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Unformatted text preview: CH201
SI:
CHAPTER
4
PRACTICE
TEST
SOLUTIONS
 For
the
following
reaction
at
298K:
 2NO2(g)
 
N2O4(g)
 1.
Determine
the
standard
enthalpy
and
standard
entropy
for
the
reaction
from
 tabulated
values.
 ΔH0

=
ΔH
0f
(N2O4(g))
–
2*
ΔH
0f
(NO2(g))
=
9.66
kJ
–
2(33.84
kJ)
=
­58.02
kJ
 ΔS0
=
S0(N2O4(g))
–
2*
S0(NO2(g))
=
304.3
J*K­1
–
2(240.45
J*K­1)
=
­176.6
J*K­1
 ΔH0
=
 ‐58.02
kJ
 ΔS
0
=
‐176.6
J*K‐1
 2.
Use
standard
enthalpy
of
reaction
and
standard
entropy
of
reaction
to
determine
 the
standard
free
energy
of
the
reaction.

Find
Keq
for
the
reaction.
 ΔG0
=
ΔH0
­
TΔS0

=
­58.02
kJ
–
298
K(­0.1766
kJ*K­1)
=


­5.39
kJ
 Keq
=
e^(­ΔG0/RT)
=
e­(­5.39
kJ)/(0.0083
kJ*mol­1*K­1
*
298K)
=
8.84
 ΔG
0
=
‐5.39
kJ
 Keq
=
8.84
 3.
Circle
All
correct
statements
about
this
reaction:
 
 
 
 a.
The
reaction
is
exothermic.
 
 The
sign
of
ΔH0
is
negative.
Thus,
the
reaction
is
exothermic
 b.
The
free
energy
of
the
products
is
greater
than
the
reactants.
 ΔG0
is
defined
as
the
reaction
above
with
initial
amounts
1.0
atm
NO2
 and
1.0
atm
N2O4.

This
effectively
takes
concentration
out
of
 consideration
allowing
direct
comparison
of
the
free
energies
between
 products
and
reactants.

In
this
case
ΔG0
is
negative,
so
the
free
energy
of
 the
products
is
lower
than
the
reactants.
 
 c.
As
temperature
increases
the
reaction
becomes
less
extensive.
 This
is
assuming
ΔH0
and
ΔS0
are
independent
of
temperature.

From
the
 equation
ΔG0
=
ΔH0
­
TΔS0
it
should
be
apparent
that
ΔG0
increases
with
 increasing
temperature
(i.e.
becomes
more
positive).

The
more
positive
 ΔG0,
the
less
extensive
the
reaction.
 
 d.
As
temperature
decreases
Keq
decreases.
 Using
the
same
reasoning
as
c,
as
temperature
decreases
ΔG0
becomes
 more
negative.

As
ΔG0
becomes
more
negative,
Keq
increases.
 
 
 
 
 
 4.
Determine
Q
for
this
reaction
if
PNO2
=
0.24
atm
and
PN2O4
=
1.73
atm.
 Q
=
(PN2O4/1
atm)/(PNO2/1
atm)2
=
(1.73
atm/1
atm)/(0.24
atm/1
atm)2

=
30.03
 
 Q
=
30.03
 5.
Determine
ΔG
for
the
reaction
from
#4.
 ΔG
=
RT
ln(Q/K)
=
(0.0083
kJ*mol­1*K­1)(298
K)
ln
(30.03/8.84)
=
3.02
kJ
 
 ΔG
=
3.02
kJ
 6.
Circle
the
correct
answer.

This
reaction
is
spontaneous
to
the:
 
 LEFT
 
 /
 RIGHT

 /
 NEITHER
 e.
At
298K
equilibrium
favors
products.
 
 See
b.

A
negative
ΔG0
tells
us
equilibrium
favors
products.
 f.
All
of
the
above
statements
are
incorrect.
 
 I’m
not
this
cruel.
 Note
that
not
only
does
the
sign
of
ΔG
tell
you
which
direction
the
reaction
is
 spontaneous,
but
the
relationship
between
Q
and
K
does
as
well.
 
 7.
See
previously
emailed
practice
test.
 a.
Note
the
reaction
must
be
spontaneous
to
the
left.

This
allows
c
to
be
 eliminated.

The
difference
between
a
and
b
is
where
equilibrium
lays.

From
 ΔG0
it
is
obvious
equilibrium
lays
towards
the
products
which
is
consistent
with
 a.

Realize
the
free
energy
of
the
products
in
a
is
lower
than
the
free
energy
of
 the
reactants
as
discussed
in
question
3.
 8.
Determine
the
equilibrium
pressures
of
NO2
and
N2O4
at
equilibrium
for
the
 reaction
from
#4.
(This
type
of
problem
will
be
covered
in
Chapter
5,
but
I’ve
 included
it
here
for
completeness.

After
we
cover
Chapter
5
material
look
back
over
 this
problem
and
see
how
it
fits
into
the
‘scheme’
of
this
worksheet)
 
 
 
 I
will
go
over
this
problem
after
we
discuss
chapter
5.

For
now
you
should
be
able
to
 determine
which
direction
the
reaction
will
proceed
(i.e.
whether
products
or
 reactants
are
consumed
and
whether
products
or
reactants
are
produce
to
reach
 equilibrium).
 
 PNO2
=
__________
atm
 PN2O4
=
__________
atm
 ...
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