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
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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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CHAPTER 15 Chemical Equilibrium
Dynamic
Equilibrium
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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 (
)
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Dynamic Eq
Equilibrium
N 2O 4
2 NO2
• Initially have only N2O4
– Only forward reaction
– As N2O4 reacts NO2 forms
• As NO2 forms
– Reverse reaction begins to occur
– NO2 collide more frequently as concentration of NO2
increases
• Eventually, equilibrium is reached
– Concentration of N2O4 does not change
– Concentration of NO2 does not change
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Dynamic Eq
Equilibrium
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Dynamic Eq
Equilibrium
N 2O 4
2NO2
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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
Dynamic Eq
Reactants
Equilibrium
Equilibrium
N2O4
Products
2NO2
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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 ]
= mass action expression
[A]
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Dynamic Eq
Reaction Quotient
• Uses stoichiometric coefficients as exponent for each
reactant
• For reaction: aA + bB
cC + dD
c
d
[C ] [D ]
Q=
b
a
[ A] [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
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Ex. 1 H2(g) + I2(g)
Exp’t
I
Initial
Amts
1.00 mol H2
10 L 1.00 mol I2
0.00 mol HI
2HI(g) 440˚C
Equil’m
Amts
0.222 mol H2
Equil’m
[M]
0.0222 M H2
0.222 mol I2
1.56 mol HI
0.0222 M I2
0.156 M HI
0.00 mol H2
0.350 mol H2
0.0350 M H2
10 L 0.100 mol I2
0.450 mol I2
0.0450 M I2
3.50 mol HI
2.80 mol HI
0.280 M HI
II
12
Ex. 1 H2(g) + I2(g)
Exp’t Initial Amts
III
0.0150 mol
H2
10 L 0.00 mol I2
1.27 mol HI
IV
0.00 mol H2
10 L 0.00 mol I2
4.00 mol HI
13
2HI(g) 440 ˚C
Equil’m
Amts
Equil’m
[M]
0.150 mol H2
0.0150 M H2
0.135 mol I2
1.00 mol HI
0.0135 M I2
0.100 M HI
0.442 mol H2
0.0442 M H2
0.442 mol I2
0.0442 M I2
3.11 mol HI
0.311 M HI
Mass Action Expression
2
[HI]
Q=
= same for all data sets at equilibrium
[H2 ][I2 ]
Equilibrium Concentrations (M )
Exp’t
[H2]
[I2]
[HI]
I
0.0222
0.0222
0.156
II
0.0350
0.0450
0.280
III
0.0150
0.0135
0.100
IV
0.0442
0.0442
0.311
[HI]2
Q=
[H2 ][I2 ]
(0.156 )2
 49.4
(0.0222 )(0.0222 )
(0.280 )2
 49.8
(0.0350 )(0.0450 )
(0.100 )2
 49.4
(0.0150 )(0.0135 )
(0.311)2
 49.5
(0.0442 )(0.0442 )
Average = 49.5
14
Group
Problem
Write mass action expressions for the following:
• 2NO2(g)
N2O4(g)
Q=
éN O ù
ë 2 4û
2
éNO ù
ë 2û
• 2CO(g) + O2(g)
Q=
2CO2(g)
éCO ù 2
ë 2û
2
éCOù éO ù
ë û ë 2û
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Group
Problem
Which of the following is the correct mass action expression
for the reaction:
Cu2+(aq) + 4NH3(aq)
[Cu(NH3)42+](aq)?
A.
B.
C.
Q =
Q=
[Cu(NH3 )2+
]
4
[Cu2+ ][NH3 ]4
[Cu(NH3 )2+
]
4
[Cu2+ ][NH3 ]
Q =
[Cu2+ ][NH3 ]4
[Cu(NH3 )2+
]
4
D. none of these
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Molecular Nature of Matter, 6E
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CHAPTER 15 Chemical Equilibrium
Equilibrium Laws
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Molecular Nature of Matter, 6E
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Equilibrium
Equilibrium Laws
• For reaction
H2(g) + I2(g)
2HI(g) at 440 ˚C
at equilibrium write the following equilibrium law
[HI]2
Kc =
= 49.5
[H2 ][I2 ]
• Equilibrium constant = Kc = constant at given T
• Use Kc since usually working with concentrations in
mol/L
• For chemical equilibrium to exist in reaction mixture,
reaction quotient Q must be equal to equilibrium
constant, Kc
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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
• Mass action expression is
[F ]f [G ]g
[D ]d [E ]e
• Note: Exponents in mass action expression are
stoichiometric coefficients in balanced equation.
f
g
• Equilibrium law is:
[F ] [G ]
Kc =
d
e
[D ] [E ]
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Predicting Equilibrium Laws
Equilibrium
• 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
Kc =
f
[products]
is scientists’ convention
d
[reactants]
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Equilibrium
Example
3H2(g) + N2(g)
2NH3(g)
Kc = 4.26 × 108 at 25 °C
What is equilibrium law?
2
Kc =
[NH3 ]
3
[H2 ] [N2 ]
= 4.26 ´ 10
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21
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
C+D
A+B
[C ][D ]
Kc =
[ A][B ]
[ A][B ] 1
K c¢ =
=
[C ][D ] K c
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Molecular Nature of Matter, 6E
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Operations
Equilibrium
1. When direction of equation is reversed, new equilibrium
constant is reciprocal of original
3H2(g) + N2(g)
Kc 
2
[NH3 ]2
3
[H2 ] [N 2 ]
K c¢ =
[NH3 ]2
at 25˚C
 4.26  10 8
3H2(g) + N2(g) at 25 ˚C
2NH3(g)
[H2 ]3 [N2 ]
NH3(g)
=
1
Kc
=
1
4.26 ´ 108
Jesperson, Brady, Hyslop. Chemistry: The
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= 2.35 ´ 10-9
23
Operations
Equilibrium
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
3A + 3B
[C ][D ]
Kc =
[ A][B ]
3C + 3D
[C ] [D ]
[C ][D ] [C ][D ] [C ][D ]
3
K c¢¢ =
=
´
´
=
K
c
3
3
[A][B ] [ A][B ] [A][B ]
[A] [B ]
3
3
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Operations
Equilibrium
2. When coefficients in equation are multiplied by factor,
equilibrium constant is raised to power equal to that factor
2NH3(g) at 25 ˚C
3H2(g) + N2(g)
Kc =
[NH3 ]2
[H2 ]3 [N2 ]
Multiply by 3
9H2(g) + 3N2(g)
K c¢¢ =
= 4.26 ´ 108
6NH3(g)
[NH3 ]6
[H2 ]9 [N2 ]3
= K c3
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Equilibrium
Operations
3. When chemical equilibria are added, their equilibrium
constants are multiplied
A+B
C+D
C+E
F+G
A+B+E
Kc
Kc
3
1
[C ][D ]
=
[ A][B ]
Kc
2
[F ][G ]
=
[C ][E ]
D+F+G
[C ][D ] [F ][G ] [D ][F ][G ]
=
´
=
= Kc ´Kc
1
2
[A][B ] [C ][E ] [A][B ][E ]
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Operations
Equilibrium
3. When chemical equilibria are added, their equilibrium
constants are multiplied
2 NO2(g)
NO3(g) + CO(g)
[NO][NO 3 ]
Therefore
1
NO2(g) + CO2(g)
NO2(g) + CO(g)
[NO 2 ]2
Kc =
NO3(g) + NO(g)
NO(g) + CO2(g)
Kc =
2
Kc =
3
[NO][NO3 ]
[NO2 ]2
[NO2 ][CO2 ]
[NO3 ][CO]
[NO][CO2 ]
[NO2 ][CO]
[NO 2 ][CO 2 ] [NO][CO2 ]


[NO 2 ][CO]
[NO 3 ][CO]
Kc ´Kc = Kc
1
2
3
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Molecular Nature of Matter, 6E
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Group
Problem
For: N2(g) + 3H2(g)
2NH3(g)
Kc = 500 at a particular temperature.
What would be Kc for following?
• 2NH3(g)
N2(g) + 3H2(g)
1
1
K c¢ =
=
=
K c 500
0.002
• 1/2N2(g) + 3/2H2(g)
NH3(g)
12
¢¢
K c = K c = 500 = 22.4
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Molecular Nature of Matter, 6E
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CHAPTER 15 Chemical Equilibrium
Equilibrium
Constant
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Molecular Nature of Matter, 6E
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Equilibrium
Constant Kc
• Most often Kc 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
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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. N2(g) + 3H2(g)
KP =
2NH3(g)
P
2
NH3
PN P
2
3
H2
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Equilibrium
Relationship between Kp and Kc
• Start with ideal gas law
PV = nRT
• Rearranging gives
æn ö
P = ç ÷ RT = MRT
èV ø
• Substituting P/RT for molar concentration into Kc results in
pressure-based formula
• ∆n = moles of gas in product – moles of gas in reactant
Kp = Kc(RT )
Dn
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Group
Problem
Consider the reaction: 2NO2(g)
N2O4(g)
If Kp = 0.480 for the reaction at 25 ˚C, what is value of Kc at
same temperature?
n = nproducts – nreactants = 1 – 2 = –1
Kp = Kc(RT )
Dn
Kp
0.480
Kc =
=
Dn
-1
(RT )
(0.0821 ´ 298 K)
Kc = 11.7
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Group
Problem
Consider the reaction A(g) + 2B(g)
4C(g)
If the Kc for the reaction is 0.99 at 25 ˚C, what would
be the Kp?
Δn = (4 – 3) = 1
A. 0.99
B. 2.0
C. 24
D. 2400
E. None of these
Kp = Kc(RT)Δn
Kp= 0.99 × (0.082057 × 298.15)1
Kp = 24
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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 Kc
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Equilibrium
Heterogeneous
2NaHCO3(s)
Na2CO3(s) + H2O(g) + CO2(g)
• Equilibrium Law =
[Na2CO3 ( s )][H 2O ( g )][CO2 ( g )]
K =
[NaHCO3 (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 NaHCO3 occupies 38.9 cm3
2 mol NaHCO3 occupies 77.8 cm3
M=
1 mol NaHCO3
0.0389 L
= 25.7 M
M=
2 mol NaHCO3
0.0778 L
= 25.7 M
Jesperson, Brady, Hyslop. Chemistry: The
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Equilibrium
Heterogeneous
2NaHCO3(s)
Na2CO3(s) + H2O(g) + CO2(g)
– Ratio (n/V ) or M of NaHCO3 is constant
(25.7 mol/L) regardless of sample size
– Likewise can show that molar concentration of Na2CO3
solid is constant regardless of sample size
• So concentrations of pure solids and liquids can be
incorporated into equilibrium constant, Kc
Kc = K
[Na2CO3 (s )]
[NaHCO3 (s )]2
= [H2O(g )][CO2 (g )]
• Equilibrium law for heterogeneous system written without
concentrations of pure solids or liquids
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Heterogeneous
Equilibrium
Write equilibrium laws for the following:
Ag+(aq) + Cl–(aq)
AgCl(s)
1
Kc =
+
[Ag ][Cl ]
H3O+(aq) + H2PO4–(aq)
H3PO4(aq) + H2O
+
-
[H3O ][H2PO4 ]
Kc =
[H3PO4]
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Interpreting KC
• Large K (K >>1)
– Means product rich mixture
– Reaction goes far toward
completion
e.g.
2SO2(g) + O2(g)
2SO3(g)
Kc = 7.0  1025 at 25 °C
[SO3 ]2
7.0 ´ 1025
Kc =
=
1
[SO2 ]2 [O2 ]
39
Interpreting KC
• Small K (K << 1)
– Means reactant rich
mixture
– Only very small amounts
of product formed
e.g.
H2(g) + Br2(g)
2HBr(g)
Kc = 1.4  10–21 at 25 °C
2
-21
[HBr]
1.4 ´ 10
Kc =
=
[H2 ][Br2 ]
1
40
Interpreting KC
• K  1
– Means product and
reactant
concentrations close to
equal
– Reaction goes only
about halfway
41
• Size of K gives measure of how
reaction proceeds
• K >> 1
• K=1
• K << 1
42
[products] >> [reactants]
[products] = [reactants]
[products] << [reactants]
CHAPTER 15 Chemical Equilibrium
Le Chatelier’s
Principle
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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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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Molecular Nature of Matter, 6E
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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
Jesperson, Brady, Hyslop. Chemistry: The
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
Jesperson, Brady, Hyslop. Chemistry: The
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?
Jesperson, Brady, Hyslop. Chemistry: The
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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
Jesperson, Brady, Hyslop. Chemistry: The
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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)
Jesperson, Brady, Hyslop. Chemistry: The
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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
Jesperson, Brady, Hyslop. Chemistry: The
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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
Jesperson, Brady, Hyslop. Chemistry: The
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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
Jesperson, Brady, Hyslop. Chemistry: The
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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
Jesperson, Brady, Hyslop. Chemistry: The
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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
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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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
Jesperson, Brady, Hyslop. Chemistry: The
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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
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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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
Jesperson, Brady, Hyslop. Chemistry: The
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
Jesperson, Brady, Hyslop. Chemistry: The
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59
Le Chatelier
How To Use Le Chatelier’s Principle
1.
Write mass action expression for reaction
1.
Examine relationship between affected
concentration and Q (direct or indirect)
1.
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
Jesperson, Brady, Hyslop. Chemistry: The
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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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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
Jesperson, Brady, Hyslop. Chemistry: The
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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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CHAPTER 15 Chemical Equilibrium
Calculating
Equilibrium
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Calculations
Overview
•
For gaseous reactions, use either KP or KC
•
For solution reactions, must use KC
•
Either way, two basic categories of calculations
1. Calculate K from known equilibrium
concentrations or partial pressures
2. Calculate one or more equilibrium
concentrations or partial pressures using known
KP or KC
Jesperson, Brady, Hyslop. Chemistry: The
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Calculations
Kc with Known Equilibrium Concentrations
•
When all concentrations at equilibrium are known
– Use mass action expression to relate
concentrations to KC
•
Two common types of calculations
A. Given equilibrium concentrations, calculate K
B. Given initial concentrations and one final
concentration
• Calculate equilibrium concentration of
all other species
• Then calculate K
Jesperson, Brady, Hyslop. Chemistry: The
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Kc with Known Equilibrium Concentrations
Calculations
Ex. 3 N2O4(g)
2NO2(g)
• If you place 0.0350 mol N2O4 in 1 L flask at
equilibrium, what is KC?
• [N2O4]eq = 0.0292 M
• [NO2]eq = 0.0116 M
2
K c=
[NO2 ]
2
[0.0116]
K c=
[0.0292]
[N2O 4 ]
KC = 4.61  10–3
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Group
Problem
For the reaction: 2A(aq) + B(aq)
3C(aq)
the equilibrium concentrations are: A = 2.0 M, B =
1.0 M and C = 3.0 M. What is the expected value of
Kc at this temperature?
A. 14
B. 0.15
3
[C ]
C. 1.5
Kc =
2
[
A
]
[B ]
D. 6.75
Kc =
Jesperson, Brady, Hyslop. Chemistry: The
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[3.0]
2
[2.0] [1.0]
69
Kc with Known Equilibrium Concentrations
Calculations
Ex. 4 2SO2(g) + O2(g)
2SO3(g)
At 1000 K, 1.000 mol SO2 and 1.000 mol O2 are placed
in a 1.000 L flask. At equilibrium 0.925 mol SO3 has
formed. Calculate K C for this reaction.
• First calculate concentrations of each
– Initial
1.00 mol
[SO2 ] = [O2 ] =
= 1.00 M
1.00 L
– Equilibrium
0.925 mol
[SO3 ] =
= 0.925 M
1.00 L
Jesperson, Brady, Hyslop. Chemistry: The
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Calculations
Example Continued
• Set up concentration table
– Based on the following:
• Changes in concentration must be in same ratio
as coefficients of balanced equation
• Set up table under balanced chemical equation
– Initial concentrations
• Controlled by person running experiment
– Changes in concentrations
• Controlled by stoichiometry of reaction
– Equilibrium concentrations
Equilibrium
Change in
Initial
=
–
Concentration
Concentration
Concentration
Jesperson, Brady, Hyslop. Chemistry: The
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Calculations
Example Continued
2SO2(g) + O2(g)
Initial Conc. (M)
1.000
Changes in Conc. (M) –0.925
Equilibrium Conc. (M) 0.075
1.000
–0.462
0.538
2SO3(g)
0.000
+0.925
0.925
[SO2] consumed = amount of SO3 formed
= [SO3] at equilibrium = 0.925 M
[O2] consumed = ½ amount SO3 formed
= 0.925/2 = 0.462 M
[SO2] at equilibrium = 1.000 – 0.975 = 0.075
[O2] at equilibrium = 1.00 – 0.462 = 0.538 M
Jesperson, Brady, Hyslop. Chemistry: The
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Calculations
Example Continued
• Finally calculate KC at 1000 K
Kc 
Kc 
[SO 3 ]2
[SO 2 ]2 [O 2 ]
2
[0.925]
2
[0.075] [0.538]
Kc = 2.8 × 102 = 280
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Calculations
ICE Table Summary
ICE tables used for most equilibrium calculations:
1.
Equilibrium concentrations are only values used in mass
action expression

2.
Initial value in table must be in units of mol/L (M)


3.
4.
Values in last row of table
[X]initial = those present when reaction prepared
No reaction occurs until everything is
mixed
Changes in concentrations always occur in same ratio as
coefficients in balanced equation
In “change” row be sure all [reactants] change in same
directions and all [products] change in opposite direction.


If [reactant]initial = 0, its change must be an increase (+) because
[reactant]final cannot be negative
If [reactants] decreases, all entries for reactants in change row should
have minus sign and all entries for products should be positive
Jesperson, Brady, Hyslop. Chemistry: The
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Calculations
Calculate [X ]equilibrium from Kc and [X ]initial
• When all concentrations but one are known
– Use mass action expression to relate Kc and known
concentrations to obtain missing concentrations
Ex. 5 CH4(g) + H2O(g)
CO(g) + 3H2(g)
• At 1500 °C, Kc = 5.67. An equilibrium mixture of
gases had the following concentrations: [CH4] = 0.400
M and
[H2] = 0.800 M and [CO] = 0.300 M.
What is [H2O] at equilibrium ?
Jesperson, Brady, Hyslop. Chemistry: The
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Calculate [X ]equilibrium from Kc and [X ]initial
Calculations
Ex. 5 CH4(g) + H2O(g)
CO(g) + 3H2(g) Kc = 5.67
[CH4] = 0.400 M; [H2] = 0.800 M; [CO] =0.300 M
• What is [H2O] at equilibrium?
• First, set up equilibrium
Kc =
[CO][H2 ]3
[CH4 ][H2O]
[H2O] =
[CO][H2 ]3
[CH4 ]K c
• Next, plug in equilibrium concentrations and Kc
[0.300][0. 800]3 0.154
[H2O] 

[0.400](5.67)
2.27
[H2O] = 0.0678 M
Jesperson, Brady, Hyslop. Chemistry: The
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Calculations
Calculating [X ]Equilibrium from Kc
When Initial Concentrations Are Given
• Write equilibrium law/mass action expression
• Set up Concentration table
– Allow reaction to proceed as expected, using
“x” to represent change in concentration
• Substitute equilibrium terms from table into mass
action expression and solve
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Ex. 6 H2(g) + I2(g)
2HI(g) at 425 ˚C
KC = 55.64
If one mole each of H2 and I2 are placed in a 0.500 L
flask at 425 °C, what are the equilibrium
concentrations of H2, I2 and HI?
Step 1. Write Equilibrium Law
[HI]2
Kc =
= 55.64
[H2 ][I2 ]
Jesperson, Brady, Hyslop. Chemistry: The
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Step 2: Construct an ICE table
Conc (M)
H2(g) + I2(g)
Initial
2.00
–x
Change
Equilibrium 2.00 – x
2HI (g)
2.00
0.000
–x
+2x
2.00 – x
+2x
• Initial [H2] = [I2] = 1.00 mol/0.500 L =2.00 M
• Amt of H2 consumed = Amt of I2 consumed = x
• Amount of HI formed = 2x
(2x )
(2x )
55.64 =
=
(2.00 - x )(2.00 - x ) (2.00 - x )2
2
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79
Calculations
Calculate [X]equilibrium from [X]initial and KC
Step 3. Solve for x
• Both sides are squared so we can take square root of both
sides to simplify
(2x)2
K = 55.64 =
2
(2.00 - x)
2x
7.459 
7.459(2.00  x )  2x
(2.00  x )
14.918  7.459x  2x
14.918  9.459x
14.918
x 
 1.58
9.459
Jesperson, Brady, Hyslop. Chemistry: The
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Step 4. Equilibrium Concentrations
Conc (M)
H2(g) +
Initial
2.00
– 1.58
Change
Equilibrium 0.42
I2(g)
2.00
2HI (g)
0.00
– 1.58
+3.16
0.42
+3.16
• [H2]equil = [I2]equil = 2.00 – 1.58 = 0.42 M
• [HI]equil = 2x = 2(1.58) = 3.16
Jesperson, Brady, Hyslop. Chemistry: The
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Ex. 7 H2(g) + I2(g)
2HI(g) at 425 ˚C
KC = 55.64
• If one mole each of H2, I2 and HI are placed in a
0.500 L flask at 425 ˚C, what are the equilibrium
concentrations of H2, I2 and HI?
• Now have product as well as reactants initially
Step 1. Write Equilibrium Law
2
[HI]
Kc =
= 55.64
[H2 ][I2 ]
Jesperson, Brady, Hyslop. Chemistry: The
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Calculate [X]equilibrium from [X]initial and KC
Calculations
Step 2. Concentration Table
Conc (M)
H2(g) +
I2(g)
2HI (g)
Initial
Change
Equil’m
2.00
2.00
2.00
–x
–x
+2x
2.00 – x
2.00 – x
2.00 + 2x
(2.00  2x ) 2
(2.00  2x ) 2
55.64 

(2.00  x )(2.00  x )
(2.00  x ) 2
K  55.64 
(2.00  2x ) 2
(2.00  x ) 2
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Calculate [X]equilibrium from [X]initial and KC
Calculations
Step 3. Solve for x
2.00  2x
7.459 
(2.00  x )
7.459(2.00  x )  2.00  2x
14.918  7.459x  2.00  2x
 [H2]equil = [I2]equil = 2.00 – x
= 2.00 – 1.37 = 0.63 M
12.918  9.459 x
12.918
x 
 1.37
9.459
 [HI]equil = 2.00 + 2x
= 2.00 + 2(1.37)
= 2.00 + 2.74
= 4.74 M
Jesperson, Brady, Hyslop. Chemistry: The
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Group
Problem
N2(g) + O2(g)
2NO(g)
Kc = 0.0123 at 3900 ˚C
If 0.25 moles of N2 and O2 are placed in a 250 mL
container, what are the equilibrium concentrations of
all species?
A.
B.
C.
D.
0.0526 M, 0.947 M, 0.105 M
0.947 M, 0.947 M, 0.105 M
0.947 M, 0.105 M, 0.0526 M
0.105 M, 0.105 M, 0.947 M
Jesperson, Brady, Hyslop. Chemistry: The
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Group
Problem
Conc (M) N2(g) + O2(g)
• Initial
1.00
1.00
• Change
–x
–x
• Equil
1.00 – x 1.00 – x
2NO (g)
0.00
+ 2x
+ 2x
0.250 mol
[N2 ] = [O2 ] =
= 1.00 M
0.250 L
(2x )2
2x
0.0123 =
0.0123 =
2
1- x
(1 - x )
x = 0.0526 M [NO] = 2x = 0.105 M
Jesperson, Brady, Hyslop. Chemistry: The
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation
Ex. 8
CH3CO2H(aq) + C2H5OH(aq)
acetic acid
ethanol
KC = 0.11
CH3CO2C2H5(aq) + H2O(l)
ethyl acetate
An aqueous solution of ethanol and acetic acid, each with initial
concentration of 0.810 M, is heated at 100 °C. What are the
concentrations of acetic acid, ethanol and ethyl acetate at
equilibrium?
Jesperson, Brady, Hyslop. Chemistry: The
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation
Step 1. Write equilibrium law
[CH3CO2C2H5 ]
Kc 
 0.11
[C 2H5OH][CH3CO2H]
• Need to find equilibrium values that satisfy this
Step 2: Set up concentration table using “x” for unknown
– Initial concentrations
– Change in concentrations
– Equilibrium concentrations
Jesperson, Brady, Hyslop. Chemistry: The
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation
Step 2 Concentration Table
(M) CH3CO2H(aq) + C2H5OH(aq)
I
C
E
•
•
•
•
0.810
–x
0.810 – x
CH3CO2C2H5(aq) + H2O(l)
0.810
0.000
0.810 – x
+x
–x
+x
Amt of CH3CO2H consumed = Amt of C2H5OH consumed = – x
Amt of CH3CO2C2H5 formed = + x
[CH3CO2H]eq and [C2H5OH ] = 0.810 – x
[CH3CO2C2H5] = x
x
0.11 =
(0.810 - x )(0.810 - x )
Jesperson, Brady, Hyslop. Chemistry: The
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation
Step 3. Solve for x
• Rearranging gives
0.11  (0.6561  1.62x  x 2 )  x
• Then put in form of quadratic equation
ax2 + bx + c = 0
2
0.07217  0.1782 x  0.11x
0.11x
2
x 0
 1.1782x  0.07217  0
• Solve for the quadratic equation using
 b  b 2  4ac
x 
2a
Jesperson, Brady, Hyslop. Chemistry: The
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation
Step 3. Solve for x
 (1.1782)  (1.1782) 2  4(0.11)(0.07217)
x 
2(0.11)
1.1782  (1.388)  (0.032) 1.1782  1.164
x 

0.22
0.22
• This gives two roots: x = 10.6 and x = 0.064
• Only x = 0.064 is possible
– x = 10.6 is >> 0.810 initial concentrations
– 0.810 – 10.6 = negative concentration,
which is impossible
Jesperson, Brady, Hyslop. Chemistry: The
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation
Step 4. Equilibrium Concentrations
CH3CO2H(aq) + C2H5OH(aq)
I 0.810
C –0.064
E 0.746
0.810
– 0.064
0.746
CH3CO2C2H5(aq) + H2O
0.000
+0.064
+0.064
[CH3CO2C2H5]equil = x = 0.064 M
[CH3CO2H]equil = [C2H5OH]equil = 0.810 M – x
= 0.810 M – 0.064 M
= 0.746 M
Jesperson, Brady, Hyslop. Chemistry: The
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Example: Cubic
When KC is very small
Ex. 9 2H2O(g)
2H2(g) + O2(g)
• At 1000 °C, KC = 7.3  10–18
• If the initial H2O concentration is 0.100 M, what will the H2
concentration be at equilibrium?
Step 1. Write Equilibrium Law
Kc 
2
[H2 ] [O 2 ]
[H2 O]2
 7.3  10
Jesperson, Brady, Hyslop. Chemistry: The
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93
Calculate [X]equilibrium from [X]initial and KC
Example: Cubic
Calculations
Step 2. Concentration Table
Conc (M ) 2H2O(g)
Initial
0.100
– 2x
Change
Equil’m 0.100 – 2x
7.3  10 18 
2H2(g) + O2(g)
0.00
0.00
+2x
+x
+2x
+x
(2x )2 x
2

4x 3
(0.100  2x )
(0.100  2x )2
• Cubic equation – tough to solve
• Make approximation
– KC very small, so x will be very small
– Assume we can neglect x
Jesperson, Brady, Hyslop. Chemistry: The
– Must prove valid later
Molecular Nature of Matter, 6E
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Calculate [X]equilibrium from [X]initial and KC
Example: Cubic
Calculations
Step 3. Solve for x
• Assume (0.100 – 2x)  0.100
Conc (M)
2H2O (g)
2H2 (g) +
Initial
Change
Equil’m
0.100
0.00
+2x
– 2x
0.100
O2 (g)
0.00
+2x
+x
+x
• Now our equilibrium expression simplifies to
2
3
(2x ) x
4x
18
7.3  10


2
0.010
4x 3
(0.100)
 0.010(7.3  10 18 ) = 7.3 × 10–20
Jesperson, Brady, Hyslop. Chemistry: The
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Example: Cubic
Step 3. Solve for x
x
3
7.3  10

4
20
 1.8  10 20
• Now take cube root
3
x  1.8  10
•
•
•
•
 20
 2.6  10
7
x is very small
0.100 – 2(2.6  10–7) = 0.09999948
Which rounds to 0.100 (3 decimal places)
[H2] = 2x = 2(2.6  10–7)
= 5.2  10–7 M
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Calculations
Simplifications: When Can You Ignore x
In Binomial (Ci – x)?
• If equilibrium law gives very complicated mathematical
problems and if K is small
– Then the change (x term) will also be small and we can
assume it can be ignored when added or subtracted
from the initial concentration, Ci.
• How do we check that the assumption is correct?
– If the calculated x is so small it does not change the
initial concentration
(e.g. 0.10 Minitial – 0.003 Mx-calc = 0.10)
– Or if the answer achieved by using the assumption
differs from the true value by less than five percent.
This often occurs when Ci > 100 x Kc
Jesperson, Brady, Hyslop. Chemistry: The
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Group
Problem
For the reaction 2A(g)
B(g)
given that Kp = 3.5 × 10–16 at 25 ˚C, and we place 0.2 atm A
into the container, what will be the pressure of B at
equilibrium?
PB
Q = KP = 2
2A  B
PA
I
0.2
0 atm
x
-16
3.5 ´ 10 =
C
–2x
+x
2
(0.2)
E
0.2 – 2x
x
x = 1.4 × 10–17
≈0.2
[B]= 1.4 × 10–17 atm
Proof: 0.2 - 1.4 × 10–17 = 0.2
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