CHAPTER 15
Chemistry: The Molecular Nature of Matter, 6 th 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)
 K c vs K p
 ICE Tables
 Quadratic Formula vs Simplifying Assumptions
 LeChatelier’s Principle
 van’t Hoff Equation
Jesperson, Brady, Hyslop. Chemistry: The
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
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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CHAPTER 15 Chemical Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Dynamic Eq Equilibrium
• Chemical equilibrium exists when
– Rates of forward and reverse reactions are equal
– Reaction appears to stop
– Concentration of reactants and products do not change over time
• Remain constant
• Both forward and reverse reaction never cease
• Equilibrium signified by double arrows ( )
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Dynamic Eq Equilibrium
N
2
O
4
2 NO
2
• Initially have only N
2
O
4
– Only forward reaction
– As N
2
O
4 reacts NO
2 forms
• As NO
2 forms
– Reverse reaction begins to occur
– NO
2 collide more frequently as concentration of NO increases
2
• Eventually, equilibrium is reached
– Concentration of N
2
O
4
– Concentration of NO
2 does not change does not change
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Dynamic Eq Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Dynamic Eq Equilibrium
N
2
O
4
Closed system
• Equilibrium can be reached from either direction
• Independent of whether it starts with
“ reactants ” or
“ products ”
• Always have the same composition at equilibrium under same conditions
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
2NO
2
8

Dynamic Eq
Equilibrium
N
2
O
4
2NO
2
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Dynamic Eq Mass Action Expression
• Simple relationship among [ reactants ] and
[ products ] for any chemical system at equilibrium
• Called the mass action expression
– Derived from thermodynamics
• Forward reaction: A

B
• Reverse reaction: A

B
• At equilibrium: A B
[ B ]
[ A ]
= mass action expression
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Dynamic Eq Reaction Quotient
• Uses stoichiometric coefficients as exponent for each reactant
• For reaction: aA + bB cC + dD
Q
[
[
C ]
A ] a c [ D ]
[ B ] d b
Reaction quotient
– Numerical value of mass action expression
– Equals “ Q ” at any time, and
– Equals “ K ” only when reaction is known to be at equilibrium
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Ex. 1 H
2
( g )
+ I
2
( g )
Exp ’ t Initial
Amts
I 1.00
mol H
2
10 L 1.00
mol I
2
0.00 mol HI
2HI
( g )
440˚C
Equil ’ m
Amts
Equil ’ m
[ M ]
0.222 mol H
2
0.0222 M H
2
0.222 mol I
2
0.0222
1.56 mol HI 0.156
M I
M HI
2
II 0.00 mol H
2
10 L 0.100 mol I
2
3.50
mol HI
12
0.350 mol H
2
0.0350 M H
2
0.450 mol I
2
0.0450 M I
2.80 mol HI 0.280 M HI
2

Ex. 1 H
2
( g )
+ I
2
( g )
Exp ’ t Initial Amts
III 0.0150
mol
H
2
10 L 0.00 mol I
2
1.27
mol HI
2HI
(
Equil ’ m
Amts
0.150 mol H
2 g )
440 ˚C
Equil ’ m
[ M ]
0.0150 M H
2
0.135 mol I
2
0.0135
1.00 mol HI 0.100
M I
M HI
2
IV 0.00 mol H
2
10 L 0.00 mol I
2
4.00
mol HI
0.442 mol H
2
0.442 mol I
2
0.0442 M H
2
0.0442 M I
3.11 mol HI 0.311 M HI
2
13

Q =
Exp ’ t
Mass Action Expression
[HI] 2
= same for all data sets at equilibrium
[H
2
][I
2
]
Equilibrium Concentrations ( M )
Q
[HI] 2
[H
2
] [I
2
] [HI]
[H
2
][I
2
]
I
II
III
IV
0.0222
0.0222
0.156
0.0350
0.0450
0.280
0.0150
0.0135
0.100
0.0442
0.0442
0.311
( 0 .
156 )
2

( 0 .
0222 )( 0
( 0 .
280
.
0222
)
2
)
( 0 .
0150 )( 0 .
0135 )
( 0 .
311 )
2

( 0 .
0350 )( 0 .
0450 )
( 0 .
100 )
2


( 0 .
0442 )( 0 .
0442 )
49 .
4
49
49
49 .
.
.
8
4
5
Average = 49.5
14

Group
Problem
Write mass action expressions for the following:
• 2NO
2
( g ) N
2
O
4
( g )
Q
N
2
O
4
NO
2
2
• 2CO( g ) + O
2
( g )
Q =
ë
é
CO
2
û
ù 2
ë
é
CO
ù
û
2 é
ë
O
2
û
ù
2CO
2
( g )
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Group
Problem
Which of the following is the correct mass action expression for the reaction:
Cu 2+ ( aq ) + 4NH
3
( aq )
A. Q =
[Cu(NH
3
) 2 +
4
]
[Cu 2 + ][NH
3
] 4
[Cu(NH
3
)
4
2+ ]( aq )?
B. Q =
[Cu(NH
3
) 2 +
4
]
[Cu 2 + ][NH
3
]
C. Q =
[Cu 2 + ][NH
3
[Cu(NH
3
) 2 +
4
] 4
]
D. none of these
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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CHAPTER 15 Chemical Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Equilibrium Equilibrium Laws
• For reaction
H
2
( g ) + I
2
( g ) 2HI( g ) at 440 ˚C at equilibrium write the following equilibrium law
K
[HI] 2 c
=
[H
2
][I
2
]
= 49.5
• Equilibrium constant = K c
• Use K c mol/L
= constant at given T since usually working with concentrations in
• For chemical equilibrium to exist in reaction mixture, reaction quotient Q must be equal to equilibrium constant, K c
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Equilibrium Predicting Equilibrium Laws
For general chemical reaction:
• dD + eE fF + gG
– Where D , E , F , and G represent chemical formulas
– d , e , f , and g are coefficients
[ F ] f [ G ] g
• Mass action expression is
[ D ] d [ E ] e
• Note: Exponents in mass action expression are stoichiometric coefficients in balanced equation.
• E quilibrium law is:
K
[ F ] f c
=
[ D ] d
[ G ] g
[ E ] e
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Equilibrium Predicting Equilibrium Laws
• Only concentrations that satisfy this equation are equilibrium concentrations
• Numerator
– Multiply concentration of products raised to their stoichiometric coefficients
• Denominator
– Multiply concentration reactants raised to their stoichiometric coefficients
K c
=
[products] f
[reactants] d is scientist s’ convention
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Equilibrium Example
3H
2
( g ) + N
2
( g )
K c
2NH
= 4.26
10 8 at 25
C
3
( g )
What is equilibrium law?
K c
[NH
3
] 2
[H
2
] 3 [N
2
]
4.26
10 8
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Equilibrium Operations
Various operations can be performed on equilibrium expressions
1.
When direction of equation is reversed, new equilibrium constant is reciprocal of original
A + B C + D
K c
=
[ C ][ D ]
[ A ][ B ]
C + D A + B
K c
¢ =
[ A ][ B ]
[ C ][ D ]
=
1
K c
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Equilibrium Operations
1.
When direction of equation is reversed, new equilibrium constant is reciprocal of original
3H
2
( g ) + N
2
( g ) 2 NH
3
( g ) at 25 ˚C
K c

[NH
[H
2
] 3
3
]
[N
2
2
]
 4 .
26  10 8
2NH
3
( g )
K ¢ c
=
[H
2
] 3 [N
2
]
[NH
3
] 2
=
3H
2
( g ) + N
2
( g ) at 25 ˚C
1
K c
=
1
4.26
´ 10 8
= 2.35
´ 10 9
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Equilibrium Operations
2.
When coefficients in equation are multiplied by a factor, equilibrium constant is raised to a power equal to that factor.
A + B C + D
K c
=
[ C ][ D ]
[ A ][ B ]
3 A + 3 B 3 C + 3 D
K c
[ C ] 3 [ D ] 3
[ A ] 3 [ B ] 3
[ C ][ D ]
[ A ][ B ]
[ C ][ D ]
[ A ][ B ]
[ C ][ D ]
[ A ][ B ]
K c
3
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Equilibrium Operations
2. When coefficients in equation are multiplied by factor, equilibrium constant is raised to power equal to that factor
3H
2
( g ) + N
2
( g )
K c
=
2NH
3
( g ) at 25 ˚C
[NH
3
] 2
[H
2
] 3 [N
2
]
= 4.26
´ 10 8
Multiply by 3
9H
2
( g ) + 3N
2
( g ) 6NH
3
( g )
K c
¢¢ =
[NH
3
] 6
[H
2
] 9 [N
2
] 3
= K c
3
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Equilibrium Operations
3. When chemical equilibria are added, their equilibrium constants are multiplied
A + B C + D
K c
1
=
[ C ][ D ]
[ A ][ B ]
C + E F + G
K c
2
=
[ F ][ G ]
[ C ][ E ]
A + B + E D + F + G
K c
3
=
[ C ][ D ]
[ A ][ B ]
´
[ F ][ G ]
[ C ][ E ]
=
[ D ][ F ][ G ]
[ A ][ B ][ E ]
= K c
1
´ K c
2
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Equilibrium Operations
3. When chemical equilibria are added, their equilibrium constants are multiplied
2 NO
2
( g )
NO
3
( g )
NO
2
( g )
+ CO ( g )
+ CO ( g )
NO
3
( g )
[NO][NO
3
[NO
2
] 2
Therefore
K
] c
1

´
[NO
[NO
K c
2
+ NO ( g )
NO
2
( g )
K c
3
+ CO
2
( g )
NO ( g ) + CO
2
( g )
2
3
][CO
2
][CO]
]
=

K c
1
K c
2
K c
3
=
=
=
[NO][NO
3
]
[NO
2
[NO
2
] 2
][CO
2
[NO
3
][CO]
[NO][CO
2
]
]
[NO
2
][CO]
[NO][CO
2
[NO
2
]
][CO]
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Group
Problem
For: N
2
( g ) + 3H
2
( g )
K c
2NH
3
( g )
= 500 at a particular temperature.
What would be K c
• 2NH
3
( g ) for following?
N
2
( g ) + 3H
2
( g )
K c
¢ =
1
K c
=
1
500
• 1/2N
2
( g ) + 3/2H
2
( g )
=
0.002
NH
3
( g )
K c
¢¢ = K 1 2 c
= 500 =
22.4
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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CHAPTER 15 Chemical Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Equilibrium Constant Kc
• Most often K c is expressed in terms of a ratio of concentrations of products and reactants as shown on previous slides
• Sometimes partial pressures, in atmospheres, may be used in place of concentrations
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Equilibrium Kp
• Based on reactions in which all substances are gaseous
• Gas quantities are expressed in atmospheres in mass action expression
• Use partial pressures for each gas in place of concentrations e.g. N
2
( g ) + 3H
2
( g )
K
P
=
P 2
NH
3
P
N
2
P
H
2
3
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
2NH
3
( g )
31

Equilibrium Relationship between Kp and Kc
• Start with ideal gas law
PV = nRT
• Rearranging gives
P
n
V
RT
MRT
• Substituting P / RT for molar concentration into K c pressure-based formula results in
• ∆ n = moles of gas in product – moles of gas in reactant
K p
K c
( RT ) D n
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Group
Problem
Consider the reaction: 2NO
2
( g ) N
2
O
4
( g )
If K p
= 0.480 for the reaction at 25 ˚C, what is value of same temperature?
K c
 n = n products
– n reactants
= 1 – 2 = –1 at
K p
K c
( RT ) D n
K c
=
K p
( RT ) D n
=
0.480
(0.0821
´ 298 K) 1
K c
= 11.7
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Group
Problem
Consider the reaction A( g ) + 2B( g )
If the K c
4C( g ) for the reaction is 0.99 at 25 ˚C, what would be the K p
?
Δn = (4 – 3) = 1
A. 0.99
B. 2.0
C. 24
D. 2400
K p
= K c
(RT) Δn
K p
= 0.99 × (0.082057 × 298.15) 1
K p
= 24
E. None of these
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Equilibrium Homogeneous and Hetergeneous
Homogeneous reaction/equilibrium
– All reactants and products in same phase
– Can mix freely
Heterogeneous reaction/equilibrium
– Reactants and products in different phases
– Can ’ t mix freely
– Solutions are expressed in M
– Gases are expressed in M
– Governed by K c
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Equilibrium Heterogeneous
2NaHCO
3
( s )
• Equilibrium Law =
Na
2
CO
3
( s ) + H
2
O( g ) + CO
2
( g )
K =
[ Na
2
CO
3
( s )
][ H
2
O
( g )
][ CO
2
( g )
]
[ NaHCO
3
( s )
] 2
• Can write in simpler form
• For any pure liquid or solid, ratio of moles to volume of substance ( M ) is constant
– e.g. 1 mol NaHCO
2 mol NaHCO
3
3 occupies 38.9 cm 3 occupies 77.8 cm 3
M =
1 mol NaHCO
3
0.0389 L
= 25.7 M
M =
2 mol NaHCO
3
0.0778 L
Jesperson, Brady, Hyslop. Chemistry: The
= 25.7 M
Molecular Nature of Matter, 6E
36

Equilibrium Heterogeneous
2NaHCO
3
( s ) Na
2
CO
3
( s ) + H
2
O( g ) + CO
2
( g )
– Ratio ( n / V ) or M of NaHCO
3 is constant
(25.7 mol/L) regardless of sample size
– Likewise can show that molar concentration of Na
2
CO
3 solid is constant regardless of sample size
• So concentrations of pure solids and liquids can be incorporated into equilibrium constant, K c
K c
= K
[Na
2
CO
[NaHCO
3
3
(
( s s )]
)] 2
= [H
2
O( g )][CO
2
( g )]
• Equilibrium law for heterogeneous system written without concentrations of pure solids or liquids
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Equilibrium Heterogeneous
Write equilibrium laws for the following:
Ag + ( aq ) + Cl – ( aq ) AgCl( s )
K c
=
1
[Ag
+
][Cl
-
]
H
3
PO
4
( aq ) + H
2
O H
3
O + ( aq ) + H
2
PO
4
–
( aq )
K c
=
[H
3
O
+
][H
2
PO
4
[H
3
PO
4
]
-
]
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Group
Problem
Given the reaction:
3Ca 2+ ( aq ) + 2PO
4
3
–
( aq ) Ca
3
(PO
4
)
2
( s )
What is the mass action expression?
A. Q =
[Ca 2 +
[Ca
3
] 3 [PO
(PO
4
3 -
4
)
2
]
] 2
B. Q =
[Ca 2 + ] 3 [PO
4
3 -
[1]
] 2
C. Q =
D. Q =
[Ca
3
(PO
4
)
2
]
[Ca 2 + ] 3 [PO
4
3 ] 2
[1]
[Ca 2 + ] 3 [PO 3 -
4
] 2
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Group
Problem
Given the reaction:
3Ca 2+ ( aq ) + 2PO
4
3 – ( aq ) Ca
3
(PO
4
)
2
( s )
What is mass action expression for the reverse reaction?
A. Q =
[Ca 2 +
[Ca
3
] 3 [PO
(PO
4
)
3 -
4
2
]
] 2
B. Q =
[Ca 2 + ] 3 [PO 3 -
4
[1]
] 2
C. Q =
[Ca
3
(PO
4
)
2
]
[Ca 2 + ] 3 [PO 3 -
4
] 2
D. Q =
[Ca 2 + ]
[1]
3 [PO 3 -
Jesperson, Brady, Hyslop. Chemistry: The
4
] 2
Molecular Nature of Matter, 6E
40