
M
PQ-21.
For
the
reaction:
LF,(g)
=
F(g),
a
reaction
mixture
initially
contains
equal
amounts
of
Fx(g)
and
F(g)
in
their
standard
states.
If
K,
=
7.55x1072
at
this
temperature,
which
statement
is
true?
A)
Q
<K,
and
the
reaction
proceeds
towards
the
reactants.
®)
2<K,
and
the
reaction
proceeds
towards
the
products.
(©)
Q
=K,
and
the
reaction
is
at
equilibrium.
D)
Q>
K,
and
the
reaction
proceeds
towards
the
reactants.
PQ-22.
Consider
this
reaction
at
equilibrium.
w“""
250,(g)
+
0,(2)
=
250,(2)
AH=-198KJ
‘Which
change
would
cause
an
increase
in
the
SO;
/
SO
mole
ratio?
(A)
adding
a
catalyst
(B)
removing
Ox(g)
(C)
decreasing
the
temperature
(D)
decreasing
the
pressure
147
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Equilibrium
ACS
Exams
PQ-23.
At298
K,
the
equilibrium
constant
for
this
reaction
AGe
(kJ-mol™")
H,(2)
+
$0,(g)
=
H,0())
H;0(l)
-237
H:0(g)
-229
(A)
has
a
value
of
1.0
at
equilibrium.
(B)
is
larger
than
the
Kq
for
H,(g)
+
+0,(g)
=
H,0(2)
.
(C)
cannot
be
computed
since
data
on
O,
and
H:
are
not
provided.
(D)
will
have
the
same
value
as
the Keq
for
H,(g)
+
+0,(g)
=
H,0(g)
M“@
PQ-24.
Which
factors
will
affect
both
the
position
of
equilibrium
and
the
value
of
the
equilibrium
constant
for
this
reaction?
N,(g)
+3H,(g)
=
2NH,(g)
AH=-92kJ
(A)
increasing
the
volume
of the
container
(B)
adding
more
nitrogen
gas
(C)
removed
ammonia
gas
(D)
lowering
the
temperature
PQ-25.
Consider
this
reaction
carried
out
at
constant
volume.
™
250,(8)
+
O.(e)
=
250,(5)
AH=-198KJ
The
concentration
of
Ox(g)
at
equilibrium
increases
if
(A)
SO
is
added
to
the
system.
(B)
SO
is
added
to
the
system.
(C)
the
temperature
of
the
system
is
lowered.
(D)
aninert
gas
is
added
to
the
system.
M@
PQ-26.
Oxygen
and
ozone
are
allowed
to
come
to
equilibrium
in
the
exozhermic
reaction
20,(2)
=30,(g)
‘Which
change
will
increase
the
numerical
value
of
the
equilibrium
constant,
K?
(A)
adding
more
Ox(g)
(B)
adding
a
suitable catalyst
(C)
decreasing
the
temperature
(D}
increasing
the
volume
of
the
container
A
PQ-27.
This
reaction
reaches
equilibrium
in
a
closed
container.
What
happens
if
the
volume
of
the
container
vfl"’»’
is
decreased
at
constant
temperature?
CH,(g)
=
H,0(g)
=
COg)
+
2H.(g)
(A)
No change
occurs.
(B)
The
equilibrium
constant
increases.
(C)
The
reaction
shifts
towards
products.
(D)
The
reaction
shifts
towards
reactants.
M
PQ-28.
For
this
equilibrium,
what
changes
will
favor
formation
of
products?
Ny(@)+3H,(9)
=
2NH,(g)
AH=-92k)
(A)
decrease
temperature,
increase
pressure
(B)
decrease
temperature,
decrease
pressure
(C)
increase
temperature,
increase
pressure
(D)
increase
temperature,
decrease
pressure
A
PQ-29.
The
table
to
the
right
gives
the
relative
concentration
of
reactants
X
+
7
¥
Xz
0.#”"
and
products
at
equilibrium
for
the
generic
reactionat
two
.
(300K
07
03
08
temperatures.
The
equilibrium
constant
is
500K
1.0
0.6
0.5
(A)
negative
at
500
K.
(B)
larger
at
500
K
than
at
300
K.
(C)
smaller
at
500
K
than
at
300
K.
(D)
the
same
at
500
K
and
at
300
K.
148
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:ACS
Equilibrium
%
Exams
A
PQ-30.
A
catalyst
is
added
to
a
system
at
equilibrium.
Which
statement
is
TRUE?
M“
(A)
The
temperature
will
decrease.
(B)
The
equilibrium
constant
will
increase.
(C)
The
concentration
of
products
will
decrease.
(D)
If
the
system
is
disturbed,
it
will
return
to
equilibrium
faster.
149
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Equilibrium
cACS
Exams
Answers
to
Study
Questions
1.
A
5.
A
2.C
6.
C
3.
A
7.
B
4.
B
Answers
to
Practice
Questions
1.
D
11.C
2.C
12.B
3.
D
13.A
4.C
14.D
5.
A
15.D
6.
D
16.D
7.C
17.D
8.
C
18.B
9.D
19.D
0.
C
20.B
150
21.D
22.C
23.B
24.D
25.B
26.C
27.D
28.A
29.C
30.D
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Chapter
12
—
Acids
and
Bases
Chapter
Summary:
This
chapter
will
focus
on
acids
and
bases
including
acid-base
models,
determining
an
acid
dissociation
constant,
and
calculating
pH
and
pOH
for
a
given
acid
dissociation.
Specific
topics
covered
in
this
chapter
are:
e
Acid-base
models
(Arrhenius,
Bronsted-Lowry,
Lewis)
*
pHand
pOH
«
Conjugate
acid-base
pairs
*
Titrations
and
neutralizations
e
Buffers
Previous
material
that
is
relevant
to
your
understanding
of
questions
in
this
chapter
include:
*
Solutions
and
Aqueous
Reactions,
Part
1
(Chapter
5)
*
Solutions
and
Aqueous
Reactions,
Part
2
(Chapter
9)
*
Equilibrium
(Chapter
11)
Common
representations
used
in
questions
related
to
this
material:
Name
Example
Used
in
questions
related
to
Titration
curves
N
titrations
and
buffers
n
10
z
8
s
4
2
o
3
20
a0
6
volume
NaOH
added
(mL)
Where
to
find
this
in
your
textbook:
The
material
in
this
chapter
typically
aligns
to
“Acid-Base
Equilibria”
in
your textbook.
The
name
of
your
chapter(s)
may
vary.
Practice
exam:
There
are
practice
exam
questions
aligned
to
the
material
in
this
chapter.
Because
there
are
a
limited
number
of
questions
on
the
practice
exam,
a
review
of
the
breadth
of
the
material
in
this
chapter
is
advised
in
preparation
for
your
exam.
How
this fits
into the
big
picture:
The
material
in
this
chapter
aligns
to
the
Big
Idea
of
Equilibrium
(8)
as
listed
on
page
12
of
this
study
guide.
Study
Questions
(SQ)
SQ-1.
Which
set
represents
a
conjugate
acid/base
pair?
(A)
HCIO/HCIO,
(B)
H:PO,/HPO
(C)
H;O"/OH-
(D)
NCI/NCl,
Knowledge
Required:
(1)
Definition
of
conjugate
acid/base
pairs
from
the
Bronsted-Lowry
acid-base
model.
Thinking
it
Through:
You
are
asked
in
the
question
to
evaluate
each
answer
option
to
determine which
set
is
a
conjugate
acid/base
pair.
Recall
from
the
Bronsted-Lowry
acid-base
model
that
conjugate
acid/base
pairs
differ
in
151
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Acids
and
Bases
.
ACS
W
Exams
one
H".
For
choice
(A),
the
two
species
differ
by
an
oxygen
atom;
these
are
a
set
of
analogous
acids.
For
choice
(B),
the
two
species
differ
by
an
H™;
therefore,
choice
(B)
is
correct.
For
choice
(C),
the
two
species
differ
by
two
H™
ions:
while
these
species
are
related
by
H™,
the
difference
in
H
is
one
too
many.
For
Choice
(D),
the
two
species
differ
by
a
chlorine
atom.
You
can
also
conceptualize
this
question
using
the
reaction
of
each
species
in
water
to
find
the
conjugate
base (and
this
is
good
practice):
Choice
(A):
HCIO(aq)+H,0(1)
=
CIO™
(aq)+H,0"
(aq)
acid
conjugate
base
(not
HCIO:
as
given
in
the
problem)
Choice
(B):
H.PO,
(aq)
+H,0(1)
=
HPO}"
(aq)
+
H,0"
(aq)
acid
conjugate
base
Choice
(C):
H,0"
(aq)+
H,0(1)
=
H,0(1)+
H,0"
(aq)
acid
conjugate
base
(nor
OH-
as
given
in
the
problem)
Choice
(D):
NCI;
would
not
behave
as
a
Bronsted-Lowry
acid
Practice
Questions
Related
to
This:
PQ-1,
PQ-2,
PQ-3,
PQ-4,
PQ-5, PQ-6, PQ-7,
and
PQ-8
5Q-2.
What
is
the
pH
of
a
0.820
M
aqueous
NH;
solution?
Ky,
(NHs)
=
1.8x10
(A)
242
(B)
925
(©)
1158
®)
1391
Knowledge
Required:
(1)
How
to
write
a
balanced
equation
for
an
aqueous
acid-base
reaction.
(2)
How
to
write
the
base
dissociation
constant
expression
(K4)
for
a
balanced chemical
equation.
(3)
How
to
use
the
K,
expression
to
determine
equilibrium
concentrations.
(4)
How
to
convert
between
pH
and
pOH.
Thinking
it
Through:
You
are
asked
to
determine
the
pH
of
a
given
aqueous
solution
for
a
known
base
and
a
known
K.
This
problem
has
several
steps:
First,
you
write
a
balanced
equation
for
the
acid-base
reaction;
recall
that
you
are
told
that
the
reaction
is
in
water,
and
thus
water
will
act
as
an
acid
with
the
given
base:
NH,
(aq)
+
H,0(1)
=
NH;
(aq)
+
OH'(aq)
Next,
you
use
the
balanced
acid-base
equation
to
determine
the
Ky,
expression:
concentrations
of
products
raised
to
power
of their
stoichiometric
coefTicients
divided
by
the
concentrations
of
reactants
raised
to
power
of
their
stoichiometric
coefficients.
Note:
pure
liquids
and
solids
are
not
included
in
the
K,
expression
(omitting
water).
NH]J[OH"
PR
[NH,]
Next,
you
construct
an
initial-change-cquilibrium
(ICE)
table.
Initial
concentrations
are
provided
in
the
question.
Stoichiometric
cocfficients
from
the
balanced
chemical
equation
are
used
to
determine
the
changes
in
concentrations.
Equilibrium
values
are
determined
by
the
addition/subtraction
of the
initial
and
change
values.
NHjaq)
=
NHi(ag)
+
OH-(aq)
Initial
0.820
0
0
Change
—x
+x
+x
Equilibrium
0.820
-
x
x
x
Becausc
acid
and
base
ionization
constants
are
only
about
95%
accurate,
you
can
compare
the
ratio
of
the
equilibrium
constant
to
the
initial
concentration
of
the
acid
or
base.
If
the ratio
is
less
than
2.5%x1073,
you
can
.
K.
3
te
or:
-
2.5x10
approximate
or:
facid],
<2.5x%
oase],
K,(NH;)
1.8x10"
[NH,],
0820M
Therefore,
you
can
assume
that
x
will
be
sufficiently
small
compared
to
the
initial
concentration
of
NHj;
therefore,
the
change
to
[NH;]y
is
small
and
the
approximation
is
[NH3Jo
=
[NH:]equitibrivm
01
0.820—x
~0.820
.
For
this
question:
=2.2x10""
which
is
less
than
2.5x10*
(so
you
can
approximate)
You now
substitute
the
equilibrium
concentrations
into
the
K,
expression:
152
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o
ACS
Acids
and
Bases
18x10°
=%
0.820
The
expression
is
then
further
simplified
and solved:
15x107
=x*
x=%384x10"
Because
x
is
the
final
concentration
of
each
of the
products
and
because
a
concentration
cannot
be
negative,
then
x
is
equal
to
+3.84x10*
M
which
is
[NH:"Lequioriun
and
[OH
Jequiitriom.
Next,
you
determine
the
pOH
from
the
concentration
of
[OH]:
POH
=
—log([OH"])
=
~log(3.84x10
")
=
2.42
Finally,
you convert
from
pOH
to
pH
knowing
that the
sum
of
pH
and
pOH
is
equal
to
14
for
an
aqueous
solution.
pOH
+pH
=
14
pH
=
14—
pOH
pH=14-242=11.58
Which
is
choice
(C)
Choice
(A)
is
not
correct
because
it
is
the
pOH
of
the
solution.
Choice
(B)
is
not
correct
because
it
was
determined
using
the
K,
value
as
the
concentration
of
OH™
when
determining
pOH.
Choice
(D)
is
not
correct
because
it
was
determined
using
the
initial
concentration
of
NH;
as
the
concentration
of
OH~
when
determining
pOH.
Practice
Questions
Related
to
This:
PQ-9,
PQ-10,
PQ-11, PQ-12,
and
PQ-13
SQ-3.
‘Which
acid
is
the
weakest
in
aqueous
solution?
(A)
acetic
acid
(K,
=
1.8x10°%)
(B)
formic
acid
(K,
=
1.8x10%)
(C)
hydrocyanic
acid
(K;
=
6.2x10
')
(D)
nitrous
acid
(K,
=
4.5x107)
Knowledge
Required:
(1)
How
to
write
an
acid
dissociation
equilibrium
expression.
(2)
Relationship
between
acid
dissociation
constants
(K)
values
and
acid
strength.
Thinking
it
Through:
You
arc
asked
to
evaluate
the
given
acids
and
determine
which
acid
is
the
weakest.
You
recall
that
K
is
expressed
as
the
concentration
of
product(s)
raised
to
the
power
of
their
stoichiometric
coefficients
(from
the
balanced
chemical
equation)
divided
by
the
concentration
of
reactant(s)
raised
to
power
of
their
stoichiometric
coefficients.
Thus,
larger
values
of
K,
are
associated
with
a
larger
product
to
reactant
ratio.
Conversely,
smaller
values
of K,
are
associated
with
a
smaller
product
to
reactant
ratio.
For
all
of
the
choices,
the
values
of
K,
are
less
than
one
(so for
all
of
these,
starting
with only
reactants,
at
equilibrium,
[reactants]
>
[products]).
However,
you
know
you
can
still
determine
relative
acid
strength
when
all
K,
values
are
less
than
one
by
determining which
value
is
the
largest
and
how
that
corresponds
to
acid
strength.
Generically:
A”
||
H,0"
HA (aq)+H,0(1)
=
A™(aq)
+H,0"
(aq)
K,
:%
The
larger
the
value
of K,
,
the
highér
[HSO‘],
the
stronger
the
acid
Thus,
small
values
of
K,
are
associated
with
weak
acids
with
the
smallest
K,
value
associated
with
the
weakest
acid.
You
consider
the
answer
options
and
note
that
hydrocyanic
acid
has
the
smallest
K,
(6.2x10
%)
and
conclude
that
it
is
the
weakest
of
all
the
acids
given
as
answer
options;
therefore,
choice
(C)
is
correct.
Practice
Questions
Related
to
This:
PQ-14
and
PQ-15
SQ-4.
Which
anion
is
the
most
basic?
(A)
Clo0”
(B)
CIO;
(€)
Cios
(D)
ClOs
Knowledge
Required:
(1)
How
to
determine
acidity
and
basicity
from
an
analogous
series
of
oxoacids.
Thinking
it
Through:
You
are
asked
in
the
question
to
evaluate
a
series
of
bases
to
determine
which
is
the
most
basic.
The
bases
are
analogous,
considering
the
answer
options
from
(A)
to
(D),
each
has
one
more
oxygen
than
the
previous;
thus,
this
a
series
of
oxoacids.
The
general
trend
for
a
series
of
oxoacids
is
that
as
the
number
of
153
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Acids
and
Bases
="
ACS
@
Exams
oxygens
increases
(or
more
oxygen
atoms
in
the
acid)
the
acid
gets
stronger.
This
is
due
to
the
increase
in
electron
withdrawing
power
of
the
added
oxygens,
which
causes
the
O-H
bond
to
weaken
and
the
acid
to
get
stronger.
Increasing
number
of
oxygen
atoms
weakest
Increasing
electron
withdrawing
————>
strongest
acid
:
G—g—H
o=G—0—H
Weakening
of
O-H
bond
Increasing
acid
strength
You
can
use
the
relative
strength
of the
acid
to
determine
the
relative
strengths
of the
conjugate
bases
as
you
know
the
value
of
K,
for
the
conjugate
base
is
related
to
the
K,
value
of
the
acid
through:
Kix
Ky
=K.
Because
K.,
is
a
constant,
at
a
given
temperature,
the
larger
the
K,
value
(stronger
the
acid)
the
smaller
the
K
value
(weaker
the
conjugate
base).
Therefore,
the
conjugate
base
of
the
weakest
acid
(fewest
oxygen
atoms)
will
be the
strongest
base
—
choice
(A).
The
next
strongest
base
would
be
the
conjugate
base
of
the
next
weakest
acid
—
choice
(B).
The
pattern
would
continue.
the
ranking
of
the
base strength
is:
strongest
base
weakest
base
CIO~>
CIO;
>
CIO;~
>
ClO4~
Although
not
part
of
this
question,
when
considering
the
structure
of
acids
and
relative
strength,
you
can
also
be
provided
with
the
same
number
of
oxygen
atoms,
but
differing
halogens,
for
example
comparing
HC1O4
and
HBrO.
Here,
the
difference
in
acid
strength
is
due
to
the
higher
electronegativity
of
chlorine
compared
to
bromine
—
with
HCIOj
as
the
stronger
acid.
Practice
Questions
Related
to
This:
PQ-16
and
PQ-17
SQ-5.
What
is
Ky
of
F?
(K.
of HF
is
6.8x107%)
(A)
6.8x10"°
(B)
1.5x10°
(C)
6.8x10%
D)
1.5%x10°"
Knowledge
Required:
(1)
Relationship
between
K,
and
K
for
a
conjugate
acid-base
pair.
Thinking
it
Through:
You
are
asked
in
the
question
to
determine
the
Ky,
for
a
species
given
a
known
K,
for
the
conjugate
acid
of the
species.
To
solve
this
problem,
it
is
necessary
to
recall
the
relationship
between
K.
and
K}
for
an
acid
and
its
conjugate
base
dissociated
in
water:
HF(aq)
<=
H'
(aq)
+
F
(aq)
X,
F"(aq)+H,0(1)
==
HF(aq)
+OH
(aq)
K,
H,0(1)
=
H*
(aq)+OH"
(aq)
K,
=K,
xK,
"
Therefore:
K,
=
;-
-
E%i_?o‘*
~1.5x107"
which
is
choice
(D).
Choice
(A)
is
not
correct
because
it
is
determined
by
dividing
K,
by
K.,
which
is
the
inverse
of
K.
Choice
(B)
is
not
correct
because
it
is
determined
by
taking
the
inverse
of
K.
Choice
(C)
is
not
correct
because
it
is
the
value
of
K,.
Practice
Questions
Related
to
This:
PQ-18
SQ-6.
‘Which
substance
will
dissolve
in
water
to
produce
an
acidic
solution?
‘p‘af"
(A)
FeCly
(B)
Na;0
(C)
NaC:Hs0,
(D)
NH;
Knowledge
Required:
(1)
Dissociation
of
ions
in
aqueous
solution.
(2)
Relative
acidity/basicity
of
common
species and
ions.
154
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=
ACS
Acids
and Bases
@
Exams
Thinking
it
Through:
You
are
asked
in
the
question
to
evaluate
a
series
of
compounds
to
determine
which
will
produce
an
acidic
solution
upon
dissolution
in
water.
The
method
you
will
use
to
do
this
is
(in
turn
for
each
compound):
1.
Consider
the
ions
formed
when
the
compound
dissolves
in
water
2.
Consider
the
reaction
of
each
ion
with
water
(hydrolysis)
3.
Consider
the
acidic/basic
ions
Choice
(A)
-
FeCl3
Step
1
FeCly(s)
>
Fe¥*(aq)
+
3Cl(aq)
Step2
Fe¥(aq)
+
6H,0(1)
=
Fe(H:0)s(aq)
=
Fe(H:0):0H
(aq)
+
H'(aq)
Cl(aq)
+
H2O(1)
=
no
reaction
(HCl
is
a
strong
acid)
Step
3
Fe'"
is
an
acidic
cation
and
CI"
is
a
neutral
anion
FeCl;
produces
an
acidic
solution
Choice
(B) —
Na,O
Step
1
Na;O(s)
>
2Na*(aq)
+
O*(aq)
Step2
Na*(aq)
+
HyO()
=
no
reaction
(NaOH
is
a
strong base)
0%(aq)
+
H0(l)
=
20H~
Step
3
Na”
is
neutral
cation
and
O
is
a
basic
anion
Na;O
produces
a
basic
solution
Choice
(C)
—
NaCH:0»
Step
1
NaC;H:0x(s)
=
Na*(aq)
+
C;H305
(aq)
Step2
Na(aq)
+
H;O(l)
=
no
reaction
(NaOH
is
a
strong
base)
C2H;0:7(aq)
+
HO(I)
=
HC;H;0x(aq)
+
OH(aq)
Step
3
Na”
is
neutral
cation
and
C,H3O;"
is
a
basic
anion
NaC>H;0;
produces
a
basic
solution
Choice
(D)
—
NH;
is
a
molecular
compound
and
not
a
salt.
Instead
for
this
compound,
you
will
consider
the
reaction
of
NHjs
is
water
directly
(omitting
the
dissociation
into
ions):
NHs(aq)
+
H20(aq)
=
NHs'(aq)
+
OH(aq)
which produces
a
basic
solution.
Choices
(B),
(C),
and
(D)
all
lead
to
the
formation
of
hydroxide
anions
and
basic
solutions.
Therefore,
Choice
(A)
is
correct
because
it
leads
to
the
formation
of
hydronium
ion
(H;0").
Practice
Questions
Related
to
This:
PQ-19,
and
PQ-20
SQ-7.
An
acetate
buffer
contains
equal
volumes
of
0.35
M
HC;H;0;
(pKa
=
4.74) and
0.55
M
NaC,H:0,.
What
is
the
pH
of the
buffer?
(A)
454
(B)
4.74
(©)
494
()
7.00
Knowledge
Required:
(1)
How
to
use
the
Henderson-Hasselbalch
equation.
(2)
How
a
buffer
works.
Thinking
it
Through:
You
are
asked
in
the
question
to
determine
the
pH
of
a
buffer
solution
given
the
pK,
of
the
acid,
concentrations
of
the
acid
and
its
conjugate
base.
The
Henderson-Hasselbalch
equation
is
used
to
relate
pH,
pK:,
and
concentrations
of
conjugate
acid-base
pairs.
A7)
H=pK,
+
I
pH=pK,
08[[“
A
You
are
given
the
concentration
of
the
acid
(HC:H;O3,
0.35
M)
and
the
conjugate
base
(NaC2H:O,,
0.55
M)
in
the
question.
In
addition,
you
are
given
the
pK,
of
the
acid
(HC;H30,,
4.74).
These
values
are
entered
into the
Henderson-Hasselbalch
equation
and
pH
is
determined:
PH
=474+
k,g(%]
pH=474+
log(1.57)
155
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Acids
and
Bases
ACS
V
Exams
pH=4.74+020=494
which
is
choice
(C)
Finally,
you
can
check
this
answer
to
make
sure
it
is
reasonable.
Because
you
have
a
higher
concentration
of
conjugate
base
compared
to
the
acid
([NaC;H;0:]
>
[HC:H:0,)),
you
would
predict
the
pH
would
be
higher
than
the
pK;
(which
it
is).
Choice
(A)
is
not
correct
because
it
is
determined
using
the
inverse
of
the
conjugate
base
to
acid
molar
ratio
in
the
Henderson-Hasselbalch.
Choice
(B)
is
not
correct
because
its
value
is
the
value
of
the
pK,
for
the
given
acid.
Choice
(D)
is
not
correct
because
its
value
is
the
value
of
the
pH
for
a
neutral
aqueous
solution.
Practice
Questions
Related
to
This:
PQ-21,
PQ-22,
PQ-23,
PQ-24,
and
PQ-25
SQ-8.
What
volume
(in
mL)
of
0.150
M
NaOH(aq)
is
required
to
neutralize
25.0
mL
of
0.100
M
HS0x(aq)?
(A)
16.7mL
(B)
333mL
(C)
66.7
mL
(D)
75.0mL
Knowledge
Required:
(1)
How
to
balance
the
chemical
equation
for
a
Brensted-Lowry
acid-base
reaction.
(2)
Solution
Stoichiometry.
Thinking
it
Through:
You
are
asked
in
the
question
to
determine
what
volume
of
a
given
solution
with
a
known
molar
concentration
is
needed
to
neutralize
a
given
volume
of
a
solution
with
a
known
molar
concentration.
To
begin,
it
is
necessary
to
write
a
balanced equation
for
the
acid-base
reaction:
2NaOH(aq)
+
H.S0,
(ag)
—2H.0(1)
+Na.SO,
(aq)
You
note
that
from
the
stoichiometric
coefficients
in
the
balanced
equation,
two
moles
of
NaOH(aq)
are
necessary
to
neutralize
one
mole
of
H,SO..
Next,
you
recall
the
relationship
between
concentration
(M),
moles
(niue).
and
volume
of
solution
(L):
mole(n,,,,.
)
=
molar
concentration
x
volume
Therefore,
based
on
the
information
in
the
question,
you
determine
the
number
of
moles
of
H2SOs
present.
25.0
mL
W‘.}(L):zso“o-‘
mol
H,S0,
L
1000
mL
h
To
determine
the
volume
of
NaOH
needed
you
usc
the
stoichiometry
from
the
balanced
chemical
equation:
N
2
(
250x10°
mol
H,30,
|
2
mNeOH
(
L
]
1000
"'L)=33.3
mL
Choice
(B)
2%\
Vol
H,50,
|
0.150
mol
NaOH
|
1L
Choice
(A)
is
not
correct
becausc
it
was
determined
using
the
inverse
mole
ratio
{1:2)
instead
of
(2:1).
Choice
(C)
is
not
correct
because
it
was
determined
using
a
1:1
mole
ratio
or
omitted
the
mole
ratio.
Choice
(D)
is
not
correct
because
it
was
determined
using
the
concentration
of
NaOH
with
the
volume
of
H2S0;
and
the
concentration
of
H2SO4
when
calculating
the
volume
of
NaOH.
Practice
Questions
Related
to
This:
PQ-26, PQ-27,
PQ-28,
PQ-29,
and
PQ-30
Practice
Questions
(PQ)
M
PQ-1.
Which
substance
is
acting
as
a
Lewis
acid?
BFs(g)
+
F(g)=
BF:i(g)
(A)
BF:
(B)
F-
(C)
BF:
(D)
This
is
not
a
Lewis
acid-base
reaction.
156
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2
ACS
Acids
and
Bases
@
Exams
PQ-2.
What
is
a
true
statement
about
Lewis
acids
and
bases?
(A)
A
Lewis
acid
must have
a
nonbonding
(lone)
pair of
electrons.
(B)
A
Lewis
base
must have
a
nonbonding
(lone)
pair
of
electrons.
(C)
In
a
Lewis
acid-base
reaction,
a
pair
of
electrons
is
donated
from
the
acid
to
the
base.
(D)
In
a
Lewis
acid-base
reaction,
one
species
always
goes
from
having
a
charge
to
being
electronically
neutral.
PQ-3.
What
are
the
Brensted-Lowry
bases
in
this
reaction?
NH,
(aq)
+
H,0(aq)
=
NH;
(aq)
+
OH
(aq)
(A)
NH;and
OH~
(B)
H:0
and
NH4~
(C)
NH;and
H;O
(D)
NH;"
and
OH~
PQ-4.
What
is
the
hydroxide
ion
concentration
in
an
aqueous
solution
with
a
pH
of
7.0
at
25
°C?
(A)
0.0M
(B)
1.0x107*M
(C)
1.0x107"M
D)
7.0M
PQ-5.
What
is
the
pH
of
a
0.0050
M
solution
of
Ba(OH)(aq)
at
25
°C?
(A)
2.00
(B)
230
©)
11.70
(D)
12.00
PQ-6.
The
formation
of
a
complex
ion
such
as
Cu(NHs)s*"(aq)
can
best
be
categorized
as
a(n)
___
reaction.
Cu**(aq)
+
4NHs(aq)
—
Cu(NH:)s*(aq)
(A)
Arrhenius
acid-base
(B)
Bronsted-Lowry
acid-base
(C)
Lewis
acid-base
(D)
oxidation-reduction
PQ-7.
A
10.0
mL
portion
of
0.010
M
HCl
is
added
to
100.0
mL
of
water.
What
is
the
pH
of
the
resulting
solution?
(A)
between
3.02
and
3.10
(B)
between
2.90
and
3.01
(C)
between
2.02
and
2.10
(D)
between
1.90
and
2.01
PQ-8.
The
pOH
of
pure
water
at
40
°C
is
6.8.
What
is
the
hydronium
concentration,
[H;0"],
in
pure
water
at
this
temperature?
(A)
1.0x10*M
(B)
63x10°M
(۩)
1.0x10"M
D)
1.6x107M
PQ-9.
What
is
the
pH
of
a
0.053
M
solution
of
KOH
at
25
°C?
(A)
0.89
(B)
128
©)
1272
(D)
13.11
PQ-10.
What
is
the
equilibrium
concentration
of
nitrous
acid,
HNO;
(K,
=
4.5%10),
in
a
solution
that
has
a
PH
of
1.657
(A)
0.0032M
(B)
0.022M
(€)
049M
@)
1.1M
PQ-11.
The
pain
reliever
codeine
is
a
weak
base
with
a
K,
equal
to
1.6%10°%.
What
is
the
pH
of
a
0.05
M
aqueous
codeine
solution?
(A)
7.1
(B)
105
©)
i
)
127
157
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Acids
and
Bases
"
ACS
Exams
A
PQ-12.
Besides
water
molecules,
what
species
is/are
present
at
the
greatest
concentration
when
NHs(g)
is
a,.d"
bubbled
into
water?
(K
for
NHa(aq)
is
1.8x10°9)
(A)
NHs(aq)
(B)
NH:OH(aq)
(C)
NH.'(aq)
and
OH-(aq)
(D)
NH;(aq)
and
H;O"(aq)
PQ-13.
Which
0.10
M
solution
will
have
the
largest
concentration
of
hydroxide
ion?
(A)
NH;
(Ky
of
NH;
=
1.8x107)
(B)
NaCN
(K,
of
HCN
=4.9x107'%)
(©)
NaClO:
(Ka
of
HCIO;
=
1.7%10°%)
(D)
NaHCO:;
(Kb
of
HCO5
=
2.3x10°%)
Mv‘,\
PQ-14.
Which
is
the
strongest
acid
in
aqueous
solution?
(A)
acetic
acid
(K,
=
1.8x107)
(B)
benzoic
acid
(K,
=
6.3x10°%)
(C)
formic
acid
(K,
=
1.7x107%)
(D)
hydrofluoric
acid
(K,
=
7.1x107%)
M,&
PQ-15.
Which
aqueous
acid
has
the
largest
K,
value?
(A)
HBrO
(B)
HBrO,
(€)
HCIO
(D)
HCIO;
s
PQ-16.
In
non-aqueous
solution,
the
trend
in
acid
strength
is
observed
to
be:
K
(HI)
>
K,
(HBr)
>
K,
(HCl)
M
‘Which
periodic
trend
best
explains
this
observed
pattern?
(A)
Atomic
radius:
as
the
bond
between
H and
the
halogen
becomes
shorter,
acid
strength
decreases
(B)
Electron
affinity:
as
the
halogen
becomes
more
attracted
to
electrons,
acid
strength
increases
(C)
Electronegativity:
as
the
bond
between
the
two
atoms
becomes
less
polar,
acid
strength
decreases
(D)
lonization
energy:
as
it
becomes
harder
to
remove
an
electron
from
the
halogen,
acid
strength
increases
s
PQ-17.
The
reaction
is
observed
to
have
a
Keg
>
1.
Il
HSO,
(ag)
+
HPO,"
(ag)
=
SO,
(ag)
~
H,PO,
(aq)
What
is
the
strongest
acid
present
in
this
equilibrium?
(A)
HzPO:
(B)
HPOZ
(C)
HSOy
(D)
SO
PQ-18.
What
is
the
pH
of
a
0.400
M
sodium
formate
(NaCHO;)
solution?
K,
(HCHO,)
=
1.8x107*
(A)
2.07
(B)
533
©)
867
D)
11.93
w0
PQ-19.
Which
salt
will
form
a
basic
aqueous
solution?
o
(A)
NaF
(B)
KBr
(©
Licl
(D)
NH.NO3
PQ-20.
The
pH
of
a
0.050
M
aqueous
solution
of
ammonium
chloride
(NH4Cl)
falls
within
what
range?
(A)
0to2
(B)
2to7
(C)
Tto
12
D)
12to0
14
PQ-21.
What
is
the
pH
of
a
buffer
solution
containing
equal
volumes
of
0.11
M
NaCH;COO
and
0.090
M.
CH;COOH?
K,
(CH;COOH)
=
1.8x107°
(A)
242
(B)
483
(©)
11.58
(D)
1391
158
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s
A
CS
Acids
and
Bases
Exams
PQ-22.
Which
pair
of
compounds
would
be the
best
choice
to
make
a
buffer solution with
a
pH
around
5?
(A)
HF
and
NaF
K.
=6.9%10*
(B)
C;HsCO;H
and
C;HsCO:Na
Ky
=1.3x107*
(C)
HCIO
and
NaClO
K,=238x10%
(D)
NHClandNH;
K,
=5.6x107"°
PQ-23.
Which
pair
could
be
used
to
make
a
buffer
solution?
(A)
acetic
acid
and
sodium
chloride
(B)
acetic
acid
and
sodium
hydroxide
(C)
hydrochloric
acid
and
potassium
chloride
(D)
hydrochloric
acid
and
sodium
hydroxide
PQ-24.
What
is
the
pH
of
a
buffer
solution
made
by
adding
0.010
mole
of solid
NaF
to
50.
mL
of
0.40
M HF?
Assume
no
change
in
volume.
K,
(HF)
=
6.9x107*
(A)
1.6
®)
19
©)
29
(D)
32
PQ-25.
An
aqueous
buffer
solution
contains
only
HCN
(pK,
=
9.31) and
KCN
and
has
a
pH
of 8.50.
What
can
be
concluded
about
the
relative
concentrations
of
HCN
and
KCN
in
the
buffer?
(A)
[HCN]>[KCN]
(B)
[HCN]
<[KCN]
(€)
[HCN]=[KCN]
(D)
nothing
can
be
concluded
about
the
relative
concentrations
PQ-26.
Methyl
orange
is
an
indicator
that
changes
color
from
red
to
yellow-orange
over
the
pH
range
from
2.9
to
4.5.
Methyl
orange
would
be
the
most
appropriate
indicator
for
which
type
of
acid
-base
titration?
(A)
A
weak
acid
titrated
with
a
strong base
(B)
A
weak
base
titrated
with
a
strong
acid
(C)
A
strong
acid
titrated
with
a
strong
base
(D)
Methy!
orange
would
not
be an
appropriate
indicator
for
any
acid-base
titration.
f
PQ-27.
Which
graph
best
represents
the
titration
of
ammonia
with
hydrochloric
acid?
e
h
£
i,
(A)
(B)
o
o
volume
HCI added
‘volume
HC1
added
14|
"
7
27
©)
(D)
N
o
volume
HC1
added
volume
HCI
added
159
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Acids
and
Bases
-
ACS
W
Exams
PQ-28.
In
the
laboratory,
50
mL
of
0.1
M
HCl
is
mixed
with
50
mL
of
0.1
M
NaOH
and
the
solution
is
stirred
gently.
At
equilibrium,
what
ionic
species,
if
any,
will be
present
in
large
amounts
in
the
reaction
mixture?
(A)
Na'and
CI-
only
(B)
H;O'and
OH
only
(C)
H;O0',OH-,Na',
and
CI-
(D)
No
ions
will
be
present
PQ-29.
What
is
the
approximate
pK,
of the
weak
1
acid
in
the
titration
curve?
N
-
z
¢
s
P
2
o
0
2
W0
e
volume
NaOH
added
(mL)
(A)
2.0
(B)
48
(©)
88
®)
11.0
PQ-30.
What
is
the
pH
of
a
solution
that
results
from
mixing
25.0
mL
of
0.200
M
HA
with
12.5
mL
of
0.400
M
NaOH?
(Ka
=
1.0x107)
(A)
294
(B)
4.94
(©)
9.06
(D)
11.06
160
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[k
ol
M
rOoOw
POXIANB
LS
TODPATN>E
>
~"ACS
Acids
and
Bases
Qvamm
Answers
to
Study
Questions
RS
wO»T
Answers
to
Practice
Questions
11.B
21.B
12.A
22.B
13.B
23.B
14.D
24.C
15.D
25.A
16.
A
26.B
17.C
27.A
18.C
28.A
19.A
29.B
20.B
30.C
161
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Acids
and
Bases
162
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Chapter
13
—
Solubility
Equilibria
Chapter
Summary:
This
chapter
will
focus
on
the
solubility
of
solids
in
aqueous
media
and
other
solvents.
Specific
topics
covered
in
this
chapter
are:
*
Molar
solubility
*
Equilibrium
expressions
for
solubility
reactions
*
Solubility
product
constants
(Ks;)
and
related
calculations
e
Common
ion
effect
«
Formation
reactions,
expressions
and
constants
(Kr)
Previous
material
that
is
relevant
to
your
understanding
of
questions
in
this
chapter
includes:
*
Solutions
and
Aqueous
Reactions,
Part
1
(Chapter
5)
*
Solutions
and
Aqueous
Reactions,
Part
2
(Chapter
9)
e
Equilibrium
(Chapter
11)
Common
representations
used
in
questions
related
to
this
material:
Name
Example
Used
in
questions
related
to
Compound
units
mol-L™!
molar
solubility
Macroscopic
diagrams
ccommon
ion
effect
Where
to
find
this
in
your
textbook:
The
material
in
this
chapter
typically
aligns
to
“Aqueous
Equilibria”
in
your textbook.
The
name
of
your
chapter(s)
may
vary.
Practice
exam:
There
are
practice
exam
questions
aligned
to
the
material
in
this
chapter.
Because
there
are
a
limited
number
of
questions
on
the
practice
exam,
a
review
of the
breadth
of the
material
in
this
chapter
is
advised
in
preparation
for
your
exam.
How
this
fits
into the
big
picture:
The
material
in
this
chapter
aligns
to
the
Big
Idea
of
Equilibrium
(8)
as
listed
on
page
12
of
this
study
guide.
Study
Questions
(SQ)
SQ-1.
What
is
the
K,
expression
for
MgCOs
in
water?
_
Mg
J[cor]
x
—Me"Iicor]
@A)
"
[MgCOH,0]
®
0]
©
K,=Mg"][coi]
()
K»=IMgCO,][H,0]
Knowledge
Required:
(1)
How
to
write
a
solubility
equilibrium
expression
for
a
given
species
in
water.
Thinking
it
Through:
Y
ou
are
asked
in
the
question
to
write
the
equilibrium
expression
that
corresponds
to
the
Kep,
solubility
product
constant,
for
a
given
species.
The
K
expression
is
the
concentrations
of
product(s)
raised
163
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Solubility
Equilibria
c:ACS
Exams
to the
power
of
their
respective
stoichiometric
coefficients
from
a
balanced
chemical
equation
divided
by
the
|
concentrations
of
reactant(s)
raised
to
the
power
of their
respective
stoichiometric
coefficients.
Solids
and
liquids
are
not
included
in
the
expression
because
their
concentrations
are
assumed
to
be
constant.
You,
therefore,
begin
by
constructing
the
balanced
equation
for the
dissociation
of
MgCOsx(s)
in
water:
MgCO;
(s)
=
Mg
(aq)
+
CO3"(aq)
From
the
balanced
equation,
you
can
construct
the
K,
expression:
K,
=
[Mg"][CO>"]
Choice
(C)
Choice
(A)
is
not
correct
because
it
includes
the
concentrations
of the
solid
and
the
liquid.
Choice
(B)
is
not
correct
because
it
includes
the
concentration
of the
liquid.
Choice
(D)
is
not
correct
because
it
includes
the
concentrations
of
the
solid
and
the
liquid,
and
does
not
include
the
concentrations
of
the
dissolved
ions.
Practice
Questions
Related
to
This:
PQ-1
and
PQ-2
SQ-2.
Given
the
two
reactions:
Cr(OH),(s)
==
Cr'"
(aq)
+
30H
(aq)
K=
1.6¥10°°
Cr**(aq)
+
4OH™(aq)
=
[Ct(OH),
T
(aq)
K;=8.0x10%
What
is
the
value
of
X
for
the
reaction?
Cr(OH),(s)
+
OH™
(aq)
=
[Cr(OH),
]"
(aq)
(A)
—8x10%
(B)
2.0x10€
(©)
13x10°
D)
5.0%10°
Knowledge
Required:
(1)
How
to
write
an
equilibrium
expression.
(2)
How
to
combine
multiple
chemical
reactions
and
their
respective
equilibrium
constants
to
form
a
new
reaction
and
equilibrium
constant.
Thinking
it
Through:
You
are
asked
to
determine
the
value
of the
equilibrium
constant
for
a
chemical
reaction
using
two
reactions
with
known
constants,
K,
and
K.
You
begin
by
writing
the
two
equations
and
corresponding
equilibrium
constants.
You
observe
that
the
target
equation
is
the
sum
of
the
two
given
equations.
You
also
then
know
that
when
two
equations
are
summed,
the
corresponding
equilibrium
constant
(K)
is
the
product
of the
two
equilibrium
constants
(K,
and
K):
Cr(OH),
(s)
=
"
(a0)
+30H
(aa)
K,
=[Cr"
J[on
|
_[[ertomT
]
[
Jlon
]
[[er(on).T]
Cr'* (aq)
+40H"
(aq)
=
[Cr(OH),
]
(aq)
K;
Cr(OH),
(s)
+OH"
(aq)
=
[
Cr(OH),
|
(aq)
K
[[CI(OH)‘]-]
.
K=[crJ[on
|
xe——
"=k
_xkK,
[
][
]
[Cri‘][OH‘]‘
»
oo
When
reactions
are
summed,
the
new
K
=
K,
x
Kr=
1.3
or
1.3x10°,
Choice
(C).
Choice
(A)
is
not
correct
because
it
is
the
value
obtained
when
K
is
substracted
from
Kip.
Choice
(B)
is
not
correct
because
it
is
the
value
obtained
when
Kyp+
Kr.
Choice
(D)
is
not
correct
because
it
is
the
value
obtained
when
K¢+
K.
Practice
Questions
Related
to
This:
PQ-3
SQ-3.
What
is
the
solubility
product
constant,
Ks,,
of
Mg(OH):
if
its
molar
solubility
in
water
is
1.6x10*
mol-L1?
(A)
2.6x10°%
(B)
1.6x107!
(C)_1.0x1072
(D)
4.1x10°"2
164
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*ACS
Solubility
Equilibria
@'
Exams
Knowledge
Required:
(1)
Definition
of
molar
solubility.
(2)
How
to
write
a
solubility
equilibrium
expression.
(3)
How
to
determine
a
solubility
product
constant
(Kp)
given
molar
solubility.
Thinking
it
Through:
You
are
asked
in
the
question
to
determine
the
solubility
product
constant
()
for
a
given
species
with
a
known
solubility
in
water.
To begin,
you
construct
a
balanced chemical
equation
for the
dissociation
and
a
corresponding
Ky,
expression:
Mg(OH),
(s)
=
Mg"
(aq)
+
20H"(aq)
K,
=
[Mg"][OH
|
You
see
that
you
are
given
the
“molar
solubility”
of
Mg(OH);
in
the
question.
To
understand
what
this
is,
you
proceed
to
set
up
the
initial-change-equilibrium
(ICE)
table
for
this
process
where
you
have
purposely
included
the
states
to
remind
yourself
that
Mg(OH)x(s)
is
not
included
in
the
K
expression:
Mg(OH)y(s)
=
Mg*(aq)
+
20H(aq)
Initial
-
0
0
Change
-
+x
+2x
Equilibrium
-
x
2x
From
here
you
see
that
the
molar
solubility
(the
value
of
“x”
in
the
table)
corresponds
to
the
number
of
moles
solute
in
1.0 L
of
a
saturated
solution
or
[Mg(OH)z].
Note:
You
know
this
is
used
to
standardize
the
comparison
of
solubility
of
various
ionic
compounds
where
the
fon
ratios
are
not
equivalent
(such
that
you
can
compare
[AI(OH)s]
or
the
molar
solubility
of
AI(OH)s
to
[AgOH]
or
the
molar
solubility
of
AgOH
but
cannot
compare
the
respective
concentrations
of
the
hydroxide
ion
to
predict
which
compound
is
more
soluble).
Based
on
the
coefficients
from
the
balanced
chemical
equation,
the
solubility
concentration
has
a
1:1
ratio
with
the
concentration
of
Mg**
and
a
1:2
ratio
with
the
concentration
of
OH".
You
can
thus
enter
these
values
into the
Ks,
expression
and
solve:
K,
=[Mg*
J[OH™
|
=(x)(2x)
=4x
K,
=[16x10*][2x(1.6x10%)]
=1.6x10""
Choice
(A)
is
not
correct
because
the
value
is
obtained
using
a
Ky,
expression
that
does
not
account
for
the
coefficient
of
hydroxide
ion
in
the
balanced chemical
equation.
Choice
(C)
is
not
correct
because
the
value
is
obtained
using
the
inverse
of
the
1:2
mole
ratio
when
calculating
the
concentration
of
hydroxide
ion.
Choice
(D)
is
not
correct
because
the
value
is
obtained
without
using
the
1:2
mole
ratio
when
calculating
the
concentration
of
hydroxide
ion.
Choice
(B)
Practice
Questions
Related
to
This:
PQ-3,
PQ-4,
PQ-5,
PQ-6
and
PQ-7
$Q-4.
_
Which
compound
has
the
highest
molar
solubility?
(A)
Agl,
Ky
=8.52x10"7
(B)
BaCOy,
Kyp=2.58x10°
(C)
Fe(OH);,
Kip=2.79%10
(D)
ZnS,
Ky
=3.00x10°%
Knowledge
Required:
(1)
Definition
of
molar
solubility.
(2)
How
to
write
a
sol\fiaili}y
Eqv
ibrium
ex}:réssion.
(3)
How
to
determine
molar
solubility
given
a
solubility
product
constant
(Kp).
Thinking
it
Through:
You
are
asked
in
the
question
to
evaluate
a
series
of
species
and
their
provided
K,
values
to
determine
which
species
has
the
highest
molar
solubility.
To
make
this
evaluation, you
consider
each
species.
To
begin,
you
write
the
balanced
equation
and
the
corresponding
K.,
expression
for
each.
Choice
(A):
Agl(s)
=
Ag"(aq)+1"(aq)
KXP=[Ag'][I’]
Choice
(B):
BaCO,
(s)
=
Ba®"
(aq)+
CO:"
(aq)
K,=[Ba*][cor]
Choice
(C):
Fe(OH),
(s)
=
Fe"
(aq)+30H™
(aq)
K,,=[Fe*
JJOH'
]’
165
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Solubility
Equilibria
ACS
¥
Exams
Choice
(D):
ZnS(s)
=
Zn*
(aq)
+S*
(aq)
K,,=[zn™
|[s*
]
Next,
you
assume
that
equilibrium
is
reached
and
the
solution
is
saturated.
Additionally,
you
assume
there
is
not
a
starting
concentration
of
any
species.
The
variable
x
will be
used
to
represent
one
mole
equivalent
of the
solid
(for
example,
Agl)
when
expressing
the
concentrations
of
each
species,
so
then
x
represents
the
molar
solubility.
Therefore,
when
x
values
and
the
K,
values
are
substituted
for
each,
you
obtain
these
expressions:
Choice
(A):
8.52x107
=
x-x
Choice
(B):
2.58x107
=
x-x
Choice
(C):
2.79x10™"
=
x-(3x)’
Choice
(D):
3.00x107
=
x-x
For each
answer
option,
you
solve
for
x;
because
x
cannot
be
negative,
only positive
values
of
x
are
retained.
Choice
(A):
9.2x107°
Choice
(B):
x=
5.1x107
Choice
(C):
1.0x10™"°
Choice
(D):
x=
5.5x10™*
Therefore,
the
species with
the
highest
molar
solubility
(the
highest
value
of
x)
is
BaCOs,
Choice
(B).
Practice
Questions
Related
to
This:
PQ-8, PQ-9,
PQ-10, PQ-11, PQ-12, PQ-13,
PQ-14,
PQ-15,
PQ-16,
and
PQ-17
SQ-5.
A
solution
contains 0.002
M
Pb*"
and
0.002
M
Ag”.
What
happens
when
Ksp_
NaCl(s)
is
added
to
bring
the
chloride
ion
concentration
to
0.01
M?
PbCly
1.2x10°°
AgCl_
1.8x10°
(A)
Neither
PbCL
nor
AgCl
will
precipitate.
(B)
Both
PbCl:
and
AgCl
precipitate.
(C)
Only
PbCl;
precipitates.
(D)
Only
AgCl
precipitates.
Knm’vledige
Required:
(1)
How
to
write
a
solubility
equilibfium
expression.
(2)’
How
to
determine
a
reaction
quotient,
Q.
(3)
How
to
predict
precipitation
by
comparing
O
with
K.
Thinking
it
Through:
Y
ou
are
asked
in
the
question
to
determine
what
occurs
when
a
concentration
of
a
given
ion
is
brought
to
a
certain
level.
To
do
this
you
will
calculate
the
value
of
O
for
each
set
of
conditions
and
compare
to
the
value
of
K.
The
criteria
for
the
comparison
are:
If
Qsp
<
Kip
the
reaction
will
proceed
to
form
more
products:
no
precipitate
forms.
Using
AgCl
as
the
example,
the
concentration
of the
ions
is
sufficiently small
that the
solution
is
unsaturated
and
the
reaction
will
shift
to
the
right:
AgCl(s)
=
Ag”
(aq)+Cl
"
(aq)
If
Qsp
>
Kip
the
reaction
will
proceed
to
form
more
reactants:
a
precipitate
forms.
Using
AgCl
as
the
example,
the
concentration
of the
ions
is
sufficiently
large
that the
solution
is
saturated
and
the
reaction
will
shift
to
the
left:
AgCl(s)
<
Ag’
(aq)+Cl
(aq)
1f
Qsp
>
Kip
the
reaction
is
at
equilibrium:
the
solution
is
just
saturated.
Using
AgCl
as
the
example,
the
concentration
of
the
ions
is
at
the
saturation
point
such
that the
solution
is
saturated
and
the
reaction
is
at
equilibrium:
AgCl(s)
=
Ag’
(aq)+Cl
(aq)
Therefore,
for
this
question,
the
relevant
O
expressions
are:
AgCl(s)
=
Ag’
(aq)+Cl
(aq)
Q=
[Ag'
]0
[Cl'
]0
PbCI,
(5)
=
Pb*"
(aq)+
2C1"
(aa)
0
=[Pb*
]
[CI
|}
For
PbCly:
0=(0.002)(0.01)°
=2x10"
For
AgCl:
0=(0.002)(0.01)=2x10"
The
Q
for
PbCl,
is
less
than
Ky,
for
PbCls,
so
PbCl>
will
net
precipitate.
166
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ACS
Solubility
Equilibria
Exams
The
Q
for
AgCl
is
greater
than
K,
for
AgCl,
so
AgCl
will
precipitate.
Choice
(D)
is
the
correct
answer.
Practice
Questions
Related
to
This:
PQ-18
and
PQ-19
SQ-6.
A
sample
of
“hard”
water contains about
2.0x102
mol
of
Ca?*
ions
per
liter.
What
is
the
maximum
concentration
of
fluoride
ion
that
could
be
present
in
hard
water?
Assume
fluoride
is
the
only
anion
present
that will
precipitate
calcium
ion.
(Ksp
of
CaFa(s)
at
25
°C
=
4.0x10°1)
(A)
2.0<10°M
(B)
22x10°M
(©)
44x10°M
(D)
2.0x102M
Knowledge
Required:
(1)
How
to
write
a
solubility
equilibrium
expression.
(2)
How
to
determine
the
concentration
of
one
ion
given
the
solubility
product
constant
(Kp)
and
concentration
of
a
common
ion.
Thinking
it
Through:
You
are
asked
to
determine
the
maximum
amount
of
a
species
that
can
be
dissolved
into
a
solution
that
contains
a
common
ion
with
the
species
being
dissolved.
You
begin
by
constructing
a
balanced
chemical
equation
and
K,
expression
for
the
dissolution:
CaF,(s)
=
Ca™
(aq)
+
2F"(aq)
=
[Ca™J[FJ
Next,
you
consider
what
this
looks
like
if
CaF;
was
first
in
pure
water.
By
slowly
adding
solid
CaF;
to
water,
initially
(when
[Ca®"]
and
[F-]
are
very
small),
the
solution
would
remain
homogeneous.
After
some
point,
the
concentration
of
the
ions
reaches
a
concentration
such
that the
solution
is
saturated
and
no
more
solid
CaF;
will
dissolve, and
a
heterogeneous
solution
has
formed.
You
know
this
is
called
the
“molar
solubility”
and
determine
this
using
the
process
outlined
in
SQ-3. Using
this,
you
find
the
molar
solubility
for
CaF;
in
pure
water
is
equal
to
2.2x10~*
M
or
in
pure water,
[Ca?"]
=
2.2x10*
M
and
[F]
=
4.4x10*
M.
Pure
water
[Ca*]
<22x10%4M
[Ca*]>22x10%
M
[Ca¥]=0M
[F]<44x104M
[F]>44x10*
M
[F1=0M
solution
is
unsaturated
solution
is
saturated
and
is
homogeneous
and
is
heterogeneous
&
>
Add
Can(s)
Contmue
to
slowly
add
CaFy(s)
slowly
Now,
you
consider
what
this
looks
like
for
CaF;
in
tap
water
containing
[Ca’*}
=
2.0x102
M.
By
slowly
adding
solid
CaF
to
tap
water,
initially
(when
[F]
is
still
very
small),
the
solution
would
remain
homogeneous.
After
some
point,
the
concentration
of
the
fluoride
reaches
a
concentration
such
that the
solution
is
saturated
and
no
more
solid
CaF;
will
dissolve, and
a
heterogeneous
solution
has
formed.
However,
this
is
different
because
you
started
with
[Ca>*}o.
So
to
determine
the
concentration
of
the
fluoride
ion,
[F-],
you
will
include
the
initial
concentration
of
the
calcium
ion:
Tap
water
[Ca*]=2.0x102M
[Ca**]=2.0x102M
[Ca*]o=2.0x102M
|
[F]<?
[F]>7M
[F]=0M
solution
is
unsaturated
solution
is
saturated
and
is
homogeneous
and
is
heterogeneous
S
-
Add
CaF
2(s)
Continue
to
slowly
add
CaFa(s)
slowly
167
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Solubility
Equilibria
.
ACS
CaF(s)
=
Ca*(aq)
+
2F
(aq)
Initial
-
2.0%102
0
Change
-
+x
2x
Equilibrium
-
2.0%107
+x
I
Because
the
concentration
of
the
calcium
ion
is
higher,
you
can
assume
the
change
in
the
concentration
of
the
calcium
ion
(from
the
added
CaFs)
will
be
sufficiently
small
(or
assume
2.0x107
+x
=
2.0x10™)
to
be
unchanged
from
the
initial
concentration:
k,=[ca
J[F
T
=[ca"](2x)
=[Ca*
Jaxr®
ar=—2=
[Fl=2x=2(22x10")=44x10"
M
Corresponding
to
choice
(C).
Choice
(A)
is
not
correct
because
it
is
the
value
of
x
if
the
coefficient
of
“2”
for
F~
was
not
considered
in
the
K
expression
and
initial-change-equilibrium
table.
Choice
(B)
is
not
correct
because
it
is
the
value
of
x;
however,
the
final
concentration
of
F~
is
2-x.
Choice
(D)
is
not
correct
because
it
is
the
initial
concentration
of
Ca’* and
the
approximate
equilibrium
concentration
of
Ca®".
Practice
Questions
Related
to
This:
PQ-20
What
will
change
the
value
of
K,
for
silver
azide
(AgN;)?
A3
M
(A)
Adding
water
to
the
solution.
(B)
Adding
silver
ions
to
the
solution.
(C)
Removing
azide
ions
from
the
solution.
(D)
Increasing
the
temperature
of
the
solution.
Knowledge
Requirt;l:
(1)
What
influences
the
value
of
Ksp
Thinking
it
Through:
Y
ou
are
asked
in
the
question
to
evaluate
what
changes
to
a
system
at
equilibrium
would
result
in
a
change
in
the
value
of
Ks,.
Choice
(A)
would
result
in
decreasing
concentrations
and
thus
increases
in
the
amount
of
silver
azide
dissolved.
Choice
(B)
would
result
in
a
common
ion
effect
whereby
the
addition
of
silver
ions
would
lead
to
more
silver
azide
precipitation.
Choice
(C)
would
have
the
opposite
effect
whereby
the
removal
of
azide
ions
would
result
in
more
silver
azide
being
dissolved.
Choices
(A),
(B),
and (C)
impact
Q
for
the
dissolution,
but
do
not
impact
the
value
of
K;.
Choice
(D),
though,
does
have
a
direct
impact
on
the
value
of
K.,
whereby
an
increase
in
temperature
results
in
a
different
K,
value.
Practice
Questions
Related
to
This:
PQ-21,
PQ-22,
PQ-23,
PQ-24,
PQ-25, PQ-26,
PQ-27,
PQ-28
and
PQ-29
8.
SQ
When
comparing
O
and
K5,
of
an
unsaturated
solution,
.
(A)
O<Ky
B)
0=Ky
©)
0>Ks
(D)
0=Ky=0
Knowledge
Required:
(1)
The
definition
of
0.
(2)
How
O and
K.
are
related.
Thinking
it
Through:
Y ou
are
asked
in
the
question
to
determine
the
relationship
between
Q
and
Ky,
for
an
unsaturated
solution.
Recall
that
for
an
unsaturated
solution,
more
starting
material
can
be
dissolved
into
solution.
Therefore,
the
molar
solubility
has
not
been
reached.
The
solubility
product
constant,
Ks,,
is
defined
as
the
product
of the
maximum
concentration
of
dissolved
species,
i.e.
the
molar
solubility.
The
value
of
O
is
defined
as
the
product
of
the
actual
concentrations
of
dissolved
species.
Therefore,
if
O
is
greater
than
K,
there
are
concentrations
of the
dissolved
species
greater
than
the
maximum
allowable
concentrations
and
thus
species
will
precipitate
out
until
the
concentrations
of
dissolved
species
matches
Ks,,
Choice
(C).
If
Q
is
equal
to
K,
there
are
exactly
the
concentration
of
dissolved
species
that
match
Ks;,
Choice
(B).
If
Q
is
less
than
K,
there
are
concentrations
of
the
dissolved
species
that are
less
than
the
maximum
concentrations
(or
molar
solubility
values)
and
thus
more
species
could
be
dissolved
into
solution,
Choice
(A).
Choice
(D)
would
only
be true
if
the
species
is
168
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cACS
Solubility
Equilibria
entirely
insoluble.
An unsaturated
solution
is
a
solution
whereby
more
species
could
be
dissolved;
therefore,
Choice
(A)
where
O
is
less
than
K,
is
correct
for
an
unsaturated
solution.
Practice
Questions
Related
to
This:
PQ-30
Practice
Questions
(PQ)
PQ-1.
What
is
the
K,
expression
for
CoCOs(s)
in
water?
@)
g
-
[ComNCOT]
®)
g
-[CoMICOT]
¥
[CoCO,][H,0]
N
[H.0]
©)
K,
=[Co™][COT]
D)
K,
=[CoCO;]H,0]
PQ-2.
What
is
the
K,
expression
for
Nis(POxs)a(s)
in
water?
A)
K,
=INiT[POIF
®)
K,
=INi"J[POIT
(©
K,
=(3xINi")(2x(POL
)
®
K,
=(3xNi)
(2x(POS
)
PQ-3.
Given
the
two
reactions:
wn"“‘
CuCl(s)
+
CI"
(aq)
=
[CuCl,
]
(aq)
K=52x10"
Cu’(aq)
+2CI"
(aq)
=
[CuCl,
]
(aq)
‘What
is
the
value
of
K,
for
CuCI?
(&)
KxKr
®
£
©
K
o
!
K,
X
KxK,
.0x10°
PQ-4.
A
saturated
solution
of
MgF;
contains 1.6x10~*
mol
of
MgF;
per
liter
at
a
certain
temperature.
What
is
the
K,
of
MgF>
at
this
temperature?
(A)
2.7x10°¢
(B)
1.6x10°¢
(©)
3.1x107°
(D)
62x10°
PQ-5.
A
saturated
solution
of
AI(OH);
has
a
molar
solubility
of
2.9x10
M
at
a
certain
temperature.
What
is
the
solubility
product
constant,
Ky,
of
AI(OH);
at
this
temperature?
(A)
2.1x10%
(B)
1.9x10%
(C)
84x107®
(D)
2.5x107"7
PQ-6.
Silver
chromate
(Ag:CrOs;
Kp
=
1.1%107'2)
has
a
molar
solubility
of
6.5x10"
M.
What
calculation
provides
the
concentration
of
silver
ions
in
a
saturated
solution
of
Ag,CrO4?
(A)
2%
(65%109M
(B)
1%
(65<109M
(C)
[2%(65x10%FM
(D)
{1x(1.1x107)
M
PQ-7.
The
solubility
of
lead(Il)
iodide
in
water
is
0.62
grams
per
1.0
L of
solution.
What
is
the
solubility
product
constant,
Ky,
for
the
dissolution?
Pbl,
(s)
=
Pb*
(aq)
+
2I"(aq)
(A)
2.4x10°
(B)
9.7x107
(©€)
1.8x10¢
(D)
4.0x10°
PQ-8.
For
Ag:S
the
solubility
product
constant,
Ks,
is
6x107*.
What
is
the
molar
solubility
of
Ag.S?
(A)
1x107"°M
(B)
2x10"°M
(©€)
210
M
(D)
2x10°°M
PQ-9.
For
BaF;
Ky,
=
1.0x10°.
What
is
the
molar
solubility
of
BaF:
in
a
solution
containing
0.10
M
NaF?
(A)
63x10°M
(B)
1.0x10*M
(©)
2.5x10°M
(D)
1.0x10°
M
169
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Solubility
Equilibria
s
ACS
Exams
M
PQ-10.
Which
saturated
solution
has
the
highest
concentration
of
F-
ions
at
25
°C?
(A)
BaFy,
Ky,
=
1.8x107
(B)
CaFy,
Kyp=3.5x107"
(€)
MgFy,
Kip=7.4x107"1
(D)
StFa,
Kip=25%10°
PQ-11.
What
is
the
molar
solubility
of
AgsPO4
in
water?
K,
(AgsPOs)
=
1.4x107'6
(A)
L1x10*M
(B)
4.8x10°M
(€)
52x10°M
(D)
6.8x10°M
PQ-12.
Given
Ki;
of
CaF3
is
3.5%107"".
The
molar
solubility
of
CaF
in
0.200
M NaF
is
(A)
2.1x10*
M
(B)
1.8x107°M
(C)
8.8x107"°M
D)
35x10""M
PQ-13.
What
is
the
maximum
mass
of
solid
barium
sulfate
(233
g-mol™')
that
can
be
dissolved
in
1.00
L
of
0.100
M
Na>SOx
solution?
Ky,
(BaSOs)
=
1.5x10°
(A)
1.5%10%¢g
(B)
2.1x10%g
(C)
3.5%x10%¢g
(D)
9.0x10%¢g
PQ-14.
Rank
the
following
species
from most
to
least
soluble
in
water.
I
Ba(NOs),,
Kip
=
4.64x10°
IL
Cd(OH);,
Kip
=
7.20x1075
.
FeCOs,
Kp=3.13x10""
(A)
1>
11>11
®)
1>
m>1n
(©)
nm>m>1
D)
1>1>Mm
PQ-15.
Rank
the
following
species
from
least
to
most
soluble
in
water.
L
PbBr;,
Ky
=6.60x10"
IL
PbCly,
Kyp=
1.70%10°
M.
Pbl,
Ky
=9.80x107
(A)
I<Hm<I
®B)
m<m<i
©
m<I<un
(D)
HI<II<I
PQ-16.
For LiF
K,
=1.84x107.
What
is
the
molar
solubility
of
LiF?
(A)
6.78x10"
M
(B)
429x102M
(©)
8.78x10°M
(D)
3.00x10°M
c#d,d
PQ-17.
Which
saturated
solution
has
the
highest
concentration
of
OH™
ions
at
25
°C?
(A)
AI(OH);,
Ky
=3.0x107
(B)
Ca(OH),
K
=
5.02x10
(C)
Cd(OHY;,
Kyp=7.2x107
(D)
Pr(OH);,
Ky
=3.4x107
PQ-18.
The
solubility
of solid
silver
chromate,
Ag.CrOs,
which
has
a
K;;
equal
to
1
pure
water
9.0x107'2
at
25
°C,
is
determined
in
water
and
in
two
different
aqueous
Il
0.1
MAgNO;
solutions.
Predict
the
relative
solubility
of
silver
chromate
in
the
three
I
0.1
M
Na,CrOq
solutions.
(A)
I=n=11
(B)
I<H<uI
©)
m=m<I
(D)
m<u<t
PQ-19.
What
will
be
the
result
if
100
mL
of
0.06
M
Mg(NO5)
is
added
to
50
mL
of
0.06
M
NaxC,04?
Assume
the
reaction
is
taking
place
at
25
°C.
(K,
of
MgCy04(s)
at
25
°C
=
8.6x10°%)
(A)
No
precipitate
will
form.
(B)
A precipitate
will
form
and
an
excess
of
Mg?*
ions
will
remain
in
solution.
(C)
A precipitate
will
form
and
an
excess
of
C;04>"
ions
will
remain
in
solution.
(D)
A
precipitate
will
form
and
both
Mg?*
and
C,04?"
ions
are
present
in
excess.
170
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§
ACS
Solubility
Equilibria
Exams
PQ-20.
A solution
contains
0.002
M
Be**
and
0.002
M
Ca?'.
What
happens
when
Ky
NaOH(s)
is
added
to
bring
the
hydroxide
ion
concentration
to
0.10
M?
Be(OH):
6.9%10%
Ca(OH):
5.0x10°
(A)
Neither
Be(OH);
nor
Ca(OH),
will
precipitate.
~
(B)
Both
Be(OH),
and
Ca(OH),
precipitate.
(C)
Only
Be(OH),
precipitates.
(D)
Only
Ca(OH);
precipitates.
w‘d’,\
PQ-21.
The
addition
of
solid
Na;SOx
to
an
aqueous
solution
in
equilibrium
with
solid
BaSOx
will
cause
(A)
no
change
in
[Ba®"]
in
solution.
(B)
more
BaSO:
to
dissolve.
(C)
precipitation
of
more
BaSOx.
(D)
an
increase
in
the
K.y
of
BaSOx.
PQ-22.
If
solid
NasPOx
is
added
to
a
saturated
aqueous
solution
of
Cas(POa)z,
what
will
happen
to
the
"afid‘\
concentration
of
Ca**
in
solution?
(A)
It
will
decrease.
(B)
It
will
increase.
(C)
It
will
be
unchanged.
(D)
It
will
become
equal
to
the
PO~
concentration.
A
PQ-23.
AgBr
has
very
low
solubility
in
water.
Silver
ions
form
a
complex
ion
with
thiosulfate.
o
AZ
(1)
25,0}
1)
=
[AR(5.0,),
"
19)
If
sodium
thiosulfate
is
added
to
a
saturated
solution
of
AgBr
in
equilibrium
with
solid
AgBr,
how
will the
concentrations
of
free
Ag'(aq)
and
Br-(aq)
change?
(A)
Ag'(aq)
decreases,
Br(aq)
decreases
(B)
Ag’(aq)
decreases,
Br(aq)
increases
(C)
Ag'(aq)
increases,
Br(aq)
decreases
(D)
Ag'(aq)
increases,
Br(aq)
increases
A
PQ-24.
For
which
slightly
soluble
substance
will the
addition
of
perchloric
acid
to
its
solution
have
no
effect
M
on
its
solubility?
(A)
AgBr(s)
(B)
Cu(OH)(s)
(€)
MgCOs(s)
(D)
PbFay(s)
PQ-25.
The
solubility
of
solid
copper(I)
chloride,
CuCl,
which
has
a
K,
L
pure
water
equal
to
1.72x107
at
25
°C,
is
determined
in
water
and
in
two
1L
0.1
MNaCl
different
aqueous
solutions.
Predict
the
relative
solubility
of
0L
0.1
M
MgCly
copper(I)
chloride
in
the
three
solutions
(A)
I<I<In
(B)
1<mI<n
©)
m<m<I1
D)
m<m<r
A
PQ-26.
The
addition
of
solid
sodium
hydroxide
(NaOH)
to
an
aqueous
solution
in
equilibrium
with
solid
o™
Me(OH):
will
cause
(A)
no
change
in
[OH"]
in
solution.
(B)
more
Mg(OH),
to
dissolve.
(C)
adecrease
in
[Mg*"]
in
solution.
(D)
an
increase
in
the
Ky,
of
Mg(OH),.
A
PQ-27.
Which
will
change
the
value
of
K,
for
mercury(I)
sulfate
(HgSO4)?
M
(A)
adding
Hg"
ions
to
the
solution.
(B)
removing
SO4*
ions
from
the
solution.
(C)
decreasing
the
temperature
of the
solution.
(D)
decreasing
the
pH
of
the
solution.
171
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Solubility
Equilibria
oACS
@
Exams
PQ-28.
For
Agl
K,
=8.5%107
and
for
[Ag(CN):]
K;=
1.0<10%'.
Which
statement
is
true?
M
(A)
Agl
is
completely
insoluble
in
a
solution
that
contains
CN~.
(B)
Agl
s
less
soluble
in
water
than
in
a
solution
that
contains
CN
.
(C)
Agl
is
more
soluble
in
water
than
in
a
solution
that
contains
CN-.
(D)
Aglis
equally
soluble
in
water
and
in
a
solution
that
contains
CN
.
PQ-29.
Ammonia
(NH;)
is
slowly
added
to
a
saturated
solution
of
Ag:CO;
(Kip
=
8.46x107'2).
Which
statement
s
true
if
K;
for
[Ag(NH3)z]"
is
1.6%107
(A)
Ag>CO;
is
completely
miscible
in
a
solution
that
contains
NHa.
(B)
AgCOs
is
more
soluble
in
water
than
in
a
solution
that
contains
NH;.
(C)
Ag;COs
is
less
soluble
in
water
than
in
a
solutjon
that
contains
NHs.
(D)
Ag:CO;s
is
equally
soluble
in
water
and
in
a
solution
that
contains
NHs.
M‘d
PQ-30.
When
comparing
O
and
K,
of
a
saturated
solution,
.
(A)
0<Kqp
(B)
O=Ks
©)
Q=K
D)
9=K3=0
172
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El
ol
wEAN
Sexnpmsprn
>EFREPTIE
SO
>
ACS
Solubility
Equilibria
@
Exams
Answers
to
Study
Questions
i
»oN0D
Answers
to
Practice
Questions
11.B
21.C
12.C
22.A
13.C
23.B
14.A
24.A
15.C
25.D
16.B
26.C
17.B
27.C
18.D
28.B
19.D
29.C
20.B
30.B
173
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Solubility
Equilibria
174
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Chapter
14
—
Thermodynamics
Chapter
Summary:
This
chapter
will
focus
on
entropy
and
free
energy,
energy,
and
enthalpy.
The
transfer
of
energy
during
chemical
and
physical
changes
will be
calculated.
The
usefulness
of
entropy
and
free
energy
being
state
functions
will be
used
to
calculate
the
values
of
entropy
and
free
energy
changes
using
Hess’s
law.
Specific
topics
covered
in
this
chapter
are:
Entropy
and
trends
in
molar
entropy
values
Predicting
the
sign
of the
entropy
change
for
a
process
Calculating
the
value
of
AS
for
a
chemical
reaction
Definition
of
a
spontaneous
process
Calculating
AG
for
a
reaction
using
Hess’s
law
Predicting
the
temperature
dependence
of
spontaneity
Relating
values
of
AG®
and
equilibrium
coefficients
Previous
material
that
is
relevant
to
your
understanding
of
questions
in
this
chapter
include:
Significant
figures
(Toolbox)
Scientific
notation
(Toolbox)
Balancing
chemical
reactions
(Chapter
3)
Heat
and
Enthalpy
(Chapter
6)
Equilibrium
(Chapter
11)
Common
representations
used
in
questions
related
to
this
material:
Name
Example
Used
in
questions
related
to
compound
units
kJ-mol™'
and
J-mol
K™
density,
enthalpy
changes
for
reactions
(AH),
Gibbs
free
energy
changes
for
reactions
(AG),
entropy
changes
for
reactions (AS)
Particulate
diagrams
ol
.
°
©
g
entropy
changes
©
ol
®
o
«©
O
o
Pred
Where
to
find
this
in
your
textbook:
The
material
in
this
chapter
typically
aligns
to
“Thermodynamics”.
The
name
of
your
chapter
may
vary.
Practice
exam:
There
are
practice
exam
questions
aligned
to
the
material
in
this
chapter.
Because
there
are
a
limited
number
of
questions
on
the
practice
exam,
a
review
of the
breadth
of
the
material
in
this
chapter
is
advised
in
preparation
for
your
exam.
How
this
fits
into the
big
picture:
The
material
in
this
chapter
aligns
to
the
Big
Ideas
of
Energy
and
Thermodynamics
(6)
and
Equilibrium
(8)
as
listed
on
page
12
of
this
study
guide.
175
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Thermodynamics
o
ACS
¥
Exams
Study
Questions
(SQ)
SQ-1.
Of
the
four
listed
compounds,
three
have
correct
values
for
molar
entropy,
S°,
associated
with
them.
M
For
which
species
is
the
entropy
value
incorrect
at
298
K?
Substance
§°(J-mol-K™')
Ay
Cas)
202
(B)
Br(l)
152
©
COAe)
214
(D)
AlICly(s)
109
Knowledge
Required:
(1)
Trends
in
molar
entropy
values.
Thinking
it
Through:
You
need
to
identify
the
incorrect pairing
of
species
and
molar
entropy
values.
You
remember
the
general
trends
for
molar
entropy
values:
(1)
The
more
restricted
the
motion
of
the
physical
state,
the
lower
the
entropy
value,
$°%(s)
<
S%(1)
<
S%(g)
(2)
The
more
complex
the
species,
more
atoms/ions
present,
the
larger
the
value
of
S°.
(3)
As
temperature
increases,
the
value
of
§°
increases.
You
would
expect
the
values
of
the
solids
to
be
less
than
the
values
of
the
liquids
and
these
would
be
less
than
the
values
of
the
gases.
Using
this
you
predict
that
the
S°
values
for
Ca(s) and
AIClx(s)
should
be
smaller
than
the
value
of
Bra(1)
and
the
S°
value
for
CO:
should
be
the
largest.
You
also
know
that the
value
of
$°
for
Ca(s)
should
be
smaller
than
the
value
for
S°
for
AICI(s).
You
predict
the
values
should
go
in
the
order
S%(Ca(s))
<
SYAICI(s))
<
S%Bra(1))
<
S(COx(g))
‘When
you
substitute
the
given
values
into
the
prediction
you
get:
202<109<152<214
It
is
clear
that the
S°
value
for
Ca(s)
is
the
incorrect
one.
Choice
(A)
is
the
incorrect
pair,
and
therefore
the
correct
answer.
The
other
species
and
molar
entropy
values
are
correctly paired
up.
Practice
Questions
Related
to
This:
PQ-1
and
PQ-2
SQ-2.
Which
change
is
likely
to
be
accompanied
by
an
increase
in
entropy?
(A)
Na(g)
+3Ha(g)
—
2NHy(g)
(B)
Ag'(aq)
+Cl(aq)
—
AgCI(s)
©)
COx(s)
—
COx(g)
(D)
H:0(g)
—
H:0(1)
Knowledge
Requifed:
(li
Definition
of
entropy.
(2)
Trends
in
molar
entropy
values.
Thinking
it
Through:
You
recall that the
entropy
change
for
a
process
is
equal
to:
AS
=
§°(products)
—
S°(reactants)
You
also
know
that
for
an
increase
in
entropy
you
are
looking
for
a
process
with
a
positive
value
of
AS
(AS
>
0),
this
occurs
when
the
entropy
of
the
products
is
greater
than
the
entropy
of
the
reactants:
AS>0
§°
(products)
-
S°
(reactants
)
>
0
§°
(products)
>
$°
(reactants)
You
consider
each
choice
and
predict
the
sign
of
AS
for
each.
In
choice
(A)
you
have:
Reactants
(4
moles
gas)
—
Products
(2
moles
gas)
You
predict
the
sign
of
AS
for
this
process
to
be
(-)
because
the
Angas
=~2
or
a
decrease
in
entropy.
176
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.
ACS
Thermodynamics
Exams
I*n"ch;ic?(fi)
i'ou
have:
Reactants
(2
moles
éolufion)»
Products
(l-mole
solid)
You
predict
the
sign
of
AS
for
this
process
to
be
()
because
the
transition
from
solution
to
solid
results
in
a
decrease
in
entropy.
In
choice
(C)
you
have:
Reactants
(1
mole
solid)
-
Products
(I
mole
gas)
You
predict
the
sign
of AS
for
this
process
to
be
(+)
because
Angs
=
+1
or
an
increase
in
entropy.
In
choice
(D)
you
have:
~
Reactants
(1
mole
gas)
—
Products
(1
mole
liquid)
You
predict
the
sign
of
AS
for
this
process
to
be
()
because
Az
=—1
or
a
decrease
in
entropy.
The
correct
choice
is
(C).
Practice
Questions
Related
to
This:
PQ-3,
PQ-4,
P-5,
PQ-6,
PQ-7,
and
PQ-8
$Q-3.
For
this
reaction
at
25
°C,
AH®
=
~1854
kJ'mol™'
and
AS®=-236
J-mol™""K™'
.
CH;COCHj(g)
+
402(2)
—
3COu(g)
+
3H:0()
What
is
the
value
of
AG®
for
this
reaction?
(A)
-1784kJ'mol"
(B)
-1848kI'mol!
(C)
-1924kJ'mol'
(D)
68500
kJ-mol"
Knowledge
Required:
(1)
The
calculation
of free
energy
changes
from
enthalpy
and
entropy
data.
(2)
The
correct
use
of
units.
Thinking
it
Through:
You
recognize
that
you
are
given
a
temperature
and
values
for
AH°
and
AS®
and
are
being
asked
to
calculate
AG®.
You
know
to
use
the
relationship:
AG®
=
AH®—TAS®.
You
also
recall
that the
units
of
AH®
are
in
k]
and
the
units
of
AS®
are
in
J,
so
you
must
convert.
The
temperature
you
are
given
is
in
°C and
you
must
use
the
value
expressed
in
kelvin.
You
begin
by
doing
the
necessary
conversions:
T=25+273=298K
AS°=-236
J'mol
"K' =-0.236
kJ'mol""K"'
Substituting
these
values
into the
expression
for
AG®:
AG®
=—1854
kJ-mol
'
(298
K)(~0.236
kJ-mol
'
-K')=—1784kJ
-
mol”
which
is
choice
(A).
Choice
(B)
is
incorrect
because
the
calculation
was
done
without
changing
the
temperature
from
°C
to
K.
Choice
(C)
is
incorrect
because
the
calculation
was
done
making
a
sign
error
in
the
second
part
of the
expression.
Choice
(D)
is
incorrect
because
the
calculation
was
done
without
converting
the
value
of
the
AS®
to
kJ.
Practice
Questions
Related
to
This:
PQ-9
and
PQ-10
SQ-4.
‘When
NH4NO;
dissolves
in
water,
the
temperature
of
the
solution
decreases.
What
describes
the
w"p‘
enthalpy
and
entropy
changes
of
the
system
and
which change
drives
the
process?
(A)
AH
= (~)
and
AS
=
()
and
the
process
is
driven
by
the
enthalpy
change.
(B)
AH=(-)and
AS=(+)
and
the
process
is
driven
by
the
enthalpy
change.
(C)
AH=(+)and
AS
=
()
and
the
process
is
driven
by
the
entropy
change.
(D)
AH
=
(+)
and
AS
=
(+)
and
the
process
is
driven
by
the
entropy change.
Knowledge
Required:
(1)
Definition
of
endothermic
and
exothermic
processes.
(2)
Prediction
of
entropy
change
for
a
process.
(3)
The
sign
of
AG
for
a
spontaneous
process.
177
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Thermodynamics
Thinking
it
Through:
You
are
asked
to
use
an
observation
to
determine
the
signs
of
the
enthalpy
and
entropy
change
for the
dissolving
of
NH:NO;
in
water.
The
first
thing
you
notice
is
the
salt
dissolves
so
the
process
is
spontaneous.
The
sign
of
AG
for
a
spontaneous
process
is
negative.
You
recall
that
endothermic
processes
take
energy
in
from
the
surroundings,
this
results
in
a
(+)
sign
for
AH.
Exothermic
processes
release
energy
into the
surroundings
and
the
sign
of
AH
would
be
negative
(~).
The
question
tells
you
that the
solution
gets
cold
when
the
NH4NOs
dissolves
in
water.
You
recognize
that
this
means
the
process
is
endothermic,
AH
> 0
because
the
dissolution
process
is
absorbing
energy
from
the
solution
(part of
the
surroundings).
You
also
know
that
when
the
solid
NHsNO;
dissolves
in
water
the
ions
become
hydrated:
NH:NOs(s)
=
NHy'(aq)
+
NOs(aq)
This
results
in
an
increase
of
entropy,
AS
>0
You
recall that
for
a
spontaneous
process,
the
value
of
AG
<
0
and
the
relationship
for
AG
is:
AG
=AH-TAS.
Putting
these
all
together:
AG
<
0,
so
AH
—
TAS
<
0
and
therefore,
AH
<
TAS.
Recall
that,
AH
>
0
and
AS>
0
and
the
process
is
driven
by
the
entropy change.
The
correct
answer
is
choice
(D).
Choice
(A)
is
incorrect
because
the
signs
of
AH
and
AS
are
incorrect and
the
process
is
being
driven
by
the
entropy change.
Choice
(B)
is
incorrect
because
the
sign
of
AH
is
incorrect
and
the
process
is
being
driven
by
the
entropy
change.
Choice
(C)
is
incorrect
because
the
sign
of
AS
is
incorrect.
Practice
Questions
Related
to
This:
PQ-
11,
PQ-12,
and
PQ-13
SQ-5.
What
is
the
free
energy
change
for
the
formation
of
one
mole
of
ammonia
from
its
elements
under
standard
conditions?
2NHs(g)
—
Na(g)
+
3Ha(g)
AG®=33.0
kJ'mol”!
(A)
~66.0
kJ-mol™!
(B)
~33.0
kJ-mol”!
(€)
-16.5kJ-mol”!
(D)
66.0
kJ-mol!
Knowledge
Required:
(1)
How
to
manipulate
thermochemical
equations.
Thinking
it
Through:
You
are
being
asked
to
find
the
free
energy
of
formation
for
a
substance
from
a
given
reaction.
You
recall
that the
definition
of
a
formation
reaction:
A reaction
that
produces
one
mole
of the
substance
from
its
elements
in
their
standard
states.
You
write
the
formation
reaction
for
ammonia:
%2
Na(g)
+*2Ha(g)
—
NHi(g)
You now
go
back
to
the
equation
you
were
given
in
the
question:
2NHs(g)
—
Na(g)
+
3Ha(g)
AG®=33.0
kJ-mol”!
You
reverse
the
reaction,
remembering
to
change
the
sign
of
AG®:
Na(g)
+
3H(g)
—
2NHs(g)
AG®=-33.0
kJ-mol”!
You
recognize
that
this
reaction
produces
ammonia
from
its
elements,
but
it
produces
2
moles
of
ammonia.
To
get
the
formation
reaction,
you
need
to
multiply
the
reaction
by
%:
V2
Na(g)
+
*2Ha(g)
—
NHi(g)
AG®
=
¥(-33.0
kJ-mol™')
=
~16.5
kJ-mol
!
The
correct
choice
is
(C).
Choice
(A)
is
incorrect
because
to
get
this
answer
the
reaction
was
multiplied
by
2
instead
of by
%2.
Choice
(B)
is
incorrect
because
to
get
this
answer
the
reaction
has not
been
multiplied
by
¥
.
Choice
(D)
is
incorrect
because
to
get
this
answer
the
sign
on
AG
was
not
inverted
and
was
multiplied
by
2
instead
of
V2.
Practice
Questions
Related
to
This:
PQ-14
and
PQ-15
178
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-
ACS
Thermodynamics
Exams
$Q-6.
Calculate
the
standard
Gibbs
free
energy
change
at
298
K
for
the
AG°1/
kJ-mol™!
reaction:
CO(g)
~137.2
Fe304(s)
+
4CO(g)
=
3Fe(s)
+
4COx(g)
COx(g)
-394.4
Fe;04(s)
-1014
A)
—757
kJ-mol™*
(B)
~14.8
kJ-mol™!
©
14.8
ki-mol™
(D)
757
kJ-mol!
Knowledge
Required:
(1)
Hess’s
law.
(2)
Rules
for
manipulating
free
energy
values.
Thinking
it
Through:
You
recall
that
Hess’s
law
allows
you
to
calculate
the
AG
of
a
reaction using
tabulated
free
energies
of
formation
values.
This
works
because
we
can
imagine
any
reaction
as
consisting
of
first
converting
the
reactants
back
into their
elements
and
then
forming
the
products
from
these
elements.
This
is,
in
effect
reversing
the
formation
reactions
of
the
reactants
and
then
using
the
formation
reactions
of the
products.
You
recall
that
this
process
is
represented
in
the
relationship:
AG’,
=
3"
nAG!
(products)
—
)
mAG?
(reactants)
You
recall
that the
symbol
3.
means
to
sum
and
7
and
m
are the
stoichiometric
coefficients
of
the
reactants
and
products.
You
also
notice
that
there
is
no
standard
free
energy
of
formation
for
Fe(s)
given,
this
is
because
you
know
the
standard
free
energy
of
formation
of
an
element
in
its
standard
state
is
zero.
The
value
of
AG°x,
can
be
calculated
as:
G,
=[4AG}(CO.
(@)
+
3AG;
(Fe(s))
|-[4AG;
(CO(@))
+
AG;
(Fe,0,(s))
|
.
~394
kJ
0kJ
—137.2
kJ
—~1014kJ
AG’,
=|
4
mol
CO,
|
—————
|+3
mol
Fe
~|
4
mol
CO|
———
|
+1
mol
Fe,0,
|
———
mol
CO,
mol
Fe
mol
CO
.
mol
Fe
O,
AG;,
=-148k)-mol”
Choice
(B)
is
the
correct
choice.
Choice
(A)
is
incorrect
because
it
did
not
use
the
coefficients
in
the
balanced equation
and used reactants
minus
products.
Choice
(C)
is
incorrect
because
it
calculated
the
answer
using
reactants
minus
products.
Choice
(D)
is
incorrect
because
it
did
not
use
the
coefficients
in
the
balanced
equation.
Practice
Questions
Related
to
This:
PQ-16, PQ-17, PQ-18,
and
PQ-19
SQ-7.
"An
equilibrium
mixture
of
Lx(g),
Cla(g),
and
ICI(g)
at
298
K
has
partial
pressures
of
0.0100
atm,
0.0100
atm,
and
0.0900
atm,
respectively.
What
is
AG°
at
298
K
for the
reaction:
L(g)
+
Ch(g)
=
2ICl(g)
(A)
-169kJ'mol!
(B)
-10.9
kJmol!
©
109
Kl-mol!
D)
169
ki
mol”
Knowledge
Required:
(1)
Relationship
between equilibrium
and
free
energy.
(2)
How
to
write
an
equilibrium
expression
for
a
reaction.
(3)
How
to
calculate
the
value
of
K.
Thinking
it
Through:
You
are
given
the
equilibrium
pressures
of
a
system
at
equilibrium
and
need
to
calculate
the
value
of
AG®.
You
recall
the
relationship
between
free
energy
and
equilibrium,
AG
=
AG®
+
RTInQ
where
0
is
the
reaction
quotient.
At
equilibrium
the
value
of
AG
is
zero
and
0
=
K
The
relationship
becomes:
AG®
=-RTInK
You
know
you
can
calculate
the
value
of
K
from
the
given
information
using
the
equilibrium
expression:
179
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Thermodynamics
*ACS
()
+
Clg)
=2iClg)
K=—ho)
(0007,
(P
)(E)
(0.0100)0.0100)
AG®
=-RTInK
=—~(8.314
J-mol™'-K~")(298K)
In(81)
=—10851
J-mol"!
Converting
to
kJ
this
value
becomes
~10.9
kJ-mol™'. This
is
choice
(B).
You
also
know
you
can
check
your
answer
to
see
if
it
is
reasonable
because
you
have
more
products
than
reactants,
so
you
predict
a
larger
value
of
K
(K>
1)
and
this
would
then
be
considered
a
spontaneous
process,
so
you
would
expect
a
negative
free
energy
(AG
<0)
as
you
calculated.
Choice
(A)
is
not
correct
because
the
value
of
K
was
calculated
without
the
exponent
for
ICI.
Choice
(C)
is
not
correct
because
the
negative
sign
was
dropped
and
the
value
of
K
was
calculated
without
the
exponent
for
ICI.
Choice
(D)
is
not
correct
because
the
negative
sign
was
dropped.
Practice
Questions
Related
to
This:
PQ-20,
PQ-21,
PQ-22,
PQ-23,
and
PQ-24
SQ-8.
‘Which
statement
is
correct
for
the
reaction
below
under
standard
conditions?
2Clx(g)
+
2NO(g)
—
Na(g)
+
2CLO(g)
AH®
=-22.0
kJ-mol™
(A)
The
reaction
is
spontaneous
at
all
temperatures.
(B)
The
reaction
is
spontaneous
only
at
low
temperatures.
(©)
The
reaction
is
spontaneous
only
at
high
temperatures.
(D)
The
reaction
is
not
spontaneous
at
any
temperature.
Knowledge
Required:
(1)
Relationship
between
AG, AH,
AS,
and
7.
(2)
Definition
of
spontaneous.
Thinking
it
Through:
You
are
being
asked
to
predict
the
spontaneity
of
a
chemical
reaction.
You
notice
that
you
are
only
given
a
value
of
AH°
and
not
for
AS®.
You
also
remember
that
a
reaction
will
be
spontaneous
under
standard
conditions
when
AG®
is
negative.
You
also
recall
the
relationship
for
AG°, AH®,
AS°, and
T.
AG®=AH®
-
TAS°.
The
sign
of
AG°
depends
on
the
signs
and
magnitudes
of
AH°
and
AS®
and
the
value
of
the
temperature.
These
relationships
are
summarized
below:
AH°
|
AS°
|
-TAS°
|
AG°
|
Result
-
+
-
-
Spontaneous
at
all
temperatures
+
-
+
+
Nonspontaneous
at
all
temperatures
-~
-
+
+or
—_|
Spontaneous
at
low
temperature/nonspontaneous
at
high
temperature
+
+
-
+or—
|
Spontaneous
at
high
temperature/nonspontaneous
at
low
temperature
You
know
the
sign
of
AH®
is
negative.
You
need
to
determine
the
sign
of
AS®.
Recalling
what
you
know
about
molar
entropy
values,
you
notice
that
the
reaction
produces
3
moles
of
gas
from
4
moles
of
gas.
You
predict
that
the
reaction
will
have
a
negative
AS°.
The
table
tells
you
that
this
type
of
reaction
will be
spontaneous
at
low
temperature.
This
is
choice
(B)
and
the
other
choices
are
incorrect.
Practice
Questions
Related
to
This:
PQ-25, PQ-26,
and
PQ-27
SQ-9.
For
a
reaction
where
K
>
1
at
all
temperatures,
which
L
AH?
is
negative
statement(s)
must
be
true?
1.
AS®
is
positive
L.
AG©
is
positive
(A)
Onlyl
(B)
Onlylll
(C)
Bothlandll
(D)
Bothland
Il
180
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.>ACS
Thermodynamics
&
Exams
f{fiowledge
Requiréd:
(1)A}i'elafil;i'm;s'};ip
between
'AGM,AAH.
AS
and
T.
(Z)Vkeylalionship
between
AG
and
K.
Thinking
it
Through:
You
are
told
that the
reaction
has
a
value
of
K
greater
than
1
at
all
temperatures.
The
relationship
between
AG®and
K
is:
AG"=-RTlmK
or
K=¢
When
AG?
is
a
negative
number
the
value
of
K
is
always
greater
than
1,
and
when
AG®
is
positive,
the
value
K
is
less
than
one.
Because
the
value
of
K
>
1
at
all
temperatures,
the
value
of
AG®
must
be
negative.
This
can
happen
at
all
temperatures
when
AH®
is
negative
AND
AS?
is
positive.
See
the
table
in
SQ-8
for
ali
the
possible
combinations.
The
correct
choice
is
(C).
Choice
(A)
is
incorrect
because
a
negative
AH®
is
not
sufficient
to
make
a
reaction
spontaneous
at
all
temperatures.
Choice
(B)
and
(D)
are
not
correct
because
if
AG®
were
positive.
K
would
be
less
than
one
at
all
temperatures.
Practice
Questions
Related
to
This:
PQ-28, PQ-29,
and
PQ-30
Practice
Questions
(PQ)
M,\a‘
PQ-1.
Which
gas has
the
largest
molar
entropy
at
298
K
and
1
atm?
(A)
Ar
(B)
CsHs
(©)
CO,
(D)
HCI
A
PQ-2.
Which
best
explains
why
the
entropy
of
a
sample
of
water
increases
as
the
temperature
changes
from
M
25°C1075
°C?
(A)
There
is
more
hydrogen
bonding
at
the
higher
temperature.
(B)
More
possible
energy
states
are
available
at
the
higher
temperature.
(C)
There
are
fewer
possible
ways
to
distribute
energy
at
the
higher
temperature.
(D)
More
water
molecules
can
be
broken
into
hydrogen
and
oxygen
at
the
higher
temperature.
M
PQ-3.
In
which
reaction
will
the
entropy
of
the
system
increase?
(A)
Na(g)
+3H:(g)
—
2NHs(g)
(B)
2Mg(s)
=
Oa(g)
—
2MgO(s)
(C)
CaCOs(s)
—
CaO(s)
~
COx(g)
(D)
H0(1)
+
COx(g)
—
H2COx(aq)
M\
PQ-4.
Which
process
is
associated
with
a
negative
entropy
change
(AS®w,
<
0)?
(A)
NaCl(s)
—
Na*(aq)
+
Ci(aq)
(B)
Ni(g)~
3Hx(2)
—
2NHs(g)
(©)
2BrCl(l)
—
Brx(D)
+
Cla(g)
(D)
H20(1)
—
H:20(g)
o
PQ-5.
For
which
of
these
processes
is
the
value
of
AS
expected
to
be
negative?
vfl"‘
I
Sugar
is
dissolved
in
water.
1L
Steam
is
condensed.
1L
CaCOs
is
decomposed
in
to
CaO
and
CO,.
(A)
Tonly
(B)
¥and
Ik only
(C)
Monly
(D)
1land
Nl
only
s
PQ-6.
For
which
process
is
the
entropy
change
per
mole
the
largest
at
constant
temperature?
(&)
H0()
—
Ha0(g)
(B}
Hx0(s)
—
H:0(g)
(C)
H:0(s)
—
H20()
(D)
H0(l)
—
H20(s)
181
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Thermodynamics
."ACS
Mp\
PQ-7.
In
which
process
is
entropy
decreased?
(A)
dissolving
sugar
in
water
(B)
expanding
a
gas
(C)
evaporating
a
liquid
(D)
freezing
water
PQ-8.
Which
figure
represents
a
process
with
a
positive
entropy
change?
o
©
°
e
o
o
05
©
©
9
—
o
—»
@w|
o
©
®|®
o
©
To
o
L
L]
e
&
&
¢
®o
g
©
?&%’
-
&cc’;og
®)
‘56’\
—>
?d%
PQ-9.
For
the
reaction:
2PCIs(l)
+
Ox(g)
—
20PCls(g)
AH®=-508
kJ'mol~'
and
AS®=—178
J-mol"K~"
at
298
K.
What
is
AG®
for
this
reaction?
(A)
-561
kJ-mol’
(B)
—504
kJ'mol'’
(©)
455
kJ'mol”!
(D)
52,500
kJ-mol™!
PQ-10.
What
is
AG°na
for
2CO(g)
+
Ox(g)
—
2COs(g)
if
AH
%
=
—566
kI
mol™!
and
AS°n,
=—173
J-mol-K-!
at298
K.
(A)
319K
mol
!
(B)
-514kJ-mol’’
(©)
739
kl'mol!
(D)
5099
kImol”!
PQ-11.
What
are
the
signs
of
the
entropy
change
and
enthalpy
change
for
the
combustion
reaction
of
magnesium?
2Mg(s)
+
Oz(g)
—
2MgO(s)
(A)
AH>0,A5>0
(B)
AH>0,AS<0
(C)
AH<0,AS5>0
D)
AH<0,AS<0
w‘f\
PQ-12.
Which
of
the
relationships
are
true
about water
boiling
in
a
container
that
is
open
to
the
atmosphere?
(A)
AH>0,AS>0
(B)
AH>0,AS<0
(C)
AH<0,AS>0
(D)
AH<0,AS<0
PQ-13.
What
are
the
signs
of
A//
and
AS
for
the
deposition
of
Zn?
AH
AS
(A)
positive
positive
(B)
positive
negative
(C)
negative
positive
(D)
negative
negative
PQ-14.
Given:
C3Hi(g)
+
504(g)
—
3COx(g)
+
4H:0(2)
AG®nn=-2074
kJ-mol
'
What
is
the
value
of
AG°xn
for
the
reaction:
6COs(g)
+
8H,0(g)
—
2C;Hs(g)
+
100a(g)
(A)
1037
kJ-mol™!
(B)
1037
kJ-mol”!
(C)
2074
kJ'mol
!
(D)
4148
kJ-mol™
182
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g
ACS
Thermodynamics
Exams
PQ-15.
What
is
AG®
for
the
reaction
2N:0(g)
+
30s(g)
—
4NOs(g)?
4NO(g)
—
2N:0(g)
+
Ox(g)
AG°=-139.56
kJ'mol”!
2NO(g)
+
Oa(g)
»2NOa(g)
AG°==69.70
ki-mol!
(A)
0.16
kJ-mol™
(B)
2.09
kJ-mol™!
(C)
69.86
kJ-mol™
(D)
—209.26
kJ-mol™!
PQ-16.
Which
has
a
value
of
zero
at
25
°C?
(A)
AG?
of
O(g)
(B)
AG®r
of
Oa(g)
(C)
AG®
of
Os(g)
(D)
all
of
these
PQ-17.
What
is
the
value
of
AG°x
for
the
reaction
below?
AG®t/
kJ-mol!
4PCly(g)
—
6Clx(g)
+
Pa(g)
PCli(g)
~269.6
Py(g;
24.1
(A)
1103
kJ-mol™
(B)
-1054
kJ-mol!
(C)
-321kJ'mol™
(D)
1103
kJ-mol™
PQ-18.
What
is
the
value
AG®
for
this
reaction?
AG®t/
kJ-mol™!
COy(g)
+
2H20(g)
—
CHa(g)
+20:(g)
COxg)
394
H:0(g)
-229
CHai(g)
~50.8
(A)
801
ki'mol™!
(B)
-118
kJ:mol™
(C)
572KkJ'mol™!
(D)
801
kJ'mol!
PQ-19.
What
is
always
true
for
a
spontaneous
process
at
constant
pressure?
(A)
AGggem
<0
(B)
ASgsen
>0
(C)
ASiysien
>
0
and
AGsysiem
>0
(D)
ASiysem
>0
and
AGiysem
=0
PQ-20.
Consider
the
ionization
of
nitrous
acid
at
25
°C:
HNO:(aq)
+
H:0(1)
=
H;0%(aq)
+
NOx(aq)
~
AG°=19.1
kI'mol™
‘What
is
the
value
of
K
for
this
equilibrium?
(A)
20x107
(B)
45x10+
©
22
®
22x10°
PQ-21.
The
ionization
of
aqueous
hydrofluoric
acid
has
an
equilibrium
constant
6.31x10*
at
25
°C.
What
is
the
value
of
AG®n,
for
this
reaction?
(A)
0.665
kJ-mol™
!
(B)
1.53kJ-mol™
(©)
7.93
kI'mol”!
(D)
18.3
ki-mol™
PQ-22.
For
the
generic
reaction
shown,
K
=
0.355
(at
950
K).
What
can
you
conclude
about
AG
and
AG®
in
a
reaction
mixture
for
which
Q
=
0.205
at
950
K?
2AZ:(g)
+
Za(g)
=
2AZ:(g)
(A)
AG
and
AG®
are
both
negative
(B)
AG
and
AG®
are
both
positive
(C)
AG
is
positive and
AG®
is
negative
(D)
AG
is
negative
and
AG®
is
positive
PQ-23.
What
is
the
value
of the
equilibrium
constant
X
for
a
reaction
for
which
AG®
is
equal
to
5.20
kJ-mol™
at
50
°C?
(A)
0.144
(B)
0287
©)
693
(D)
86.4
PQ-24.
For
the
reaction
NH.CI(s)
=
NHs(g)
+
HCI(g)
AH®
=176
kJmol™
and
AG®
=91.2
kJ'mol™'
at
298
K.
What
is
the
value
of
AG
at
1000
K?
(A)
-109
kJ'mol"!
(B)
64
kI-mol!
(©)
64Kk
mol”
()
109
ki'mol”!
183
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Thermodynamics
."ACS
w’v‘,\d
PQ-25.
Thereactionis
Fe:05(s)
+
3C(s)
—
2Fe(s)
+
3CO(g)
AH®
=+490.7
kJ-mol”!
AS®=+541
J-mol
K-
(A)
never
spontaneous.
(B)
always
spontaneous.
(C)
spontaneous
only
below
a
certain
temperature.
(D)
spontaneous
only
above
a
certain
temperature.
v‘”,p\
PQ-26.
A
spontancous
reaction
has
a
negative
AS.
What
must
be
true
about
the
reaction?
(A)
The
reaction
is
exothermic
and
the
temperature
is
sufficiently
low.
(B)
The
reaction
is
exothermic
and
the
temperature
is
sufficiently
high.
(C)
The
reaction
is
endothermic
and
the
temperature
is
sufficiently
low.
(D)
The
reaction
is
endothermic
and
the
temperature
is
sufficiently
high.
A
PQ-27.
For
the
process
Ox(g)
—
20(g),
AH°
=
498
kI-mol~'.
What
would
be the
predicted
sign
of
AS®y,
and
o\@"’fl
the
conditions
under
which
this
reaction
would
be
spontaneous?
AS°n
Spontaneous
(A)
positive
at
low
temperatures
only
(B)
positive
at
high
temperatures
only
(C)
negative
at
high
temperatures
only
(D)
negative
at
low
temperatures
only
“‘,\
PQ-28.
Which
gcid
h_ss
amore
positive
value
of
AG®
for
its
ionization
in
water,
Acid
Ko
c,#’
and
which
acid
is
stronger?
HNO,
3x10*
HCN
5x10710
more
positive
AG®
stronger
acid
(A)
HNO,
HCN
(B)
HNO:;
HNO:
(C)
HCN
HCN
(D)
HCN
HNO:
f"
PQ-29.
Given
that
the
AG®
value
for
a
particular
reaction,
A
=
B,
is
negative,
what
can
be
said
about
the
o
relative
portions
of
reactants and
products
when
the
reaction
reaches
equilibrium?
(A)
There
are
fewer
reactant
molecules
than
product
molecules.
(B)
There
are
equal
numbers
of
reactant
and
product
molecules.
(C)
There
are
more
reactant
molecules
than
product
molecules.
(D)
A
relationship
cannot
be
drawn
without
calculating
K.
and
AG®.
M
PQ-30.
When
the
reaction
A
=
B
reaches
equilibrium,
which
quantity
equals
zero?
(A)
The
concentrations
of
A
and
B.
(B)
The
enthalpy
change,
AH.
(C)
The
equilibrium
constant,
K.
(D)
The
free
energy
change
AG.
184
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W=
0
PREIAAN
SR
WS
EOATUROAOWAT®
>
ACS
Thermodynamics
@
Exams
Answers
to
Study
Questions
4.D
7.
B
5
C
8.
B
6.
B
9.C
Answers
to
Practice
Questions
11.D
21.D
12.A
22.D
13.D
23.A
14.D
24.A
15.A
25.D
16.B
26.A
17.D
27.B
18.D
28.D
19.A
29.A
20.B
30.D
185
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Thermodynamics
186
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Chapter
15
—
Electrochemistry
Chapter
Summary:
This
chapter
will
focus
on
concepts
and
reactions
that
involve
the
transfer
of
electrons.
Specific
topics
covered
in
this
chapter
are:
Galvanic
cells
and
cell
diagrams
Calculating
cell
potentials
for
galvanic
reactions
Relating
values
of
£%e
and
AG®,
and
Keq
Using
the
Nernst
equation
Calculating
quantities
in
electrolytic
cells
Calculations
using
concentration
cells
Previous
material
that
is
relevant
to
your
understanding
of
questions
in
this
chapter
include:
«
Significant
figures
(Toolbox)
e
Scientific
notation
(Toolbox)
*
Balancing
redox
reactions
(Chapter
5)
*
Definitions
of
anode,
cathode,
reduction,
and
oxidation
(Chapter
5)
«
Equilibrium
(Chapter
11)
*
Thermodynamics
(Chapter
14)
Common
representations
used
in
questions
related
to
this
material:
Name
Example
Used
in
questions
related
to
Cell
notation
Ni(s)
|
Ni**(aq)
||
Ag*(aq)
|
Ag(s
Galvanic
cells
Cell
diagrams
Redox
reactions
Fe
metal
1.0M
Sn(NO3):(aq)
10M
Fe(NOs)s(aq)
Where
to
find
this
in
your
textbook:
The
material
in
this
chapter
typically
aligns
to
“Electrochemistry™
in
your
textbook.
The
name
of
your
chapter
may
vary.
Practice
exam:
There
are
practice
exam
questions
aligned
to
the
material
in
this
chapter.
Because
there
are
a
limited
number
of
questions
on
the
practice
exam,
a
review
of
the
breadth
of
the
material
in
this
chapter
is
advised
in
preparation
for
your
exam.
How
this
fits
into
the
big
picture:
The
material
in
this
chapter
aligns
to
the
Big
Idea
of
Energy
and
Thermodynamics
(6)
and
Equilibrium
(8)
as
listed
on
page
12
of
this
study
guide.
187
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Electrochemistry
cACS
W
Exams
Study
Questions
(SQ)
SQ-1.
What
is
the
coefficient
for
the
nitrite
ion
(NO,")
when
the
redox
reaction
is
balanced
in
basic
solution?
MnOs
(aq)
+
NO;(aq)
—
MnOx(s)
+
NOs(aq)
@
1
®)
2
©
3
o
4
Knowledge
Required:
(1)
Ability
to
write
half-reactions.
(2)
Rules
for
balancing
redox
equations.
Thinking
it
Through:
Y
ou
are
given
a
redox
reaction and
are
asked
to
determine
the
coefficient
of
a
species
when
the
equation
is
balanced
in
basic
solution.
You
recall
that the
rules
for
balancing
in
acidic
and
basic
solution
are
the
same
until
the
second
to
last
step
(where
you
balance
charge
—
using
H
for
acidic
solutions
and
OH~
for
basic
solutions).
The
first
step
is
to
separate
the
reaction
into
two
half-reactions
and
proceed
as
shown
in
the
table
for
each
half
reaction:
Step
Rxn
1
Rxn2
Asi;m
E
45
47
"
?s’:o:,"‘f:bi'?‘gf
NO;
(ag)
—
NO;
(aq)
MnO;
(aq)
-
MnO,
(s)
.
Oxidation
reduction
Assignas
ox/red
(2oing
from
+3
to
+5
is
a
loss
of
electrons)
(going
from
+7
to
+4
is
a
gain
of
electrons)
Balance element
3o
S
7
4
undergoing
NO;
(aq)
>
NO;
(aq)
MnO;
(aq)
>
MnO,
(s)
ox/red
N
already
halanced
Mn
already
balanced
‘Add
e
/balance
5
45
B
M
ox
stae Sw"h
NO;
(aq)
=
NOj
(aq)+2e
MnOj;
(aq)+3e”
—
MnO,
(s)
Balance
charg
winot
|
20H"(aq)+NO;
(aq)
>
NO;
(aq)
+2¢
|
MnO;
(aq)+3¢”
—>
MnO,
(aq)+
4OH
(aq)
(charge
below)"
'
N
-
-
’
Balance
Hwitn
|
2OH
™
(aa)+NO;3
(ag)
—
2H,0(1)+MnO;
(ag)+3¢”
—
H0
NO;
(aq)
+2e”
+H,0(1)
MnO,
(aq)+40H"
(aq)
*As
a
reminder,
you
know
you
would
use
H*
if
this
was
in
acidic
solution
rather
than
basic
solution.
The
next
step
is
to
multiple
the
half-reactions
so
that
the
number
of
electrons
produced
equals
the
number
of
electrons
used.
The
overall
equation
is
then:
Ox:
[20H"(aq)+NO3
(aq)
>
NO;
(aq)
+
2¢"
+
H,0(1)|x3
Red:
[2H,0(1)+MnO;
(ag)+3e
—
MnO,
(aq)+
4OH"
(aq)
2
Overall:
3NO;
(aq)+2MnOj
(aq)
+
H,0(1)
-
3NO;
(aq)+
2MnO,
(aq)
+
20H™
(ag)
Where
the
electrons
are
balanced
and
canceled
by
multiplying
each
(or
both
as
here)
half
reaction
by
a
factor.
Finally,
you
canceled
the
extra
water
molecules
(from
the
right
or
products
side)
and
the
extra
hydroxide
ions
(from
the
left
or
reactants
side).
You
can
now
find the
coefficient
for
the
nitrite
ion
(listed
first)
which
is
3.
Choice
(C)
is
the
correct
answer.
Choice
(A)
is
incorrect
because
it
is
the
result
of
not
multiplying
the
half-reactions
to
make
the
number
of
electrons
cancel.
Choice
(B)
and
(D)
are
incorrect
because
they
are the
result
of
incorrectly
multiplying
the
half-reactions.
Practice
Questions
Related
to
This:
PQ-1,
PQ-2,
PQ-3
188
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>ACS
Electrochemistry
@
Exams
SQ-2
A
spontaneous
electrochemical
cell
is
set
up
as
shown.
Which
statement
is
true?
Noltmeter
Salt
Bridge
Sn**+2¢—
Sn
E°=-0.136V
Fe*
+3e—>
Fe
E°=-0036V
b
metal
]
Sn(NO3):
(aq)
(A)
The
tin
electrode
is
the
cathode
and
electrons
move
from
left
to
right.
(B)
The
tin
electrode
is
the
cathode
and
electrons
move
from
right
to
left.
(C)
The
tin
electrode
is
the
anode
and
electrons
move
from
left
to
right.
(D)
The
tin
electrode
is
the
anode
and
electrons
move
from
right
to
left.
Knowledge
Required:
(1)
Definition
of
spmfuc{aous
reaction.
(2)
Definition
of
anode
and
cathode.
Thinking
it
Through:
You
are
being asked
to
evaluate
some
statements
about
a
spontaneous
electrochemical
cell.
You
begin
by
using
the
given
cell
potentials
to
determine
the
spontaneous
reaction.
Recall
that
for
a
reaction
to
be
spontaneous
the
value
of
E°
is
positive
because
reactions
with
positive
cell
potentials
have
negative
free
energy
changes,
and
are
spontaneous.
To
get the
spontaneous
reaction
you
need
to
reverse
the
reaction
involving
Sn.
You
also
have
to
multiply
the
half-reactions
so
that
electrons
cancel
(here
6
electrons
are
being
transferred
as
written).
Remember
that
the
oxidation
half-reaction
provides
the
electrons
for
the
reduction
half-reaction.
You
also
remember
to
NOT
multiply
the
values
of
E°
since these
cell
potentials
are
intensive
properties
and
so
their
value
does
not
depend
on
the
number
of
times
the
half-reaction
occurs.
Sn
—
Sn?"
+2e”
E°=0136V
oxidation
Fe*
+3¢”
—
Fe
E°=-0.036
V
reduction
3(Sn
—
Sn**
+2¢)
E°=0.136V
2(Fe*"
+
3¢~
—
Fe)
E°=-0.036
V
3Sn
+
2Fe™
—
38n**
+
2Fe
E°=0.100
V
You
also
know
that the
oxidation
reaction
occurs
at
the
anode
and
the
reduction
occurs
at
the
cathode.
For
the
spontaneous
reaction,
the
anode
is
the
Sn
and
the
Fe
is
the
cathode.
You
know
that
electrons
will
move
from
where
they
are
produced
(the
oxidation
half-reaction
or
the
reaction
at
anode)
to
where
they
are
consumed
(the
reduction
half-reaction
or
the
reaction
at
the
cathode).
The
source
of
electrons
for the
reduction
is
the
anode.
Therefore,
the
correct
choice
is
(C).
Choices
(A)
and
(B)
are
not
correct
because
tin
is
the
anode.
Choice
(D)
is
not
correct
because
the
electrons
move
from
the
anode
to
the
cathode.
Practice
Questions
Related
to
This:
PQ-4,
PQ-5,
PQ-6
SQ-3.
Given
the
standard
reduction
potentials
at
25
°C,
Half
Reaction
E°
what
is
the
standard
cell
potential
of
a
Aflg'(aq)
+e
—
Ag(.s)
079V
Ni(s)
|
Ni?*(aq)
||
Ag'(aq)
|
Ag(s)
galvanic
cell?
Ni**(aq)
+
2¢”
—
Ni(s)
023V
(A)
0.56
V
(B)
102V
©)
135V
(D)
181V
189
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Electrochemistry
.
ACS
W
Exams
|
Knowledge
Required:
(1)
Ability
to
interpret
electrochemical
cell
notation.
(2)
Definition
of
a
galvanic
cell.
Thinking
it
Through:
You
are told
the
cell
is
galvanic
and
you
remember
that
galvanic
cells are
spontaneous
reactions.
For
a
reaction
to
be
spontaneous
under
standard
conditions
the
value
of
AG°
must
be
negative.
You
also
recall
that
when
AG®
is
negative,
then
£°..s
positive.
You
are
given
the
reaction
in
electrochemical
cell
notation.
You
know
that
in
this
notation,
the
first
pair
represents
the
reaction
at
the
anode
(oxidation)
and
the
second
pair
is
the
reaction
at
the
cathode
(reduction).
The
double
lines
(*}|)
represent
the
salt
bridge.
oxidation
half-reaction||reduction
half-reaction
oxidation
reactant
|
oxidation
product
||
reduction
reactant
|
reduction product
Converting
the
given
notation
Ni(s)
|
Ni*(aq)
||
Ag'(aq)
|
Ag(s)
to
half-reactions
you
write:
Ni(s)
—»
Ni*"(aq)
+2¢°
E°=023V
(Note
you
changed
the
sign
of
£°
from
the
reduction
in
the
table)
Ag(aq)+¢
—
Ags)
E°=079V
Multiplying
the
Ag"
half-reaction
by
2
to
make
the
electrons
cancel,
and
remembering
to
NOT
multiply
the
values
of
the
cell
potentials,
since
they
are
intensive
properties,
you
get.
Ni(s)
+
2Ag*(aq)
—
2Ag(s)
+
Ni**(aq)
E°=102V
whichis
choice
(B).
Choice
(A)
is
not
correct
because
the
half-reaction
for Ni
was
not
reversed.
Choice
(C)
is
not
correct
because
the
half-reaction
reaction
for
silver
was
multiplied
by
2
and
the
Ni
reaction
was
not
reversed.
Choice
(D)
is
not
correct
because
the
half-reaction
reaction
for
silver
was
multiplied
by
2
and
the
Ni
reaction
was
reversed.
Practice
Questions
Related
to
This:
PQ-7,
PQ-8, PQ-9,
PQ-10,
PQ-11
SQ-4.
An
oxidation-reduction
reaction
in
which
3
electrons
are
transferred
has
a
AG°
=
18.55
kI-mol~*
at
25°C.
What
is
the
value
of
£°7
A
0192V
B
0064V
©
0192V
D)
0577V
[Knowledge
Required:
(1)
The
relationship
of
fiee
encrgy
to
cell
potential.
Thinking
it
Through:
You
are
given
a
value
of
AG®
and
are
asked
for
the
cell
potential.
You
recall
the
relationship
between
AG®
and
£°1
AG®
=-nFE°
[Note:
It
isn’t
needed
in
this
problem
but
the
value
of
£°
is
also
related
to
the
equilibrium
constant.}
AG°=-RTInK
so
-RTInK
=-nFE®
E°=(RT/nF)ink
where
»
is
the
number
of
electrons
transferred
and
F
is
the
Faraday
constant.
Note:
The
value
of
the
Faraday
constant
will
typically
be
given
on
the
reference page
of
the
exam
with
other constants.
Solving
for
E°
you
get:
E°=-AG/nF
Substituting
values
results
in:
1000)
—(13.55kJ)(——~)
1kJ
96485
C
(3
mol
electrons)|
-
———————-
1
mol
electrons
)
0.06421-C”'
=~
0.0642V
Where
you
have
used
the
definition
1
V=1
J-C"'.
The
correct
choice
is
(B).
Choice
(A)
is
not
correct
because
it
omitted
the
number
of
electrons
and
dropped
the
negative
sign.
Choice
(C)
is
not
correct
because
it
omitted
the
number
of
electrons.
Choice
(D)
is
not
correct
because
it
multiplied
by
the
number
of
electrons
rather
than
dividing.
Practice
Questions
Related
to
This:
PQ-12,
PQ-13
190
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Electrochemistry
Consider
the
reaction.
Cu?(aq)
+
Fe(s)
—
Cu(s)
+
Fe*(aq)
E°=0.78V
What
is
the
value
of
E
when
[Cu?"]
is
equal
to
0.040
M
and
[Fe?]
is
equal
to
0.40
M?
(A)
072V
(B)
075V
(©)
081V
(D)
0.84V
Knowledge
Required:
(1)
Use
the
Nemst
equation
to
calculate
cell
potential
under
nonstandard
conditions.
Thinking
it
Through:
You
recall
the
relationship
between
£
and
£°
is
given
by
the
Nernst
equation.
[Note:
the
equation
will
be
provided
either
in
the
problem
or
on
the
reference
sheet.]
There
are
two
common
versions
of the
Nernst
equation:
og
O
The
0.0592
comes
from
using
the
values
for
R, F,
and
25
°C
(298
K)
and
the
conversion
between
the
natural
and
base-10
log.
You
then
use
the
reaction
to
write
an
expression
for
Q;
recall
from
chapter
11
that
Q
is
an
expression
written
exactly
like
an
equilibrium
expression,
but
the
values
substituted
into
the
expression
are
not
necessarily
equilibrium
values.
Once
the
expression
is
written and
the
appropriate
values
are
used
you
calculate
Q.
RT
E=E"~—1
nF nQ
Reactants
Cn:']
T
00a0M
Now,
using
the
value
of
n
=2
because
2
mole
of
electrons
are
transferred
for
every
mole
of
reaction:
0.0592
E=078V-
1og(10)=0.75V
which
is
choice
(B).
Choice
(A)
is
incorrect
because
it
omits
n.
Choice
(C)
is
incorrect
because
it
added
the
nonstandard
cell
potential
term,
(0.0592/2)logQ.
Choice
(D)
is
incorrect
because
it
added
the
nonstandard
cell
potential
term
and
didn’t
use
n
=2.
Practice
Questions
Related
to
This:
PQ-14,
PQ-15,
PQ-16,
PQ-17,
PQ-18,
PQ-19,
PQ-20
$Q-6.
‘What
half-reaction
occurs
at
the
anode
during
the
electrolysis
of
molten
sodium
iodide?
A)
A
—bL+2e
B)
L+2e
20
©
Na—Na"+e"
(D)
Na"+e
—Na
Knowledge
Required:
(1)
The
definition
of
electrolysis.
(2)
Definitions
of
anode
and
cathode
in
electrochemical
cells.
Thinking
it
Through:
You
are
being
asked
to
identify
the
reaction
occurring
at
the
anode
of
an
electrolytic
cell.
You
remember
that
electrolysis
reactions
are
reactions
that
use
electrical
energy
to
drive
a
chemical
process.
Specifically,
these
reactions
have
a
positive
AG
and
can
be
driven
by
the
application
of
electrical
energy.
This
is
different
from
the
spontaneous
reactions
used
in
galvanic
cells,
which
have
negative
AG
values
and
the
excess
free
energy
can
be
used
to
do
electrical
work.
You
also
remember
that
no
matter
if
the
cell
is
a
galvanic
or
electrolytic
cell,
oxidation
occurs
at
the
anode
and
reduction
occurs
at
the
cathode.
Because
the
electrolysis
involves
molten
sodium
iodide,
you
do
not
have
to
consider
the
oxidation
or
reduction
of
water.
Remember
that
in
aqueous
solutions,
these
reactions
are
often
the
ones
that
occur.
Recall
that
in
molten
salts,
the
species
of
interest
are
the
ions.
For
molten
sodium
iodide
the
ions
are:
Na*
and
[".
These
are the
only
options
for
reactants
for the
half-reactions.
Because
the
process
at
the
anode
will be
oxidation,
you
eliminate
Na“
because
it
is
in
its
highest
oxidation
state.
Therefore,
the
only
possible
reaction
is
the
oxidation
of
I:
21"
—
I,
+
2e~
which
is
choice
(A).
Choice
(B)
is
not
correct
because
this
is
the
reduction
of
I,.
Choice
(C)
is
not
correct
because
it
is
the
oxidation
of
Na,
a
species
not
present
in
the
molten
salt.
Choice
(D)
is
not
correct
because
it
is
the
reduction
of
Na'.
Practice
Questions
Related
to
This:
PQ-21, PQ-22,
PQ-23
191
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Electrochemistry
>
ACS
=
W
Exams
How
many
minutes
are
required
to
plate
2.08
g
of
copper
at
a
Constants
constant
current
flow
of
1.26
A?
Molar
mass,
Cu
|
63.5
g'mol'
Cu(aq)
+
2&”
—
Cu(s)
1
faraday
(F)
96485
C
(A)
41.8min
(B)
83.6
min
(©
133
min
®)
5016
min
Km)wlzdée
Req;:ired:
1)
The
stoichiometry
of
electrolytic
processes.
i2)
The
relat}onships
among
an"\i)eres:
faradays,
time,
and
moles
of
electrons.
Thinking
it
Through:
You
are
given
a
mass
and
current.
You
recall
that an
Ampere
is
defined
as
1
C
of
charge
per
second:
1A=
l_
.
Therefore,
the
1.26
A
can
be
written
as
1.26
C's™
or
as
a
conversion
factor;
1.26
C=1s.
s
The
stoichiometric
relationship
involving
electrons
and
copper
is:
2
mol
¢™
=
1
mol
Cu
The Faraday
is
also
a
conversation
between
coulomb
and
moles
of
electrons:
1
mol
e”
=
96485
C
You
can
use
these
relationships
to
find
the
time
required.
You
start
with
the
mass
of
copper
desired:
1
mol
Cu
2mole”
{96485
C
1s
1
min
N
L
B
2.08
g
Cu
=836
min
which
is
choice
(B).
63.5gCu/\
1
mol
Cu
/\1mole”
/\1.26
C/\
60s
Choice
(A)
is
incorrect
because
it
didn’t
use
the
2
mol
of
electrons
per
1
mol
Cu
relationship.
Choice
(C)
is
incorrect
because
it
incorrectly
multiplied
by
1.26.
Choice
(D)
is
incorrect
because
it
is
the
number
of
seconds,
and
did not
convert
the
time
to
minutes.
Practice
Questions
Related
to
This:
PQ-24,
PQ-25,
PQ-26, PQ-27,
PQ-28
SQ-8.
For
the
concentration
cell
Fe(s)
|
Fe*(aq),
0.100
M
||
Fe**(aq),
1.00
M
|
Fe(s)
what
is
Ecar
at
25
°C
(in
mV)?
(A)
9.87
mV/
(B)
19.7
mV/
©)
29.6
mV
D)
59.2mV
Knowledge
Required:
(1)
Definition
of
a
concentration
cell.
(2)
Ability
to
interpret
electrochemical
cell
notation.
|
Thinking
it
Through:
You
are
being asked about
a
concentration
cell.
You
recall that
in
a
concentration
cell
the
source
of
the
electrical
potential,
and
the
current
flow,
is
the
differences
in
the
concentrations
of
the
species.
The
given
cell
notation
can
be
converted
to
half-reactions:
Fe(s)
|
Fe**(aq),
0.100
M
||
Fe¥(aq),
1.00
M
|
Fe(s)
oxidation
/
\lmduction
Fe(s)
—
Fe¥*(aq,
0.100
M)
+
3e~
Fe¥*(aq,
1.00
M)
+
3¢~
—
Fe(s)
After
adding
the
half-reactions,
you
get:
Fe(s)
+
Fe**(aq,
1.00
M)
—
Fe**(aqg,
0.100
M)
+
Fe(s)
The
Nernst
equation
takes
into
account
the
dependence
of
cell
potential
on
concentrations:
5
[0.0592
o
=
Foa
Jlog
O
remembering
that
Q
is
the ratio
of
the
concentrations
of
products
to
reactants.
n
You
also
recall
that the
E°;
for
a
concentration
cell
is
zero,
because
the
two
half-reactions
are the
same:
192
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ACS
Electrochemistry
Exams
00592)
(0.100
M
E,
=
o-(
jlog
~00197V
3
1.00M
which
is
choice
(B).
1000
mv
0.0197
v[ool%]
=19.7
mV
Choice
(A)
is
not
correct
because
it
used
6
for
the
number
of
electrons
in
the
Nernst
equation.
Choice
(C)
is
not
correct
because
it
used
2
for
the
number
of
electrons
in
the
Nernst
equation.
Choice
(D)
is
not
correct
because
it
did not
use
the
number
of
electrons
in
the
Nernst
equation.
Practice
Questions
Related
to
This:
PQ-29,
PQ-30
Practice
Questions
(PQ)
PQ-1.
What
is
the
coefficient
of
the
hydroxide
ion for
the
half
reaction
when
the
hypobromite
ion
reacts
to
form
the
bromide
ion
in
basic
solution?
BrO(aq)
>
Bri(aq)
unbalanced
)1
(B)
2
©
3
D)
4
PQ-2.
What
is
the
balanced
half
reaction
of
the
nitrate
ion
reacting
to
form
nitrogen
monoxide
in
acidic
M
solution?
(A)
NOs™(aq)
=
HoO(l)
>
NO(g)
+
3e”
+
2H
(aq)
(B)
NOy(aq)
+3¢”
~
4H'(aq)
>
NO(g)
+
2H:0(1)
(C)
NOs(aq)
+
3¢+
2H0(l)
—
NO(g)
+
40H(aq)
(D)
NOs(aq)
+
4™~
5H'(aq)
=
NO(g)
+
2H,0(1)
PQ-3.
What
is
the
coefficient
for
water
when
the
reaction
is
balanced
in
acidic
solution?
Cr;07*(aq)
+
Mn**(aq)
-
Cr*(aq)
+
MnOx(s)
unbalanced
@A)
1
(B)
2
©)
4
@)
5
PQ-4.
Which
reaction
occurs
at
the
cathode
of
a
galvanic
cell
constructed
from
these
two
half-ceils?
The
standard
reduction
potential
of
gold(III)
is
+1.50
V.
The
standard
reduction
potential
of
tin(II)
is
-0.14
V.
(A)
Au*(ag)
=
3e”—
Au(s)
(B)
Sn**(aq)
+2¢”
—
Sn(s)
(©)
Au(s)
—
Au**(ag)
+
3¢
(D)
Sn(s)
—
Sn**(aq)
+
2
PQ-5.
Which
reaction
will
occur
if
each
substance
is
Standard
Reduction
Potentials
E°
in
its
standard
state?
Ni*(aq)
+2e”
—
Ni(s)
-028
V
Assume
potentials
are
given
in
water
at
25
°C.
Sn*(aq)
+
2e”
—
Sn**(aq)
0.15V
Bry(l)
+
2e”
—
2Br7(aq)
1.06
V
(A)
Ni**
will
oxidize
Sn**
to
give Sn**
(B)
Sn**
will
oxidize
Br~
to
give
Bry
(C)
Br,
will
oxidize
Ni(s)
to
give
Ni*~
(D)
Ni*"
will
oxidize
Br;
to
give
Br-
193
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Electrochemistry
*ACS
Exams
PQ-6.
Which
represents
a
spontaneous
reaction
and
what
is
|
Standard
Reduction
Potentials
_E°
the
correct
E°u?
Fe'"(aq)
+
3¢
—
Fe(s)
004V
Clyg)
+
2¢”
—
2Cl(aq)
136V
Reaction
E%an
(A)
Fe(s)
+
Cly(g)
—
Fe''(ag)
+
2CI(aq)
140V
(B)
Fe¥(aq)
+
Cly(g)
—
Fe(s)
+
2CI(aq)
140V
(C)
2Fe(s)
+
3Ch(g)
—
2Fe**(aq)
+
6Cl(ag)
140V
(D)
2Fe(s)
+3Cla(g)
—
2Fe™"(aq)
+
6CI(aq)
~
4.16
V
PQ-7.
Which
combination
of
reactants
will
produce
the
Standard
Reduction
Potentials
E°
greatest
voltage
based
on
these
standard
electrode
Cu'*(aq)
+
e~
—
Cu(s)
0.52V
potentials?
Sn*(ag)
=
2¢-
—
Sn’*(aq)
015V
Cr'(aq)
+
e
—
Cr*(aq)
041V
(A)
Cu’and
Sn*
(B)
Cu*and
Cr**
(C)
CuandSn*
(D)
Sn**
and
Cr**
PQ-8.
What
is
the
standard
electrode
potential
for
a
voltaic
|
Standard
Reduction
Potentials
£E°
cell
constructed
in
the
appropriate
way
from
these
two
|
Cr**(aq)
+
3¢~
—
Cr(s
—0.74V
half-cells?
Co*(aq)
+
2"
—
Cos,
028V
(A)
-1.02V
(B)
046V
(C)
064V
D)
1.02V
PQ-9.
What
is
the
value
of the
missing
standard
reduction
Standard
Reduction
Potentials
E°
potential?
~
Pd?*(aq)
+
2e~
—
Pd(s)
?
Cu(s)
+
Pd**(aq)
—
Cu?(aq)
+
Pd(s)
£
=
0.650
V
Cu*(aqg)
+
2e”—
Cu(s).
0337V,
(A)
0987V
(B)
-0.987V
(©)
0313V
D)
0313V
PQ-10.
The
standard
cell
potential,
£°,
for
this
reaction
is
0.79
V.
61"
(aq)
~
Cr.O7*(aq)
+
14H'(aq)
~»
31x(aq)
~
2Cr'(ag)
+
7H0(1)
What
is
the
standard
potential
for
Ix(aq)
being
reduced
to
I
(aq)
given
that
the
standard
reduction
potential
for
Cr,O-*<(aq)
changing
to
Cr’~(aq)
is
1.33
V2
(A)
054V
(B)
—0.54V
(©
018V
(D)
018V
PQ-11.
What
i
the
standard
cell
potential,
£°,
for
this
[Standard
Reduction
Potentials
_E°
M
reaction?
Ce*(aq)
+
&
—
Ce>*(aq)
161V
Bra()
+2Ce¥(aq)
—
2Br
(aq)
+
2Ce*'(aa)
Bry(l)
+
2
—
JBraq)
106V
(A)
267V
(B)
-2.16V
(€)
055V
D)
267V
PQ-12.
Which
statement
is
true
for
an
electrochemical
cell
built
from
an
oxidation-reduction
reaction
if
K
for
the
reaction
is
greater
than
1?7
(A)
AG®is
negative,
E°
is
negative
(B)
AG®isnegative,
E°
is
positive
(C)
AG®is
positive,
E°
is
negative
(D)
AG?is
positive,
£°
is
positive
PQ-13.
What
is
the
equilibrium
constant
at
298
K
for the
spontaneous
|
Reduction
Potentials
E°
reaction
that
occurs
when
a
Cu®"|Cu
half-cell
is
connected
toa
|
Cu?|Cu
034V
Ag™
Ag
half-cell?
A
AL
0.80
V
(A)
2.8x107
(B)
6.0x10
(€)
3.6x10"
194
D)
3.7x10°®
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.*ACS
Electrochemistry
W
Exams
PQ-14.
Consider
the
reaction
at
298
K.
Sn**(aq)
+2Fe**(aq)
—
Sn*(aq)
+
2Fe**(aq)
E°=0617V
What
is
the
value
of
E
when
[Sn**]
and
[Fe**]
are
equal
to
0.50
M
and
[Sn**]
and
[Fe**]
are
equal
to
0.10
M?
(A)
0.069V
(B)
0679V
(C)
0658V
(D)
0576
V
A
PQ-15.
Consider
this
reaction.
w\w“p
2Al(s)
+
6HCl(aq)
==2AICl(aq)
+
3Hx(g)
E°=1.66
V
Which
statement
is
true
if
the
hydrogen
ion
concentration
is
initially
at
1.0
M
and
the
initial
pressure
of
hydrogen
gas
is
1.0
atm?
(A)
Addition
of
a
base
should
result
in
a
value
of
£
which
is
less
than
1.66
V.
(B)
n
=3,
because
three
moles
of
hydrogen
is
being
produced.
(C)
E*is
independent
of
the
pH
of
the
solution.
_[mJ[acL]
A
[Ha)
w‘"@
PQ-16.
According
to
the
Nernst
equation,
when
Ecai
=
0
then
A)
0=K
B)
K=1
(€)
AG°=0
(D)
Ei=0
PQ-17.
What
is
£
at
25
°C
if
[Ag']
=
0.200
M
and
[Cu?']
=
0.200
M?
2Ag
(aq)
+
Cu(s)
—
Cu*(aq)
+
2Ag(s)
E’e=0.460
V
(A)
0419V
(B)
0439V
(C)
0.460V
(D)
0481V
PQ-18.
A
voltmeter
connected
to
the
cell
under
M
standard
state
conditions
reads
+0.25
V.,
and
the
standard
free
energy
(AG®)
is
—44
kJ-mol™'.
Salt
Bridge
)
Tan
If
the
concentration
of
Ni*"
is
increased
to
(e
2.0
M,
the
cell
potential
will
and
N
AG
will
be
than
AG®.
Pi(s)
Ni**(aq)
+
2
—Ni(s)
E°=-025V
(A)
decrease,
less
negative
(B)
decrease,
more
negative
(C)
increase,
less
negative
(D)
increase,
more
negative
PQ-19.
What
is
the
potential
of
a
galvanic
cell
made
from
Standard
Reduction
Potentials
E°
Mn?'[Mn
and
Ag*|Ag
half-cells
at
25
°C
if
[Ag']
=
|
Mn®
(aq)
+
2
—
Mn(s)
“1LI8V
0.10
M
and
[Mn?]
=
0.10
M?
Ag'(aq)
+
e
—
Ag(s)
0.80
V
(A)
192V
(B)
195V
(€)
198V
D)
201V
PQ-20.
What
is
the
cell
potential
of
the
voltaic
zinc-copper
cell
at
25
°C?
Zn(s)|Zn**(1.00%107°
M)[[Cu?*(1.00x107%)|Cu(s)
The
standard
cell
potential
of
the
cell
is
1.10
V.
(A)
092V
(B)
128V
©
1.01v
D)
119V
195
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Electrochemistry
o
ACS
¥
Exams
PQ-21.
During
the
electrolysis
of
an
aqueous
solution
of
CuSOx
using
inert
electrodes
M
(A)
the
anode
loses
mass
and
the
cathode
gains
mass.
(B)
the
mass
of the
anode
decreases
but
the
mass
of
the
cathode
remains
constant.
(C)
the
mass
of the
anode
remains
the
same
but
the
cathode
gains
mass.
(D)
the
anode
and
the
cathode
neither
gain
nor
lose
mass.
M
PQ-22.
Which
products
are
formed
during
the
electrolysis
of
a
concentrated
aqueous
solution
of
sodium
chloride?
1Clg)
11
NaOH(aq)
1
Hy(g)
(A)
Tonly
(B)
Iand
Il
only
(C)
1and
Il
only
(D)
LT
and
I
wfd
PQ-23.
The
half-reaction
occurring
at
the
anode
during
the
electrolysis
of
molten
sodium
bromide,
NaBr,
is
(A)
Na*+e
—Na
(B)
Na—Na"+e
(C)
Brz+2e—2Br
(D)
2Br — Bry+2e”
PQ-24.
What
mass
of
platinum
could
be
plated
on
an
electrode
from
the
electrolysis
of
a
PtNOs),
solution
with
a
current
of
0.500
A
for
55.0
s?
The
molar
mass
of
platinum
is
195.1
g-mol~.
(A)
27.8mg
(B)
455mg
(C)
53.6mg
(D)
91.0mg
PQ-25.
A current
of
5.00
A
is
passed
through
an
aqueous
solution
of
chromium(III)
nitrate
for
30.0
min.
How
many
grams
of
chromium
metal
will be
deposited
at
the
cathode?
The
molar
mass
of
chromium
is
52.0
gmol™!.
(A)
0.027¢
®)
162g
(©)
485g
)
633g
PQ-26.
A
vanadium
electrode
is
oxidized
electrically.
If
the
mass
of
the
electrode
decreases
by
114
mg
during
the
passage
of
650
coulombs,
what
is
the
oxidation
state
of
the
vanadium
product?
A
-1
(B)
+2
©)
+3
(D)
+4
PQ-27.
Given
|
amp
of
current
for
1
hour,
which
solution
would
Molar
mass
/
g'mol-1
deposit
the
smallest
mass
of
metal?
Fe
559
Ni_58.7
Cu
63.5
Ag
1079
(A)
Fe
from
aqueous
FeCly
(B)
Ni
from
aqueous
NiCly
(C)
Cu
from
aqueous
CuSO4
(D)
Ag
from
aqueous
AgNO;
PQ-28.
How
many
minutes
will be
required
to
deposit
1.00
g
of
chromium
metal
from
an
aqueous
CrO>
solution
using
a
current
of
6.00
A?
(A)
186
min
(B)
30.9
min
(C)
15.4
min
(D)
5.15
min
PQ-29.
What
is
the
[Cu®"]
in
the
oxidation
half-cell
if
the
observed
cell
potential
is
0.068
V?
Cu(s)
|
Cu*'(?
M)
|
Cu?"(2.0
M)
|
Cu(s)
(A)
0.043M
(B)
0.010M
(©)
0.14M
D)
2.0M
PQ-30.
Which
statement
best
describes
a
concentration
cell
when
the
solutions
in
each
compartment
are
0.50
M?
(A)
The
cell
potential
is
zero.
(B)
The
cell
is
an
electrolytic
cell.
(C)
Anions
will
flow
away
from
the
anode.
(D)
Electrons
will
flow
away
from
the
cathode.
196
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.ACS
Electrochemistry
we
wO0
B
R
ol
ol
PErEEOONPEPER
Y
Exams
Answers
to
Study
Questions
EELFS
rE®
%
ww
Answers
to
Practice
Questions
11.C
12.B
13.C
14.B
15.A
16.
A
17.B
18.A
19.B
20.D
SRENERERERE
PROBRINN
AL
PEITDAEPOON
197
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Electrochemistry
198
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Chapter
16
—
Nuclear
Chemistry
Chapter
Summary:
This
chapter
will
focus
on
expressing
and
predicting
nuclear
reactions.
Specific
topics
covered
in
this
chapter
are:
*
Radioactivity
*
Radioactive
decay
*
Balancing
nuclear
reactions
Previous
material
that
is
relevant
to
your
understanding
of
questions
in
this
chapter
include:
*
Atomic
Structure
(Chapter
1)
*
Electronic
Structure
(Chapter
2)
o
Stoichiometry
(Chapter
4)
Common
representations
used
in
questions
related
to
this
material:
Name
Example
Used
in
questions
related
to
Graphs
B
nuclear
binding
energies
AL
T
off
¢
o
Binding energy
per
nucleon
(Me'
o
4
w20
w0
w0
Mass
number
(A)
Where
to
find
this
in
your
textbook:
The
material
in
this
chapter
typically
aligns
to
“Nuclear
Chemistry”
in
your textbook.
The
name
of
your
chapter(s)
may
vary.
Practice
exam:
There
are
practice
exam
questions
aligned
to
the
material
in
this
chapter.
Because
there
are
a
limited
number
of
questions
on
the
practice
exam,
a
review
of
the
breadth
of the
material
in
this
chapter
is
advised
in
preparation
for
your
exam.
How
this
fits
into the
big
picture:
The
material
in
this
chapter
aligns
to
the
Big
Idea
of
Atomic
Structure
(1)
as
listed
on
page
12
of
this
study
guide.
Study
Questions
(SQ)
SQ-1.
What
is
the
missing
particle
in
the
nuclear
equation?
ZiTh—
¥
Ra+
@
(B)
iHe
©
H
™)
n
Knowledge
Required:
(1)
How
to
balance
a
nuclear
equation.
Thinking
it
Through:
You
are
asked
in
the
question
to
determine
the
missing
particle
from
a
balanced
nuclear
equation.
You
can
assume
that
all
mass
units
are
accounted
for
in
the
missing
particle.
Therefore,
you
note
that the
starting
material
(Th)
undergoes
the loss
of
4
mass
units
in
becoming
the
product
shown
(Ra)
as
4
(remember,
4
is
199
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Nuclear
Chemistry
ACS
w
Exams
‘mass
number
or
the
number
of
protons
plus
the
number
of
neutrons)
goes from
230
(**°Th)
to
226
(***Ra).
The
missing
particle
will
have
a
mass
number
of
4
(*X).
In
addition,
you
note
that the
starting
material
(Th)
is
2
atomic
numbers
(or
two
protons)
more
than
the
product
(Ra).
as
Z
goes
from
90
(50Th)
to
88
(ssRa).
The
missing
particle
will
have
an
atomic
number
of
2
(X).
Four
mass
units
with
two
of
those
as
protons
is
a
He
particle.
Therefore,
you
can
conclude
that
this
is
an
example
of
alpha
decay
and
that
;He
is
the
missing
particle,
choice
(B).
Choice
(A)
is
not
correct
because
it
is
the
particle
lost
for
a
beta
decay
equation.
Choice
(C)
is
not
correct
because
it
is
a
proton
particle.
Choice
(D)
is
not
correct
because
it
is
a
neutron
particle.
Practice
Questions
Related
to
This:
PQ-1,
PQ-2, PQ-3,
and
PQ-4
SQ2.
A
U
atom
undergoes
nuclear
fission
by
being
bombarded
with
a
neutron
to
produce
a
total
of
four
neutrons,
a
'%,Cs
atom,
and
@
U
®
Np
(©)
ZRb
(D)
Rb
Knowledge
Required:
(1)
Definition
of
nuclear
fission.
(2).
How
to
write
and
balance
a
nuclear
equation.
Thinking
it
Through:
You
are
asked
in
the
question
to
determine
the
missing
particle
for
a
given
nuclear
fission
reaction.
The
reaction
is
expressed
in
words;
it
is
helpful
in
this
situation
to
express
the
reaction
as
an
equation:
BU4In
o
MCs+dlns
Next,
you
balance
the
protons;
there
are
92 on
the
reactant
side
(52U
and
on)
and
55
on
the
product
side
(s5Cs
and
4
x
on)
or
92-55
=
37
protons
missing
for
the
unknown
atom.
You
then
balance
the
mass
numbers;
there
are
235
+
1
on
the
reactant
side
(***U
and
'n)
and
140
+
4x1
on
the
product
side
(**'Cs
and
4
*
'n)
or
236
—
144
=
92
mass
numbers
missing
for
the
unknown
atom.
Thus,
X’
is
missing
in
order
to
balance
the
reaction;
this
is
the
atom
?
Rb,
Choice
(C).
Choice
(A)
is
not
correct
because
it
assumes
no
change
in
the
starting
material.
Choice
(B)
is
not
correct
because
it
is
the
result
of
positron
emission,
i.e.
a
neutron
becomes
a
proton.
Choice
(D)
is
not
correct
because
it
does
not
account
for
the
four
neutrons
that are
produced
in
the
nuclear
fission
reaction.
Practice
Questions
Related
to
This:
PQ-5
SQ-3.
An
atom
of the
element
of
atomic
number
53
and
mass
number
131
undergoes
beta
decay.
The
residual
atom
after
this
change
has
an
atomic
number
of
and
a
mass
number
of
.
(A)
54,131
(B)
53,131
(C)
53,131
D)
51,127
emission
occur.
(2)
How
to
write
and
balance
a
nuclear
reaction.
Thinking
it
Through:
You
are
asked
in
the
question
to
determine
the
atomic
number
and
mass
number
for
an
atom
after
a
particular radiation
emission
has
occurred.
In
this
instance,
the
atom
undergoes
beta
decay
(a
particle);
this
emission
results
in
a
net
gain
of
one
atomic
number
(or
transformation
such
that the
nucleus
now
has
one
additional
proton).
During
beta
decay,
there
is
no
overall
change
in
mass
number.
You
identify
the
element
in
the
question
as
iodine,
I
(Z=
53).
i
0g
4
1
sl
B+
5Xe
200
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“
ACS
Nuclear
Chemistry
Exams
Therefore,
if
a
particle
that
is
"1
undergoes
beta
decay,
the
resulting
atom
is
',
Xe
(Z
=
54,
4
=
131),
choice
(A).
Choice
(B)
is
not
correct
because
it
would
be the
residual
atom
after
a
positron
emission.
Choice
(C)
is
not
correct
because
it
would
be the
residual
atom
after
a
gamma
emission.
Choice
(D)
is
not
correct
because
it
would
be the
residual
atom
after
alpha
decay.
Practice
Questions
Related
to
This:
PQ-6,
PQ-7,
PQ-8,
PQ-9,
PQ-10, PQ-11,
and
PQ-12
Practice
Questions
(PQ)
PQ-1.
Cobalt-60
undergoes
beta
decay.
What
is
the
balanced
nuclear
reaction
for
this
process?
(@A)
#Co—p+5Fe
®)
5Co——
Jp+ENi
(©
#Co——ja+iMn
@)
#Co+
je——fFe
PQ-2.
What
is
the
missing
particle
in
the
balanced
nuclear
equation?
BMg—
TNa+_
+
v
@
®)
B
©
H
®)
o
PQ-3.
What
is
the
missing
product
of
this
reaction?
iR
—
@
B
®
B
©
oY
®
PQ-4.
What
is
the
missing
particle
in
the
balanced
nuclear
equation?
TAl+
jHe—>
P+
3
3
@)
e
®
e
©
H
®
o
PQ-5.
When
2;sz
captures
an
alpha
particle,
it
produces
a
neutron
and
which
isotope?
@)
“cCm
®
Cf
©
P
®
Pu
M
PQ-6.
Which
particle,
if
lost
from
the
nucleus,
will
nof
result
in
a
change
in
the
atomic
mass
number?
(A)
proton
(B)
alpha
particle
()
beta
particle
(D)
neutron
PQ-7.
An
atom
of
the
element
of
atomic
number
84
and
mass
number
199
emits
an
alpha
particle.
The
residual
atom
after
this
change
has
an
atomic
number
of
and
a
mass
number
of
(A)
82,195
(B)
84,203
(C)
85,195
(D)
86,199
PQ-8.
What
is
the
product
when
*;Po
undergoes
positron
emission?
(@A)
‘uPb
(B)
Rn
©
D)
8Bi
PQ-9.
Which
equation
qualifies
as
electron
capture?
@
SLiton—>
H+la
®)
Mgt
p
>
[Natia
(©
[Ar+
e
o
JClthy
®
Ko
Cat
g
201
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Nuclear
Chemistry
s
A
CS
Exams
PQ-10.
Which
nuclear
process
does
not
necessarily
result
in
nuclear
transmutation
(the
transformation
of
one
M
element
into
another)?
(A)
alpha
(o)
emission
(B)
beta
(B)
emission
(C)
gamma
(y)
emission
(D)
positron
emission
PQ-11.
Medical
imaging
is
used
when
a
patient
ingests
a
radioactive
emitter,
called
a
tracer,
so
a
doctor
can
view
an
internal
system.
Which
tracer
emission
will
penetrate
through
the
body?
(A)
alpha
particle
(B)
beta
particle
(C)
gamma
particle
(D)
helium
nucleus
PQ-12.
Based
on
their
binding
energies,
what
nuclear
5
10
transformations
are
predicted
for
F
and
Br?
g
.
PN,
e
F
Br
s
g4
=
8
g2
£
K
A
0
4
s
120
160
200
‘Mass
number
(
A)
(A)
Both
could
undergo
nuclear
fission.
(B)
Both
could
undergo
nuclear
fusion.
(C)
F
could
undergo
nuclear
fission
and
Br
could
undergo
nuclear
fusion.
(D)
Br
could
undergo
nuclear
fission
and
F
could
undergo
nuclear
fusion.
202
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bl
o=
ol
orww
~*ACS
Nuclear
Chemistry
¥
Exams
Answers
to
Study
Questions
Answers
to
Practice
Questions
[adal
2k
4
crOw
=
gann
203
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