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the laws of thermodynamics
·
first
law
states: energy cant
be created or destroyed
0 DE
=
q
W-work
mechanical
A
heat
surroundings
a
energy
internal
w
G
energy
->
->
System
W
G
& transfer of heat energy
inlout of the
System
to the surroundings
·
second
law: for a spontanous process
entropy
of the universe increases
entropy:
like randomness
(disorder
a
randomness entropy4 A
opposite
·
IS universe
so
change
in
entropy
of the universe is
a
positive
#so
30
·
Asystem
u
two
of
the combined
are
the universe
· 3rdlaw: a perfectcrystal
ezepokelvin has zero entropy
tmp associated with
randomness
I disorder
Remp4 disorderentropy A
opposite
entropy
S
=
KIn IW)
OR S
=
KB((-)
k
=
x 10-235/K
w
=
A microstates
~
more microstates,
more
entropy
·
mixtures
have more
entropy
than
pure
substances
mixing
is
a spontanous process
a
leads to
inc.
entropy
FACTORS AFFECTING ENTROPY
1
phase
a
solids:
mostorder=
least entropy
gas=
least order:
most
entropy
temperature
high
tempt
entropy temps entropy b
volume
a volumet entropy
particles
more particles,
more
entropy/disorder
or more moles
predicting
the
sign of IS
moles
of gas
look
reaction
to
see if
youre getting
move or less moves
of
gas
4Oas
DSCO,
bGas
DS
no
change
in
to
guess,
somewhatclose
to
zero
Phase changes
S-1-goenotherher,
entro
is
St-
--
g DHC
exother mic
DSCO -
less
disorder,
entropy
NaCIs)
Nalag)
for
salts
=
DSSO
dissolving
is associated with
more
disorder,
entropy
- Complexity
simple or
more
complex molecules?
use method iffirst
dont
work
ex4MPICS:
molgas
2
moles
gas
122(x
=>
more
gas,
entropy
so
DSO
H20,1)
H
is)
las less
entropy SO
DSCO
kBris)
kBr,nN)
dissolving,
ASJO
calculating
DS, DH,
DS
·
DGA DHP
in
k5/mol
B6ixn
=
nD61, pro
mas
·
DS in
51mol
A,
reac
DMn
=
EnBHE, prod-EmDti,
reac
->
ntm mean multiply
Sim
=
SnSpron-SmS" reac
by coeff
·
elements in their standard
stale
have
a DCoo
DH
of 0
if DJ is not provided, use
D6=DH-TDS
to calculatein
u
units must
match!
F
ORMATION REACTIONS
form 1 mol
ofa single product
Reactants
are
elements
in standard
state
6 It,
tC
H2119)
3(9)
must
balance but still keep the product
as I move
18.5-Gibbs free energy
&
eg.
constant
D
A
Keq
DO
=
D60+
RTIn(a)
o
f
spartial
pressure
proal
R
=
8.1395/mol D8: given
in"
mol
Q=
coeft
IPpveact
f
2
Pp
react
(
D6" = 0
Gequilibrium
C
notthe same
AG
=
usually NOT
=
o
Keg
is same
thing
as
p
160
=
RT(n((eq)
Ab
Keg
=
1 RT
2
Relationship between K, $6,
as QCK
1630 al 3600
Keq)
-ASK
ACSO
b)
b6°>
Keq<
CI
a
= IC
=
0
xD
=
0
Keg
=
Review:
to
know
predicting
the
sign of IS
1) # moles
of
gas
look reaction
to see ifyoure
getting
move or less moves
ofgas
4Oas
DSCO,
bGas DS
no
change in -
unable
to
guess,
somewhatclose
to
zero
Phase
changes
S-1-g
DHSO
endothermic
DSJO w
more disorder, entropy
S-l-9 DHCO
exothermic
DSCO -
less
disorder,
entropy
NaCIs)
Nalag)
for
salts
=
DSSO
dissolving is associated with
more
disorder,
entropy
- Complexity
simple or
more complex molecules?
use
method
if first
dont
work
16.3-Gibbs
free energy
& relationship
between C, DH,
DS
* get
higher test scores
DG
=
DH
TDS
Surroundings
spontanous
system
D6s
nonspontanous
b
=
0 e equilibrium
surroundings
DG vs D
~
means a standard conditions:
a)
I
mocare concentration for
ag.
Species
b)
a
m for gasses
C)
250 on
298 Kelvin what the
universe doesnt
want
A
what
the universe wants
DG
=
DH-TDS
BH DS
A
B
un
s
I
spontanious
= -
-a
=
A
--
+ -honspontanious:
-a
b
=
+#
AS>
plug
values
I
t -Spontanious
I
a
b
=
into
high
temp
a
=
equation---
spontanious
&
low
Rrup
18.5-Gibbs free energy
&
eg.
constant
D
A
Keq
DO
=
D60+
RTIn(a)
o
f
spartial
pressure
proal
R
=
8.1395/mol D8: given
in"
mol
Q=
coeft
IPpveact
(uett
2
Pp
react
(
D6" = 0
Gequilibrium
C
notthe same
AG
= usually NOT
=
o
Keg
is same
thing
as
p
160
=
RT(n((eq)
Ab
keg
=
~RT
2
Relationship between K, $6,
as QCK
1630 al 3600
Keq)
-ASK
ACSO
b)
b6°>
Keq<
CI
a
=
IC
=
0
xD
=
0
Keg
=
