Equilibrium © 2009, Prentice-Hall, Inc. Chapter 14 Chemical Equilibrium.
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Transcript of Equilibrium © 2009, Prentice-Hall, Inc. Chapter 14 Chemical Equilibrium.
Equilibrium
© 2009, Prentice-Hall, Inc.
Chapter 14Chemical Equilibrium
Equilibrium
© 2009, Prentice-Hall, Inc.
The Concept of Equilibrium
Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.
Equilibrium
© 2009, Prentice-Hall, Inc.
The Concept of Equilibrium• As a system
approaches equilibrium, both the forward and reverse reactions are occurring.
• At equilibrium, the forward and reverse reactions are proceeding at the same rate.
Equilibrium
© 2009, Prentice-Hall, Inc.
A System at Equilibrium
Once equilibrium is achieved, the amount of each reactant and product remains constant.
Equilibrium
© 2009, Prentice-Hall, Inc.
Depicting Equilibrium
Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow.
N2O4 (g) 2 NO2 (g)
Equilibrium
© 2009, Prentice-Hall, Inc.
The Equilibrium Constant
Equilibrium
© 2009, Prentice-Hall, Inc.
The Equilibrium Constant
• Forward reaction:N2O4 (g) 2 NO2 (g)
• Rate Law:Rate = kf [N2O4]
Equilibrium
© 2009, Prentice-Hall, Inc.
The Equilibrium Constant
• Reverse reaction:2 NO2 (g) N2O4 (g)
• Rate Law:Rate = kr [NO2]2
Equilibrium
© 2009, Prentice-Hall, Inc.
The Equilibrium Constant
• Therefore, at equilibrium
Ratef = Rater
kf [N2O4] = kr [NO2]2
• Rewriting this, it becomeskf
kr [NO2]2
[N2O4]=
Equilibrium
© 2009, Prentice-Hall, Inc.
The Equilibrium Constant
The ratio of the rate constants is a constant at that temperature, and the expression becomes
Keq =kf
kr [NO2]2
[N2O4]=
Equilibrium
© 2009, Prentice-Hall, Inc.
The Equilibrium Constant
• Consider the generalized reaction
• The equilibrium expression for this reaction would be
Kc = [C]c[D]d
[A]a[B]b
aA + bB cC + dD
Equilibrium
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The Equilibrium Constant
Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written
Kp =(PC
c) (PDd)
(PAa) (PB
b)
Equilibrium
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Relationship Between Kc and Kp
• From the Ideal Gas Law we know that
• Rearranging it, we get
PV = nRT
P = RTnV
Equilibrium
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Relationship Between Kc and Kp
Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes
where
Kp = Kc (RT)n
n = (moles of gaseous product) - (moles of gaseous reactant)
Equilibrium
© 2009, Prentice-Hall, Inc.
Equilibrium Can Be Reached from Either Direction
As you can see, the ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.
Equilibrium
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Equilibrium Can Be Reached from Either Direction
This is the data from the last two trials from the table on the previous slide.
Equilibrium
© 2009, Prentice-Hall, Inc.
Equilibrium Can Be Reached from Either Direction
It doesn’t matter whether we start with N2 and H2 or whether we start with NH3: we will have the same proportions of all three substances at equilibrium.
Equilibrium
© 2009, Prentice-Hall, Inc.
What Does the Value of K Mean?
• If K>>1, the reaction is product-favored; product predominates at equilibrium.
Equilibrium
© 2009, Prentice-Hall, Inc.
What Does the Value of K Mean?
• If K>>1, the reaction is product-favored; product predominates at equilibrium.
• If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium.
Equilibrium
© 2009, Prentice-Hall, Inc.
Manipulating Equilibrium Constants
The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.
Kc = = 0.212 at 100 C[NO2]2
[N2O4]N2O4 (g)
QuickTime™ and aPhoto - JPEG decompressor
are needed to see this picture. 2 NO2 (g)
Kc = = 4.72 at 100 C[N2O4][NO2]2
N2O4 (g)
QuickTime™ and aPhoto - JPEG decompressor
are needed to see this picture.2 NO2 (g)
Equilibrium
© 2009, Prentice-Hall, Inc.
Manipulating Equilibrium ConstantsThe equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number.
Kc = = 0.212 at 100 C[NO2]2
[N2O4]N2O4(g)
QuickTime™ and aPhoto - JPEG decompressor
are needed to see this picture. 2 NO2(g)
Kc = = (0.212)2 at 100 C[NO2]4
[N2O4]22 N2O4(g)
QuickTime™ and aPhoto - JPEG decompressor
are needed to see this picture. 4 NO2(g)
Equilibrium
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Manipulating Equilibrium Constants
The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps.
Equilibrium
© 2009, Prentice-Hall, Inc.
Heterogeneous Equilibrium
Equilibrium
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The Concentrations of Solids and Liquids Are Essentially Constant
Both can be obtained by multiplying the density of the substance by its molar mass — and both of these are constants at constant temperature.
Equilibrium
© 2009, Prentice-Hall, Inc.
The Concentrations of Solids and Liquids Are Essentially Constant
Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression.
Kc = [Pb2+] [Cl-]2
PbCl2 (s) Pb2+ (aq) + 2 Cl-(aq)
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are needed to see this picture.
Equilibrium
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As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same.
CaCO3 (s) CO2 (g) + CaO(s)
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are needed to see this picture.
Equilibrium
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Writing Equilibrium Constant Expressions
• The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm.
• The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions.
• The equilibrium constant is a dimensionless quantity.
• In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature.
• If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.
14.2
Equilibrium
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Equilibrium Calculations
Equilibrium
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An Equilibrium Problem
A closed system initially containing 1.000 x 10-3 M H2 and 2.000 x 10-3 M I2 at 448 C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10-3 M. Calculate Kc at 448 C for the reaction taking place, which is
H2 (g) + I2 (s) 2 HI (g)
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are needed to see this picture.
Equilibrium
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What Do We Know?
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change
At equilibrium 1.87 x 10-3
Equilibrium
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[HI] Increases by 1.87 x 10-3 M
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change +1.87 x 10-3
At equilibrium 1.87 x 10-3
Equilibrium
© 2009, Prentice-Hall, Inc.
Stoichiometry tells us [H2] and [I2] decrease by half as much.
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3
At equilibrium 1.87 x 10-3
Equilibrium
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We can now calculate the equilibrium concentrations of all three compounds…
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3
At equilibrium 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3
Equilibrium
© 2009, Prentice-Hall, Inc.
…and, therefore, the equilibrium constant.
Kc =[HI]2
[H2] [I2]
= 51
=(1.87 x 10-3)2
(6.5 x 10-5)(1.065 x 10-3)
Equilibrium
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Example• Sulfuryl chloride, SO2Cl2, reacts to
produce sulfur dioxide, and chlorine gas.
• 3.509 grams of the compound decomposes in a 1.00 L flask, the temp is raised to 375K, determine the pressure in the flask.
• At equilibrium the total pressure is 1.43 atm, calculate the partial pressure of all the substances.
• Calculate the Kp value for the reaction.
Equilibrium
© 2009, Prentice-Hall, Inc.
The Reaction Quotient (Q)
• Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.
• To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.
Equilibrium
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If Q = K,
the system is at equilibrium.
Equilibrium
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If Q > K,there is too much product, and the
equilibrium shifts to the left.
Equilibrium
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If Q < K,there is too much reactant, and the
equilibrium shifts to the right.
Equilibrium
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The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression.
IF
• Qc > Kc system proceeds from right to left to reach equilibrium
• Qc = Kc the system is at equilibrium
• Qc < Kc system proceeds from left to right to reach equilibrium
14.4
Equilibrium
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Le Châtelier’s Principle
Equilibrium
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Le Châtelier’s Principle
“If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”
Equilibrium
© 2009, Prentice-Hall, Inc.
If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.
Le Châtelier’s Principle
• Changes in Concentration
N2 (g) + 3H2 (g) 2NH3 (g)
AddNH3
Equilibrium shifts left to offset stress
14.5
Equilibrium
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Le Châtelier’s Principle
• Changes in Concentration continued
Change Shifts the Equilibrium
Increase concentration of product(s) left
Decrease concentration of product(s) right
Decrease concentration of reactant(s)
Increase concentration of reactant(s) right
left14.5
aA + bB cC + dD
AddAddRemove Remove
Equilibrium
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Le Châtelier’s Principle
• Changes in Volume and Pressure
A (g) + B (g) C (g)
Change Shifts the Equilibrium
Increase pressure Side with fewest moles of gas
Decrease pressure Side with most moles of gas
Decrease volume
Increase volume Side with most moles of gas
Side with fewest moles of gas
14.5
Equilibrium
© 2009, Prentice-Hall, Inc.
Le Châtelier’s Principle
• Changes in Temperature
Change Exothermic Rx
Increase temperature K decreases
Decrease temperature K increases
Endothermic Rx
K increases
K decreases
14.5colder hotter
Equilibrium
© 2009, Prentice-Hall, Inc.
uncatalyzed catalyzed
14.5
Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift equilibrium.
• Adding a Catalyst• does not change K• does not shift the position of an equilibrium system• system will reach equilibrium sooner
Le Châtelier’s Principle
Equilibrium
© 2009, Prentice-Hall, Inc.
Le Châtelier’s Principle
Change Shift EquilibriumChange Equilibrium
Constant
Concentration yes no
Pressure yes no
Volume yes no
Temperature yes yes
Catalyst no no
14.5
Equilibrium
© 2009, Prentice-Hall, Inc.
Catalysts
Equilibrium
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Catalysts
Catalysts increase the rate of both the forward and reverse reactions.
Equilibrium
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Catalysts
When one uses a catalyst, equilibrium is achieved faster, but the equilibrium composition remains unaltered.
Equilibrium
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Gibbs Free Energy and Chemical Equilibrium
G = G0 + RT lnQ
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature (K)
Q is the reaction quotient
At Equilibrium
G = 0 Q = K
0 = G0 + RT lnK
G0 = RT lnK
18.6
Equilibrium
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G0 = RT lnK
18.6
Equilibrium
© 2009, Prentice-Hall, Inc.
Ch 16: Solubility Equilibria
16.6
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = [Ag+][Cl-] Ksp is the solubility product constant
MgF2 (s) Mg2+ (aq) + 2F- (aq) Ksp = [Mg2+][F-]2
Ag2CO3 (s) 2Ag+ (aq) + CO32- (aq) Ksp = [Ag+]2[CO3
2-]
Ca3(PO4)2 (s) 3Ca2+ (aq) + 2PO43- (aq) Ksp = [Ca2+]3[PO3
3-]2
Dissolution of an ionic solid in aqueous solution:
Q = Ksp Saturated solution
Q < Ksp Unsaturated solution No precipitate
Q > Ksp Supersaturated solution Precipitate will form
Equilibrium
© 2009, Prentice-Hall, Inc.16.6
Equilibrium
© 2009, Prentice-Hall, Inc.
Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a saturated solution.
Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution.
16.6
Equilibrium
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What is the solubility of silver chloride in g/L ?
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = [Ag+][Cl-]Initial (M)
Change (M)
Equilibrium (M)
0.00
+s
0.00
+s
s s
Ksp = s2
s = Ksps = 1.3 x 10-5
[Ag+] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M
Solubility of AgCl = 1.3 x 10-5 mol AgCl
1 L soln143.35 g AgCl
1 mol AgClx = 1.9 x 10-3 g/L
Ksp = 1.6 x 10-10
16.6
Equilibrium
© 2009, Prentice-Hall, Inc.16.6
Equilibrium
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If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will a precipitate form?
16.6
The ions present in solution are Na+, OH-, Ca2+, Cl-.
Only possible precipitate is Ca(OH)2 (solubility rules).
Is Q > Ksp for Ca(OH)2?
[Ca2+]0 = 0.100 M [OH-]0 = 4.0 x 10-4 M
Ksp = [Ca2+][OH-]2 = 8.0 x 10-6
Q = [Ca2+]0[OH-]02 = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8
Q < Ksp No precipitate will form
Equilibrium
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What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br- and Cl- at a concentration of 0.02 M?
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = [Ag+][Cl-]
Ksp = 1.6 x 10-10
16.7
AgBr (s) Ag+ (aq) + Br- (aq) Ksp = 7.7 x 10-13
Ksp = [Ag+][Br-]
[Ag+] = Ksp
[Br-]7.7 x 10-13
0.020= = 3.9 x 10-11 M
[Ag+] = Ksp
[Br-]1.6 x 10-10
0.020= = 8.0 x 10-9 M
3.9 x 10-11 M < [Ag+] < 8.0 x 10-9 M
Equilibrium
© 2009, Prentice-Hall, Inc.
The Common Ion Effect and Solubility
The presence of a common ion decreases the solubility of the salt.
What is the molar solubility of AgBr in (a) pure water and (b) 0.0010 M NaBr?
AgBr (s) Ag+ (aq) + Br- (aq)
Ksp = 7.7 x 10-13
s2 = Ksp
s = 8.8 x 10-7
NaBr (s) Na+ (aq) + Br- (aq)
[Br-] = 0.0010 M
AgBr (s) Ag+ (aq) + Br- (aq)
[Ag+] = s
[Br-] = 0.0010 + s 0.0010
Ksp = 0.0010 x s
s = 7.7 x 10-10
16.8