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Subject
Chemistry
Date
Nov 24, 2024
Type
Pages
7
Uploaded by ksusanto
Week
7
Discussion
+
Tutorial
S22
CHEM
110A
May.
9**
—
May.
15*
1
Things
learnt
after
MT1
Entropy
=
2A0F
(')’1-'
LT
1=
RT=
HREK
2
Microscopic
picture
2.1
PM1-Q1
aH
=
-[ootm™
Tertiary
amines
can
form
a
bond
with
certain
metals
and
metalfxide
sur-
faces.
The
formation
of
this
bond
is
exothermic
and
releases
100
e~
of
heat
~(00
&h"'
per
bond.
You
design
a
molecule
with
four
tertiary
amine
limbs
to
adsorb
to
L
'.-b.,l
surfaces.
This
molecule
can
adsorb
either
with
four
(1
way),
three
(4
way),
two
-
e
e
(6
way),
or
one
(4
way)
“limbs".
The
head
does
not
stick
to
the
surflace,
(a)
Calculate
the
partition
function
Z
for
the
four
energy
states
shown
in
the
figure.
Use
kgT
=
207
emn™'
You
are
provided
with
the
possible
ways
to
achieve
each
state
in
the
figure,
—_
(b)
What
percentage
of
molecules
will
be
attached
with
one
imb?
What
(
percentage
will
be
attached
with
four
limbs?
<)
-1
*
(c)
Calculate
the
equilibrium
constant
for
a
molecule
going
from
one
limb
-
(’000"
attached
to
four
limbs
attached.
s
(d)
Without
doing
any
calculations,
explain
what
will
happen
to
this
equi-
4.-
é’&
librium
constant
if
temperature
is
raised?
E:
(a)
2=
i;AASe
A
~
-6,
o,
TR
o~
BOAR
(e
o
ot
+
G
et/
|
s
(b)
/
twb
D
_
4
oo
categ
S
(=
1"%op
.
Ne”
4-€
=
0%
E’
—
-
“).
—
,6‘
0
*(e)
IS4
>
K=
'}}72
%21-0-7
d)
AG=
A"'/'%—é-@w
48
s
Jowored
!
v::a:
#
mrywstodes
2.2
PM2-Q1
The
molecules
have
[thioll-like
functional
groups
that
allow
them
to
cova-
lently
bind
to
gold
on
a
2-D
surface.
Consider
a
2-by-2
portion
of
this
surface,
«
0
hydrogen
bonding
=
2
configurations
(highest
E)
«
2
hydrogen
bonding
=
12
configurations
(intermediate
E)
-’@4
hydrogen
bonding
=
2
configurations
(lowest
E)
A
(a)
Shown
above
are
all
possible
arrangements
in
this
2
x
2
grid.
If
there
is
no
enthalpy
associated
with
the
hydrogen
bonds,
what
is
the
probability
of
finding
2
x
2
system
evenly
distributed
(the
“four
hydrogen
bonds®™
section)?
(b)
In
reality,
formation
of
one
hydrogen
bond
is
exothermic
and
releases
0’
300
em~".
Assuming
“no
hydrogen
bonds”
corresponds
to
0
joules,
calculate
-
the
partition
function
at
T
=
208
K
for
hydrogen bonding
in
the
2
x
2
grid.
%\
vo
{(c)
At
T
=
298
K,
what
is
the
probability
of
achieving
no
hydrogen
bonds?
s’
w‘
What
is
the
probability
of
achieving
four
hydrogen
bonds”
(d)
No
calculation
necessary.
To
increase
the
probability
of
the
“evenly
distributed”
state
with
four
hydrogen
bonds,
should
you
increase
or
decrease
temperature’
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2.3
PM1-Q6
and
Q7
Consider
the
following
endothermic
reaction:
(‘a('();u'.'
=
(‘flol.)
+
("02!_91'
Decreasing
the
amount
of
CaO
in
the
reaction
vessel
will
(a)
Shift
the
reaction
towards
the
products
(L)
Shift
the
reaction
towards
the
reactants
@lnw
no
effect
on
the
reaction
é——
9'
&&3
Mc
‘)
Decreasing
the
temperature
of
the
reaction
vessel
will
(a)
Shift
the
reaction
towards
the
products
Shift
the
reaction
towards
the
reactamnts
(c)
Have
no
effect
on the
reaction
N
a\ow-b
AH
>0
R,:
P
AHP"AHR,
>0
T
it
~
o
P
yeotk
/fie@
e/@
.
¥
3
Macroscopic
picture
3.1
PM1-Q3
5kJ
of
hea
t;w
through
a
lm!
n
¢
where
the
hot
source
is
at
300
°C
and
the
»Il
l
25
('
The
l
gine
n;wmlm
ul
muimmn
efficiency.
How
mm'!
ropy
is
generated
a
l
ll
ink,
assuming
the
temperature
of
he
cold
xink
'us
constant?
~
T
]
30
AQ
=
éé:“_c
000
c
T
‘g"
=+
L
3
205
-w-3.=
.w-%j
8%
=
—2.68]
=-
g
=
5c>;+t‘i‘3)k
(=-4%5)
3-3%
3.2
PM1-Q4
The
bond
between
carbon
and
iodine
in
organic
molecules
is
quite
weak
in
comparison
to
carbon-hydrogen
or
carbon-carbon
bonds,
The
reported
bond
dissociation
enthalpy
for
the
cleavage
of
the
C-1
bond
in
iodomethane
is
AH®
=
240k
/mol,
CH3yl
—
CH,
+
I,
This
enthalpy
change
is
the
standard
enthalpy
and
therefore
only
applicable
at
room
temperature.
What
is
the
enthalpy
change
at
1000
K
under
constant
pressure
conditions?
Use:
5
cecuyt
=
4R,
cpen,
=4R,
cpy
=
ER'
Ctote
et
&F
H.
AH((wok)
=
AH
+
4ArH
=
AHO
¢+
(A”r“’-‘”fl)
=
AHO
+
(M
(g
8T*
N
CaT
=
Nerg
Corgz
4T)
AHO
+
(Con
+
G
=
Congr
)
AT
-
349%
+
[
gRt
Le-
u).((ooo—lfif)k
=
Jm;‘%
+
l@é%
Q58
RS
fnel
]
1\
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3.3
PM1-Q8
Standard
entropies
at
room
temperature
can
be
Yositive
only
5)
Negative
only
0
(c)
Positive
or
zero
S.—-—
(d)
Negative
or
zero
a'b
T:OK
(e)
Positive,
negative,
or
zero
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с
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mp: 113-15°C mp = 14.2°C
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© Macmillan Learning
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achieve.macmillanlearning.com
H-C-OH
reduction
H-C-OH
H₂C-OH
OCT
11
MacBook Pro
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<6
Reaction B
Ċ
Chapter 6 HW - General, Organic, and Biological Chemistry for Health Sciences - A-
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? Hint
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C
0 H
G
8*
(
H
0
H
H➡C
OH
| |
A
OH
H₂C
―OH
9
0
Q2Q
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