Page 1 1 CHEMICAL EQUILIBRIUM Chapter 13 2 Chemical Equilibrium • The state where the concentrations of all reactants and products remain constant with time. • On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation. 3 Properties of an Equilibrium Equilibrium systems are • DYNAMIC (in constant motion) • REVERSIBLE • can be approached from either direction Pink to blue Co(H O) Cl → Co(H O) Cl + 2 H O Blue to pink Co(H 2 O) 4 Cl 2 + 2 H 2 O ---> Co(H 2 O) 6 Cl 2 Co(H 2 O) 6 Cl 2 qe Co(H 2 O) 4 Cl 2 + 2 H 2 O 4 Chemical Equilibrium Fe 3+ + SCN - qe FeSCN 2+ + Fe(H 2 O) 6 3+ + SCN - qe Fe(SCN)(H 2 O) 5 2+ + H 2 O qe 5 Chemical Equilibrium Fe 3+ + SCN - qe FeSCN 2+ • After a period of time, the concentrations of reactants and products are constant. • The forward and reverse reactions continue after equilibrium is attained. 6 Examples of Chemical Equilibria Phase changes such as H 2 O(s) qe H 2 O(liq)
Transcript
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CHEMICAL EQUILIBRIUM
Chapter 13
2
Chemical Equilibrium
• The state where the concentrations of all reactants and products remain constantwith time.
• On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.
3Properties of an EquilibriumEquilibrium systems are• DYNAMIC (in constant
motion)• REVERSIBLE • can be approached from
either direction
Pink to blueCo(H2O)6Cl2 → Co(H2O)4Cl2 + 2 H2O
Blue to pinkCo(H2O)4Cl2 + 2 H2O ---> Co(H2O)6Cl2
Co(H2O)6Cl2 q e Co(H2O)4Cl2 + 2 H2O
4
Chemical EquilibriumFe3+ + SCN- q e FeSCN2+
+
Fe(H2O)63+ + SCN- q e Fe(SCN)(H2O)5
2+ + H2O
q e
5
Chemical EquilibriumFe3+ + SCN- q e FeSCN2+
• After a period of time, the concentrations of reactants and products are constant.
• The forward and reverse reactions continue after equilibrium is attained.
• At a given T and P of CO2, [Ca2+] and [HCO3-] can
be found from the EQUILIBRIUM CONSTANT.
Chemical Equilibria
9Example• Remember we are looking at a net change.• Reactions at equilibrium still react. It is just that the
forward and reverse reactions are equal.H2O(g) + CO(g) q e H2(g) + CO2(g)
• Figure 13.2: The changes in concentrations with time for the reaction H2O(g) + CO(g) H2(g) + CO2(g) when equimolarquantities of H2O(g) and CO(g) are mixed.
10
• Figure 13.3: (a) H2O and CO are mixed in equal numbers and begin to react (b) to form CO2 and H2. After time has passed, equilibrium is reached (c) and the numbers of reactant and product molecules then remain constant over time (d).
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• Figure 13.4: The changes with time in the rates of forward and reverse reactions for H2O(g) + CO(g) q e H2(g) + CO2(g) when equimolar quantities of H2O(g) and CO(g) are mixed. The rates do not change in the same way with time because the forward reaction has a much larger rate constant than the reverse reaction.
12Characteristics of Chemical Equilibrium
N2(g) + 3H2(g) q e 2NH3(g)• When we mix N2, H2, and
NH3 in a closed container at 25°C, we do no see a change in the concentrations over time regardless of the original amounts of gas.
• Two possibilities exist– The system is at chemical
equilibrium– The forward and reverse
reactions are so slow that the system moves towards equilibrium at a rate that cannot be detected.
• The second reason applies.• Remember
– N2 triple bond = 941 kJ/mol– H2 single bond = 432 kJ/mol
• How can we speed up the reaction?
– Catalyst
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13The Equilibrium Constant• NOT TO BE CONFUED WITH THE RATE CONSTANT!!!!!• Equilibrium Constant: The value obtained when
equilibrium concentrations of the chemical species are substituted in the equilibrium expression.
• Equilibrium Expression: The expression (from the law of mass action) obtained by multiplying the product concentrations and dividing by the multiplied reactant concentrations, with each concentration raised to a power represented by the coefficient in the balanced equation.
• Law of Mass Action: A general description of the equilibrium condition; it defines the equilibrium constant expression.
• WHAT THE *%#@!
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The Law of Mass Action• Two Norwegian chemist
– Cato Maximilian Guldberg (1836-1902)– Peter Waage (1833-1900)
• Proposed in 1864 a general description of an equilibrium condition.
– Based upon observations of many chemical reactions.
• For some reactionjA + kB ↔ lC + mD
• The law of mass action is represented by the equilibrium expression & K is the constant:
Kl m
j k=C DA B
15
Equilibrium Expression Example
4NH3(g) + 7O2(g) q e 4NO2(g) + 6H2O(g)
K= NO H ONH O
2
2
24 6
34 7
16
Example• [NH3] = 3.1×10-2 M• [N2] = 8.5×10-2 M• [H2] = 3.1×10-3 M• N2(g) + 3H2(g) q e 2NH3(g)• What is the value of K?
4331
22
322
23 103.8
)10)(3.110(8.5)10(3.1
]][H[N][NHK ×=
×××
== −−
−
17
Example• [NH3] = 3.1×10-2 M• [N2] = 8.5×10-2 M• [H2] = 3.1×10-3 M• 2NH3 (g)q e N2(g) + 3H2(g)• Just the reverse of the previous
example• What is the value of K’?
52
3
322 102.6
K1
][NH]][H[NK' −×===
18
Example• [NH3] = 3.1×10-2 M• [N2] = 8.5×10-2 M• [H2] = 3.1×10-3 M• ½ N2(g) + 3/2 H2(g) q e NH3(g)• Just ½ of the original example• What is the value of K’’?
21/2 101.9(K)'K' ×==
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Notes on Equilibrium Expressions (EE)
The Equilibrium Expression for a reaction is the reciprocal of that for the reaction written in reverse.When the equation for a reaction is multiplied by n, EEnew = (EEoriginal)n
Knew = (Korginal)n
The units for K depend on the reaction being considered.
Customarily written without units.
20Equilibrium with Pressure• We have been talking about
equilibrium of gases in terms of concentration.
• Why not pressure?• You should know
RTPC & CRTP
RTVnP
Vol(L)molsM & nRTPV
==
=
==
21Equilibrium with Pressure
))(P(PP
K
K))(C(C
C]][H[N
][NHK
3HN
2NH
P
c3HN
2NH
322
23
22
3
22
3
=
∴
===
22
But• There must be some sort of relationship
between K and KP.• So we have the following reaction.
jA + kB q e lC + mD
))(P(P))(P(PK &
[B][A][D][C]K
KNOWWe
kB
jA
mD
lC
Pkj
ml
==
CRTP =
23
KP = K(RT)∆n
∆n
k)(jm)(lkj
ml
kB
jA
mD
lC
kB
jA
mD
lC
kB
jA
mD
lC
P
K(RT)
K(RT)(RT)(RT)
)(C)(C)(C)(C
RT)(CRT)(CRT)(CRT)(C
))(P(P))(P(PK
=
=×=
××××
==
+−++
+
24
Example
KP = 1.9×103 @ 25 °C2NO(g) + Cl2(g) q e 2NOCl(g)
Calculate K from KP
KP = K(RT)∆n
1.9×103 = K(RT)∆n
∆n = 2 – (2+1) = -11.9×103 = K(RT)-1
1.9×103(RT) = K1.9×103(0.082057)(298) = K = 4.6×104
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Heterogeneous Equilibria• Heterogeneous Equilibrium: an equilibrium involving
reactants and/or products in more than one phase.• Example: CaCO3(s) q e CaO(s) + CO2(g)• You would think that the equilibrium constant would
be
• But heterogenous equilibrium does not depend on the amounts of pure solid or liquids present.
• Therefore• K=[CO2]
][CaCO][CaO][COK'
3
2=
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Writing and Manipulating K Expressions
Solids and liquids NEVERappear in equilibrium expressions.S(s) + O2(g) q e SO2(g)
K = [SO2][O2]
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Writing and Manipulating K Expressions
Solids and liquids NEVERappear in equilibrium expressions.
NH3(aq) + H2O(liq) q e
NH4+(aq) + OH-(aq)
K = [NH4+][OH-]
[NH3]
28The Reaction Quotient, QIn general, all reacting chemical systems
are characterized by their REACTION QUOTIENT, Q.
• If Q = K, then system is at equilibrium.• If Q > K, then conc. of products too large shifts
towards reactants.• If Q < K, then conc. of reactants too large shifts
towards products.
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The Meaning of K1. Can tell if a reaction is product-
favored or reactant-favored.For N2(g) + 3 H2(g) q e 2 NH3(g)
Kc = [NH3]2
[N2][H2]3 = 3.5 x 108
Conc. of products is much greaterthan that of reactants at equilibrium. The reaction is strongly product-favored.
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The Meaning of KFor AgCl(s) q e
Ag+(aq) + Cl-(aq)Kc = [Ag+] [Cl-] = 1.8 x 10-5
Conc. of products is much less than that of reactants at equilibrium.
The reaction is stronglyreactant-favored.
Ag+(aq) + Cl-(aq) q e AgCl(s)
is product-favored.
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Product- or Reactant Favored
Product-favored Reactant-favored
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The Meaning of KK comes from thermodynamics.See Chapter 19, page 812-813
∆G˚ < 0: reaction is product favored∆G˚ > 0: reaction is reactant-favored
If K > 1, then ∆G˚ is negativeIf K < 1, then ∆G˚ is positive
∆Go = -RT ln K
33
The Meaning of K2. Can tell if a reaction is at equilibrium.
If not, which way it moves to approach equilibrium.
H
H
H
H
H
H
H
H
H H H
CH
HHH H
H—C—C—C—C—H H—C—C—C—H
K =
n-butane iso-butane
[iso][n]
= 2.5
H
H
H
H
H
H
H
H
H H H
CH
HHH H
H—C—C—C—C—H H—C—C—C—H
K =
n-butane iso-butane
[iso][n]
= 2.5
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The Meaning of K
If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium?
If not, which way does the reaction “shift” to approach equilibrium?
H
H
H
H
H
H
H
H
H H H
CH
HHH H
H—C—C—C—C—H H—C—C—C—H
K =
n-butane iso-butane
[iso][n]
= 2.5
35The Meaning of KAll reacting chemical systems are
characterized by their REACTION QUOTIENT, Q.
Q = product concentrationsreactant concentrations
Q = conc. of isoconc. of n
= 0.350.15
= 2.3
Q (2.33) < K (2.5) Reaction is NOT at equilibrium, so [iso] must become ________ and [n] must ____________.
If Q = K, then system is at equilibrium.
36
The Extent of a Reaction• The inherent tendency for a reaction to occur it
indicated by the magnitude of the equilibrium constant.
• A value of K much larger than 1 mean?– At equilibrium the reaction system will consist of mostly
products.– Lies to the right.
• A very small value of K– At equilibrium the reaction system will consist of mostly
reactants.– Lies to the left.
• Remember the size of K and the time required to reach equilibrium are NOT directed related.
• The time is dependent on the Activation Energy.
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MANY MANY EXAMPLES• For the reaction
N2(g) + 3H2(g)q e 2NH3(g)• K=6.0×10-2 @ 500 °C• Predict the direction in which the system will shift
to reach equilibrium if• [NH3] = 1.0×10-5 M• [N2] = 1.0×10-5 M• [H2] = 2.0×10-3 M
7333
23
322
23 1031
)10)(2.010(1.0)10(1.0
]][H[N][NHQ ×=
×××
== −−
−
.
Therefore, since Q>K the direction goes towards the reactants
38N2O4(g)q e 2NO2(g)• N2O4 used as one of the
fuels on the lunar lander.• Closed container of
N2O4(g) allowed to reach equilibrium.
• KP = 0.133• PN2O4 at eq = 2.71 atm• PNO2 = ?
0.6000.360X
2.71atmX0.133
0.133PPK
2
ON
2NO2
P42
==
=
∴
==Apollo II lunar
landing module at Tranquility Base,
1969.
39 40
Determining K2 NOCl(g) q e 2 NO(g) + Cl2(g)Place 2.00 mol of NOCl is a 1.00 L flask. At
equilibrium you find 0.66 mol/L of NO. Calculate K.
• In an experiment in which 1.0 mol NOClis placed in a 2.0 L flask, what are the equilibrium concentrations?
522
2
106.1]NOCl[
]Cl[]NO[K −×==
x2x0.5 - 2xE
+x+2x-2xC
000.50I
Cl2(g)2NO(g) +2NOCl(g)q e
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67
• The problem is you end up with an x3 component, which are not fun to solve for
• But, since K is so small the reaction will not proceed that far to the right.
• Which means that x is a relatively small number.
• So 2x does nothing to 0.50
2
25
22
2
)x250.0()x()x2(106.1
]NOCl[]Cl[]NO[K
−=×== −
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2
25
22
2
)x250.0()x()x2(106.1
]NOCl[]Cl[]NO[K
−=×== −
2
3
2
25
22
2
)50.0(x4
)50.0()x()x2(106.1
]NOCl[]Cl[]NO[K ==×== −
Therefore, x = 1.0×10-2
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x2x0.5 - 2xE
+x+2x-2xC
000.50I
Cl2(g)2NO(g) +2NOCl(g)q e
x = 1.0×10-2
• Therefore at eq.– [NOCl]I = 0.50 M– [NO] = 2.0×10-2 M– [Cl] = 1.0×10-2 M
70
EQUILIBRIUM AND EXTERNAL EFFECTS
• Temperature, catalysts, and changes in concentration affect equilibria.
• The outcome is governed by LE CHATELIER’S PRINCIPLE
• “...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.”
71
EQUILIBRIUM AND EXTERNAL EFFECTS
Henri Le Chatelier1850-1936Studied mining
engineering.Interested in glass
and ceramics.
72
EQUILIBRIUM AND EXTERNAL EFFECTS
• Temperature change ---> change in K• Consider the fizz in a soft drink
CO2(aq) + HEAT q e CO2(g) + H2O(liq) • K = P (CO2) / [CO2]• Increase T. What happens to equilibrium
position? To value of K?• K increases as T goes up because P(CO2)
increases and [CO2] decreases.• Decrease T. Now what?• Equilibrium shifts left and K decreases.
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Temperature Effects on Equilibrium
N2O4 (colorless) + heat
q e 2 NO2 (brown)
∆Ho = + 57.2 kJ
Kc = [NO2]2
[N2O4]
Kc (273 K) = 0.00077Kc (298 K) = 0.0059
74
EQUILIBRIUM AND EXTERNAL EFFECTS
• Add catalyst ---> no change in K• A catalyst only affects the RATE of
approach to equilibrium.
Catalytic exhaust system
75
N2(g) + 3 H2(g) q e 2 NH3(g) + heatK = 3.5 x 108 at 298 K
NH3Production
76
Haber-Bosch Ammonia Synthesis
Fritz Haber1868-1934Nobel Prize, 1918
Carl Bosch1874-1940Nobel Prize, 1931
77
EQUILIBRIUM AND EXTERNAL EFFECTS
• Concentration changes–no change in K –only the position of equilibrium changes.
78
Le Chatelier’s Principle
Adding a “reactant” to a chemical system.
CH13_slide78.mov
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Le Chatelier’s Principle
Removing a “reactant” from a chemical system.
CH13_slide79.mov
80
Le Chatelier’s Principle
Adding a “product” to a chemical system.
CH13_slide80.mov
81
Le Chatelier’s Principle
Removing a “product” from a chemical system.
CH13_slide81.mov
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Butane-Isobutane
Equilibrium
K = [isobutane][butane]
= 2.5
butane
isobutane
83
Butane Isobutanebutane
isobutane
• At equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. K = 2.5.
• Add 1.50 M butane. • When the system comes to equilibrium
again, what are [iso] and [butane]? CH13_slide83.mov
84
Assume you are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]? K = 2.5
SolutionCalculate Q immediately after adding more
butane and compare with K.
Q = [isobutane][butane]
= 1.250.50 + 1.50
= 0.63
Q is LESS THAN K. Therefore, the reaction will shift to the ____________.
Butane q e Isobutane
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You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane.
SolutionQ is less than K, so equilibrium shifts right —
away from butane and toward isobutane.Set up ICE table
[butane] [isobutane]InitialChangeEquilibrium
0.50 + 1.50 1.25- x + x2.00 - x 1.25 + x
Butane q e Isobutane86
You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane.
Solution
K = 2.50 = [isobutane][butane]
= 1.25 + x2.00 - x
x = 1.07 MAt the new equilibrium position,
[butane] = 0.93 M and [isobutane] = 2.32 M. Equilibrium has shifted toward isobutane.
Butane e Isobutane
87
EQUILIBRIUM AND EXTERNAL EFFECTS
• Temperature, catalysts, and changes in concentration affect equilibria.
• The outcome is governed by LE CHATELIER’S PRINCIPLE
• “...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.”
88
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”
89
Nitrogen Dioxide EquilibriumN2O4(g) q e 2 NO2(g)
Increase P in the system by reducing the volume (at constant T).
Kc = [NO2]2
[N2O4] = 0.0059 at 298 K
e
CH13_slide89.mov
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Nitrogen Dioxide EquilibriumN2O4(g) q e 2 NO2(g)
Increase P in the system by reducing the volume.
In gaseous system the equilibrium will shift to the side with fewer molecules (in order to reduce the P).
Therefore, reaction shifts LEFT and P of NO2decreases and P of N2O4 increases.
