kp and kc
TRANSCRIPT
Chapter 14 Notes
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14Chemistry: A Molecular Approach by Nivaldo J. TroChemistry: A Molecular Approach by Nivaldo J. TroCHEM 1411 General ChemistryCHEM 1411 General Chemistry
Mr. Kevin A. BoudreauxAngelo State Universitywww.angelo.edu/faculty/kboudrea
Mr. Kevin A. BoudreauxAngelo State Universitywww.angelo.edu/faculty/kboudrea
Chemical EquilibriumChemical EquilibriumObjectives:• Understand the concept of equilibrium, the significance of
the equilibrium constant K, how to manipulate values of K, and how to assess the direction in which an equilibrium proceeds.
• Understand how to perform equilibrium calculations using reaction tables.
• Use Le Châtelier’s principle to predict the direction in which equilibria can be shifted.
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Chemical EquilibriumChemical Equilibrium• Kinetics tells us how quickly a chemical reaction
takes place; equilibrium tells us how far the reaction will go — that is, how much product there will be when the reaction is “done.”
• Most chemical reactions proceed until they reach a state of chemical equilibrium, in which the concentrations of the reactants and products no longer change.– Many chemical reactions proceed in both a
forward and reverse direction.– When the rate in the forward direction equals the
rate in the reverse direction, the concentrations of the products and any remaining reactants remains constant.
Chapter 14 Notes
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The Equilibrium The Equilibrium Constant, Constant, KK
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Dynamic EquilibriumDynamic Equilibrium• H2 and I2 react to form HI:
H2(g) + I2(g) → 2HI(g)• However, the reaction is reversible,
and can also run backwards:2HI(g) → H2(g) + I2(g)
– At the beginning of the reaction (14.2a), the concentrations of H2 and I2 are high, and the concentration of HI is zero. The rate in the forward direction is high.
– As the reaction proceeds (14.2b-d), the no. of H2’s and I2’s decrease, and the number of HI’s increase, slowing down the forward reaction but speeding up the reverse reaction.
– Eventually, the the rates of the forward and reverse reactions are equal, and no further change in concentration takes place. Movie
(1st half)Figure 14.2
Chapter 14 Notes
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Equilibrium ConcentrationsEquilibrium Concentrations• There is still stuff happening at the molecular level;
but the change in one direction equals exactly the change in the opposite direction — we have a dynamic equilibrium:
H2(g) + I2(g) h 2HI(g)
Figure 14.2
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Reversible ReactionsReversible Reactions• In a dynamic equilibrium, the rate of the forward
reaction is equal to the rate of the reverse reaction.– This (apparent) cessation of the reaction arises
because the change in the forward direction cancels out the change in the reverse direction.
– The same equilibrium state could be reached by starting with HI instead of H2and I2.
• To indicate an equilibrium reaction, we use two half-arrows, one pointing in each direction:
• All chemical reactions are in principle reversible; “irreversible” reactions are those that proceed nearly to completion, so that the final equilibrium mixture contains almost entirely products.
h
Chapter 14 Notes
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The Law of Chemical EquilibriumThe Law of Chemical Equilibrium• For the reaction
N2O4(g) h 2NO2(g)• the rate in the forward and reverse directions are:
ratefwd = kfwd[N2O4]raterev = krev[NO2]2
• At equilibrium, ratefwd = raterev
kfwd[N2O4] = krev[NO2]2
]O[N][NO 42
22
rev
fwd ==kkK
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The Law of Chemical EquilibriumThe Law of Chemical Equilibrium• Law of chemical equilibrium or law of mass
action — at a given temperature, a chemical system reaches a state in which a particular ratio of reactant and product concentration terms is a constant.– For a particular system and temperature, the same
equilibrium state is reached no matter what the starting concentrations are.
Equilibrium Constant
Equilibrium Concentration
Initial Concentrations
4.63×10-30.0141 M0.0429 M0.0600 M0.0200 M54.65×10-30.0107 M0.0246 M0.0600 M0.0000 M44.66×10-30.0156 M0.0522 M0.0000 M0.0600 M34.64×10-30.0125 M0.0337 M0.0800 M0.0000 M24.64×10-30.0125 M0.0337 M0.0000 M0.0400 M1[NO2]2/[N2O4][NO2][N2O4][NO2][N2O4]Expt.
Chapter 14 Notes
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The Equilibrium Constant, KThe Equilibrium Constant, Kcc• For any general reversible reaction
aA + bB h cC + dD• we can define an equilibrium constant, Kc, using
the following equilibrium equation:
ba
dc
K[B][A][D][C] c =
Equilibrium equation:
Equilibrium constant
Equilibrium constant expression or mass action expression
Reactants
Products
coefficients
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The Equilibrium Constant, The Equilibrium Constant, KKcc• No matter what the individual equilibrium
concentrations may be for a particular experiment, the equilibrium constant for a reaction at a particular temperature is always the same.
• We will write values of Kc as unitless numbers.
Chapter 14 Notes
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The Significance of the Equilibrium ConstantThe Significance of the Equilibrium Constant• The numerical value of the equilibrium constant
indicates the extent to which reactants are converted to products:
Significant amounts of both reactant and product present at equilibrium.K ≈ 1
Very little product formedK << 1
Little reactant left at equilibrium (reaction “goes to completion”)K >> 1
Figure 14.4 Figure 14.5
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Examples: Equilibrium ConstantsExamples: Equilibrium Constants1. Which of the following reactions will tend to
proceed farthest toward completion?
a. H2(g) + Br2(g) h 2HBr(g); Kc = 1.4×10-21
b. 2NO(g) h N2(g) + O2(g); Kc = 2.1×1030
c. 2BrCl h Br2 + Cl2; Kc = 0.145
Chapter 14 Notes
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Examples: Equilibrium ConstantsExamples: Equilibrium Constants2. Write the equilibrium equation for the following
reactions. (sim. to Ex. 14.1)a. N2(g) + H2(g) h NH3(g) [unbalanced]
b. C3H8(g) + O2(g) h CO2(g) + H2O(g) [unbal.]
c. 2H2(g) + O2(g) h 2H2O(g)
d. PCl3(g) + Cl2(g) h PCl5(g)
e. PCl5(g) h PCl3(g) + Cl2(g)
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Manipulating K ExpressionsManipulating K Expressions• If all of the coefficients in a balanced equation are
multiplied by a number, each term in the Kexpression must be raised to that power.
• The K for a reverse reaction is the reciprocal of the K for the forward reaction. (This is equivalent to multiplying the equation by -1 and using -1 as an exponent).
Kc(rev) = 1 / Kc(fwd)
• For a reaction which is the sum of two or more reactions, the overall K is the product of the expressions for K for the individual steps:
Koverall = K1 × K2 × K3 • • •
Chapter 14 Notes
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Examples: Manipulating Examples: Manipulating KK’’ss3. Given the following reaction and Kc,N2(g) + 3H2(g) h 2NH3(g); Kc = 2.4×10-3 at 1000K
calculate the value of Kc for the following equations (similar to Ex. 14.2).a. 2NH3(g) h N2(g) + 3H2(g)
b. NH3(g) h 1/2N2(g) + 3/2H2(g)
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Examples: Manipulating Examples: Manipulating KK’’ss4. At 25°C, the following reactions have the
equilibrium constants shown:2CO(g) + O2(g) h 2CO2(g); Kc1 = 3.3 × 1091
2H2(g) + O2(g) h 2H2O(g); Kc2 = 9.1 × 1080
Use these data to calculate Kc for the reactionCO(g) + H2O(g) h CO2(g) + H2(g)
Answer: 1.9×105
Chapter 14 Notes
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The Equilibrium Constant, KThe Equilibrium Constant, Kpp• The equilibrium constant obtained when all species
are present at their equilibrium partial pressures is Kp, the equilibrium constant based on pressures.
2NO(g) + O2(g) h 2NO2(g)
• Kp and Kc are related by the following equation:
Δngas = moles of gas product - moles of gas reactants and R is the gas constant (0.0821 L atm K-1 mol-1).
• When there is no change in the number of moles of gas, Δngas = 0, and Kp = Kc.
2
2
O2
NO
2NO
p PP
PK =
gas)( cpnRTKK Δ=
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Examples: Relating Examples: Relating KKcc and and KKpp5. Calculate Kc for the following reaction: (sim. to
Ex. 14.3)CaCO3(s) h CaO(s) + CO2(g); Kp = 2.1×10-4 at 1000K
Answer: 2.6×10-6
Chapter 14 Notes
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Heterogeneous EquilibriaHeterogeneous Equilibria• Homogeneous equilibrium — all reactants and
products are in the same phase.• Heterogeneous equilibrium — more than one phase
exists in a reaction mixture. (One or more of the components is a liquid or solid.)
CaCO3(s) h CaO(s) + CO2(g); Kc = ?• A pure solid always has the same concentration at a
given temperature, and its volume does not change much.
• Since we’re only concerned with quantities that are changing, we eliminate the terms for pure liquids and solids from the reaction quotient:
Kc = [CO2] Kp = PCO2
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Examples: Heterogeneous EquilibriaExamples: Heterogeneous Equilibria6. Write equilibrium laws for the following
heterogeneous reactions. (sim. to Ex. 14.4)a. 2CO(g) h CO2(g) + C(s)
b. 2Hg(l) + Cl2(g) h Hg2Cl2(s)
c. NH3(g) + HCl(g) h NH4Cl(s)
d. Ag2CrO4(s) h 2Ag+(aq) + CrO42–(aq)
e. 2NaHCO3(s) h Na2CO3(s) + H2O(g) + CO2(g)
Chapter 14 Notes
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The Reaction The Reaction Quotient, Quotient, QQ
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The Reaction Quotient, QThe Reaction Quotient, Q• At conditions other than equilibrium, the equilibrium
constant expression can be evaluated to obtain a value called the reaction quotient, Q:
aA + bB h cC + dD
• The numerical value of Q changes during the course of the reaction until the system reaches equilibrium. At this point, there is no further change and
Q = K (at equilibrium)
ba
dc
Q[B][A][D][C] c = ba
dc
QBA
DCp P P
P P =
Chapter 14 Notes
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Predicting the Direction of ReactionPredicting the Direction of Reaction• By comparing the value of Q at a particular time
with the value of K, we can determine if the reaction is at equilibrium, or if not, in which direction it is progressing:
No net reaction occurs; the reaction is at equilibrium:
reactants h productsQ = K
The net reaction goes to the left:reactants ← productsQ > K
The net reaction goes to the right:reactants → productsQ < K
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Predicting the Direction of ReactionPredicting the Direction of Reaction
Chapter 14 Notes
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Predicting the Direction of ReactionPredicting the Direction of Reaction
Figure 14.7
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Examples: The Direction of ReactionExamples: The Direction of Reaction7. For the reaction
N2O4(g) h 2NO2(g)Kc = 0.21 at 100°C. At a point during the reaction, [N2O4] = 0.12 M and [NO2] = 0.55 M. (a) Calculate the value of Qc for the reaction. (b) Is the reaction at equilibrium, and if not, in what direction is it progressing? (sim. to Ex. 14.7)
Answer: Qc = (a) 2.5; (b) left
Chapter 14 Notes
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Examples: Minding Your Examples: Minding Your KK’’ss and Qand Q’’ss8. Consider the reaction
I2(g) + Cl2(g) h 2ICl(g), Kp = 81.9A reaction mixture contains 0.114 atm of I2, 0.102 atm of Cl2, and 0.355 atm of ICl. (a) Calculate the value of Qp for the reaction. (b) Is the reaction at equilibrium, and if not, in what direction is it progressing? (Ex. 14.7)
Answer: Qc = (a) 10.8; (b) right
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Examples: Minding Your Examples: Minding Your KK’’ss and Qand Q’’ss9. For the reaction
CO(g) + H2O(g) h CO2(g) + H2(g)Kc = 4.06 at 500°C. If [CO] = 0.0331 M, [H2O] = 0.0331 M, [CO2] = 0.0667 M, and [H2] = 0.0667 M, calculate the value of Qc and state in what direction will the reaction progress in?
Answer: Qc = 4.06, at equilibrium
Chapter 14 Notes
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Equilibrium Equilibrium CalculationsCalculations
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How To Solve Equilibrium ProblemsHow To Solve Equilibrium Problems• Most equilibrium problems can be grouped into two
types:– Equilibrium quantities (concentrations or partial
pressures) are known, and the value of Kc or Kpmust be determined.
– Initial quantities and the value of K are known, and the equilibrium concentrations must be determined.
• Many of these types of problems require the use of an reaction table, or “ICE” table which will be illustrated in the following examples.
Chapter 14 Notes
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Examples: Equilibrium ProblemsExamples: Equilibrium Problems10. Calculating Kc from Equilibrium Concentrations
At a certain temperature, a mixture of H2 and I2was prepared by placing 0.200 mol of H2 and 0.200 mol of I2 into a 2.00 liter flask. After a period of time the equilibrium
H2(g) + I2(g) h 2HI(g)was established. The purple color of the I2 vapor was used to monitor the reaction, and from the decreased intensity of the purple color it was determined that, at equilibrium, the I2concentration had dropped to 0.020 mol/L. What is the value of Kc for this reaction at this temperature? (sim. to Ex. 14.9)
Answer: Kc = 64
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Examples: Equilibrium ProblemsExamples: Equilibrium Problems11. Calculating Kc from Equilibrium Concentrations
A student placed 0.200 mol of PCl3(g) and 0.100 mol of Cl2(g) into a 1.00 L container at 250°C. After the reaction
PCl3(g) + Cl2(g) h PCl5(g)came to equilibrium it was found that the flask contained 0.120 mol of PCl3. What were the initial and equilibrium concentrations of the reactants and products? What is the value of Kc for this reaction at this temperature?
Answer: Kc = 33
Chapter 14 Notes
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Examples: Equilibrium ProblemsExamples: Equilibrium Problems12. Determining an Equilibrium Concentration from
Kc and some Equilibrium ConcentrationsIn a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous CH4 and H2O in a 0.32 L flask at 1200 K. At equilibrium, the flask contains 0.26 mol CO, 0.091 mol H2, and 0.041 mol CH4. What is [H2O] at equilibrium? Kc= 0.26 for this reaction. (sim. to Ex. 14.8)
CH4(g) + H2O(g) h CO(g) + 3H2(g)
Answer: [H2O] = 0.53 M
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Examples: Equilibrium ProblemsExamples: Equilibrium Problems13. Using Kc to Calculate Equilibrium Concentrations
The equilibrium constant Kc for the reaction of H2with I2 is 57.0 at 700 K:H2(g) + I2(g) h 2HI(g); Kc = 57.0 at 700K
If 1.00 mol of H2 is allowed to react with 1.00 mol of I2 in a 10.0 L reaction vessel at 700 K, what are the concentrations of H2, I2, and HI at equilibrium?
Answer: [HI] 0.158 M, [H2] = [I2] = 0.021 M
Chapter 14 Notes
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Examples: Equilibrium ProblemsExamples: Equilibrium Problems14. Using Kc to Calculate Equilibrium Concentrations
The reactionCO(g) + H2O(g) h CO2(g) + H2(g)
has Kc = 4.06 at 500°C. If 0.100 mol of CO and 0.100 mol of H2O(g) are placed in a 1.00 L reaction vessel at this temperature, what are the concentrations of the reactants and products when the system reaches equilibrium?
Answer: [CO2]=[H2]=0.067 M, [CO]=[H2O]=0.033 M
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Examples: Equilibrium ProblemsExamples: Equilibrium Problems15. Using Kc to Calculate Equilibrium Concentrations
At a certain temperature Kc = 4.50 for the reactionN2O4(g) h 2NO2(g)
If 0.300 mol of N2O4 is placed into a 2.00 L container at this temperature, what will be the equilibrium concentrations of both gases?
Answer: [NO2] 0.268 M, [N2O4] = 0.016 M
Chapter 14 Notes
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Examples: Equilibrium ProblemsExamples: Equilibrium Problems16. Using Kp to Calculate Equilibrium Concentrations
One reaction that occurs in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and carbon dioxide. The equilibrium constant Kp for the reaction at 1000 K is 0.259.
FeO(s) + CO(g) h Fe(s) + CO2(g)Kp=0.259 at 1000 K
What are the equilibrium partial pressures of CO and CO2 at 1000 K if the initial partial pressures are PCO = 1.000 atm and PCO2 = 0.500 atm?
Answer: [CO2] 0.309 atm, [N2O4] = 1.191 atm
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Examples: Equilibrium ProblemsExamples: Equilibrium Problems17. Simplifying Equilibrium Calculations for Reactions
with Small Kc
Hydrogen, a potential fuel, is found in great abundance in water. Before the hydrogen can be used as a fuel, however, it must be separated from the oxygen; the water must be split into H2 and O2. One possibility is thermal decomposition, but this requires very high temperatures. Even at 1000°C, Kc = 7.3×10-18 for the reaction
2H2O(g) h 2H2(g) + O2(g)If the H2O concentration in a reaction vessel is set initially at 0.100 M, what will the H2 concentration be at equilibrium? (sim. to Ex. 14.12)
Answer: [H2] = 5.3×10-7 M
Chapter 14 Notes
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Examples: Equilibrium ProblemsExamples: Equilibrium Problems18. Simplifying Equilibrium Calculations for Reactions
with Small Kc
In air at 25°C and 1 atm, the N2 concentration is 0.033 M and the O2 concentration is 0.00810 M. The reaction
N2(g) + O2(g) h 2NO(g)has Kc = 4.8×10-31 at 25°C. Taking the N2 and O2concentrations given above as initial values, calculate the equilibrium NO concentration that should exist in our atmosphere from this reaction at 25°C.
Answer: [NO] = 1.1×10-17 M
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Chapter 14 Notes
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Disturbing the Disturbing the Equilibrium:Equilibrium:
Le ChâtelierLe Châtelier’’s s PrinciplePrinciple
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Le ChâtelierLe Châtelier’’s Principles Principle• There are a number of parameters that can be
adjusted in order to maximize the yield of products in a chemical reaction:– the concentration of the reactants or products.– the pressure and the volume.– the temperature.
• The effect that changing any of these parameters has on a reaction can be predicted using a principle first described by Henri-Louis Le Châtelier:
When a system at equilibrium is disturbed, the system shifts in a direction that minimizes the disturbance.
Le Châtelier’s Principle
Chapter 14 Notes
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Le ChâtelierLe Châtelier’’s Principles Principle• “Disturbance” in this context means a change in
concentration, pressure, volume, or temperature that disturbs the original equilibrium.– Reaction then occurs until a new state of
equilibrium is reached; the direction that the reaction takes is one that reduces the stress.
– The system, when disturbed, changes in a way that reduces the disturbance, and attains a new equilibrium.
MOV: Le Châtelier’s Principle
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Disturbing the Equilibrium Disturbing the Equilibrium —— ConcentrationConcentrationA + B h C + D
• Changing concentration:– If the concentration of a substance increases, the
equilibrium shifts to consume some of it.– If the concentration of a substance decreases, the
equilibrium shifts to produce some of it.• In terms of reactants and products:
– The equilibrium position shifts to the right if a reactant is added or a product is removed.
– The equilibrium position shifts to the left if a reactant is removed or a product is added.
Movie(2nd half)
Chapter 14 Notes
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Disturbing the Equilibrium Disturbing the Equilibrium —— ConcentrationConcentration
Figure 14.9
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Disturbing the Equilibrium Disturbing the Equilibrium —— ConcentrationConcentration
Figure 14.10
Chapter 14 Notes
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Disturbing the Equilibrium Disturbing the Equilibrium —— Volume/PressureVolume/Pressure• Changing pressure by changing volume: From
Boyle’s Law, we know that changing the volume of a reaction that involves gases changes the pressure:– Decreasing the volume of a system increases the
pressure, and favors the side of the reaction with the smaller number of moles of gas.
– Increasing the volume of a system decreases the pressure, and favors the side of the reaction with the larger number of moles of gas.
– If the two sides have the same number of moles of gas, changing the volume has no effect.
• Adding an inert gas increases the pressure, but does not change the partial pressure of any of the species, and does not affect the equilibrium.
MOV: N2O4NO2 Equilibrium
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Disturbing the Equilibrium Disturbing the Equilibrium —— Volume/PressureVolume/Pressure
Figure 14.11
Chapter 14 Notes
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Disturbing the Equilibrium Disturbing the Equilibrium —— TemperatureTemperature• The value of Kc is not affected by changing
concentration, pressure, or volume. Changing the temperature does change the value of Kc:
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Disturbing the Equilibrium Disturbing the Equilibrium —— TemperatureTemperature• Changing temperature:
– Raising temperature increases Kc if ΔHrxn > 0– Raising temperature decreases Kc if ΔHrxn < 0
It’s easier if you think about where heat is located:• Exothermic: A + B h C + D + heat
– adding heat (increasing temp.) drives the reaction to the left, while removing heat (decreasing temp.) drives the reaction to the right.
• Endothermic: A + B + heat h C + D– adding heat (increasing temp.) drives the reaction
to the right, while removing heat (decreasing temp.) drives the reaction to the left.
Chapter 14 Notes
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Disturbing the Equilibrium Disturbing the Equilibrium —— CatalystsCatalysts• Adding a catalyst:
– The rates of the forward and reverse reactions speed up by the same amount.
– A catalyst shortens the time a system needs to reach equilibrium, but has no effect on the position of the equilibrium.
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Chemical Equilibrium and KineticsChemical Equilibrium and Kinetics• The relative values of the rate constants for the
forward and the reverse reactions determine the composition of the equilibrium mixture:– When kf is much larger than kr, Kc is very large
and the reaction goes to completion; the reaction is essentially irreversible.
– When kr is much larger than kf, Kc is very small and the reaction does not produce much product.
– When kr is comparable to kf, Kc has a value near 1, and significant amounts of both reactants and products are present at equilibrium.
– Addition of a catalyst increases kr and kf, by the same amount, and does not affect the equilibrium composition.
Chapter 14 Notes
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Examples: Le ChâtelierExamples: Le Châtelier’’s Principles Principle19. (sim. to Ex. 14.14-14.16) The reaction
N2O4(g) h 2NO2(g)is endothermic, with ΔH° = +56.9 kJ. How will the amount of NO2 at equilibrium be affected by:a. adding N2O4
b. increasing the volume of the container
c. raising the temperature
d. adding a catalyst to the system?
e. Which of these changes alters the value of Kc?
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Examples: Le ChâtelierExamples: Le Châtelier’’s Principles Principle20. Iron(III) oxide reacts with carbon monoxide in a
blast furnace, reducing it to iron metal:Fe2O3(s) + 3CO(g) h 2Fe(l) + 3CO2(g)
Use Le Châtelier’s principle to predict the direction of net reaction when an equilibrium mixture is disturbed by:a. removing CO2
b. removing COc. adding COd. adding Fe2O3
e. decreasing the volume of the containerf. Which of these changes alters the value of Kc?
Chapter 14 Notes
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Examples: Le ChâtelierExamples: Le Châtelier’’s Principles Principle21. Does the number of moles of reaction products
increase, decrease, or remain the same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?a. PCl5(g) h PCl3(g) + Cl2(g)
b. H2(g) + I2(g) h 2HI(g)
c. C(s) + H2O(g) h CO(g) + H2(g)
d. CaO(s) + CO2(g) h CaCO3(s)
e. 3Fe(s) + 4H2O(g) h Fe3O4(s) + 4H2(g)
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Examples: Le ChâtelierExamples: Le Châtelier’’s Principles Principle22. How would you change the volume of each of the
following reactions to increase the yield of the products?a. CaCO3(s) h CaO(s) + CO2(g)
b. S(s) + 3F2(g) h SF6(g)
c. Cl2(g) + I2(g) h 2ICl(g)
Chapter 14 Notes
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Examples: Le ChâtelierExamples: Le Châtelier’’s Principles Principle23. How would an increase in temperature affect the
position of the equilibrium and the value of Kc for the following reactions?a. CaO(s) + H2O(l) h Ca(OH)2(aq); ΔH° = -82 kJ
b. CaCO3(s) h CaO(s) + CO2(g); ΔH° = +178 kJ
c. SO2(g) h S(s) + O2(g); ΔH° = +297 kJ
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Example: The Haber Synthesis of AmmoniaExample: The Haber Synthesis of Ammonia• Nitrogen is a vital element in many essential natural
and synthetic compounds. Eighty percent of the air we breathe is nitrogen, but N2 is very unreactive.– Natural nitrogen fixation of “unusable” N2 to
usable ammonia, NH3, occurs in bacteria on the root nodules of leguminous plants (peas, beans), and in lightning.
– Industrially, nitrogen is converted to ammonia by the Haber process (Fritz Haber, 1913):
N2(g) + 3H2(g) h 2NH3(g); ΔH°rxn = -91.8 kJ– Over 80% of the 110 million tons of ammonia
produced each year is used in fertilizers.• How would we maximize the yield of NH3 using Le
Châtelier’s Principle?
Chapter 14 Notes
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The Haber Synthesis of AmmoniaThe Haber Synthesis of Ammonia
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Equation SummaryEquation SummaryaA + bB h cC + dD
gas)( cpnRTKK Δ=
ba
dc
K[B][A][D][C] c = ba
dc
KBA
DCp P P
P P =
ba
dc
Q[B][A][D][C] c = ba
dc
QBA
DCp P P
P P =