slide 1 of 33 chapter 15: principles of chemical equilibrium
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Slide 1 of 33
Chapter 15: Principles of Chemical Equilibrium
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Contents
15-1 Dynamic Equilibrium
15-2 The Equilibrium Constant Expression
15-3 Relationships Involving Equilibrium Constants
15-4 The Magnitude of an Equilibrium Constant
15-5 The Reaction Quotient, Q: Predicting the Direction of a Net Change
15-6 Altering Equilibrium Conditions:
Le Châtelier’s Principle
15-7 Equilibrium Calculations: Some Illustrative Examples
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15-1 Dynamic Equilibrium
Equilibrium – two opposing processes taking place at equal rates.
H2O(l) H2O(g)
I2(H2O) I2(CCl4)
NaCl(s) NaCl(aq)H2O
CO(g) + 2 H2(g) CH3OH(g)
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Dynamic Equilibrium
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15-2 The Equilibrium Constant Expression
Methanol synthesis is a reversible reaction.
CO(g) + 2 H2(g) CH3OH(g)k1
k-1
CH3OH(g) CO(g) + 2 H2(g)
CO(g) + 2 H2(g) CH3OH(g)k1
k-1
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Three Approaches to the Equilibrium
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Three Approaches to Equilibrium
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Three Approaches to Equilibrium
CO(g) + 2 H2(g) CH3OH(g)k1
k-1
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The Equilibrium Constant Expression
Forward: CO(g) + 2 H2(g) → CH3OH(g)
Reverse: CH3OH(g) → CO(g) + 2 H2(g)
At Equilibrium:
Rfwrd = k1[CO][H2]2
Rrvrs = k-1[CH3OH]
Rfwrd = Rrvrs
k1[CO][H2]2 = k-1[CH3OH]
[CH3OH]
[CO][H2]2=
k1
k-1
= Kc
CO(g) + 2 H2(g) CH3OH(g) k1
k-1
k1
k-1
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General Expressions
a A + b B …. → g G + h H ….
Equilibrium constant = Kc= [A]m[B]n ….
[G]g[H]h ….
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15-3 Relationships Involving the Equilibrium Constant
Reversing an equation causes inversion of K. Multiplying by coefficients by a common factor
raises the equilibrium constant to the corresponding power.
Dividing the coefficients by a common factor causes the equilibrium constant to be taken to that root.
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Combining Equilibrium Constant Expressions
N2O(g) + ½O2 2 NO(g) Kc= ?
Kc= [N2O][O2]½
[NO]2
=[N2][O2]½
[N2O][N2][O2]
[NO]2
Kc(2)
1Kc(3)= = 1.710-13
[N2][O2]
[NO]2
=
[N2][O2]½
[N2O]=N2(g) + ½O2 N2O(g) Kc(2)= 2.710-18
N2(g) + O2 2 NO(g) Kc(3)= 4.710-31
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KP = Kc(RT)Δn
:
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Pure Liquids and Solids
Equilibrium constant expressions do not contain concentration terms for solid or liquid phases of a single component (that is, pure solids or liquids).
Kc = [H2O]2
[CO][H2]
C(s) + H2O(g) CO(g) + H2(g)
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Worked Examples Follow:
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15-1 Practice Example B
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CRS Questions Follow:
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H2
CO
CH4 = H2O
time
mol
es o
f su
bsta
nce
0
1
2
3
f
r
k
2 4 2kCO(g) + 3H (g) CH (g) + H O(g) Which of the following statements is
correct?
1. At equilibrium the reaction stops.
2. At equilibrium the rate constants for the forward and reverse reactions are equal.
3. At equilibrium the rates of the forward and reverse reactions are equal.
4. At equilibrium the rates of the forward and reverse reactions are zero.
Slide 21 of 33
H2
CO
CH4 = H2O
time
mol
es o
f su
bsta
nce
0
1
2
3
f
r
k
2 4 2kCO(g) + 3H (g) CH (g) + H O(g) Which of the following statements is
correct?
1. At equilibrium the reaction stops.
2. At equilibrium the rate constants for the forward and reverse reactions are equal.
3. At equilibrium the rates of the forward and reverse reactions are equal.
4. At equilibrium the rates of the forward and reverse reactions are zero.
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