Transcript AP Notes Chapter16

AP Notes Chapter 16
Equilibrium
Dynamic chemical system in
which two reactions, equal and
opposite, occur
simultaneously
Properties
1. Appear from outside to
be inert or not functioning
2. Can be initiated in both
directions
Pink to blue
Co(H2O)6Cl2  Co(H2O)4Cl2 + 2 H2O
Blue to pink
Co(H2O)4Cl2 + 2 H2O  Co(H2O)6Cl2
Equilibrium achieved
Product conc. increases
and then becomes constant
at equilibrium
Reactant conc. declines
and then becomes
constant at equilibrium
At any point in the reaction
H2 + I2 2 HI
Q  reaction quotient =
2
[HI]
[H2 ][I2 ]
Equilibrium achieved
In the equilibrium region
[HI] 2
= 55.3 = K
[H2 ][I2 ]
K = equilibrium constant
Kinetics Definition
Rf = Rr
At equilibrium, the
rates of the forward
and reverse
reactions are equal.
aA + bB  cC
a
[A]
b
[B]
Rf(eq) = kf
c
Rr(eq) = kr [C]
Rf = Rr
a
b
c
kf [A] [B] = kr [C]
By convention
K is a concentration
quotient for a system
at equilibrium.
K=Q
For a system NOT at
equilibrium
Q≠K
if Q > K
The reverse reaction will
occur until equilibrium
is achieved.
if Q < K
The forward reaction
will occur until
equilibrium is
achieved.
Achieving equilibrium
is a driving force in
chemical systems and
will occur when
possible. It cannot be
stopped (spontaneous)
1. For the equilibrium system,
2NO2 (g)  N2O4 (g) , the
equilibrium constant, KC , is 8.8
0
at 25 C. If analysis shows that
2.0 x 10-3 mole of NO2 and 1.5 x
10-3 mole of N2O4 are present in
a 10.0 L flask, is the reaction at
equilibrium?
2NO2 (g)  N2O4 (g) , the equilibrium
constant, KC , is 8.8 at 250C. If analysis shows
that 2.0 x 10-3 mole of NO2 and 1.5 x 10-3 mole
of N2O4 are present in a 10.0 L flask, is the
reaction at equilibrium?
2NO2 (g)  N2O4 (g)
KC = 8.8 so… Q = [product]m =
[reactant]n
2NO2 (g)  N2O4 (g) , the equilibrium
constant, KC , is 8.8 at 250C. If analysis shows
that 2.0 x 10-3 mole of NO2 and 1.5 x 10-3 mole
of N2O4 are present in a 10.0 L flask, is the
reaction at equilibrium?
2NO2 (g)  N2O4 (g)
KC = 8.8 so… Q = [product]m = (2.0 x 10-3 mole NO2/ 10.0L)
[reactant]n
(1.5 x 10-3 mole of N2O4 /10.0L)2
=
2NO2 (g)  N2O4 (g) , the equilibrium
constant, KC , is 8.8 at 250C. If analysis shows
that 2.0 x 10-3 mole of NO2 and 1.5 x 10-3 mole
of N2O4 are present in a 10.0 L flask, is the
reaction at equilibrium?
2NO2 (g)  N2O4 (g)
KC = 8.8 so… Q = [product]m = (2.0 x 10-3 mole NO2/ 10.0L)
[reactant]n
(1.5 x 10-3 mole of N2O4 /10.0L)2
= 8888
KC = 8.8 so… Q = 8888
Is K = Q?
Concentrations and orders must be used!!
NO Q > K so [product] is BIGGER than is should be to be at
Equilibrium with given K value so products will convert to reactants
to reach equilibrium.
Types of Reactions
1. one way
(goes to completion)
NaOH(s) 
+
Na (aq) +
OH (aq)
Types of Reactions
2. Equilibrium
(two opposite reactions at
same time)
a. dimerization
2NO2(g)  N2O4(g)
b. dissociation of a
weak electrolyte
CH3COOH + H2O 
CH3COO- + H3O+
c. saturated aqueous
solutions
AgCl(s) 
+
Ag (aq)
+
Cl (aq)
C6H12O6(s)  C6H12O6(aq)
EQUILIBRIUM CONSTANT Keq
By convention
if Keq > 1
[products]coeff > [reactants]coeff
the forward reaction
proceeded to a greater extent
than the reverse reaction to
achieve equilibrium (i.e. the
products predominate at
equilibrium)
if Keq < 1
[products]coeff < [reactants]coeff
the forward reaction
proceeded to a lesser extent
than the reverse reaction to
achieve equilibrium (i.e. the
reactants predominate at
equilibrium)
How are kf and kr related
to temperature?
kf and kr are temperature
dependent
thus, Keq is temperature
dependent
N2O4 + heat 2 NO2
(colorless)
(brown)
∆Ho = + 57.2 kJ
2
[NO 2 ]
Kc 
[N 2 O 4 ]
Kc (273 K) = 0.00077
Kc (298 K) = 0.0059
Examples of Equilibrium
Expressions
N2O4(g)  2NO2(g)
2
K eq
[NO2 ]

[N2O 4 ]
CH3COOH + H2O CH3COO- +H3O+

K eq

[CH3COO ][H3O ]

[CH3COOH]
AgCl(s) 
+
Ag (aq)

+
Cl (aq)

K eq  [ Ag ][Cl ]
Concentrations of pure
liquids and solids are
NOT included in
equilibrium expressions,
as their concentrations
are themselves
constants.
The value of Keq
may appear to
change based on
way equation is
balanced.
1
2
N2O 4  NO2
K
'
eq

[NO2 ]
[N2O 4 ]
1
2
K
'
eq
 K eq
A value that is
mathematically related
to another (eg. temp) is
NOT considered a new
value
Multiple Equilibria
H3PO4 + 3 H2O 
3+
PO4 + 3 H3O
H3PO4 + H2O  H2PO4- + H3O+
H2PO4- + H2O  HPO42- + H3O+
HPO42- + H2O  PO43- + H3O+
Keq =
.
.
K1 K2 K3
for the complete
dissociation of
phosphoric acid
So far, Keq has been
studied as a function of
concentration, or
expressed with
appropriate notation, Kc
But, what about
equilibrium systems
where all components
are gases?
Partial pressures 
mole distribution
PV  nRT
n
P
 [gas] 
V
RT
where
V = a container
parameter
(constant for all gases)
T = constant for given
values of K
R = constant
aA(g) + bB(g)  cC(g)
c
[C]
Kc 
a
b
[ A ] [B ]
Substituting for a gas
c
 PC 


RT 

Kc 
a
b
 PA   PB 

 

 RT   RT 
c
PC
KC  a
b
PA  PB
 1 


 RT 
c  ( a b )
Let
c - (a + b) = n
where n is the change in #
of moles of gas (product reactant) for the forward
reaction.
If we express the
equilibrium constant
as a function of
partial pressures
c
PC
KP  a
b
PA  PB
Thus
KC =
-n
KP(RT)
or
KP = Kc
n
(RT)
2. When 2.0 moles of HI(g)
are placed in a 1.0 L
container and allowed to
come to equilibrium with
it’s elements, it is found
that 20% of the HI
decomposes. What is KC
and KP?
Applications of the
Equilibrium Constant
&
LeChatelier’s
Principle
3. 0.017 mol of n-butane is
placed in a 0.50 L container
and allowed to come to
equilibrium with its isomer
isobutane. KC at 250C is
2.5. What are the
equilibrium concentrations
of the two isomers?
Set up an ICE table
Initial [ ] of components
Change in [ ]
Equilibrium [ ]
n-butane  isobutane
I
C
E
0.034
-x
0.034-x
0
+x
x
[isobu tan e ]
KC 
n  bu tan e
x
2 .5 
0.034  x
solve
4. 2.0 mols Br2 are placed in
a 2.0 L flask at 1756 K, which
is of sufficient energy to split
apart some of the molecules.
If KC = 4.0 x 10-4 at 1756 K,
what are the equilibrium
concentrations of the
bromine molecules and
atoms?
Br2(g)  2 Br(g)
I
C
E
1.0
-x
1.0 - x
0
+2x
2x
2
[Br ]
KC 
[Br2 ]
4.0  10
4
2
(2x )

1 .0  x
solve
if K <<< [A]0, then can
assume amount that
dissociated to reach
equilibrium is VERY
small, thus
[ A ]eq  [ A ]0
2
[Br ]
KC 
[Br2 ]
4.0  10
4
(2x )

1 .0
solve
2
[OH ]
5. Calculate
at
equilibrium of a solution
that is initially 0.020 M
nicotine.
2H2O + Nic  NicH22+ + 2 OH-
I
0.020
0
C
-x
+x
E 0.020 - x x
0
+2x
2x
.
KC = K1 K2
KC = (7.0 x
-7
10 )(1.1
KC = 7.7 x
x
-17
10
-10
10 )

 2
[NicH2 ][OH ]
KC 
[Nic ]
7.7  10
17
2
(x )(2x )

(0.020  x )
LeChatlier’s Principle
When a stress is placed
on a system at
equilibrium, the system
will adjust so as to
relieve that stress.
Stress Factors
1. Change in
concentration of
reactants or products
2. Change in volume or
pressure (for gases)
3. Change in temperature
Responses to Stress
3 H2(g) + N2(g)  2 NH3(g)
1. Change concentration
a. add either H2 or N2
b. remove NH3
Responses to Stress
3H2(g) + N2(g)  2 NH3(g)
2. Change in volume or
pressure
a. increase volume
b. Increase pressure
c. add He
Responses to Stress
3H2(g) + N2(g)  2 NH3(g)
H = -92 kJ
3. Change in temperature
a. increase temperature
AgCl(s)  Ag+(aq) + Cl-(aq)
a. add AgCl
b. add H2O
c. add NaCl
d. add NH3(aq)
Keq is a temperature
dependent constant,
similar to the rate
constant, kf or kr
slope of line is different
ln Keq
m = -HR/R
1/T
ln K eq
HR
R
 1
 b
T
K1
 HR  1
1

ln

 
K2
R  T2 T1 
Le Chatelier’s Principle
• Change T
– change in K
– therefore change in P or
concentrations at equilibrium
• Use a catalyst: reaction comes more
quickly to equilibrium. K not changed.
• Add or take away reactant or product:
– K does not change
– Reaction adjusts to new equilibrium
“position”