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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Properties of an Equilibrium Properties of an Equilibrium Properties of an Equilibrium Properties of an Equilibrium Equilibrium systems areEquilibrium systems are
• DYNAMIC (in constant DYNAMIC (in constant motion)motion)
• REVERSIBLE REVERSIBLE
• can be approached from can be approached from either directioneither direction
Equilibrium systems areEquilibrium systems are
• DYNAMIC (in constant DYNAMIC (in constant motion)motion)
• REVERSIBLE REVERSIBLE
• can be approached from can be approached from either directioneither direction
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Properties of an Equilibrium Properties of an Equilibrium Properties of an Equilibrium Properties of an Equilibrium Equilibrium systems areEquilibrium systems are
• DYNAMIC (in constant DYNAMIC (in constant motion)motion)
• REVERSIBLE REVERSIBLE
• can be approached from can be approached from either directioneither direction
Equilibrium systems areEquilibrium systems are
• DYNAMIC (in constant DYNAMIC (in constant motion)motion)
• REVERSIBLE REVERSIBLE
• can be approached from can be approached from either directioneither direction
Pink to bluePink to blueCo(HCo(H22O)O)66ClCl22 ---> Co(H ---> Co(H22O)O)44ClCl22 + 2 H + 2 H22OO
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Properties of an Equilibrium Properties of an Equilibrium Properties of an Equilibrium Properties of an Equilibrium Equilibrium systems areEquilibrium systems are
• DYNAMIC (in constant DYNAMIC (in constant motion)motion)
• REVERSIBLE REVERSIBLE
• can be approached from can be approached from either directioneither direction
Equilibrium systems areEquilibrium systems are
• DYNAMIC (in constant DYNAMIC (in constant motion)motion)
• REVERSIBLE REVERSIBLE
• can be approached from can be approached from either directioneither direction
Pink to bluePink to blueCo(HCo(H22O)O)66ClCl22 ---> Co(H ---> Co(H22O)O)44ClCl22 + 2 H + 2 H22OO
Blue to pinkBlue to pinkCo(HCo(H22O)O)44ClCl22 + 2 H + 2 H22O ---> Co(HO ---> Co(H22O)O)66ClCl22
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Chemical EquilibriumChemical EquilibriumFeFe3+3+ + SCN + SCN-- FeSCNFeSCN2+2+
Chemical EquilibriumChemical EquilibriumFeFe3+3+ + SCN + SCN-- FeSCNFeSCN2+2+
Fe(H2O)63+ Fe(SCN)(H2O)5
3++ SCN-
+
+ H2O
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Chemical EquilibriumChemical EquilibriumFeFe3+3+ + SCN + SCN-- FeSCN FeSCN2+2+
• After a period of time, the concentrations of reactants After a period of time, the concentrations of reactants and products are constant. and products are constant.
• The forward and reverse reactions continue after The forward and reverse reactions continue after equilibrium is attained.equilibrium is attained.
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Examples of Chemical Examples of Chemical EquilibriaEquilibria
Phase changes such asPhase changes such as H H22O(s) O(s) HH22O(liq)O(liq)
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Examples Examples of of
Chemical Chemical EquilibriaEquilibria
Formation of stalactites and stalagmitesFormation of stalactites and stalagmites
Place 2.00 mol of NOCl is a 1.00 L flask. At Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. equilibrium you find 0.66 mol/L of NO. Calculate K.Calculate K.
SolutionSolution
Set of a table of concentrationsSet of a table of concentrations
[NOCl][NOCl] [NO][NO] [Cl[Cl22]]
BeforeBefore 2.002.00 00 00
ChangeChange
EquilibriumEquilibrium 0.660.66
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Place 2.00 mol of NOCl is a 1.00 L flask. At Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. equilibrium you find 0.66 mol/L of NO. Calculate K.Calculate K.
SolutionSolution
Set of a table of concentrationsSet of a table of concentrations
[NOCl][NOCl] [NO][NO] [Cl[Cl22]]
BeforeBefore 2.002.00 00 00
ChangeChange -0.66-0.66 +0.66+0.66 +0.33+0.33
EquilibriumEquilibrium 1.341.34 0.660.66 0.330.33
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
K [SO2 ]
[O2 ] K
[SO2 ]
[O2 ]
Changing directionChanging direction
S(s) + OS(s) + O22(g) (g) SO SO22(g)(g)
SOSO22(g) (g) S(s) + O S(s) + O22(g)(g)
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Changing directionChanging direction
S(s) + OS(s) + O22(g) (g) SO SO22(g)(g)
SOSO22(g) (g) S(s) + O S(s) + O22(g)(g)
K [SO2 ]
[O2 ] K
[SO2 ]
[O2 ]
Knew [O2 ][SO2 ]
Knew [O2 ][SO2 ]
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Changing directionChanging direction
S(s) + OS(s) + O22(g) (g) SO SO22(g)(g)
SOSO22(g) (g) S(s) + O S(s) + O22(g)(g)
K [SO2 ]
[O2 ] K
[SO2 ]
[O2 ]
Knew [O2 ][SO2 ]
= 1
Kold Knew
[O2 ][SO2 ]
= 1
Kold
Knew [O2 ][SO2 ]
Knew [O2 ][SO2 ]
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Concentration UnitsConcentration Units
We have been writing K in terms of mol/L. We have been writing K in terms of mol/L.
These are designated by These are designated by KKcc
But with gases, P = (n/V)•RT = conc • RTBut with gases, P = (n/V)•RT = conc • RT
P is proportional to concentration, so we can P is proportional to concentration, so we can write K in terms of P. These are designated write K in terms of P. These are designated
by by KKpp. .
KKcc and K and Kpp may or may not be the same. may or may not be the same.
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of K1.1. Can tell if a reaction is product-Can tell if a reaction is product-
favored or reactant-favored.favored or reactant-favored.
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of K2. Can tell if a reaction is at equilibrium. 2. Can tell if a reaction is at equilibrium.
If not, which way it moves to approach If not, which way it moves to approach equilibrium.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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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of K
If [iso] = 0.35 M and [n] = 0.15 M, are you If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? at equilibrium?
Which way does the reaction “shift” to Which way does the reaction “shift” to approach equilibrium?approach equilibrium?
See Screen 16.9.See Screen 16.9.
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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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of KIn general, all reacting chemical systems In general, all reacting chemical systems
are characterized by their are characterized by their REACTION REACTION QUOTIENT, QQUOTIENT, Q..
If Q = K, then system is at equilibrium.If Q = K, then system is at equilibrium.
Q = product concentrationsreactant concentrations
Q = product concentrationsreactant concentrations
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of KIn general, all reacting chemical systems In general, all reacting chemical systems
are characterized by their are characterized by their REACTION REACTION QUOTIENT, QQUOTIENT, Q..
If Q = K, then system is at equilibrium.If Q = K, then system is at equilibrium.
If [iso] = 0.35 M and [n] = 0.15 M, are you at If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium?equilibrium?
Q (2.3) < K (2.5). Q (2.3) < K (2.5).
Q = product concentrationsreactant concentrations
Q = product concentrationsreactant concentrations
Q = conc. of isoconc. of n
= 0.350.15
= 2.3Q = conc. of isoconc. of n
= 0.350.15
= 2.3
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of KIn general, all reacting chemical systems are In general, all reacting chemical systems are
characterized by their characterized by their REACTION REACTION QUOTIENT, QQUOTIENT, Q..
If Q = K, then system is at equilibrium.If Q = K, then system is at equilibrium.
Q (2.33) < K (2.5). Q (2.33) < K (2.5).
Reaction is NOT at equilibrium, so [Iso] must Reaction is NOT at equilibrium, so [Iso] must become ________ and [n] must become ________ and [n] must ____________. ____________.
Q = product concentrationsreactant concentrations
Q = product concentrationsreactant concentrations
Q = conc. of isoconc. of n
= 0.350.15
= 2.3Q = conc. of isoconc. of n
= 0.350.15
= 2.3
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
PROBLEM: Place 1.00 mol each of HPROBLEM: Place 1.00 mol each of H22 and I and I22 in a in a 1.00 L flask. Calc. equilibrium concentrations.1.00 L flask. Calc. equilibrium concentrations.
HH22(g) + I(g) + I22(g) (g) ¸̧ 2 HI(g) 2 HI(g)
Kc = [HI]2
[H2 ][I2 ] = 55.3Kc =
[HI]2
[H2 ][I2 ] = 55.3
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H2(g) + I2(g) 2 HI(g), Kc = 55.3
Step 1. Set up table to define Step 1. Set up table to define EQUILIBRIUM concentrations.EQUILIBRIUM concentrations.
[H[H22]] [I[I22]] [HI][HI]
Initial Initial 1.001.00 1.001.00 00
ChangeChange
EquilibEquilib
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H2(g) + I2(g) 2 HI(g), Kc = 55.3
Step 1. Set up table to define Step 1. Set up table to define EQUILIBRIUM concentrations.EQUILIBRIUM concentrations.
[H[H22]] [I[I22]] [HI][HI]
Initial Initial 1.001.00 1.001.00 00
ChangeChange -x-x -x-x +2x+2x
EquilibEquilib1.00-x1.00-x 1.00-x1.00-x 2x2x
where where xx is defined as am’t of H is defined as am’t of H22 and I and I22
consumed on approaching equilibrium.consumed on approaching equilibrium.
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H2(g) + I2(g) 2 HI(g), Kc = 55.3
Step 2. Put equilibrium concentrations Step 2. Put equilibrium concentrations into Kinto Kcc expression. expression.
Kc = [2x]2
[1.00 - x][1.00 - x] = 55.3Kc =
[2x]2
[1.00 - x][1.00 - x] = 55.3
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H2(g) + I2(g) 2 HI(g), Kc = 55.3
Step 3. Solve KStep 3. Solve Kcc expression - take expression - take
square root of both sides.square root of both sides.
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H2(g) + I2(g) 2 HI(g), Kc = 55.3
Step 3. Solve KStep 3. Solve Kcc expression - take expression - take
square root of both sides.square root of both sides.
7.44 = 2x
1.00 - x7.44 =
2x1.00 - x
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H2(g) + I2(g) 2 HI(g), Kc = 55.3
Step 3. Solve KStep 3. Solve Kcc expression - take expression - take
square root of both sides.square root of both sides.
x = 0.79x = 0.79
7.44 = 2x
1.00 - x7.44 =
2x1.00 - x
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H2(g) + I2(g) 2 HI(g), Kc = 55.3
Step 3. Solve KStep 3. Solve Kcc expression - take expression - take
square root of both sides.square root of both sides.
x = 0.79x = 0.79
Therefore, at equilibriumTherefore, at equilibrium
[H[H22] = [I] = [I22] = 1.00 - x = 0.21 M] = 1.00 - x = 0.21 M
7.44 = 2x
1.00 - x7.44 =
2x1.00 - x
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H2(g) + I2(g) 2 HI(g), Kc = 55.3
Step 3. Solve KStep 3. Solve Kcc expression - take expression - take
square root of both sides.square root of both sides.
x = 0.79x = 0.79
Therefore, at equilibriumTherefore, at equilibrium
[H[H22] = [I] = [I22] = 1.00 - x = 0.21 M] = 1.00 - x = 0.21 M
[HI] = 2x = 1.58 M[HI] = 2x = 1.58 M
7.44 = 2x
1.00 - x7.44 =
2x1.00 - x
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Nitrogen Dioxide EquilibriumNitrogen Dioxide EquilibriumNN22OO44(g) (g) 2 NO 2 NO22(g)(g)
If initial concentration of NIf initial concentration of N22OO44 is 0.50 M, what are the is 0.50 M, what are the equilibrium concentrations?equilibrium concentrations?
Step 1. Set up an equilibrium tableStep 1. Set up an equilibrium table
[N[N22OO44]] [NO[NO22]]
InitialInitial 0.500.50 00
ChangeChange
EquilibEquilib
Kc = [NO2 ]2
[N2O4 ] = 0.0059 at 298 KKc =
[NO2 ]2
[N2O4 ] = 0.0059 at 298 K
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Nitrogen Dioxide EquilibriumNitrogen Dioxide EquilibriumNN22OO44(g) (g) 2 NO 2 NO22(g)(g)
If initial concentration of NIf initial concentration of N22OO44 is 0.50 M, what are the is 0.50 M, what are the equilibrium concentrations?equilibrium concentrations?
Step 1. Set up an equilibrium tableStep 1. Set up an equilibrium table
[N[N22OO44]] [NO[NO22]]
InitialInitial 0.500.50 00
ChangeChange -x-x +2x+2x
EquilibEquilib 0.50 - x0.50 - x 2x2x
Kc = [NO2 ]2
[N2O4 ] = 0.0059 at 298 KKc =
[NO2 ]2
[N2O4 ] = 0.0059 at 298 K
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Nitrogen Dioxide EquilibriumNitrogen Dioxide EquilibriumNN22OO44(g) (g) 2 NO 2 NO22(g)(g)
Step 2. Substitute into KStep 2. Substitute into Kcc expression and solve. expression and solve.