ch 13 chemical equilibrium. the concept of equilibrium chemical equilibrium occurs when a reaction...
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CH 13
Chemical Equilibrium
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The Concept of EquilibriumThe Concept of Equilibrium
• Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.
• 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.
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Chemical equilibrium
• Dynamic processDynamic process • Rate of forward Rxn = Rate of reverse RxnRate of forward Rxn = Rate of reverse Rxn
HH22OO(l) (l) HH22OO((g)g)
(reactant) (product)(reactant) (product)
Dynamic
EquilibriumConcentration of reactants Concentration of reactants and products remain constantand products remain constant over time.over time.
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2SO2(g) + O2(g) 2SO3(g)
At Equilibium
SO2(g)+O2(g)
Initially
SO3(g)
Initially
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2SO2(g) + O2(g) 2SO3(g)
At Equilibium
SO2(g)+O2(g)
Initially
SO3(g)
Initially
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Depicting EquilibriumDepicting Equilibrium
• Consider the reversible reaction between N2O4 and NO2:
• In a system at equilibrium, both the forward and reverse reactions are being carried out; as a result, we write its equation with a double arrow.
N2O
4 2NO
2
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A System at EquilibriumA System at Equilibrium
• Once equilibrium is achieved, the amount of each reactant and product remains constant.
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The Equilibrium ConstantThe Equilibrium Constant
• Mathematically, the rate of a chemical reaction is described though its rate law.
• For the forward reaction:
• The forward rate law is given as:
Where: kf is the rate constant
[N2O4] is the concentration of N2O4
N2O
4 2NO
2
Rate = kf [N
2O
4]
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The Equilibrium ConstantThe Equilibrium Constant
• For the reverse reaction:
• And the rate law is:
Rate = kr [NO
2]2
2NO2 N
2O
4
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The Equilibrium ConstantThe Equilibrium Constant
• Therefore, at equilibrium
Ratef = Rater
kf [N2O4] = kr [NO2]2
• Rewriting, it becomes:
kf
kr
[NO
2]2
[N2O
4]
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The Equilibrium ConstantThe Equilibrium Constant
• The ratio of the rate constants is a constant, at that temperature, thus:
• Where Keq is the “Equilibrium Constant” for the reversible reaction.
K
eq
kf
kr
[NO
2]2
[N2O
4]
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The Equilibrium ConstantThe Equilibrium Constant
• To generalize this expression, consider the general chemical reaction:
• The equilibrium expression for this reaction would be:
• This is the generalized equilibrium constant based on concentration of the components in the system.
aA bB cC dD
K
c
[C]c [D]d
[A]a[B]b
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[C]c [D]d
[A]a [B]bKeq =
Equilibrium constant (K)
• Equilibrium expressionEquilibrium expression
(for any reaction at constant temperature)
aA + bB cC + dD
moles per litermoles per liter
coefficients
productsreactants
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[C]c [D]d
[A]a [B]bKeq =
aA + bB cC + dDproductsreactants
Equilibrium constant (K)
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N2(g) + 3 H2(g) 2 NH3(g)
Keq = [ NH3 ] 2
[ N2 ] [ H2 ] 3
Equilibrium constant (K)
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Equilibrium Expression
4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(g)
K NO H O
NH O2
2
24 6
34 7
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The Equilibrium ConstantThe Equilibrium Constant
• Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written as a pressure-dependent constant:
• Since, from the Ideal Gas Law:
K
p
pCc p
Dd
pAa p
Bb
PV nRT
p
i n
iV RT c
iRT
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Relationship between Relationship between KKcc and and KKpp
• Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes:
• Where: δn = (moles of product) – (moles of reactant)
K
pK
cRT n
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Relationship between Kc and Kp
• Relationship between concentration and pressure obtained from the ideal gas law. • Recall PV = nRT or
• Substitute for P in equilibrium expression. Consider the reaction:aA + bB cC + dD
• Use this relationship to relate KP and Kc
E.g. Determine Kp for the synthesis of ammonia at 25°C;N2(g) + 3H2(g) 2NH3(g) Kc = 4.1x108
E.g. 2 Determine Kp for the synthesis of HI at 458°C;½ H2(g) + ½ I2(g) HI(g) Kc = 0.142.
RT]A[
RTV
nP A
A
nc
)ba()dc(cP
baba
dcdc
ba
dc
bB
aA
dD
cC
P
RTKRTKK
RT]B[]A[
RT]D[]C[
RT]B[RT]A[
RT]D[RT]C[
PP
PPK
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• It does not 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.
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What Does the Value of What Does the Value of KK Mean? 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.
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The Concentrations of Solids and The Concentrations of Solids and Liquids Are Essentially ConstantLiquids Are Essentially Constant
• Both can be obtained by dividing the density of the substance by its molar mass — and both of these are constants at constant temperature.
• e.g., the concentration of water molecules can be calculated as:
H2O
MwtH2O
1000 g/L
18 g/mol55.6 mol/L
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The Concentrations of Solids and The Concentrations of Solids and Liquids Are Essentially ConstantLiquids Are Essentially Constant
• Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression – they are taken to have invariant concentrations or “activities” equal to unity (1):
• And:
PbCl
2(s) Pb2
(aq) 2Cl
(aq)
K
c
[Pb2 ][Cl ]2
[PbCl2]
[Pb2 ][Cl ]2
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Equilibrium CalculationsEquilibrium Calculations
• A closed system initially containing 0.001M H2 and 0.002 M I2 is allowed to reach equilibrium at 448oC.
• Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10−3 M.
• Calculate Kc at 448oC for the reaction taking place, which is:
H
2(g) I
2(g)2HI
(g)
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What Do We Know?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
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[HI] Increases by 1.87 x 10[HI] Increases by 1.87 x 10-3-3 MM
[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
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StoichiometryStoichiometry tells us [H tells us [H22] and [I] and [I22]]
decrease by half as much …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
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We can now calculate the equilibrium We can now calculate the equilibrium concentrations of all three compounds…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
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……and, therefore, the equilibrium and, therefore, the equilibrium constant is calculated as:constant is calculated as:
Kc
[HI]2
[H2][I
2]
1.87x10 3 26.5x10 5 1.065x10 3 51
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Heterogeneous Equilibria
. . . are equilibria that involve more than one phase.
CaCO3(s) CaO(s) + CO2(g)
K = [CO2]
The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present, so only gases and/or aqueous species are included in the expression.
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Reaction Quotient
If the law of mass action is applied with initial concentrations, the expression becomes a value called the reaction quotient (Q). This helps to determine the direction of the move toward equilibrium.
If Q > K then the reaction will favor reactants
If Q < K then the reaction will favor products
If Q=K the reaction is at equilibrium already
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Reaction Quotient (continued)
H2(g) + F2(g) 2HF(g)
QHF
H F2 2
02
0 0
Compare Q with K to determine direction :
Q > K to left Q < K to right Q = K at Equilibrium
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The Reaction Quotient (The Reaction Quotient (QQ))
• To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.
• Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.
Q
[C]ic [D]
id
[A]ia[B]
ib
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• If Q = K, then the system is at equilibrium.
• If Q > K, there is too much product and the equilibrium will shift to the left.
• If Q < K, there is too much reactant and the equilibrium will shift to the right.
aA bB cC dD
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Le Châtelier’s PrincipleLe 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.”
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Le Chatelier’s principle
• Stress causesStress causes shift shift in equilibriumin equilibrium• Adding or removing reagent
• N2(g) + 3 H2(g) 2 NH3(g)
• Stress causesStress causes shift shift in equilibriumin equilibrium• Adding or removing reagent
• N2(g) + 3 H2(g) 2 NH3(g)
Add moreN2?
NN22
Reaction shifts to the right[NH3] inc, [H2] dec
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Le Chatelier’s principleLe Chatelier’s principle
Adding or removing reagent
N2(g) + 3 H2(g) 2 NH3(g)
Adding or removing reagent
N2(g) + 3 H2(g) 2 NH3(g)
Add moreNH3?
NHNH33
Reaction shifts to the left[NN22] and [H2] inc
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Le Chatelier’s principleLe Chatelier’s principle
Adding Pressure affects an equilibrium with gases
N2(g) + 3 H2(g) 2 NH3(g)
Add PP?
Increasing pressure causes the equilibrium to shift to the side with the least moles of gas.
4 mol
of reactants
2 molof products
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Solving Equilibrium Problems From Initial Conditions
1. Balance the equation.
2. Write the equilibrium expression.
3. List the Initial concentrations.
4. Calculate Q and determine the shift (Change) to
equilibrium.
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Solving Equilibrium Problems
5. Define Equilibrium concentrations.
6. Substitute equilibrium concentrations into equilibrium expression and solve.
7. Check calculated concentrations by calculating K.
Take note of ICE calculations
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Example 13.9
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Example 13.10
• Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700K the value of K is 5.10. Find equilibrium concentrations if 1.00 mol of each component is mixed in a 1.00 L flask.
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Example 13.12
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Example 13.13—For Small Values of K