Chap15-Part5

Chap15-Part5 -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2)
Pressure
Changes:

Gas‐Phase
Equilibria
 A
reac'on
is
at
equilibrium
in
the
gas
phase.
 The
volume
is
reduced
while
keeping
the
temperature
constant.
 The
reduc'on
in
volume
causes
an
ini'al
increase
in
pressure.
 The
par'al
pressures
of
all
reactants
and
products
are
all
ini'ally

 increased
by
the
same
factor.
 The
equilibrium
constant
Kp
is
not
changed.
 The
reac'on
proceeds
un'l
the
new
reactant
and
product

 concentra'ons
agree
with
the
value
of
Kp.

 Example: 
A

+

B

↔

C
 The
reac'on
proceeds
to
the
right.
 Example: 
A

↔

B

+

C
 The
reac'on
proceeds
to
the
leE.
 Volume
and
Pressure
Changes:

Gas‐Phase
Equilibria
 In
general,
 decreasing
the
volume
shiEs
the
reac'on
towards
the
 direc'on
that
decreases
the
total
number
of
mols
of
gas;
 and
 increasing
the
volume
shiEs
the
reac'on
towards
the
 direc'on
that
increases
the
total
number
of
mols
of
gas.
 Decreasing
the
volume
shiEs
the
reac'on
to
the
leE.
 The
volume
is
decreased
for
the
following

 gas
phase
reac'ons
ini'ally
at
equilibrium.


 In
which
direc'on
will
the
reac'on
proceed

 to
re‐establish
equilibrium?
 2A

↔

B 
 
 

 A
+
3B

↔

2C
 

 A

+

B

↔

2C
 
 

 The
volume
is
decreased
for
the
following

 gas
phase
reac'ons
ini'ally
at
equilibrium.


 In
which
direc'on
will
the
reac'on
proceed

 to
re‐establish
equilibrium?
 2A

↔

B 
 
 
to
the
right
 A
+
3B

↔

2C
 

 A

+

B

↔

2C
 
 

 The
volume
is
decreased
for
the
following

 gas
phase
reac'ons
ini'ally
at
equilibrium.


 In
which
direc'on
will
the
reac'on
proceed

 to
re‐establish
equilibrium?
 2A

↔

B 
 
 
to
the
right
 A
+
3B

↔

2C
 
to
the
right
 A

+

B

↔

2C
 
 

 The
volume
is
decreased
for
the
following

 gas
phase
reac'ons
ini'ally
at
equilibrium.


 In
which
direc'on
will
the
reac'on
proceed

 to
re‐establish
equilibrium?
 2A

↔

B 
 
 
to
the
right
 A
+
3B

↔

2C
 
to
the
right
 A

+

B

↔

2C
 
no
change 

 The
volume
is
increased
for
the
following

 gas
phase
reac'ons
ini'ally
at
equilibrium.


 In
which
direc'on
will
the
reac'on
proceed

 to
re‐establish
equilibrium?
 2A

+

B
↔

2C 
 

 6A

+
B
↔

6C

+
D 
 

 A

↔

B
+
2C
 
 
 

 
 The
volume
is
increased
for
the
following

 gas
phase
reac'ons
ini'ally
at
equilibrium.


 In
which
direc'on
will
the
reac'on
proceed

 to
re‐establish
equilibrium?
 2A

+

B
↔

2C 
 
to
the
leE
 6A

+
B
↔

6C

+
D 
 

 A

↔

B
+
2C
 
 
 

 
 The
volume
is
increased
for
the
following

 gas
phase
reac'ons
ini'ally
at
equilibrium.


 In
which
direc'on
will
the
reac'on
proceed

 to
re‐establish
equilibrium?
 2A

+

B
↔

2C 
 
to
the
leE
 6A

+
B
↔

6C

+
D 
no
change
 A

↔

B
+
2C
 
 
 
 

 The
volume
is
increased
for
the
following

 gas
phase
reac'ons
ini'ally
at
equilibrium.


 In
which
direc'on
will
the
reac'on
proceed

 to
re‐establish
equilibrium?
 2A

+

B
↔

2C 
 
to
the
leE
 6A

+
B
↔

6C

+
D 
no
change
 A

↔

B
+
2C
 
 
to
the
right 
 

 2)
Pressure
Changes
 Pressure
changes
(via
volume
changes)
affect
equilibria
in
the

 gas
phase
because
the
concentra'ons
of
reactants
and
products

 are
changed.

 In
these
cases,
the
equilibrium
is
disturbed
only
if
the
overall
 stoichiometry
of
the
reac'on
involves
more
(or
less)
reactants
 or
products,
changing
the
value
of
the
reac'on
quo'ent
Q
and
 its
value
rela've
to
Kp.
 Pressure
changes
do
not
usually
appreciably
affect
equilibria

 in
the
liquid
phase
because
the
volume
changes
are
very
small

 and
the
consequent
changes
in
reactant
and
product
concentra'ons
 are
also
very
small
and
negligible.
 3)
Temperature
Changes
 Changes
in
temperature
affect
chemical
equilibria

 in
both
gas
and
liquid
phases
because
the
values
of

 Kp
and
Kc
change
with
temperature.
 A
conceptual
way
to
think
about
this
result
is
to

 treat
the
heat
(energy)
created
by
the
reac'on
as
a

 reactant
or
product.

 [email protected] In
an
endothermic
reac'on,
ΔH
>
0.
 The
system
absorbs
(endo)
heat.


 The
surroundings
become
(temporarily)
cold.
 We
can
think
of
the
reac'on
as 
reactants
+
heat
↔

products
 [ products] K= [ reac tan ts] If
the
temperature
is
increased,
more
heat
is
available,
the
equilibrium
 shiEs
to
the
right,
and
K
increases.
 If
the
temperature
is
decreased,
less
heat
is
available,
the
equilibrium
 € shiEs
to
the
leE
and
K
decreases.
 [email protected] In
an
exothermic
reac'on,
ΔH
<
0.
 The
system
releases
(exo)
heat.


 The
surroundings
become
(temporarily)
hot.
 We
can
think
of
the
reac'on
as 
reactants
↔

products
+
heat
 [ products] K= [ reac tan ts] If
the
temperature
is
increased,
more
heat
is
available,
the
equilibrium
 shiEs
to
the
leE,
and
K
decreases.
 If
the
temperature
is
decreased,
less
heat
is
available,
the
equilibrium
 € shiEs
to
the
right,
and
K
increases.
 Quan'ta'vely,
 In
an
endothermic
reac'on,
ΔH
>
0.


 Increasing
the
temperature
decreases
the
(nega've)
magnitude
of
the

 argument
of
the
exponen'al,
which
leads
to
a
larger
K.
 Example: 
ΔH
=
RT1 
& 
T2
=
2
T1
 Quan'ta'vely,
 In
an
exothermic
reac'on,
ΔH
<
0.


 Increasing
the
temperature
decreases
the
(posi've)
magnitude
of
the

 argument
of
the
exponen'al,
which
leads
to
a
smaller
K.
 Example: 
ΔH
=
‐RT1 
& 
T2
=
2
T1
 This
reac'on
is

 endothermic.
 ΔH

>

0
 An
increase
in
T
 results
in
an
increase
 in
K,
favoring
products.
 A
decrease
in
T
 results
in
a
decrease
 in
K,
favoring
reactants.
 Is
this
reac'on
endothermic
or
exothermic?

 Is
this
reac'on
endothermic
or
exothermic?

 Exothermic,
ΔH
<
0.

K
decreases
with
increasing
T.
 Effect
of
Temperature
and
Pressure
on
the

 Haber
Process
 The
percent
of
NH3
at
equilibrium
in
mixtures
star'ng
with
a
3:1
molar

 ra'o
of
H2
and
N2
varies
with
temperature
and
pressure.
 Larger
amounts
of
NH3
are
found
at
lower
temperatures
and
higher
 pressures.
 The
reac'on
is
A2

+

B
↔

A
+
AB.

The
systems
are
in
equilibrium.
 A
atoms
are
red
and
B
atoms
are
blue.

(i)

T
=
300
K,

(ii)

T
=
500
K.
 Is
the
reac'on
endothermic
or
exothermic?
 The
reac'on
is
A2

+

B

↔

A
+
AB.

The
systems
are
in
equilibrium.
 A
atoms
are
red
and
B
atoms
are
blue.

(i)

T
=
300
K,

(ii)

T
=
500
K.
 Is
the
reac'on
endothermic
or
exothermic?
 The
equilibrium
constant
decreases
with
increasing
temperature.
 The
reac'on
is
exothermic.
 Effect
of
Catalysts
 A
catalyst
increases
the
rate
of

 approach
to
equilibrium.
 A
catalyst
does
not
affect
the
 final
equilibrium
state.
 Example:

A

↔

B


 Δ E
 Without
catalyst:

kf1
&

kr1
 With
catalyst:

kf2
&
kr2
 For
the
following
reac'on,
ΔH
>
0:
 6
CO2(g)

+
6
H2O(l)

↔

C6H12O6(s)

+

6
O2(g)
 How
is
the
equilibrium
yield
of
C6H12O6
affected
by
 (a)
increasing
P(CO2)?
 

 (b)
increasing
temperature? 
 (c)
removing
CO2?
 
 

 (d)
decreasing
pressure? 
 

 (e)
adding
a
catalyst? 
 

 
 

 For
the
following
reac'on,
ΔH
>
0:
 6
CO2(g)

+
6
H2O(l)

↔

C6H12O6(s)

+

6
O2(g)
 How
is
the
equilibrium
yield
of
C6H12O6
affected
by
 (a)
increasing
P(CO2)?
 
higher
yield
 (b)
increasing
temperature? 
 (c)
removing
CO2?
 
 

 (d)
decreasing
pressure? 
 

 (e)
adding
a
catalyst? 
 

 
 

 For
the
following
reac'on,
ΔH
>
0:
 6
CO2(g)

+
6
H2O(l)

↔

C6H12O6(s)

+

6
O2(g)
 How
is
the
equilibrium
yield
of
C6H12O6
affected
by
 (a)
increasing
P(CO2)?
 
higher
yield
 (b)
increasing
temperature? 
 (c)
removing
CO2?
 
 

 (d)
decreasing
pressure? 
 

 (e)
adding
a
catalyst? 
 

 
 
higher
yield
 For
the
following
reac'on,
ΔH
>
0:
 6
CO2(g)

+
6
H2O(l)

↔

C6H12O6(s)

+

6
O2(g)
 How
is
the
equilibrium
yield
of
C6H12O6
affected
by
 (a)
increasing
P(CO2)?
 
higher
yield
 (b)
increasing
temperature? 
 (c)
removing
CO2?
 
 
higher
yield
 
lower
yield
 (d)
decreasing
pressure?
 
 

 (e)
adding
a
catalyst? 
 
 

 For
the
following
reac'on,
ΔH
>
0:
 6
CO2(g)

+
6
H2O(l)

↔

C6H12O6(s)

+

6
O2(g)
 How
is
the
equilibrium
yield
of
C6H12O6
affected
by
 (a)
increasing
P(CO2)?
 
higher
yield
 (b)
increasing
temperature? 
 (c)
removing
CO2?
 
 
higher
yield
 
lower
yield
 (d)
decreasing
pressure? 
 
no
change
 (e)
adding
a
catalyst? 
 

 
 For
the
following
reac'on,
ΔH
>
0:
 6
CO2(g)

+
6
H2O(l)

↔

C6H12O6(s)

+

6
O2(g)
 How
is
the
equilibrium
yield
of
C6H12O6
affected
by
 (a)
increasing
P(CO2)?
 
higher
yield
 (b)
increasing
temperature? 
 (c)
removing
CO2?
 
 
higher
yield
 
lower
yield
 (d)
decreasing
pressure? 
 
no
change
 (e)
adding
a
catalyst? 
 
no
change
 
 Example:

Dimeriza'on
 2A

↔

B
 K 
equilibrium
constant
 Aequil 
monomer
concentra'on
at
equilibrium
 Bequil 
dimer
concentra'on
at
equilibrium
 Aini'al 
ini'al
monomer
concentra'on
 Bini'al 
ini'al
dimer
concentra'on

 C 
total
subunit
concentra'on
(constant)
 Dimeriza'on
 Eliminate
Bequil:
 Which
root
to
choose?

When
C
=0,
Aequil
must
be
zero.
 
Dimeriza'on:

Sample
Calcula'on
 K
=
3 
 
Aini'al
=
0.60
M
 Note
that
 
Binital
=
0.05
M
 ...
View Full Document

This note was uploaded on 09/13/2011 for the course CHEM 102 taught by Professor Todd during the Spring '08 term at UNC.

Ask a homework question - tutors are online