Chapter 16 Principles of Reactivity: Chemical Equilibria
description
Transcript of Chapter 16 Principles of Reactivity: Chemical Equilibria
John C. Kotz • State University of New York, College at Oneonta
John C. KotzPaul M. TreichelJohn Townsend
http://academic.cengage.com/kotz
Chapter 16Principles of Reactivity: Chemical Equilibria
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CHEMICAL CHEMICAL EQUILIBRIUMEQUILIBRIUMChapter 16Chapter 16
PbPb2+2+(aq) + 2 Cl(aq) + 2 Cl––(aq) (aq) ee PbCl PbCl22(s)(s)
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Properties of an Properties of an Equilibrium 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 ff Co(H Co(H22O)O)44ClCl22 + 2 H + 2 H22OO
Blue to pinkBlue to pinkCo(HCo(H22O)O)44ClCl22 + 2 H + 2 H22O O ff Co(H Co(H22O)O)66ClCl22
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Chemical EquilibriumChemical EquilibriumFeFe3+3+ + SCN + SCN-- ee FeSCN FeSCN2+2+Chemical EquilibriumChemical EquilibriumFeFe3+3+ + SCN + SCN-- ee FeSCN FeSCN2+2+
+
Fe(H2O)63+ Fe(SCN)(H2O)5
2++ SCN- ee
+ H2O
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Chemical EquilibriumChemical EquilibriumFeFe3+3+ + SCN + SCN-- ee FeSCN FeSCN2+2+Chemical EquilibriumChemical EquilibriumFeFe3+3+ + SCN + SCN-- ee 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 Examples of Chemical EquilibriaChemical Equilibria
Phase changes such asPhase changes such as H H22O(s) O(s) ee HH22O(liq)O(liq)
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Examples Examples of of
Chemical Chemical EquilibriEquilibri
aa
Formation of Formation of
stalactites and stalagmitesstalactites and stalagmites
CaCOCaCO33(s) + H(s) + H22O(liq) + COO(liq) + CO22(g)(g)
ee 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)
ee Ca Ca2+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 found from the can be found from the EQUILIBRIUM EQUILIBRIUM
CONSTANTCONSTANT. .
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Reaction Quotient & Reaction Quotient & Equilibrium ConstantEquilibrium Constant
Product conc. increases and then becomes constant at equilibrium
Reactant conc. declines and then becomes constant at equilibrium
Equilibrium achieved
See Active Figure 16.2See Active Figure 16.2
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Reaction Quotient & Reaction Quotient & Equilibrium ConstantEquilibrium Constant
At any point in the reaction
H2 + I2 ee 2 HI
Q = reaction quotient = [HI]2
[H2 ][I2 ]
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Reaction Quotient & Reaction Quotient & Equilibrium ConstantEquilibrium Constant
[HI]2
[H2 ][I2 ] = 55.3 = K
K = equilibrium constant
Equilibrium achieved
In the equilibrium region
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The Reaction Quotient, The Reaction Quotient, QQ
In general, ALL reacting chemical systems are In general, ALL reacting chemical systems are
characterized by their characterized by their REACTION REACTION
QUOTIENT, QQUOTIENT, Q..
a A + b B a A + b B ee c C + d D c C + d D
If Q = K, then system is at If Q = K, then system is at equilibrium.equilibrium.
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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 ee c C + d D c 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 If K is known, then we can predict concs. of products or reactants.concs. of products or reactants.
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Determining KDetermining KDetermining KDetermining K2 NOCl(g) 2 NOCl(g) ee 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 an Set of an “ICE”“ICE” table of concentrations table of concentrations
[NOCl][NOCl] [NO][NO] [Cl[Cl22]]
IInitialnitial 2.002.00 00 00
CChangehange
EEquilibriumquilibrium 0.660.66
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Determining KDetermining KDetermining KDetermining K2 NOCl(g) 2 NOCl(g) ee 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 an Set of an “ICE”“ICE” table of concentrations table of concentrations
[NOCl][NOCl] [NO][NO] [Cl[Cl22]]
IInitialnitial 2.002.00 00 00
CChangehange -0.66-0.66 +0.66+0.66 +0.33+0.33
EEquilibriumquilibrium 1.341.34 0.660.66 0.330.33
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Determining KDetermining KDetermining KDetermining K2 NOCl(g) 2 NOCl(g) ee 2 NO(g) + Cl 2 NO(g) + Cl22(g)(g)
[NOCl][NOCl] [NO][NO] [Cl[Cl22]]
InitialInitial 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 = 0.080
K = [NO]2[Cl2 ]
[NOCl]2 =
(0.66)2(0.33)
(1.34)2 = 0.080
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Writing and Writing and Manipulating K Manipulating K ExpressionsExpressions
Writing and Writing and Manipulating K Manipulating K ExpressionsExpressions
Solids and liquids Solids and liquids NEVERNEVER appear in appear in equilibrium equilibrium expressions.expressions.
S(s) + OS(s) + O22(g) (g)
ee SO SO22(g)(g)
K = [SO2 ]
[O2 ]
K = [SO2 ]
[O2 ]
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Writing and Writing and Manipulating K Manipulating K ExpressionsExpressions
Writing and Writing and Manipulating K Manipulating K ExpressionsExpressions
Solids and liquids Solids and liquids NEVERNEVER appear appear in equilibrium expressions.in equilibrium expressions.
NHNH33(aq) + H(aq) + H22O(liq) O(liq) ee
NHNH44++(aq) + OH(aq) + OH--(aq)(aq)
K = [NH4
+ ][OH- ]
[NH3 ]
K = [NH4
+ ][OH- ]
[NH3 ]
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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) ee 2 NH 2 NH33(g)(g)
Kc = [NH3 ]2
[N2 ][H2 ]3 = 3.5 x 108
Kc = [NH3 ]2
[N2 ][H2 ]3 = 3.5 x 108
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-favoredproduct-favored..
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The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of KFor AgCl(s) For AgCl(s) ee
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 reactants than that of reactants at equilibrium. at equilibrium.
The reaction with small K is The reaction with small K is strongly strongly
reactant-reactant-favoredfavored..
AgAg++(aq) + Cl(aq) + Cl--(aq) (aq)
ee AgCl(s) AgCl(s)is product-favored.is product-favored.
AgAg++(aq) + Cl(aq) + Cl--(aq) (aq)
ee AgCl(s) AgCl(s)is product-favored.is product-favored.
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Product- or Reactant Product- or Reactant FavoredFavored
Product-favoredProduct-favoredK > 1K > 1
Reactant-favoredReactant-favoredK < 1K < 1
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The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of KK comes from thermodynamics. (K comes from thermodynamics. (See See
Chapter 19)Chapter 19)
∆∆G˚ < 0: reaction is G˚ < 0: reaction is product favoredproduct favored
∆∆G˚ > 0: reaction is G˚ > 0: reaction is reactant-favoredreactant-favored
If K > 1, then ∆G˚ is negative
If K < 1, then ∆G˚ is positive
If K > 1, then ∆G˚ is negative
If K < 1, then ∆G˚ is positive
∆Go
= -RT ln K ∆Go
= -RT ln K
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The 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
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The Meaning of KThe Meaning of K
If [iso] = 0.35 M and [n] = 0.15 M, are If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium?you at equilibrium?
If not, which way does the reaction If not, which way does the reaction “shift” to approach equilibrium?“shift” to approach equilibrium?
See Chemistry NowSee Chemistry Now
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 KAll reacting chemical systems are All reacting chemical systems are
characterized by their characterized by their REACTION QUOTIENT, REACTION QUOTIENT, QQ..
Q =
product concentrations
reactant concentrations Q =
product concentrations
reactant concentrations
Q =
conc. of iso
conc. of n =
0.35
0.15 = 2.3
Q =
conc. of iso
conc. of n =
0.35
0.15 = 2.3
Q (2.33) < K (2.5)Q (2.33) < K (2.5) Reaction is Reaction is NOTNOT at equilibrium, so [iso] must at equilibrium, so [iso] must become ________ and [n] must become ________ and [n] must ________________________..
If Q = K, then system is at equilibrium.If Q = K, then system is at equilibrium.
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ÆÆ
Typical 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.
Kc = [HI]2
[H2 ][I2 ] = 55.3
Kc = [HI]2
[H2 ][I2 ] = 55.3
HH22(g) + I(g) + I22(g) (g) ee 2 HI(g) 2 HI(g)
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HH22(g) + I(g) + I22(g) (g) ee 2 HI(g) 2 HI(g)KKcc = 55.3 = 55.3
HH22(g) + I(g) + I22(g) (g) ee 2 HI(g) 2 HI(g)KKcc = 55.3 = 55.3
Step 1. Set up Step 1. Set up ICEICE table to define 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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Step 1. Set up Step 1. Set up ICEICE table to define 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.
HH22(g) + I(g) + I22(g) (g) ee 2 HI(g) 2 HI(g)KKcc = 55.3 = 55.3
HH22(g) + I(g) + I22(g) (g) ee 2 HI(g) 2 HI(g)KKcc = 55.3 = 55.3
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HH22(g) + I(g) + I22(g) (g) ee 2 HI(g) 2 HI(g)KKcc = 55.3 = 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] = 55.3
Kc =
[2x]2
[1.00-x][1.00-x] = 55.3
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[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
[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
Step 3. Solve KStep 3. Solve Kcc expression - take expression - take
square root of both sides.square root of both sides.
x = 0.79x = 0.79Therefore, at equilibriumTherefore, at equilibrium
HH22(g) + I(g) + I22(g) (g) ee 2 HI(g) 2 HI(g)KKcc = 55.3 = 55.3
Kc =
[2x]2
[1.00-x][1.00-x] = 55.3
Kc =
[2x]2
[1.00-x][1.00-x] = 55.3
7.44 =
2x
1.00-x 7.44 =
2x
1.00-x
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Nitrogen Dioxide Nitrogen Dioxide EquilibriumEquilibrium
NN22OO44(g) (g) ee 2 NO 2 NO22(g)(g)
Nitrogen Dioxide Nitrogen Dioxide EquilibriumEquilibrium
NN22OO44(g) (g) ee 2 NO 2 NO22(g)(g)
eeee
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Nitrogen Dioxide EquilibriumNitrogen Dioxide Equilibrium
NN22OO44(g) (g) ee 2 NO 2 NO22(g)(g)
Nitrogen Dioxide EquilibriumNitrogen Dioxide Equilibrium
NN22OO44(g) (g) ee 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 ICE tableStep 1. Set up an ICE table
[N[N22OO44]] [NO[NO22]]
InitialInitial 0.500.50 00
ChangeChange
EquilibEquilib
Kc = [NO2 ]2
[N2O4 ] = 0.0059 at 298 K
Kc = [NO2 ]2
[N2O4 ] = 0.0059 at 298 K
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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 ICE tableStep 1. Set up an ICE table
[N[N22OO44]] [NO[NO22]]
InitialInitial 0.500.50 00
ChangeChange -x-x +2x+2x
EquilibEquilib 0.50 - x0.50 - x 2x2x
Nitrogen Dioxide EquilibriumNitrogen Dioxide Equilibrium
NN22OO44(g) (g) ee 2 NO 2 NO22(g)(g)
Nitrogen Dioxide EquilibriumNitrogen Dioxide Equilibrium
NN22OO44(g) (g) ee 2 NO 2 NO22(g)(g)
Kc = [NO2 ]2
[N2O4 ] = 0.0059 at 298 K
Kc = [NO2 ]2
[N2O4 ] = 0.0059 at 298 K
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Nitrogen Dioxide EquilibriumNitrogen Dioxide Equilibrium
NN22OO44(g) (g) ee 2 NO 2 NO22(g)(g)
Nitrogen Dioxide EquilibriumNitrogen Dioxide Equilibrium
NN22OO44(g) (g) ee 2 NO 2 NO22(g)(g)Step 2. Substitute into KStep 2. Substitute into Kcc expression and solve. expression and solve.
Kc = 0.0059 = [NO2 ]2
[N2O4 ]=
(2x)2
(0.50 - x)
Kc = 0.0059 = [NO2 ]2
[N2O4 ]=
(2x)2
(0.50 - x)
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 = 0a = 4a = 4 b = 0.0059b = 0.0059 c = -0.0029c = -0.0029
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Nitrogen Dioxide EquilibriumNitrogen Dioxide Equilibrium
NN22OO44(g) (g) ee 2 NO 2 NO22(g)(g)
Nitrogen Dioxide EquilibriumNitrogen Dioxide Equilibrium
NN22OO44(g) (g) ee 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 = 0.0059b = 0.0059 c = -0.0029c = -0.0029
x =
-b ± b2 - 4ac
2a x =
-b ± b2 - 4ac
2a
x =
-0.0059 ± (0.0059)2 - 4(4)(-0.0029)
2(4) x =
-0.0059 ± (0.0059)2 - 4(4)(-0.0029)
2(4)
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
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Nitrogen Dioxide EquilibriumNitrogen Dioxide Equilibrium
NN22OO44(g) (g) ee 2 NO 2 NO22(g)(g)
Nitrogen Dioxide EquilibriumNitrogen Dioxide Equilibrium
NN22OO44(g) (g) ee 2 NO 2 NO22(g)(g)
x = x = 0.0260.026 or or -0.028-0.028
But a negative value is not reasonable.But a negative value is not reasonable.
Conclusion: Conclusion: x = 0.026 Mx = 0.026 M [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)
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
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Solving Quadratic Solving Quadratic EquationsEquations
• Recommend you solve the equation exactly on a calculator or use the “method of successive approximations”
• See Appendix A.
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Writing and Manipulating Writing and Manipulating K ExpressionsK Expressions
Writing and Manipulating Writing and Manipulating K ExpressionsK Expressions
Adding equations for reactionsAdding equations for reactions
S(s) + OS(s) + O22(g) (g) ee SO SO22(g)(g)
SOSO22(g) + 1/2 O(g) + 1/2 O22(g) (g) ee SO SO33(g(g))
K1 = [SO2 ]
[O2 ]
K1 = [SO2 ]
[O2 ]
K2 = [SO3 ]
[SO2 ][O2 ]1/2
K2 = [SO3 ]
[SO2 ][O2 ]1/2
Net equationNet equation
S(s) + 3/2 OS(s) + 3/2 O22(g) (g) ee SO SO33(g)(g)
Knet = [SO3 ]
[O2 ]3/2
Knet = [SO3 ]
[O2 ]3/2 Knet = K1 x K2 Knet = K1 x K2
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Writing and Manipulating Writing and Manipulating K ExpressionsK Expressions
Writing and Manipulating Writing and Manipulating K ExpressionsK Expressions
Changing coefficientsChanging coefficients
S(s) + 3/2 OS(s) + 3/2 O22(g) (g) ee SO SO33(g)(g)
2 S(s) + 3 O2 S(s) + 3 O22(g) (g) ee 2 SO 2 SO33(g)(g)
K = [SO3 ]
[O2 ]3/2
K = [SO3 ]
[O2 ]3/2
Knew= [SO3 ]2
[O2 ]3
Knew= [SO3 ]2
[O2 ]3
Knew= [SO3 ]2
[O2 ]3 = (Kold)2
Knew= [SO3 ]2
[O2 ]3 = (Kold)2
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Writing and Manipulating Writing and Manipulating K ExpressionsK Expressions
Writing and Manipulating Writing and Manipulating K ExpressionsK Expressions
Changing directionChanging direction
S(s) + OS(s) + O22(g) (g) ÆÆ SO SO22(g)(g)
SOSO22(g) (g) ÆÆ 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 Writing and Manipulating K ExpressionsK Expressions
Writing and Manipulating Writing and Manipulating K ExpressionsK Expressions
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)But with gases, P = (n/V)··RT = conc RT = conc ·· RT RT
P is proportional to concentration, so we can write K P is proportional to concentration, so we can write K
in terms of P. These are designated by in terms of P. These are designated 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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Writing and Manipulating Writing and Manipulating K ExpressionsK Expressions
Writing and Manipulating Writing and Manipulating K ExpressionsK Expressions
K using concentration and pressure unitsK using concentration and pressure units
KKpp = K = Kc c (RT)(RT)∆n∆n
For For S(s) + OS(s) + O22(g) (g) ee SO SO22(g)(g)
∆∆n = 0 and Kn = 0 and Kpp = K = Kcc
For SOFor SO22(g) + 1/2 O(g) + 1/2 O22(g) (g) ee SO SO33(g)(g)
∆∆n = –1/2 and Kn = –1/2 and Kpp = K = Kcc(RT)(RT)–1/2–1/2
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EQUILIBRIUM AND EQUILIBRIUM AND EXTERNAL EFFECTSEXTERNAL EFFECTS
• Temperature, catalysts, and changes in Temperature, catalysts, and changes in concentration affect equilibria.concentration affect equilibria.
• The outcome is governed by The outcome is governed by LE LE CHATELIER’S PRINCIPLECHATELIER’S PRINCIPLE
• ““...if a system at equilibrium is ...if a system at equilibrium is disturbed, the system tends to shift its disturbed, the system tends to shift its equilibrium position to counter the equilibrium position to counter the effect of the disturbance.”effect of the disturbance.”
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EQUILIBRIUM AND EQUILIBRIUM AND EXTERNAL EFFECTSEXTERNAL EFFECTSEQUILIBRIUM AND EQUILIBRIUM AND EXTERNAL EFFECTSEXTERNAL EFFECTS
Henri Le ChatelierHenri Le Chatelier
1850-19361850-1936
Studied mining Studied mining engineering.engineering.
Interested in glass Interested in glass and ceramics.and ceramics.
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EQUILIBRIUM AND EXTERNAL EFFECTSEQUILIBRIUM AND EXTERNAL EFFECTSEQUILIBRIUM AND EXTERNAL EFFECTSEQUILIBRIUM AND EXTERNAL EFFECTS
• Temperature change Temperature change ff change in K change in K• Consider the fizz in a soft drinkConsider the fizz in a soft drink
COCO22(aq) + (aq) + HEATHEAT ee COCO22(g) + H(g) + H22O(liq) O(liq)
• K = P (COK = P (CO22) / [CO) / [CO22] ]
• Increase T. What happens to equilibrium Increase T. What happens to equilibrium position? To value of K?position? To value of K?
• K increases as T goes up because P(COK increases as T goes up because P(CO22) )
increases and [COincreases and [CO22] decreases.] decreases.
• Decrease T. Now what?Decrease T. Now what?
• Equilibrium shifts left and K decreases.Equilibrium shifts left and K decreases.
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Temperature Temperature Effects on Effects on EquilibriumEquilibrium
NN22OO44 (colorless) + (colorless) + heatheat
ee 2 NO 2 NO22 (brown) (brown)
∆H∆Hoo = + 57.2 kJ = + 57.2 kJ
Kc = [NO2 ]2
[N2O4 ]
Kc = [NO2 ]2
[N2O4 ]
KKc (273 K) = 0.00077 (273 K) = 0.00077
KKc (298 K) = 0.0059 (298 K) = 0.0059PLAY MOVIE
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Temperature Effects on Temperature Effects on EquilibriumEquilibrium
See Figure 16.8See Figure 16.8
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EQUILIBRIUM AND EXTERNAL EFFECTSEQUILIBRIUM AND EXTERNAL EFFECTS
• Add catalyst Add catalyst ff no change in K no change in K• A catalyst only affects the RATE of A catalyst only affects the RATE of
approach to equilibrium.approach to equilibrium.
Catalytic exhaust systemCatalytic exhaust system
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• NN22(g) + 3 H(g) + 3 H22(g) (g) ee 2 NH 2 NH33(g) + (g) + heatheat
• K = 3.5 x 10K = 3.5 x 1088 at 298 K at 298 K
Haber-Bosch Process for NHHaber-Bosch Process for NH33
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Haber-Bosch Haber-Bosch Ammonia SynthesisAmmonia Synthesis
Fritz HaberFritz Haber1868-19341868-1934Nobel Prize, 1918Nobel Prize, 1918
Carl BoschCarl Bosch1874-19401874-1940Nobel Prize, 1931Nobel Prize, 1931
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EQUILIBRIUM AND EXTERNAL EFFECTSEQUILIBRIUM AND EXTERNAL EFFECTS
• Concentration changes Concentration changes
–no change in K no change in K
–only the equilibrium only the equilibrium composition changes.composition changes.
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Le Chatelier’s PrincipleLe Chatelier’s PrincipleLe Chatelier’s PrincipleLe Chatelier’s Principle
Adding a “reactant” to a chemical Adding a “reactant” to a chemical system.system.
PLAY MOVIE
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© 2009 Brooks/Cole - Cengage
Le Chatelier’s PrincipleLe Chatelier’s PrincipleLe Chatelier’s PrincipleLe Chatelier’s Principle
Removing a “reactant” from a chemical Removing a “reactant” from a chemical system.system.
PLAY MOVIE
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© 2009 Brooks/Cole - Cengage
Le Chatelier’s PrincipleLe Chatelier’s PrincipleLe Chatelier’s PrincipleLe Chatelier’s Principle
Adding a “product” to a chemical Adding a “product” to a chemical system.system.
PLAY MOVIE
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© 2009 Brooks/Cole - Cengage
Le Chatelier’s PrincipleLe Chatelier’s PrincipleLe Chatelier’s PrincipleLe Chatelier’s Principle
Removing a “product” from a chemical Removing a “product” from a chemical system.system.
PLAY MOVIE
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© 2009 Brooks/Cole - Cengage
EQUILIBRIUM AND EQUILIBRIUM AND EXTERNAL EFFECTSEXTERNAL EFFECTS
• Temperature, catalysts, and changes in Temperature, catalysts, and changes in concentration affect equilibria.concentration affect equilibria.
• The outcome is governed by The outcome is governed by LE LE CHATELIER’S PRINCIPLECHATELIER’S PRINCIPLE
• ““...if a system at equilibrium is ...if a system at equilibrium is disturbed, the system tends to shift its disturbed, the system tends to shift its equilibrium position to counter the equilibrium position to counter the effect of the disturbance.”effect of the disturbance.”
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Butane-Butane-Isobutane Isobutane EquilibriEquilibri
umum
Butane-Butane-Isobutane Isobutane EquilibriEquilibri
umum
K =
[isobutane]
[butane]= 2.5
K =
[isobutane]
[butane]= 2.5
butanebutane
isobutaneisobutane
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ButaneButane ee IsobutaneIsobutane
• At equilibrium with [iso] = 1.25 M and At equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. K = 2.5.[butane] = 0.50 M. K = 2.5.
• Add 1.50 M butane. Add 1.50 M butane. • When the system comes to equilibrium When the system comes to equilibrium
again, what are [iso] and [butane]? again, what are [iso] and [butane]?
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Assume you are at equilibrium with [iso] = 1.25 M and [butane] = 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 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]? equilibrium again, what are [iso] and [butane]? K = 2.5K = 2.5
SolutionSolution
Calculate Q immediately after adding more Calculate Q immediately after adding more butane and compare with K.butane and compare with K.
Q =
[isobutane]
[butane] =
1.25
0.50 + 1.50 = 0.625
Q =
[isobutane]
[butane] =
1.25
0.50 + 1.50 = 0.625
Q is LESS THAN K. Therefore, the Q is LESS THAN K. Therefore, the reaction will shift to the ____________.reaction will shift to the ____________.Q is LESS THAN K. Therefore, the Q is LESS THAN K. Therefore, the reaction will shift to the ____________.reaction will shift to the ____________.
ButaneButane ee IsobutaneIsobutane
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You are at equilibrium with [iso] = 1.25 M and [butane] You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. = 0.50 M. Now add 1.50 M butane.
SolutionSolutionQ is less than K, so equilibrium shifts right — Q is less than K, so equilibrium shifts right —
away from butane and toward isobutane.away from butane and toward isobutane.Set up ICE tableSet up ICE table [butane][butane] [isobutane][isobutane]InitialInitial ChangeChangeEquilibriumEquilibrium
0.50 + 1.500.50 + 1.50 1.251.25
- x- x + x+ x
2.00 - x2.00 - x 1.25 + x1.25 + x
ButaneButane ee IsobutaneIsobutane
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You are at equilibrium with [iso] = 1.25 M and [butane] You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. = 0.50 M. Now add 1.50 M butane.
SolutionSolution
K = 2.50 =
[isobutane]
[butane] =
1.25 + x
2.00 - x K = 2.50 =
[isobutane]
[butane] =
1.25 + x
2.00 - x
x = 1.07 Mx = 1.07 MAt the new equilibrium position, At the new equilibrium position, [butane] = 0.93 M and [isobutane] = 2.32 M. [butane] = 0.93 M and [isobutane] = 2.32 M.
Equilibrium has shifted toward isobutane.Equilibrium has shifted toward isobutane.
ButaneButane ee IsobutaneIsobutane
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Le Chatelier’s PrincipleLe 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”
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Nitrogen Dioxide Nitrogen Dioxide EquilibriumEquilibrium
NN22OO44(g) (g) e 2 NO 2 NO22(g)(g)
Increase P in the system Increase P in the system by reducing the by reducing the volume (at constant T). volume (at constant T).
Kc = [NO2 ]2
[N2O4 ] = 0.0059 at 298 K
Kc = [NO2 ]2
[N2O4 ] = 0.0059 at 298 K
ee
PLAY MOVIE
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Nitrogen Dioxide Nitrogen Dioxide EquilibriumEquilibrium
NN22OO44(g) (g) e 2 NO 2 NO22(g)(g)
Increase P in the system by reducing the volume. Increase P in the system by reducing the volume.
In gaseous system the equilibrium will shift to the side In gaseous system the equilibrium will shift to the side with fewer molecules (in order to reduce the P). with fewer molecules (in order to reduce the P).
Therefore, reaction shifts Therefore, reaction shifts LEFTLEFT and P of NOand P of NO22 decreases decreases and P of Nand P of N22OO44 increases. increases.
Kc = [NO2 ]2
[N2O4 ] = 0.0059 at 298 K
Kc = [NO2 ]2
[N2O4 ] = 0.0059 at 298 K