Chemical Equilibrium
CHAPTER 15
Chemistry: The Molecular Nature of Matter, 6th edition
By Jesperson, Brady, & Hyslop
CHAPTER 15 Chemical Equilibrium
Learning Objectives:
 Reversible Reactions and Equilibrium
 Writing Equilibrium Expressions and the Equilibrium
Constant (K)
 Reaction Quotient (Q)
 Kc vs Kp
 ICE Tables
 Quadratic Formula vs Simplifying Assumptions
 LeChatelier’s Principle
 van’t Hoff Equation
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Molecular Nature of Matter, 6E
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CHAPTER 15 Chemical Equilibrium
Lecture Road Map:
① Dynamic Equilibrium
② Equilibrium Laws
③ Equilibrium Constant
④ Le Chatelier’s Principle
⑤ Calculating Equilibrium
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Molecular Nature of Matter, 6E
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CHAPTER 15 Chemical Equilibrium
Le Chatelier’s
Principle
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Le Chatelier
Definition
• Equilibrium positions
– Combination of concentrations that allow Q = K
– Infinite number of possible equilibrium positions
• Le Châtelier’s principle
– System at equilibrium (Q = K) when upset by
disturbance (Q ≠ K) will shift to offset stress
• System said to “shift to right” when forward reaction
is dominant (Q < K)
• System said to “shift to left” when reverse direction
is dominant (Q > K)
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Le Chatelier
Q & K Relationships
• Q = K reaction at equilibrium
• Q < K reactants go to products
– Too many reactants
– Must convert some reactant to product to
move reaction toward equilibrium
• Q > K products go to reactants
– Too many products
– Must convert some product to reactant to
move reaction toward equilibrium
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Molecular Nature of Matter, 6E
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Le Chatelier
Change in Concentration
Cu(H2O)42+(aq) + 4Cl–(aq)
CuCl42–(aq) + 4H2O
blue
yellow
• Equilibrium mixture is blue-green
Kc =
4
[CuCl2(aq )][H O]
4
2
4
[Cu(H2O)2+
( aq )][Cl ( aq )]
4
K c¢ =
Kc
4
[H2O]
=
[CuCl2(aq )]
4
4
[Cu(H2O)2+
(aq )][Cl (aq )]
4
• Add excess Cl– (conc. HCl)
– Equilibrium shifts to products
– Makes more yellow CuCl42–
– Solution becomes green
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Molecular Nature of Matter, 6E
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Change in Concentration
Le Chatelier
Cu(H2O)42+(aq) + 4Cl–(aq)
blue
Kc =
CuCl42–(aq) + 4H2O
yellow
[CuCl2(aq )]
4
4
[Cu(H2O)2+
( aq )][Cl ( aq )]
4
• Add Ag+
– Removes Cl–: Ag+(aq) + Cl–(aq)  AgCl(s)
– Equilibrium shifts to reactants
– Makes more blue Cu(H2O)42+
– Solution becomes increasingly more blue
• Add H2O?
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Le Chatelier
Change in Concentration: Example
For the reaction
2SO2(g) + O2(g)
Kc = 2.4 × 10–3 at 700 °C
2SO3(g)
Which direction will the reaction move if 0.125 moles of
O2 is added to an equilibrium mixture?
A.Towards the products
B.Towards the reactants
C.No change will occur
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Molecular Nature of Matter, 6E
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Le Chatelier
Change in Concentration
• When changing concentrations of reactants or products
– Equilibrium shifts to remove reactants or products that
have been added
– Equilibrium shifts to replace reactants or products that
have been removed
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Le Chatelier
Change in Pressure or Volume
• Consider gaseous system at constant T and n
3H2(g) + N2(g)
2NH3(g)
2
KP =
PNH
3
3
• If volume is reduced
PN PH
2
2
– Expect pressure to increase
– To reduce pressure, look at each side of reaction
– Which has less moles of gas
– Reactants = 3 mol + 1 mol = 4 mol gas
– Products = 2 mol gas
– Reaction favors products (shifts to right)
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Le Chatelier
Change in Pressure or Volume
Consider gaseous system at constant T and n
H2(g) + I2(g)
2HI(g)
KP =
PHI2
PH PI
2
2
• If pressure is increased, what is the effect on equilibrium?
– nreactant = 1 + 1 = 2
– nproduct = 2
– Predict no change or shift in equilibrium
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Le Chatelier
Change in Pressure or Volume
2NaHSO3(s)
NaSO3(s) + H2O(g) + SO2(g)
K P = PH OPSO
2
2
• If you decrease volume of reaction, what is the effect
on equilibrium?
– Reactants: All solids, no moles gas
– Products: 2 moles gas
– Decrease in V, causes an increase in P
– Reaction shifts to left (reactants), as this has fewer
moles of gas
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Molecular Nature of Matter, 6E
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Le Chatelier
Change in Pressure or Volume
• Reducing volume of gaseous reaction mixture
causes reaction to decrease number of molecules
of gas, if it can
– Increasing pressure
• Moderate pressure changes have negligible effect
on reactions involving only liquids and/or solids
– Substances are already almost incompressible
• Changes in V, P and [X ] effect position of
equilibrium (Q), but not K
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Le Chatelier
Change in Temperature
Boiling
water
Ice
water
Cu(H2O)42+(aq) + 4Cl–(aq)
CuCl42–(aq) + 4H2O
blue
yellow
– Reaction endothermic
– Adding heat shifts equilibrium toward products
– Cooling shifts equilibrium toward reactants
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Le Chatelier
Change in Temperature
H2O(s)
– Energy + H2O(s)
H2O(l)
Hf°=+6 kJ (at 0 °C)
H2O(l )
– Energy is reactant
– Add heat energy, shift reaction right
3H2(g) + N2(g)
– 3 H2(g) + N2(g)
2NH3(g)
Hrxn= –47.19 kJ
2 NH3(g) + energy
– Energy is product
– Add heat, shift reaction left
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Le Chatelier
Change in Temperature
• Increase in temperature shifts reaction in
direction that produces endothermic (heat
absorbing) change
• Decrease in temperature shifts reaction in
direction that produces exothermic (heat
releasing) change
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Le Chatelier
Change in Temperature
• Changes in T change value of mass action
expression at equilibrium, so K changed
– K depends on T
– Increase in temperature of exothermic reaction
makes K smaller
• More heat (product) forces equilibrium to
reactants
– Increase in temperature of endothermic reaction
makes K larger
• More heat (reactant) forces equilibrium to
products
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Le Chatelier
Change with Catalyst
• Catalyst lowers Ea for
both forward and reverse
reaction
• Change in Ea affects
rates k r and k f equally
• Catalysts have no effect
on equilibrium
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Molecular Nature of Matter, 6E
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Le Chatelier
Addition of Inert Gas at Constant Volume
Inert gas
– One that does not react with components of reaction
e.g. argon, helium, neon, usually N2
• Adding inert gas to reaction at fixed V (n and T), increase
P of all reactants and products
• Since it doesn’t react with anything
– No change in concentrations of reactants or products
– No net effect on reaction
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Le Chatelier
How To Use Le Chatelier’s Principle
1.
Write mass action expression for reaction
2.
Examine relationship between affected
concentration and Q (direct or indirect)
3.
Compare Q to K
– If change makes Q > K, shifts left
– If change makes Q < K, shifts right
– If change has no effect on Q, no shift expected
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Group
Problem
Consider:
H3PO4(aq) + 3OH–(aq)
Q=
3H2O(l) + PO43–(aq)
[PO3–
]
4
[OH– ]3[H3PO4 ]
What will happen if PO43– is removed?
 Q is proportional to [PO43–]
 Decrease [PO43–], decrease in Q
 Q < K equilibrium shifts to right
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Molecular Nature of Matter, 6E
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Group
Problem
The reaction
H3PO4(aq) + 3OH–(aq)
3H2O(aq) + PO43–(aq)
is exothermic.
What will happen if system is cooled?
Q =




heat
[PO3–
]
4
[OH– ]3 [H3PO4 ]
Since reaction is exothermic, heat is product
Heat is directly proportional to Q
Decrease in T, decrease in Q
Q < K equilibrium shifts to right
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Molecular Nature of Matter, 6E
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Group
Problem
The equilibrium between aqueous cobalt ion and the chlorine
ion is shown:
[Co(H2O)6]2+(aq) + 4Cl–(aq)
[Co(Cl)4]2–(aq) + 6H2O
pink
blue
It is noted that heating a pink sample causes it to turn violet.
The reaction is:
A. endothermic
B. exothermic
C. cannot tell from the given information
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Molecular Nature of Matter, 6E
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Group
Problem
The following are equilibrium constants for the reaction of acids
in water, Ka. Which reaction proceeds the furthest to products?
A. Ka = 2.2 × 10–3
B. Ka = 1.8 × 10–5
C. Ka = 4.0 × 10–10
D. Ka = 6.3 × 10–3
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