Post on 17-Dec-2015
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Chemical Equilibrium
Brown, LeMay Ch 15AP Chemistry
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15.1: Chemical Equilibrium
Occurs when opposing reactions are proceeding at the same rate Forward rate = reverse rate of reaction
Ex:Vapor pressure: rate of vaporization =
rate of condensationSaturated solution: rate of dissociation
= rate of crystallization
Expressing concentrations: Gases: partial pressures, PX
Solutes in liquids: molarity, [X]
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Forward reaction: A → B Rate = kforward [A]
Reverse reaction: B → A Rate = kreverse [B]
or
RT
P
V
nM
RT
PA A][
RT
PB B][
Forward reaction:
Reverse reaction:
RT
PkRate A
f
RT
PkRate B
r
nRTPV
R = 0.0821 L•atm
mol•K
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If equilibrium: A ↔ Bforward rate = reverse rate
[B]k [A]k rf
r
f
k
k
[A]
[B]
RT
Pk
RT
Pk B
rA
f or
eqr
f
A
B Kconstantk
k
P
Por
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PX or [X]
Time →
[B] or PB / RT
Equilibrium is established
Figure 1: Reversible reactions
[A] or PA / RT
[A]0 or PA0 / RT
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Reversible Reactions and Rate
Reaction Rate
Time
Backward rate
Forward rate
Equilibrium is established:
Forward rate = Backward rate
When equilibrium is achieved:[A] ≠ [B] and kf/kr = Keq
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15.2: Law of Mass Action
Derived from rate laws by Guldberg andWaage (1864) For a balanced chemical reaction
in equilibrium:
a A + b B ↔ c C + d D Equilibrium constant expression (Keq):
ba
dc
c [B] [A]
[D] [C] K b
Ba
A
dD
cC
p )(P)(P
)(P)(PK
Keq is strictly based on stoichiometry of the reaction (is independent of the mechanism).
Units: Keq is considered dimensionless (no units)
Cato Guldberg Peter Waage
(1836-1902) (1833-1900)
or
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Relating Kc and Kp
Convert [A] into PA:
ba
dc
([B]RT)([A]RT)
([D]RT)([C]RT)b
Ba
A
dD
cC
p )(P)(P
)(P )(PK
baba
dcdc
(RT)[B][A]
(RT)[D][C]
(RT) K K b)(a - d)(ccp
where x == change in coefficents of products – reactants (gases only!)= (c+d) - (a+b)
(RT) K xc
RT
P
V
nM RTAPA ][
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Magnitude of Keq
Since Keq [products]/[reactants], the magnitude of Keq predicts which reaction direction is favored:
If Keq > 1 then [products] > [reactants]and equilibrium “lies to the
right”
If Keq < 1 then [products] < [reactants]and equilibrium “lies to the
left”
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15.3: Types of Equilibria
Homogeneous: all components in same phase (usually g or aq)
N2 (g) + H2 (g) ↔ NH3 (g)
3H
1N
2NH
P )(P)(P
)(PK
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3
bB
aA
dD
cC
P )(P)(P
)(P)(PK
3 21
Fritz Haber(1868 – 1934)
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Heterogeneous: different phasesCaCO3 (s) ↔ CaO (s) + CO2 (g)
Definition: What we use:
][CaCO
)(P [CaO] K
3
COeq
22COp P K
Even though the concentrations of the solids or liquids do not appear in the equilibrium expression, the substances must be present to achieve equilibrium.
Concentrations of pure solids and pure liquids are not included in Keq expression because their concentrations do not vary, and are “already included” in Keq (see p. 548).
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15.4: Calculating Equilibrium Constants
Steps to use “ICE” table:1. “I” = Tabulate known initial and
equilibrium concentrations of all species in equilibrium expression
2. “C” = Determine the concentration change for the species where initial and equilibrium are known
• Use stoichiometry to calculate concentration changes for all other species involved in equilibrium
3. “E” = Calculate the equilibrium concentrations
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Ex: Enough ammonia is dissolved in 5.00 L of water at 25ºC to produce a solution that is 0.0124 M ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that [OH1-] is 4.64 x 10-4 M. Calculate Keq at 25ºC for the reaction:
NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1-
(aq)
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NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1-
(aq)
Initial
Change
Equilibrium
][NH
]][OH[NH K
3
-114
c
0.0124 M
- x
0.0119 M
0 M 0 M
+ x + x
4.64 x 10-4 M 4.64 x 10-4 M
5-2-4
101.81x 0.0119
)10(4.64
NH3 (aq)H2O (l)
NH41+ (aq) OH1- (aq)
XXX
x = 4.64 x 10-4 M
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Ex: A 5.000-L flask is filled with 5.000 x 10-3
mol of H2 and 1.000 x 10-2 mol of I2 at 448ºC. The value of Keq is 1.33. What are the concentrations of each substance at equilibrium?
H2 (g) + I2 (g) ↔ 2 HI (g)
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H2 (g) + I2 (g) ↔ 2 HI (g)
Initial
Change
Equilibrium
]][I[H
[HI] K
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2
c
1.000x10-3 M
- x M
(1.000x10-3 – x) M
2.000x10-3 M 0 M
- x M + 2x M
(2.000x10-3 – x) M 2x M
33.1x)-10x)(2.000-10(1.000
(2x)3-3-
2
4x2 = 1.33[x2 + (-3.000x10-3)x + 2.000x10-6]
0 = -2.67x2 – 3.99x10-3x + 2.66x10-6
Using quadratic eq’n: x = 5.00x10-4 or –1.99x10-3; x = 5.00x10-4
Then [H2]=5.00x10-4 M; [I2]=1.50x10-3 M; [HI]=1.00x10-3 M
H2 (g) I2 (g) HI (g)
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15.6: Le Châtelier’s Principle
If a system at equilibrium is disturbed by a change in: Concentration of one of the
components, Pressure, or Temperature
…the system will shift its equilibrium position to counteract the effect of the disturbance.
Henri Le Châtelier(1850 – 1936)
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4 Changes that do not affect Keq:
1. Concentration Upon addition of a reactant or product,
equilibrium shifts to re-establish equilibrium by consuming part of the added substance.
Upon removal of reactant or product, equilibrium shifts to re-establish equilibrium by producing more of the removed substance.Ex: Co(H2O)6
2+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l)
•Add HCl, temporarily inc forward rate•Add H2O, temporarily inc reverse rate
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2. Volume, with a gas present (T is constant)
Upon a decrease in V (thereby increasing P),equilibrium shifts to reduce the number of moles of gas.
Upon an increase in V (thereby decreasing P),equilibrium shifts to produce more moles of gas.
Ex: N2 (g) + 3 H2 (g) ↔ 2 NH3 (g) If V of container is decreased, equilibrium shifts
right. XN2 and XH2
dec
XNH3 inc3HN
2NH
P )(PP
)(PK
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THTN
2TNH
)P)(XP(X
)P(X
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3 4T
3HN
2T
2NH
P)(XX
P)(X
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3 2T
3HN
2NH
P)(XX
)(X
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3
23HN
2NH
P )(
)(K
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Since PT also inc, KP remains constant.
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3. Pressure, but not Volume
Usually addition of a noble gas, p. 560 Avogadro’s law: adding more non-reacting
particles “fills in” the empty space between particles.
In the mixture of red and blue gas particles, below, adding green particles does not stress the system, so there is no Le Châtelier shift.
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4. Catalysts
Lower the activation energy of both forward and reverse rxns, therefore increases both forward and reverse rxn rates.Increase the rate at which equilibrium is achieved, but does not change the ratio of components of the equilibrium mixture (does not change the Keq)
Energy
Rxn coordinate
Ea, uncatalyzed
Ea, catalyzed
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1 Change that does affect Keq:Temperature: consider “heat” as a part of the reaction
Upon an increase in T, endothermic reaction is favored (equilibrium shifts to “consume the extra heat”)
Upon a decrease in T, equilibrium shifts to produce more heat.Effect on Keq
1. Exothermic equilibria: Reactants ↔ Products + heat
• Inc T increases reverse reaction rate which decreases Keq
2. Endothermic equilibria: Reactants + heat ↔ Products
• Inc T increases forward reaction rate increases Keq
Ex: Co(H2O)62+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l);
H=+?•Inc T temporarily inc forward rate•Dec T temporarily inc reverse rate