Chap15-part3

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Unformatted text preview: EXAM‐3:
April
15th
Friday
2011
(not
April
11th)
 Syllabus:
Chapters
15,
16
and
17
 Mastering
Chemistry
Available
for
Ch‐15
 Homogeneous
equilibria
involve
reactants
and
products
in
the
same
phase.
 Examples: 
N2O4(g)

↔
2
NO2(g)

 
 
CO(g)
+
Cl2(g)
↔
COCl2(g)

 
 
CO(g)

+
2
H2(g)

↔

CH3OH(g)
 Heterogeneous
equilibria
involve
at
least
one
reactant
or
product
in
a
different
 phase.
 Examples: 
PbCl2(s)

↔

Pb2+(aq)

+

2
Cl‐(aq)
 
 
H2O(l)

+

CO32‐
(aq)

↔

OH‐(aq)

+

HCO3‐(aq)
 
 
CO2(g)

+

H2(g)

↔

CO(g)

+

H2O(l)
 
 
SnO2(s)

+

2
CO(g)

↔

Sn(s)

+
2
CO2(g)
 2
 Heterogeneous
Equilibrium
 When
a
pure
solid
or
a
pure
liquid
is
involved
in
a
heterogeneous
 equilibrium,
its
concentraJon
is
not
included
in
the
equilibrium‐
 constant
expression.
 The
concentraJon
of
a
pure
liquid
or
pure
solid
remains
constant

 during
the
reacJon.


 Doubling
the
amount
of
a
pure
liquid
or
solid
doubles
the
volume

 but
does
not
change
the
concentraJon.


 Equilibrium
constants
are
concerned
with
changes
in
concentraJons.
 Also,
there
is
the
noJon
of
acJviJes
and
reference
values.

 DecomposiJon
of
calcium
carbonate:

CaCO3(s)

↔

CaO(s)

+

CO2(g)
 The
equilibrium
pressure
of
CO2(g)
is
the
same
even
though
the
amounts
of
 CaCO3(s)
and
CaO(s)
are
much
different.

 Examples:

Homogeneous
Equilibria
 Examples:

Heterogeneous
Equilibria
 The
reacJon
 has
an
equilibrium
constant
Kc
=
7.1
x
10‐4.

What
are
the
equilibrium

 concentraJons
of
Ca2+
and
CrO42‐
in
a
saturated
soluJon
of
CaCrO4?
 The
reacJon
 has
an
equilibrium
constant
Kc
=
7.1
x
10‐4.

What
are
the
equilibrium

 concentraJons
of
Ca2+
and
CrO42‐
in
a
saturated
soluJon
of
CaCrO4?
 1)
Write
 € 2− K c = [Ca 2 + ][CrO4 ] = 7.1x10−4 The
reacJon
 has
an
equilibrium
constant
Kc
=
7.1
x
10‐4.

What
are
the
equilibrium

 concentraJons
of
Ca2+
and
CrO42‐
in
a
saturated
soluJon
of
CaCrO4?
 1)
Write
 2)
Note
that
for
each
CaCrO4
molecule
which
dissolves,
equal
amounts
of

 Ca2+
and
CrO42‐
are
created.

Thus,
 The
reacJon
 has
an
equilibrium
constant
Kc
=
7.1
x
10‐4.

What
are
the
equilibrium

 concentraJons
of
Ca2+
and
CrO42‐
in
a
saturated
soluJon
of
CaCrO4?
 1)
Write
 2)
Note
that
for
each
CaCrO4
molecule
which
dissolves,
equal
amounts
of

 Ca2+
and
CrO42‐
are
created.

Thus,
 3)
Calculate
the
concentraJons.
 CalculaJng
equilibrium
constants
when
all
reactant
and
product

 concentraJons

are
not
known:
 1)
Make
a
table
with
the
known
informaJon.
 2)
For
species
where
iniJal
and
final
concentraJons
are
known,
calculate

 the
change
in
the
concentraJon.
 3)
Use
the
stoichiometry
to
find
the
changes
in
the
other

 concentraJons.
 4)
Calculate
the
remaining
equilibrium
concentraJons.
 5)
Calculate
the
equilibrium
constant.
 Example:

Sample
Exercise
15.9
 InformaJon:
 The
iniJal
concentraJon
of
H2
is
1.00
x
10‐3
M.
 The
iniJal
concentraJon
of
I2
is
2.00
x
10‐3
M.
 The
iniJal
concentraJon
of
HI
is
0.
 The
equilibrium
concentraJon
of
HI
is
1.87
x
10‐3
M.
 1)
Make
a
table
with
the
known
informaJon.
 H2 (M) initial I2 (M) HI (M) 1.00 x 10-3 2.00 x 10-3 0 change Equilib 1.87 x 10-3 2)
For
species
where
iniJal
and
final
concentraJons
are
known,

 calculate
the
change
in
the
concentraJon.
 H2 (M) initial I2 (M) HI (M) 1.00 x 10-3 2.00 x 10-3 0 change 1.87 x 10-3 Equilib 1.87 x 10-3 3)
Use
the
stoichiometry
to
find
the
changes
in
the
other
concentraJons.

 H2 (M) initial I2 (M) HI (M) 1.00 x 10-3 2.00 x 10-3 0 change -0.935 x 10-3 -0.935 x 10-3 1.87 x 10-3 Equilib 1.87 x 10-3 4)
Calculate
the
remaining
equilibrium
concentraJons
 H2 (M) initial I2 (M) HI (M) 1.00 x 10-3 2.00 x 10-3 0 change -0.935 x 10-3 -0.935 x 10-3 1.87 x 10-3 Equilib 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3 5)
Calculate
the
equilibrium
constant.
 H2 (M) initial I2 (M) HI (M) 1.00 x 10-3 2.00 x 10-3 0 change -0.935 x 10-3 -0.935 x 10-3 1.87 x 10-3 Equilib 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3 A
mixture
of
1.374
gm
of
H2
and
70.31
gm
of
Br2
is
heated
in
a
2.00
L
 container
at
700
K.

The
reactants
combine
to

form
HBr:
 
 
H2(g)

+
Br2(g)

↔

2
HBr(g)
 At
equilibrium
the
container
has
0.566
gm
H2.

Calculate
the

 equilibrium
constant
Kc.
 A
mixture
of
1.374
gm
of
H2
and
70.31
gm
of
Br2
is
heated
in
a
2.00
L
 container
at
700
K.

The
reactants
combine
to

form
HBr:
 
 
H2(g)

+
Br2(g)

↔

2
HBr(g)
 At
equilibrium
the
container
has
0.566
gm
H2.

Calculate
the

 equilibrium
constant
Kc.
 Convert
the
masses
to
concentraJons:

 1.374 gm mol initial H 2 : ( )( ) = 0.34 M 2.00 L 2 gm 0.566 gm mol final H 2 : ( )( ) = 0.14 M 2.00 L 2 gm initial Br2 : ( 70.31 gm mol )( ) = 0.22 M 2.00 L 160 gm 1)
Make
a
table
with
the
known
informaJon.
 H2 (M) initial Br2 (M) HBr (M) 0.34 0.22 0 change final 0.14 2)
For
species
where
iniJal
and
final
concentraJons
are
known,

 calculate
the
change
in
the
concentraJon.
 H2 (M) Br2 (M) HBr (M) initial 0.34 0.22 0 change -0.20 final 0.14 3)
Use
the
stoichiometry
to
find
the
changes
in
the
other
concentraJons.

 H2(g)

+
Br2(g)

↔

2
HBr(g)
 H2 (M) Br2 (M) HBr (M) initial 0.34 0.22 0 change -0.20 -0.20 +0.40 final 0.14 4)
Calculate
the
remaining
equilibrium
concentraJons.
 H2 (M) Br2 (M) HBr (M) initial 0.34 0.22 0 change -0.20 -0.20 +0.40 final 0.14 0.02 0.40 5)
Calculate
the
equilibrium
constant.
 H2(g)

+
Br2(g)

↔

2
HBr(g)
 H2 (M) Br2 (M) HBr (M) initial 0.34 0.22 0 change -0.20 -0.20 +0.40 final 0.14 0.02 0.40 ReacMon
QuoMent
(Q)
 The
value
of
the
equilibrium
constant
(Kc
or
Kp)
can
be
used
to
predict

 the
direcJon
in
which
the
reacJon
will
spontaneously
proceed
(lec
or

 right)
to
achieve
equilibrium.
 This
noJon
is
formalized
by
defining
the
reacJon
quoJent
Q
(denoted

 as
Qc
or
Qp)
by
using
the
starJng
concentraJons
or
parJal
pressures
in

 the
expression
for
the
equilibrium
constant.
 1)  If
Q
<
K,
the
starJng
concentraJons
indicate
a
deficit
of
products
and

 the
reacJon
will
spontaneously
proceed
to
the
right.
 2)
If
Q
>
K,
the
starJng
concentraJons
indicate
a
deficit
of
reactants
and

 the
reacJon
will
spontaneously
proceed
to
the
lec.
 3)
If
Q
=
K,
the
starJng
concentraJons
are
equivalent
to

equilibrium

 concentraJons
and
no
further
change
will
occur.
 Example:

Q
=
K
(equilibrium)
 Example:

Q
<
K
(more
products
will
form)
 Example:

Q
>
K
(more
reactants
will
form)
 Consider
the
reacJon
 For
each
of
the
following
starJng
condiJons,
determine
whether
 the
system
is
in
equilibrium,
the
reacJon
will
proceed
to
the
lec,
 or
the
reacJon
will
proceed
to
the
right.
 (a)  P(N2)
=
35
atm,
P(H2)
=
495
atm,
P(NH3)
=
105
atm
 (b)
P(N2)
=
0
atm,
P(H2)
=
595
atm,
P(NH3)
=
35
atm
 (c)
P(N2)
=
202
atm,
P(H2)
=
42
atm,
P(NH3)
=
26
atm
 (a)  P(N2)
=
35
atm,
P(H2)
=
495
atm,
P(NH3)
=
105
atm
 More
products
are
needed
for
equilibrium.
 The
reacJon
will
spontaneously
proceed
to
the
right.
 (b)
P(N2)
=
0
atm,
P(H2)
=
595
atm,
P(NH3)
=
35
atm
 More
reactants
(N2!)
are
needed
for
equilibrium.
 The
reacJon
will
spontaneously
proceed
to
the
lec.
 (c)
P(N2)
=
202
atm,
P(H2)
=
42
atm,
P(NH3)
=
26
atm
 The
starJng
condiJons
are
very
close
to
equilibrium.
 The
reacJon
will
not
proceed
much
further.
 The
reacJon
A2
+
B2
↔
2
AB
has
an
equilibrium
constant
Kc
=
1.5.

Below,

 A
molecules
are
red
and
B
molecules
are
blue.

Are
the
samples
at
equilibrium

 and,
if
not,
in
which
way
will
the
reacJon
proceed?
 The
reacJon
A2
+
B2
↔
2
AB
has
an
equilibrium
constant
Kc
=
1.5.

Below,

 A
molecules
are
red
and
B
molecules
are
blue.

Are
the
samples
at
equilibrium

 and,
if
not,
in
which
way
will
the
reacJon
proceed?
 The
sample
is
not
at
equilibrium
and
 the
reacJon
will
proceed
to
the
lec.


 There
are
too
many
products
and
not

 enough
reactants.
 The
reacJon
A2
+
B2
↔
2
AB
has
an
equilibrium
constant
Kc
=
1.5.

Below,

 A
molecules
are
red
and
B
molecules
are
blue.

Are
the
samples
at
equilibrium

 and,
if
not,
in
which
way
will
the
reacJon
proceed?
 The
sample
is
at
equilibrium.
 The
reacJon
A2
+
B2
↔
2
AB
has
an
equilibrium
constant
Kc
=
1.5.

Below,

 A
molecules
are
red
and
B
molecules
are
blue.

Are
the
samples
at
equilibrium

 and,
if
not,
in
which
way
will
the
reacJon
proceed?
 The
sample
is
not
at
equilibrium
and
 the
reacJon
will
proceed
to
the
right.


 There
are
too
many
reactants
and
not

 enough
products.
 ...
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