CHEMICAL EQUILIBRIUM Chapter 16 Copyright © 1999 by Harcourt Brace & Company All rights reserved....
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Transcript of CHEMICAL EQUILIBRIUM Chapter 16 Copyright © 1999 by Harcourt Brace & Company All rights reserved....
CHEMICAL CHEMICAL EQUILIBRIUMEQUILIBRIUM
Chapter 16Chapter 16
CHEMICAL CHEMICAL EQUILIBRIUMEQUILIBRIUM
Chapter 16Chapter 16
Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida
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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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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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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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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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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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Examples Examples of of
Chemical Chemical EquilibriaEquilibria
Formation of stalactites and stalagmitesFormation of stalactites and stalagmites
CaCOCaCO33(s) + H(s) + H22O(liq) + COO(liq) + CO22(g)(g)
CaCa2+2+(aq) + 2 HCO(aq) + 2 HCO33--(aq) (aq)
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Chemical Chemical EquilibriaEquilibria
CaCOCaCO33(s) + H(s) + H22O(liq) + COO(liq) + CO22(g)(g)
CaCa2+2+(aq) + 2 HCO(aq) + 2 HCO33--(aq)(aq)
At a given T and P of COAt a given T and P of CO22, [Ca, [Ca2+2+] and [HCO] and [HCO33--] can be ] can be
found from the found from the EQUILIBRIUM CONSTANTEQUILIBRIUM CONSTANT. .
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THE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANT
For any type of chemical equilibrium of the For any type of chemical equilibrium of the typetype
a A + b B a A + b B c C + d Dc C + d D
the following is a CONSTANT (at a given T)the following is a CONSTANT (at a given T)
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THE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANT
For any type of chemical equilibrium of the For any type of chemical equilibrium of the typetype
a A + b B a A + b B c C + d Dc C + d D
the following is a CONSTANT (at a given T)the following is a CONSTANT (at a given T)
K =[C]c [D]d
[A]a [B]b
conc. of products
conc. of reactantsequilibrium constant
If K is known, then we can predict concs. of If K is known, then we can predict concs. of products or reactants. products or reactants.
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Determining KDetermining KDetermining KDetermining K2 NOCl(g) 2 NOCl(g) 2 NO(g) + Cl 2 NO(g) + Cl22(g)(g)
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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Determining KDetermining KDetermining KDetermining K2 NOCl(g) 2 NOCl(g) 2 NO(g) + Cl 2 NO(g) + Cl22(g)(g)
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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Determining KDetermining KDetermining KDetermining K2 NOCl(g) 2 NOCl(g) 2 NO(g) + Cl 2 NO(g) + Cl22(g)(g)
[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
K [NO]2[Cl2 ]
[NOCl]2 K
[NO]2[Cl2 ]
[NOCl]2
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Determining KDetermining KDetermining KDetermining K2 NOCl(g) 2 NOCl(g) 2 NO(g) + Cl2 NO(g) + Cl22(g)(g)
[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
K [NO]2[Cl2 ]
[NOCl]2 K
[NO]2[Cl2 ]
[NOCl]2
K [NO]2[Cl2 ]
[NOCl]2 =
(0.66)2(0.33)
(1.34)2 = 0.080K
[NO]2[Cl2 ]
[NOCl]2 =
(0.66)2(0.33)
(1.34)2 = 0.080
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Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Solids and liquids Solids and liquids NEVER appear in NEVER appear in equilibrium equilibrium expressions.expressions.
S(s) + OS(s) + O22(g) (g)
SO SO22(g)(g)
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Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Solids and liquids Solids and liquids NEVER appear in NEVER appear in equilibrium equilibrium expressions.expressions.
S(s) + OS(s) + O22(g) (g)
SO SO22(g)(g)
K [SO2 ][O2 ]
K [SO2 ][O2 ]
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Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Solids and liquids NEVER Solids and liquids NEVER appear in equilibrium appear in equilibrium expressions.expressions.
NHNH33(aq) + H(aq) + H22O(liq) NHO(liq) NH44++
(aq) + OH(aq) + OH--(aq)(aq)
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Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Solids and liquids NEVER Solids and liquids NEVER appear in equilibrium appear in equilibrium expressions.expressions.
NHNH33(aq) + H(aq) + H22O(liq) NHO(liq) NH44++
(aq) + OH(aq) + OH--(aq)(aq)
K [NH4
+][OH- ][NH3 ]
K [NH4
+][OH- ][NH3 ]
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Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Adding equations for reactionsAdding equations for reactions
S(s) + OS(s) + O22(g) SO(g) SO22(g) K(g) K11 = [SO = [SO22] / [O] / [O22]]
SOSO22(g) + 1/2 O(g) + 1/2 O22(g) (g) SO SO33(g)(g)
KK22 = [SO = [SO33] / [SO] / [SO22][O][O22]]1/21/2
NET EQUATIONNET EQUATION
S(s) + 3/2 OS(s) + 3/2 O22(g) (g) SO SO33(g)(g)
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Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Adding equations for reactionsAdding equations for reactions
S(s) + OS(s) + O22(g) (g) SO SO22(g) K(g) K11 = [SO = [SO22] / [O] / [O22]]
SOSO22(g) + 1/2 O(g) + 1/2 O22(g) (g) SO SO33(g)(g)
KK22 = [SO = [SO33] / [SO] / [SO22][O][O22]]1/21/2
NET EQUATIONNET EQUATION
S(s) + 3/2 OS(s) + 3/2 O22(g) (g) SO SO33(g)(g)
Knet
[SO3]
[O2 ]3/2 = K1 • K2 Knet [SO3]
[O2 ]3/2 = K1 • K2
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Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Changing coefficientsChanging coefficients
S(s) + 3/2 OS(s) + 3/2 O22(g) (g) SO SO33(g)(g)
2 S(s) + 3 O2 S(s) + 3 O22(g) (g) 2 SO2 SO33(g)(g)
K [SO3 ]
[O2 ]3/2 K [SO3 ]
[O2 ]3/2
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Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Changing coefficientsChanging coefficients
S(s) + 3/2 OS(s) + 3/2 O22(g) (g) SOSO33(g)(g)
2 S(s) + 3 O2 S(s) + 3 O22(g) (g) 2 SO2 SO33(g)(g)
K [SO3 ]
[O2 ]3/2 K [SO3 ]
[O2 ]3/2
Knew [SO3 ]2
[O2 ]3 Knew
[SO3 ]2
[O2 ]3
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Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions
Changing coefficientsChanging coefficients
S(s) + 3/2 OS(s) + 3/2 O22(g) (g) SO SO33(g)(g)
2 S(s) + 3 O2 S(s) + 3 O22(g) (g) 2 SO 2 SO33(g)(g)
Knew [SO3 ]2
[O2 ]3 = (Kold )2 Knew
[SO3 ]2
[O2 ]3 = (Kold )2
K [SO3 ]
[O2 ]3/2 K [SO3 ]
[O2 ]3/2
Knew [SO3 ]2
[O2 ]3 Knew
[SO3 ]2
[O2 ]3
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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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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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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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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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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.
For NFor N22(g) + 3 H(g) + 3 H22(g) (g) 2 NH2 NH33(g)(g)
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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.
For NFor N22(g) + 3 H(g) + 3 H22(g) (g) 2 NH2 NH33(g)(g)
Kc = [NH3 ]2
[N2 ][H2]3 = 3.5 x 108Kc =
[NH3 ]2
[N2 ][H2]3 = 3.5 x 108
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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.
For NFor N22(g) + 3 H(g) + 3 H22(g) (g) 2 NH 2 NH33(g)(g)
Conc. of products is Conc. of products is much greatermuch greater than that of reactants at equilibrium. than that of reactants at equilibrium.
Kc = [NH3 ]2
[N2 ][H2]3 = 3.5 x 108Kc =
[NH3 ]2
[N2 ][H2]3 = 3.5 x 108
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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.
For NFor N22(g) + 3 H(g) + 3 H22(g) (g) 2 NH 2 NH33(g)(g)
Conc. of products is Conc. of products is much greatermuch greater than that of reactants at equilibrium. than that of reactants at equilibrium.
The reaction is strongly The reaction is strongly product-product-favoredfavored..
Kc = [NH3 ]2
[N2 ][H2]3 = 3.5 x 108Kc =
[NH3 ]2
[N2 ][H2]3 = 3.5 x 108
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The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of KFor AgCl(s) For AgCl(s)
AgAg++(aq) + Cl(aq) + Cl--
(aq)(aq)
KKcc = [Ag = [Ag++] [Cl] [Cl--] = 1.8 x 10] = 1.8 x 10-5-5
Conc. of products is Conc. of products is much much lessless than that of than that of reactants at equilibrium. reactants at equilibrium.
The reaction is strongly The reaction is strongly
reactant-favoredreactant-favored..
AgAg++(aq) + Cl(aq) + Cl--(aq) (aq) AgCl(s)AgCl(s)is product-favored.is product-favored.
AgAg++(aq) + Cl(aq) + Cl--(aq) (aq) AgCl(s)AgCl(s)is product-favored.is product-favored.
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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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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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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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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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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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Typical CalculationsTypical CalculationsTypical CalculationsTypical Calculations
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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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
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 Nitrogen Dioxide EquilibriumEquilibrium
NN22OO44(g) (g) 2 2 NONO22(g)(g)
Nitrogen Dioxide Nitrogen Dioxide EquilibriumEquilibrium
NN22OO44(g) (g) 2 2 NONO22(g)(g)
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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
50
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 -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
51
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
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.
Rearrange: Rearrange: 0.0059 (0.50 - x) = 4x0.0059 (0.50 - x) = 4x22
0.0029 - 0.0059x = 4x0.0029 - 0.0059x = 4x22
4x4x22 + 0.0059x - 0.0029 = 0 + 0.0059x - 0.0029 = 0
This is a This is a QUADRATIC EQUATIONQUADRATIC EQUATION axax22 + bx + c = 0 + bx + c = 0
a = 4a = 4 b = 0.0059 b = 0.0059 c = -0.0029c = -0.0029
Kc = 0.0059 = [NO2 ]2
[N2O4 ]=
(2x)2
(0.50 - x) Kc = 0.0059 =
[NO2 ]2
[N2O4 ]=
(2x)2
(0.50 - x)
52
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Nitrogen Dioxide EquilibriumNitrogen Dioxide EquilibriumNN22OO44(g) (g) 2 NO 2 NO22(g)(g)
Nitrogen Dioxide EquilibriumNitrogen Dioxide EquilibriumNN22OO44(g) (g) 2 NO 2 NO22(g)(g)
Solve the quadratic equation for x.Solve the quadratic equation for x.
axax22 + bx + c = 0 + bx + c = 0
a = 4a = 4 b = 0.0059 b = 0.0059 c = -0.0029c = -0.0029
x = -b b2 - 4ac
2ax =
-b b2 - 4ac2a
53
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Nitrogen Dioxide EquilibriumNitrogen Dioxide EquilibriumNN22OO44(g) (g) 2 NO 2 NO22(g)(g)
Solve the quadratic equation for x.Solve the quadratic equation for x.
axax22 + bx + c = 0 + bx + c = 0
a = 4a = 4 b = 0.0059 b = 0.0059 c = -0.0029c = -0.0029
x = -b b2 - 4ac
2ax =
-b b2 - 4ac2a
x = -0.0059 (0.0059)2 - 4(4)(-0.0029)
2(4)x =
-0.0059 (0.0059)2 - 4(4)(-0.0029)2(4)
54
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Nitrogen Dioxide EquilibriumNitrogen Dioxide EquilibriumNN22OO44(g) (g) 2 NO 2 NO22(g)(g)
Solve the quadratic equation for x.Solve the quadratic equation for x.
axax22 + bx + c = 0 + bx + c = 0
a = 4a = 4 b = 0.0059 b = 0.0059 c = -0.0029c = -0.0029
x = -0.00074 ± 1/8(0.046)x = -0.00074 ± 1/8(0.046)1/21/2 = -0.00074 ± 0.027 = -0.00074 ± 0.027
x = -b b2 - 4ac
2ax =
-b b2 - 4ac2a
x = -0.0059 (0.0059)2 - 4(4)(-0.0029)
2(4)x =
-0.0059 (0.0059)2 - 4(4)(-0.0029)2(4)
55
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Nitrogen Dioxide EquilibriumNitrogen Dioxide EquilibriumNN22OO44(g) (g) 2 NO 2 NO22(g)(g)
x = -0.00074 ± 1/8(0.046)x = -0.00074 ± 1/8(0.046)1/21/2 = -0.00074 ± 0.027 = -0.00074 ± 0.027
x = 0.026 or -0.028x = 0.026 or -0.028
But a negative value is not reasonable.But a negative value is not reasonable.
ConclusionConclusion [N[N22OO44] = 0.050 - x = 0.47 M] = 0.050 - x = 0.47 M
[NO[NO22] = 2x = 0.052 M] = 2x = 0.052 M
x = -0.0059 (0.0059)2 - 4(4)(-0.0029)
2(4)x =
-0.0059 (0.0059)2 - 4(4)(-0.0029)2(4)