ch3sol - CH201
SI:
CHAPTER
3
PRACTICE
TEST


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Unformatted text preview: CH201
SI:
CHAPTER
3
PRACTICE
TEST
 Answer
the
following
questions
for
butane
gas
(C4H10(g))
unless
otherwise
noted:
 1.

Write
the
formation
reaction
for
butane.

Determine
the
standard
enthalpy
of
 formation.
 4C(s)
+
5H2(g)

C4H10(g)
 ΔH0f
=
ΔH0f(C4H10(g))
–
[4ΔH0f(C(s))
+
5ΔH0f(H2(g))
=
­124.73
kJ
–
4(0
kJ)
–
5(0
kJ)
=
 ΔH0f
=
­124.73
kJ/mol
 2.
Write
the
combustion
reaction
for
butane.

Determine
the
standard
heat
of
 combustion
from
standard
enthalpies
of
formation.
 C4H10(g)
+
(13/2)O2(g)

4CO2(g)
+
5H2O(l)
 ΔH0comb
=
[4ΔH0f(CO2(g))
+
5ΔH0f(H2O(l))]
–
[ΔH0f(C4H10(g))
+
(13/2)ΔH0f(O2(g))]
 =
4(­393.51
kJ)
+
5(­285.83
kJ)
–
(­124.73
kJ)
–
(13/2)(0
kJ)
=
 ΔH0COMB
=

­2990.46kJ/mol
 3.

Determine
the
heat
of
formation
of
n‐butanol
(C4H10O)
given
the
heat
of
 combustion
of
butanol
is
‐3213.23
kJ/mol.
 C4H10O(g)
+
6O2(g)

4CO2(g)
+
5H2O(l)
 ΔH0comb

=

[4ΔH0f(CO2(g))
+
5ΔH0f(H2O(l))]
–
[ΔH0f(C4H10O(g))
+
6ΔH0f(O2(g))]
 ΔH0f(C4H10O(g))
=
4ΔH0f(CO2(g))
+
5ΔH0f(H2O(l))
+
6ΔH0f(O2(g))
­
ΔH0comb
 =
4(­393.51
kJ)
+
5(­285.83
kJ)
+
6(0
kJ)
–
(­3213.23)
 ΔH0f(butanol)
=
210.04
kJ/mol
 4.
Write
the
atomization
reaction
for
butane
gas.

Use
standard
enthalpies
of
 formation
determine
the
heat
of
atomization
for
butane
gas.

Estimate
the
average
 C‐C
bond
energy
in
butane
gas.
 C4H10(g)

4C(g)
+
10H(g)
 ΔH0atom
=
4ΔH0f(C(g))
+
10ΔH0f(H(g))
‐
ΔH0f(C4H10(g))
 =
4(716.68
kJ)
+
10(217.97
kJ)
–
(­124.73
kJ)
=
5171.15
kJ
 Butane
contains
3
C­C
bonds
and
10
C­H
bonds,
so….
 ΔH0atom
=
3DC‐C
+
10DH‐H
 DC‐C
=
(ΔH0atom
–
10DH‐H)/3
=
(5171.15
kJ
–
10(413
kJ))/3
=
347.05
kJ
 
 
 ΔH0atom
=
5171
kJ/mol
 DC­C
=
347.05
kJ/mol
 5.
Use
bond
energies
to
estimate
the
heat
of
combustion
of
liquid
butane
at
298K.
 To
use
bond
dissociation
to
estimation
we
need
everything
to
be
is
gas
phase
period.

 So
choose
a
reaction
that
‘looks’
like
a
combustion
reaction
but
keep
all
species
in
the
 gas
phase:
 C4H10(g)
+
(13/2)O2(g)

4CO2(g)
+
5H2O(g)
 Draw
Lewis
structures
of
all
species
to
determine
the
number
and
types
of
bonds
being
 broken
and
formed.
 This
is
a
non­standard
reaction
so
we
are
finding
ΔH.
 ΔH
=
ΣnDBonds
Broken
­
ΣnDBonds
Formed
=
3DC­C
+
10DH­H
+
(13/2)DO=O

­
8DC=O
–
10DO­H

 =
3(347
kJ)
+
10(413
kJ)
+
(13/2)(445
kJ)
–
8(799
kJ)
–
10(463
kJ)
=
­2633.5
kJ
 Next,
simply
use
Hess’
law
and
the
careful
selection
of
reactions
to
convert
our
reaction
 to
a
standard
combustion
reaction
for
liquid
butane
at
298K.
 C4H10(g)
+
(13/2)O2(g)

4CO2(g)
+
5H2O(g)
 C4H10(l)

C4H10(g)
 
 
 
 ΔH
=
­2633.5
kJ
 ΔH
=
ΔHf(C4H10(g))
­
ΔHf(C4H10(l))
 
=
­124.73
kJ
–
(­147.6
kJ)
=
22.87
kJ
 5H2O(g)

5H2O(l)
 
 
 
 
 
 
 
 ΔH
=
5[ΔHf(H2O(l))
­
ΔHf(H2O(g))]
 =
5[­285.83
–(­241.82)]
=
­220.05
kJ
 
 
 
 C4H10(l)
+
(13/2)O2(g)

4CO2(g)
+
5H2O(l)
 

 
 
 
 
 
 ΔH0comb
=
­2633.5
kJ
+
22.87
kJ
 +
(­220.05
kJ)
=
­2830.68
kJ/mol
 ΔH0comb
=
­2830.68
kJ/mol
 Note
these
problems
are
very
similar
(some
identical)
tothose
we
did
in
S.I.

See
 CH201SIJuly6
for
further
explanation.
 Good
luck
on
the
exam!
 ...
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This note was uploaded on 01/27/2011 for the course CHEM 201 taught by Professor Wilson during the Spring '10 term at N.C. Central.

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