D_-_102_F11_Notes

D_-_102_F11_Notes - D)
Equilibrium


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Unformatted text preview: D)
Equilibrium
 Petrucci
et
al.,
10th
Edition:
15.1‐15.7,
18.1‐18.5,
+
part
of
19.7
 2.0 1.0 0.0 Mols of product & reactant Mols of product & reactant Mols of product & reactant 
 Dynamic
Equilibrium
 • Continuous
movement
of
particles
from
one
phase
to
another
 • Equilibrium
can
exist
for
both
physical
and
chemical
processes
 
 
 Chemical
Equilibrium
Constant
Expression
 Methanol
is
produced
from
syngas
according
to
the
following
reversible
reaction:
 
 CO
(g)
+
2
H2
(g)
→
CH3OH
(g)
 
 CH3OH
(g)
→
CO
(g)
+
2
H2
(g)
 When
chemical
equilibrium
is
reached,
the
concentrations
of
all
reactants
and
products
 remain
constant
 Chemical
equilibrium
is
a
dynamic
process:

 • Rate
of
forward
reaction
=
Rate
of
reverse
reaction
 • Reaction
does
not
stop
 
 
 CO
(g)
+
2
H2
(g)
↔
CH3OH
(g)
 2.0 1.0 0.0 Time Experiment 1 
 
 
 2.0 1.0 0.0 Time Experiment 2 Time Experiment 3 =
mol
CO
 =
mol
H2
 =
mol
CH3OH
 
 
 
[similar
to
Petrucci
et
al.,
Figure
15.3]
 ChE102
Fall
2011
Class
Notes
‐
Equilibrium
 
 
 
 
 
 
 
 D ‐1 
 Exp
1
 Exp
2
 Exp
3
 € [similar
to
Petrucci
et
al.,
Table
15.1]
 
 
 CO
(g)
 H2
(g)
 Initial
(mol)
 1.000
 1.000
 Equilibrium
(mol)
 0.911
 0.822
 Equilibrium
conc
(mol/L)
 
 
 CH3OH
(g)
 0.000
 0.0892
 
 Initial
(mol)
 Equilibrium
(mol)
 Equilibrium
conc
(mol/L)
 0.000
 0.753
 1.000
 1.506
 1.000
 0.247
 
 
 
 Initial
(mol)
 Equilibrium
(mol)
 Equilibrium
conc
(mol/L)
 1.000
 1.380
 1.000
 1.760
 1.000
 0.620
 
 
 
 
 
 For
the
reaction
(in
a
10L
vessel):


CO
(g)
+
2
H2
(g)
↔
CH3OH(g)

 
 the
equilibrium
constant
is
defined
as:

 
 
 
 Using
the
values
given
in
the
Table
above:

 
 (0.00892) (0.0247) (0.062) L2 
 Kc = = = = 14.5 (0.0911)(0.0822) 2 (0.0753)(0.151) 2 (0.138)(0.176) 2 mol 2 
 
 The
value
of
the
equilibrium
constant
is
constant
at
____________________________________
 
 
 Chemical
Equilibrium:
Reversible
&
Spontaneous
 
 CO
(g)
+
2
H2
(g)
↔
CH3OH
(g)
 
 What
happens
if
we
add
more
CO
(g)
after
chemical
equilibrium
is
reached?
 • [CO]
>
[CO]eq

 • Rate
of
forward
reaction
>
Rate
of
reverse
reaction
 • More
CH3OH
is
formed
→
[CH3OH]
increases
→
rate
of
reverse
reaction
increases
 • CO
and
H2
are
consumed
→
[CO]
and
[H2]
decrease
→
rate
of
forward
reaction
 decreases
 • Finally,
a
new
equilibrium
is
reached
when
the
rate
of
the
forward
reaction
equals
the
 rate
of
the
reverse
reaction

 ChE102
Fall
2011
Class
Notes
‐
Equilibrium
 
 
 
 
 
 
 
 D ‐2 
 Magnitude
of
the
Equilibrium
Constant
(Kc)
 For
the
general
reversible
chemical
reaction:
 
 a
A
+
b
B
↔
c
C
+
d
D

 The
equilibrium
constant
is
defined
by
the
equation:
 
 
 Kc
is
constant
for
a
given
temperature
 Kc
is
independent
of
the
initial
concentration
of
reactants
and
products
(path
independent)
 Kc
has
units
of
(concentration)
d+c–a–b

 • Large
Kc:
products
are
present
in
higher
concentration
than
reactants
 • Small
Kc:
products
are
present
in
lower
concentration
than
reactants
 
 
 2SO2(g)
+
O2(g)
↔
2SO3(g)
at
1000K
 
 
 
 [see
Petrucci
et
al.,
Figure
15.6]
 
 
 
 
 
 
 Question
D1
 One
mole
of
A
and
two
moles
of
B
are
placed
in
a
container.

At
equilibrium,
the
density
of
 the
gas
mixture
is
0.312
g/L
at
0.2
atm
and
500K.

What
is
Kc?
 Data:

A(g)
+
B(g)
↔
C(g)
 
 Molar
Mass
A
=
80
g/mol,
B
=
40
g/mol,
C
=
120
g/mol
 ChE102
Fall
2011
Class
Notes
‐
Equilibrium
 
 
 
 
 
 
 
 D ‐3 
 Relationships
Involving
Equilibrium
Constants
 Consider
the
reaction
2
SO2(g)
+
O2
(g)
↔
2
SO3
(g)
 
 Kc = € € [ 
 [ 
 Use
the
ideal
has
law
to
calculate
[SO3],
[SO2]
and
[O2]:
 
 P P 
 [ SO3 ] = = ; [ SO2 ] = SO 2 ; [O2 ] = O 2 
 RT RT 
 Combine
with
the
expression
for
Kc:
 
 
 K= = x RT € € [ c 
 
 Equilibrium
Constant
(Kp)
 For
the
reaction
2
SO2(g)
+
O2
(g)
↔
2
SO3
(g),
the
equilibrium
constant
Kp
is:
 
 
 
 
 Therefore,
equilibrium
constants
Kc
and
Kp
for
this
reaction
are
related
by
the
 expression:
 
 Kc = ⇒ Kp = 
 
 
 General
Relationship
between
Kp
and
Kc
 For
the
reaction:
a
A
(g)
+
b
B
(g)
↔
c
C
(g)
+
d
D
(g)
 
 n P n P n P n P 
 
 
 
 [ A] = A = A ; [ B] = B = B ; [C ] = C = C ; [ D] = D = D ; 
 V RT V RT V RT V RT 
 Kc = € 
 
 





 
 € ChE102
Fall
2011
Class
Notes
‐
Equilibrium
 
 
 
 
 
 
 
 D ‐4 
 Reactions
Involving
Pure
Solids
&
Liquids
 For
the
reaction:

CaCO3(s)
↔
CaO
(s)
+
CO2
(g)
 
 
 
 Kc
=
 
 
 and
Kp
=
 
 Concentrations
of
pure
solids
and
liquids
are
not
included
in
the
definition
of
the
 equilibrium
constants
Kc
and
Kp
because
these
concentrations
are
constant
 
 For
the
reaction
Cu2+(aq)
+
Zn
(s)
↔
Cu
(s)
+
Zn2+
(aq)
 
 KC = 
 € 
 
 CaCO3(s)
↔
CaO
(s)
+
CO2
(g)
 
 Introducing
more
solid
(CaO
and
CaCO3)
from
(a)
to
(b)
does
not
affect
the
reaction
 equilibrium
→
the
partial
pressure
of
CO2
remains
the
same
 
 [see
Petrucci
et
al.,
Figure
15.4]
 
 
 Question
D2
 Consider
the
reversible
reaction
N2O4
(g)
↔
2
NO2
(g),
with
Kp
=
0.15
atm
at
25˚C.

 Calculate
Kc
in
units
of
moles
per
litre.
 
 
 Some
Properties
of
Kc
and
Kp
 If
a
reaction
is
multiplied
by
a
constant,
Kc
must
be
raised
to
a
power
equal
to
that
 constant
 [ B] 
 A ↔ B : Kc = [ A] 
 
 [ B ]2 2 A ↔ 2B : K c = 
 [ A ]2 
 
 Kc
for
a
reaction
and
its
reverse
are
reciprocals
of
each
other
 
 [ B] A " B : Kc = 
€ [ A] 
 [ A] 
 B " A : Kc = 
 [ B] ChE102
Fall
2011
Class
Notes
‐
Equilibrium
 
 ! 
 
 
 
 
 
 D ‐5 
 When
two
reactions
or
more
are
added,
multiply
their
Kc
values
to
obtain
the
Kc
for
the
 overall
reaction.


 
 [ B] A ↔ B : K c1 = 
 [ A] 
 [C ] 
 B ↔ C : Kc2 = [ B] 
 
 [C ] A ↔ C : Kc3 = = K c1K c 2 
 [ A] 
 Question
D3
 The
equilibrium
constants
for
the
following
reactions
have
been
measured
at
823
K:

 
 € CoO
(s)
+
H2
(g)
↔
Co
(s)
+
H2O
(g)

 KP1
=
67

 
 CoO
(s)
+
CO
(g)
↔
Co
(s)
+
CO2
(g)

 KP2
=
490
 From
this
data,
calculate
the
equilibrium
constant
(Kp)
of
the
following
reaction
at
823
K:
 
 CO2
(g)
+
H2
(g)
↔
CO
(g)
+
H2O
(g)
 
 
 
 The
Reaction
Quotient:
Predicting
the
Direction
of
Change
 Consider
the
general
reversiblechemical
reaction
(NOT
at
chemical
equilibrium):
 
 a
A
+
b
B
↔
c
C
+
d
D

 
 
 with
initial
concentrations
[A]init,
[B]init,
[C]init,
[D]init

 
 The
reaction
quotients
for
this
reaction
are
defined
as:
 
 PC c PD d [C ]c [ D]d init init init Qc = ; Qp = init b a 
 a b 
 [ A]init [ B]init PA init PB init Qc
and
Qp
are
analogous
to
Kc
and
Kp
BUT
Qc
and
Qp
are
calculated
for
non­equilibrium
 concentrations
 € 
 Qc
and
Direction
of
Change
 If
Qc
=
Kc:
 • Chemical
equilibrium
 • Product
and
reactant
concentrations
do
not
change
 If
Qc
<
Kc:
 • Reactants
are
in
excess
 • Reactant
concentrations
will
decrease
and
product
concentrations
will
increase
 If
Qc
>
Kc:
 • Products
are
in
excess
 • Product
concentrations
will
decrease
and
reactant
concentrations
will
increase ChE102
Fall
2011
Class
Notes
‐
Equilibrium
 
 
 
 
 
 
 
 D ‐6 
 [see
Petrucci
et
al.,
Figure
15.5]
 
 
 Question
D4
 At
elevated
temperatures,
NO2
reacts
with
CO
according
to
the
reversible
chemical
 equation:
 
 
 NO2
(g)
+
CO
(g)
↔
NO
(g)
+
CO2
(g)
 NO2
has
a
brownish
colour.
The
other
gases
taking
place
in
the
reaction
are
colourless.

 When
a
gas
mixture
is
prepared
at
500
K,
in
which
3.4
atm
is
the
initial
partial
pressure
of
 both
NO2
and
CO,
and
1.4
atm
is
the
partial
pressure
of
both
NO
and
CO2,
the
brown
colour
 of
the
mixture
is
observed
to
fade
as
the
reaction
proceeds
towards
equilibrium.

Give
a
 condition
that
must
be
satisfied
by
the
equilibrium
constant
Kp
under
these
reaction
 conditions.


 
 
 
 Question
D5
 Consider
the
equilibrium,
with
Kc
=
2
×
1037:
 
 
 Zn
(s)
+
Cu2+
(aq)
↔
Cu
(s)
+
Zn2+
(aq)
 i) What
will
happen
to
Cu2+
and
Zn2+
ions
in
a
solution
that
has
a
0.1M
concentration
of
 each
species
but
contains
no
Zn
(s)
or
Cu
(s)?

 ii) What
will
happen
to
Cu2+
and
Zn2+
ions
in
a
solution
that
has
a
0.1M
concentration
of
 each
species
if
only
Zn
(s)
is
added
to
the
initial
solution
from
(a)?
 iii) What
will
happen
to
Cu2+
and
Zn2+
ions
in
a
solution
that
has
a
0.1M
concentration
of
 each
species
if
only
Cu
(s)
is
added
to
the
initial
solution
from
(a)?
 iv) What
will
happen
to
Cu2+
and
Zn2+
ions
in
a
solution
that
has
a
0.1M
concentration
of
 each
species
if
both
Zn
(s)
and
Cu
(s)
are
added
to
the
initial
solution
from
(a)?
 
 Altering
Equilibrium
Conditions
 Le
Chatelier’s
principle:
 


If
a
system
in
equilibrium
is
disturbed
by
changing:
 • concentration
of
reactants
and/or
products
 • pressure,
or

 • temperature,

 


it
will
react
to
minimize
the
effect
of
the
disturbance

 ChE102
Fall
2011
Class
Notes
‐
Equilibrium
 
 
 
 
 
 
 
 D ‐7 
 Effect
of
Changes
in
Reagent
Concentration
 Increasing
the
concentration
(or
pressure)
of
reactants:
 • Qc
<
Kc

 • Rate
of
forward
reaction





rate
of
reverse
reaction
 • More
_______________
are
formed
until
Qc
=
Kc
(new
equilibrium)
 Increasing
the
concentration
(or
pressure)
of
products:
 • Qc
>
Kc

 • Rate
of
forward
reaction





rate
of
reverse
reaction
 • More
_______________
are
formed
until
Qc
=
Kc
(new
equilibrium)
 
 
 Physical
Equilibrium
 Consider
the
dissolution
equilibrium:
 
 BaSO4
(s)
↔
Ba2+(aq)
+
SO42‐(aq)
 
 What
happens
if
Na2SO4
is
added
to
this
system?

 
 
 
 
 Effects
of
Change
in
Pressure
or
Volume
 Add
or
remove
a
gaseous
reactant
or
product:
 • same
effect
as
changing
the
concentration
of
a
reactant
or
product
 Change
the
pressure
by
changing
the
volume
of
the
system:

 • this
action
may
cause
a
change
in
chemical
equilibrium,
depending
on
the
type
of
 reaction
under
investigation.
 ChE102
Fall
2011
Class
Notes
‐
Equilibrium
 
 
 
 
 
 
 
 D ‐8 
 Change
the
Pressure
by
Changing
the
Volume
of
the
System
 2
SO2
(g)
+
O2
(g)
↔
2
SO3
(g)
 
 [Petrucci
et
al.,
Figure
15.7]
 
 2 n SO 3 2 2 [SO3 ] = V0 = n SO3 xV Initially:


 K c = 0
 2 2 2 [SO2 ] [O2 ] n SO2 nO2 n SO2 nO2 V0 V0 
 Volume
decreases
from
V0
=10
L
to
V1
=
1
L
 
 € 
 
 
 Since
Qc
<
Kc,
[SO3]
increases
and
[SO2]
and
[O2]
decrease
until
Qc
=
Kc
(new
equilibrium)
 
 
 
 
 Increases
10
times
to
re‐establish
the
chemical
equilibrium

 
 
 
 (a)

 2 n SO3 2 n SO2 nO2 2 (0.68) = = 28 
 2 0.32) (0.16) ( 
 (b)
 2 2 n SO3 2 n SO2 nO2 (0.83) = = 280 
 2 0.17) (0.085) ( 
 
 Additional
Notes
 € € Reducing
the
volume
of
a
reactive
mixture
(or
increasing
the
pressure):
 • chemical
equilibrium
shifts
towards
the
side
with
_______________
gas
molecules
 Adding
an
inert
gas
at
constant
pressure:

 • same
effect
as
changing
the
volume
by
changing
the
pressure
 Condensing
of
phases
(liquids
and
solids)

 • generally
________________________
because
condensed
phases
are
not
significantly
 compressible

 ChE102
Fall
2011
Class
Notes
‐
Equilibrium
 
 
 
 
 
 
 
 D ‐9 
 Temperature
Effects
 When
temperature
is
varied,
Kp
and
Kc
generally
change
 For
exothermic
reactions
(heat
is
released,
ΔH
<
0):
 • • Kp
and
Kc
______________
with
increasing
temperature
 equilibrium
shifts
towards
the
reactants’
side
 For
endothermic
reactions
(heat
is
absorbed,
ΔH
>
0):
 • • Kp
and
Kc
______________
with
increasing
temperature
 equilibrium
shifts
towards
the
products’
side

 
 
 For
Example
 CH4
(g)
+
H2O
(g)
↔
3
H2
(g)
+
CO
(g)
 
 Kp
=
2.679
atm2
@
649°C

 Kp
=
6
755
atm2
@
982°C

 ΔH
>
0

 N2
(g)
+
3H2
(g)
↔
2
NH3
(g)
 
 
 Kp
=
0.0902
atm‐2
@
227°C

 Kp
=
9.71x10‐8
atm‐2
@
827°C

 ΔH
<
0

 N2O4
(g)
↔
2
NO2
(g)
 ∆H
>
0
 
 
 
 
 [Figure
from
Petrucci
et
al.,
Figure
15.8]
 
 Question
D6
 A
vessel
is
filled
at
800K
with
pure
UF6
to
1.2
atm.

UF6
dissociates
to
form
UF4
and
F2.


At
 equilibrium,
the
pressure
in
the
vessel
is
1.23
atm.

What
is
the
value
of
Kc
(in
mol/L)?
 
 
 The
Van’t
Hoff
Equation
(19.7,
p850)
 The
van’t
Hoff
equation
can
be
used
to
estimate
Kc
or
Kp
at
a
given
temperature,
provided
 that
Kc
or
Kp
are
known
at
some
other
temperature:

 
 K ΔH rxn 1 1 ln 2 = − 
 R T1 T2 K1 
 
 € • • • T1
and
T2
are
absolute
temperatures
(K
or
°R)
 ∆Hrxn
is
the
enthalpy
of
reaction
 K1
and
K2
can
be
in
terms
of
concentration
or
pressure
 ChE102
Fall
2011
Class
Notes
‐
Equilibrium
 
 
 
 
 
 
 
 D‐10
 Question
D7
 The
equilibrium
constants
Kp
for
the
reaction

 
 2
SO2
(g)
+
O2
(g)
↔
2
SO3
(g)
 are
tabulated
at
several
temperatures:
 
 T
(K)
 Kp
(atm‐1)
 
 800
 910

 
 850
 170
 
 900
 42
 
 950
 10
 At
which
temperature
will
Kp
=
106
atm‐1
?
 
 
 Solubility
Product
Constant
(18)
 The
equilibrium
between
a
solid
ionic
solute
and
a
liquid
solvent
is
treated
in
a
similar
way
 to
the
one
used
to
study
chemical
equilibrium:
 
 
 AgI
(s)
↔
Ag+
(aq)
+
I–
(aq)
 
 
 Ksp
=

 
 
 
 
 
 
 
 
 (see
Petrucci
et
al.,
Table
18.1)
 
 
 Relationship
between
Solubility
Product
Constant
(Ksp)
and
Solubility
(S)
 • The
solubility
of
a
solid
in
a
solvent
is
the
concentration
of
the
dissolved
salt
in
a
 saturated
solution
at
a
given
temperature
 • A
saturated
solution
is
one
in
which
solid
solute
is
in
equilibrium
with
dissolved
 solute

 
 
 S
and
Ksp
for
AgCl
(s)
↔
Ag+(aq)
+
Cl–(aq)
 • S
is
the
molar
solubility
 • S
moles
of
AgCl
(s)
⇒
____
moles
of
Ag+(aq)
 • S
moles
of
AgCl
(s)
⇒
____
moles
of
Cl–(aq)

 • S
=

 • Ksp
=
[Ag+][Cl–]
=

 • 

 ChE102
Fall
2011
Class
Notes
‐
Equilibrium
 
 
 
 
 
 
 
 D‐11
 S
and
Ksp
for
CaF2(s)
↔
Ca2+(aq)
+
2
F–(aq)
 • Ksp
=
[Ca2+][F–]2

 • S
moles
of
CaF2(s)
⇒
____
moles
of
Ca2+(aq)
 • S
moles
of
CaF2(s)
⇒
____
moles
of
F–(aq)
 • Ksp
=

 • 


 
 Question
D8
 Ammonium
hexachloroplatinate,
(NH4)2(PtCl6)
has
a
Ksp
of
5.6×10–6
mol3/L3
at
20˚C.

 Compute
its
solubility
in
g/L
of
solution.
 
 
 Common
Ion
Effect
&
18.5
­
Precipitation
 a) A
clear
saturated
solution
of
PbI2

 b) When
KI
is
added,
the
common
ion
I‐
makes
part
of
the
Pb2I
precipitate
(yellow
 precipitate)
 [Petrucci
et
al.,
Figure
18.1]
 
 
 
 Question
D9
 Calculate
the
mass
of
AgCl
that
can
dissolve
in
1L
of
a
0.15M
NaCl
solution.
 Data:
Ksp
(AgCl)
=
1.6
×
10–10
mol2/L2.

 
 
 Question
D10
 Three
drops
of
0.20M
KI
are
added
to
100
mL
of
a
0.010M
solution
of
AgNO3.

Will
a
 precipitate
of
silver
iodide
form?
 Data:

 Ksp
(AgI)
=
8.5
x
10‐17
(from
Table
18.1)
 
 1
drop
has
a
volume
of
0.05
mL
 ChE102
Fall
2011
Class
Notes
‐
Equilibrium
 
 
 
 
 
 
 
 D‐12
 ...
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