Name:___________________________
Thermo
Unit
–
Entropy
of
Physical
and
Chemical
Changes
PHYSICAL
CHANGE
Let’s
look
at
a
process
with
which
we
are
all
familiar:
Ice
melting
in
a
glass
of
water.
Imagine
you
start
with
a
glass
that
has
36
g
of
solid
ice.
The
initial
temperature
of
the
ice
is
0
°C.
If
you
leave
this
glass
in
a
room
that
has
a
constant
temperature
of
25
°C,
the
ice
will
spontaneously
melt
and
you
will
end
up
with
all
liquid
water
at
25
°C.
Here
we
will
calculate
the
entropy
change
for
the
process
of
the
ice
melting
and
increasing
in
temperature
to
25
°C.
1.
First
identify
what
the
system
and
surroundings
are.
(what
are
they
composed
of,
what
is
their
temperature,
etc…)
Often
it
is
helpful
to
make
a
sketch
or
diagram.
Surroundings,
25˚C
System,
36
g
block
of
ice
at
0˚C
2.
Now
identify
what
the
initial
and
final
states
of
the
system
are.
Again
a
diagram,
equation,
or
sketch
can
be
helpful.
You
should
break
this
down
into
two
steps:
ice
melting,
then
water
warming
up.
Initial:
Middle:
Final:
0˚C,
Ice
0˚C,
Water
25˚C,
Water
3.
Now
look
at
the
energy
change
for
this
process.
Does
energy
flow
into
or
out
of
the
system?
If
so,
as
heat
or
work?
Energy
flows
in
as
heat.
Revised
SH
7/17/13
©
LaBrake
&
Vanden
Bout
2013
Name:___________________________
4.
Given
that
ΔfusH°
=
6.02
kJ
mol‐1
and
that
the
heat
capacity
for
water
is
4.184
J
g‐1
°C‐1
.
How
much
heat
flows
into
the
system
for
this
process
(in
KJ)?
There
are
two
steps
in
this
process:
melting
the
ice
and
warming
the
water.
For
the
first
step
we
have
to
convert
grams
into
moles:
36
g
H2O
1
mole
H2O
18
g
H2O
=
2
moles
H2O
Melting
the
Ice:
q1 = ΔH fus x moles
kJ
x 2moles mol
q1 = 12.04kJ
Warming
the
Water:
q2 = Cm ΔTm
J
q2 = (4.184
)(25˚C − 0˚C)(36g) g˚C
q2 = 3765.6J = 3.77kJ
Total
Heat:
qtotal
=
q1+q2
=
12.04
kJ
+
3.77
kJ
=
15.81
kJ
5.
What
is
the
change
in
entropy
for
the
melting
of
the
ice?
What
is
the
change
in
entropy
for
the
melting
of
the
water?
What
is
the
change
in
entropy
for
the
system?
Work:
q
12.04kJ
kJ
J
ΔS = rev =
= 0.0441 = 44.1 ice melting
T
273K
K
K
T
J
298K
J
ΔS = mCs ln( f ) = (36g)(4.184
)ln(
) = 13.12
water war min g Ti
g˚C
273K
K
J
J
J
ΔSsys = 44.1 + 13.12 = 57.22
K
K
K
6.
Did
heat
flow
into
or
out
of
the
surroundings
during
this
change?
Out
of
surroundings
q1 = 6.02
€
€
€
Revised
SH
7/17/13
©
LaBrake
&
Vanden
Bout
2013
€
€
€
Name:___________________________
7.
What
is
the
entropy
change
for
the
surroundings?
Work:
−q
−15.81kJ
kJ
J
ΔSsur = rev =
= −0.053 = −53 T
298K
K
K
8.
What
is
the
entropy
change
for
the
universe
for
this
process?
Work:
J
J
J
ΔSuniverse = ΔSsys + ΔSsur = 57.2 + −53 = 4.2 K
K
K
9.
The
answer
to
part
8
is
extensive
(this
for
the
36
g
of
solid
ice
that
melts
in
this
process).
What
is
the
entropy
change
per
mole
of
solid
ice
that
melts
under
these
conditions?
We
already
know
from
working
question
#4
that
36
g
of
ice
is
2
moles
of
H2O.
J
4.2
ΔSuniverse
K = 2.1 J =
moles
2moles
molK
CHEMICAL
CHANGE
The
equation
ΔS
=
qrev/T
holds
for
reversible
heat
flow.
However,
during
a
chemical
change
the
heat
flow
is
not
reversible.
Therefore,
we
need
another
way
to
determine
the
change
in
entropy
for
the
system
such
that
we
can
predict
the
spontaneity
of
chemical
change.
The
Boltzman
definition
allows
a
method
for
determining
the
standard
molar
entropy
a
compound,
because
S
=
kBlnΩ.
When
there
is
only
one
microstate,
S
=
0.
This
occurs
at
0
K
in
a
perfect
crystal,
thus
there
is
an
absolute
value
for
entropy.
The
standard
molar
entropies
for
elements
and
compounds
have
been
computed
and
are
available
in
tables.
One
can
use
these
values
to
determine
the
change
in
entropy
for
chemical
change!
Remember
that
entropy
is
a
state
function,
and
the
change
in
the
state
function
only
depends
on
the
final
condition
minus
the
initial
condition.
Revised
SH
7/17/13
©
LaBrake
&
Vanden
Bout
2013
€
€
Name:___________________________
1.
Use
the
following
tabulated
data:
S°(H2)g
=
131
J/K
mol
S°(O2)g
=
205
J/
K
mol
S°(H2O)l
=
70
J/K
mol
to
determine
the
change
in
entropy
for
the
combustion
of
1
mole
of
hydrogen.
First
we
need
to
write
out
a
chemical
equation
for
the
combustion
of
hydrogen
gas
in
which
the
stoichiometric
coefficient
for
hydrogen
gas
is
1:
H2
(g)
+
½O2
(g)

H2O
(l)
ΔS = ∑ products − ∑ reac tan ts
J
J
1
J
J
kJ ΔS = [70
] − [(131
) + (205
)] = −163.5
= −0.1635
Kmol
Kmol 2
Kmol
Kmol
Kmol
2.
Use
the
following
tabulated
data:
ΔHf°(H2O)l
=
‐286
kJ/mol
to
determine
the
change
in
entropy
for
the
surroundings
for
this
change
under
standard
conditions.
Work:
−ΔH sys
ΔSsur =
Tsurroudings
kJ
−(−286
)
mol = 0.96 kJ
ΔSsur =
298K
molK
3.
Use
your
answers
from
2
and
3
to
determine
the
change
in
entropy
for
the
universe,
and
predict
if
this
change
is
spontaneous
under
standard
conditions.
Work:
ΔSuniverse = ΔSsys + ΔSsur
kJ
kJ
kJ ΔSuniverse = −0.1635
+ 0.96
= 0.797
molK
molK
molK
Based
only
on
the
entropy
of
the
universe,
this
reaction
is
spontaneous.
€
Revised
SH
7/17/13
©
LaBrake
&
Vanden
Bout
2013