AP Notes Chapter 16 Equilibrium Dynamic chemical system in which two reactions, equal and opposite,...
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Transcript of AP Notes Chapter 16 Equilibrium Dynamic chemical system in which two reactions, equal and opposite,...
AP Notes Chapter 16Equilibrium
Dynamic chemical system in which two reactions, equal and opposite, occur simultaneously
Properties1. Appear from outside to
be inert or not functioning
2. Can be initiated in both directions
Pink to blueCo(H2O)6Cl2 Co(H2O)4Cl2 + 2 H2O
Blue to pinkCo(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 reactionH2 + I2 2 HI
Q reaction quotient = [HI]2
[H2 ][I2 ]
Equilibrium achieved
In the equilibrium region
[HI]2
[H2 ][I2 ] = 55.3 = K
K = equilibrium constant
Kinetics Definition
Rf = Rr
At equilibrium, the rates of the forward and reverse reactions are equal.
aA + bB cCRf(eq) = kf [A]a [B]b
Rr(eq) = kr [C]c
Rf = Rr
kf [A]a [B]b = kr [C]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 > KThe 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 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?
Types of Reactions1. one way
(goes to completion)
NaOH(s) Na+(aq) + OH-(aq)
Types of Reactions2. 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)
By convention EQUILIBRIUM CONSTANT Keq
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
]O[N
][NO K
42
22
c ]O[N
][NO K
42
22
c
Kc (273 K) = 0.00077
Kc (298 K) = 0.0059
N2O4(g) 2NO2(g)
]ON[
]NO[K
42
22
eq
Examples of EquilibriumExpressions
CH3COOH + H2O CH3COO- +H3O+
]COOHCH[
]OH][COOCH[K
3
33eq
AgCl(s) Ag+(aq) + Cl-(aq)
]Cl][Ag[Keq
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.
21
]ON[
]NO[K
NOON
42
2'eq
24221
eq'eq KK
A value that is mathematically related to another (eg. temp) is NOT considered a new
value
Multiple EquilibriaH3PO4 + 3 H2O
PO43- + 3 H3O+
H3PO4 + H2O H2PO4- + H3O+
H2PO4- + H2O HPO4
2- + H3O+
HPO42- + H2O PO4
3- + 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
RT
P]gas[
V
n
nRTPV
whereV = a container parameter (constant for all gases)T = constant for given values of KR = constant
aA(g) + bB(g) cC(g)
ba
c
c ]B[]A[
]C[K
Substituting for a gas
b
B
a
A
c
C
c
RTP
RTP
RTP
K
)ba(c
bB
aA
cC
C RT
1
PP
PK
Letc - (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
bB
aA
cC
PPP
PK
Thus
KC = KP(RT)-n
or
KP = Kc(RT)n
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 theEquilibrium 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 0.034 0
C -x +x
E 0.034-x x
x034.0
x5.2
etanbun
]etanisobu[KC
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 1.0 0
C -x +2x
E 1.0 - x 2x
x0.1
)x2(100.4
]Br[
]Br[K
24
2
2
C
solve
if K <<< [A]0, then can assume amount that dissociated to reach equilibrium is VERY
small, thus
0eq ]A[]A[
0.1
)x2(100.4
]Br[
]Br[K
24
2
2
C
solve
5. Calculate [OH-] at equilibrium of a solution that is initially 0.020 M nicotine.
2H2O + Nic NicH22+ + 2 OH-
I 0.020 0 0
C -x +x +2x
E 0.020 - x x 2x
KC = K1 . K2
KC = (7.0 x 10-7)(1.1 x 10-10)
KC = 7.7 x 10-17
)x020.0(
)x2)(x(107.7
]Nic[
]OH][NicH[K
217
22
C
LeChatlier’s PrincipleWhen a stress is placed
on a system at equilibrium, the system
will adjust so as to relieve that stress.
Stress Factors1. Change in
concentration of reactants or products
2. Change in volume or pressure (for gases)
3. Change in temperature
Responses to Stress3 H2(g) + N2(g) 2 NH3(g)
1. Change concentration
a. add either H2 or N2
b. remove NH3
Responses to Stress3H2(g) + N2(g) 2 NH3(g)
2. Change in volume or pressure
a. increase volume
b. Increase pressure
c. add He
Responses to Stress3H2(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
1/T
ln K
eq m = -HR/R
bT
1
R
H- K ln R
eq
12
R
2
1
T
1
T
1
R
H
K
K ln
• 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”
Le Chatelier’s Principle