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Chapter 6
Chemical Equilibrium
Chemistry: The Molecular Nature
of Matter, 6E
Jespersen/Brady/Hyslop
Dynamic Equilibrium in Chemical
Systems
 Chemical equilibrium exists when
 Rates of forward and reverse reactions are equal
 Reaction appears to stop
 [reactants] and [products] don't change over
time
 Remain constant
 Both forward and reverse reaction never cease
 Equilibrium signified by double arrows (
or equal sign (=)
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
)
2
Dynamic Equilibrium
N2O4
2 NO2
 Initially forward reaction rapid
 As some reacts [N2O4] so rate forward 
 Initially Reverse reaction slow
 No products
 As NO2 forms
  Reverse rate
 Ions collide more frequently as [ions] 
 Eventually rateforward = ratereverse
 Equilibrium
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Dynamic Equilibrium
(Fig. 15.1)




Almost all systems come to equilibrium
Where equilibrium lies depends on system
Some systems’ equilibrium hard to detect
Essentially no reactants or no products present
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Reaction Reversibility
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
Jespersen/Brady/Hyslop
N 2O 4

Chemistry: The Molecular Nature of Matter, 6E
2 NO2
5
Reactants
Equilibrium
N2O4
Products
2 NO2
 For given overall system composition
 Always reach same equilibrium concentrations
 Whether equilibrium is approached from forward or
reverse direction
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Chemistry: The Molecular Nature of Matter, 6E
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Equilibrium
 Simple relationship among [reactants] and
[products] for any chemical system at
equilibrium
 Called = mass action expression
 Derived from thermodynamics
 Forward reaction: A  B
k
 Reverse reaction: A  B
 At equilibrium:
A
B
kf
r
 rate forward = rate reverse
 rearranging:
Jespersen/Brady/Hyslop
Rate = kf[A]
Rate = kr[B]
kf[A] = kr[B]
[B] k f

 constant
[A] k r
Chemistry: The Molecular Nature of Matter, 6E
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Ex: H2(g) + I2(g)
2HI(g)
440°C
Exp’t Initial
#
Amts
I 1.00 mol H2
Equil’m
Amts
0.222 mol H2
Equil’m
[M]
0.0222 M H2
10 L 1.00 mol I2
0.00 mol HI
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
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Ex: H2(g) + I2(g)
Exp’t Initial Amts
#
2HI(g)
440°C
Equil’m
Amts
Equil’m
[M]
III
0.0150 mol H2
0.150 mol H2
0.0150 M H2
10 L
0.00 mol I2
0.135 mol I2
0.0135 M I2
1.27 mol HI
1.00 mol HI
0.100 M HI
IV
0.00 mol H2
0.442 mol H2
0.0442 M H2
10 L
0.00 mol I2
0.442 mol I2
0.0442 M I2
4.00 mol HI
3.11 mol HI
0.311 M HI
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Mass Action Expression (MAE)
 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
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Mass Action Expression
2
[HI]
Q
= same for all data sets at equilibrium
[H2 ][I 2 ]
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 ][I 2 ]
(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
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Equilibrium Law
 For reaction at equilibrium write the following
Equilibrium Law
(at 440 °C)
Kc
2
[HI]

 49.5
[H2 ][I 2 ]
 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
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Predicting Equilibrium Law
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
 Mass action expression =
[F] [G]
g
[D]d [E]e
 Note: Exponents in mass action expression
are stoichiometric coefficients in balanced
equation.
[F] f [G]g
K

c
 Equilibrium law is:
[D]d [E]e
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Ex. Equilibrium Law
3 H2(g) + N2(g)
2 NH3(g)
 Kc = 4.26 x 108 at 25 °C
 What is equilibrium law?
Kc 
[NH3 ]
3
2
[H2 ] [N 2 ]
Jespersen/Brady/Hyslop
 4.26  10
8
Chemistry: The Molecular Nature of Matter, 6E
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Learning Check
Write mass action expressions for the following:
 2 NO2 (g)
N2O4 (g)
[N2O4]
Q
2
[NO2]
 2CO (g) + O2 (g)
2 CO2 (g)
2
[CO2]
Q
2
[CO] [O2]
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Your Turn!
Which of the following is the correct mass
action expression (MAE) for the reaction:
Cu2+(aq) + 4NH3(aq)
[Cu(NH3)42+](aq)?
A.
B.
C.
Q
[Cu(NH3 ) 24  ]
[Cu2  ][NH3 ]4
Q
[Cu(NH3 ) 24  ]
Q
[Cu2  ][NH3 ]4
[Cu2  ][NH3 ]
[Cu(NH3 ) 24  ]
D. none of these
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Manipulating Equations for Chemical
Equilibria
Various operations can be performed on
equilibrium expressions
1. When direction of equation is reversed,
new equilibrium constant is reciprocal of
original
[C][D]
Kc 
A+B
C+D
[A][B]
C +D
A+B
Jespersen/Brady/Hyslop
[A][B]
1
K c 

[C][D] K c
Chemistry: The Molecular Nature of Matter, 6E
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Ex. Manipulating Equilibria 1
1. When direction of equation is reversed, new
equilibrium constant is reciprocal of original
3 H2(g) + N2(g)
Kc 
[NH3 ]
[H2 ]3 [N 2 ]
2 NH3(g)
K c 
2
 4.26  10
at 25°C
8
3 H2(g) + N2(g)
3
[H2 ] [N2 ]
[NH3 ]
2 NH3(g)
2

Jespersen/Brady/Hyslop
1
Kc

1
4.26  10
8
at 25°C
 2.35  10
Chemistry: The Molecular Nature of Matter, 6E
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Manipulating Equilibria 2
2. When coefficients in equation are
multiplied by a factor, equilibrium
constant is raised to a power equal to that
factor.
A+B
3A + 3B
K c 
Kc
C+D
3
[C][D]

[A][B]
3C + 3D
3
[C] [D]
[A]3 [B] 3
[C][D] [C][D] [C][D]
3



 Kc
[A][B] [A][B] [A][B]
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Manipulating Equilibria 2
2. When coefficients in equation are multiplied by
factor, equilibrium constant is raised to power
equal to that factor
3 H2(g) + N2(g)  2 NH3(g) at 25°C
Kc 
[NH3 ]
2
 4.26  10 8
[H2 ]3 [N 2 ]
multiply by 3
9 H2(g) + 3 N2(g)  6 NH3(g)
K c 
[NH3 ]6
9
[H2 ] [N 2 ]
Jespersen/Brady/Hyslop
3
3
 Kc
Chemistry: The Molecular Nature of Matter, 6E
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Manipulating Equilibria 3
3. When chemical equilibria are added,
their equilibrium constants are
multiplied
[C ][D ]
A+B
C + D Kc 
1
[A ][B ]
C+E
[F ][G ]
K c2 
[C ][E ]
D+F+G
F+G
A+B+E
[C ][D ] [F ][G ] [D ][F ][G ]
K c3 


 K c1  K c 2
[A ][B ] [C ][E ] [A ][B ][E ]
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Manipulating Equilibria 3
3. When chemical equilibria are added, their
equilibrium constants are multiplied
2 NO2(g)
NO3(g) + NO(g)
NO3(g) + CO(g)
NO2(g) + CO2(g)
NO2(g) + CO(g)
NO(g) + CO2(g)
[NO][NO 3 ]
K c1 
[NO][NO3 ]
[NO 2 ]2
K c2
[NO2 ][CO2 ]

[NO3 ][CO]
K c3
[NO][CO2 ]

[NO 2 ][CO]
[NO 2 ][CO 2 ] [NO][CO2 ]


[NO2 ][CO]
[NO 3 ][CO]
[NO 2 ]2
K c1  K c 2  K c 3
Therefore
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Learning Check
For: N2(g) + 3 H2(g)
2 NH3(g)
Kc = 500 at a particular temperature.
What would be Kc for following?
 2 NH3(g)
N2(g) + 3 H2(g)
K c 
1
Kc
1

 0.002
500
 ½ N2(g) + 3/2 H2(g)
K c  K c
12
NH3(g)
 500  22.4
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Equilibrium Constant, Kc
 Constant value equal to ratio of product
concentrations to reactant concentrations
raised to their respective exponents
[products]f
Kc 
d
[reactants ]
 Changes with temperature (van’t Hoff
Equation)
 Depends on solution concentrations
 Assumes reactants and products are in
solution
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Equilibrium Constant, Kp
 Based on reactions in which substances are
gaseous
 Assumes gas quantities are expressed in
atmospheres in mass action expression
 Use partial pressures for each gas in place of
concentrations
 Ex. N2 (g) + 3 H2 (g)  2 NH3 (g)
KP 
Jespersen/Brady/Hyslop
2
PNH
3
PN2 PH3
2
Chemistry: The Molecular Nature of Matter, 6E
25
How are Kp and Kc Related?
 Start with Ideal Gas Law
PV=nRT
 Rearranging gives
n 
P   RT  MRT
V 
 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 )n
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Learning Check
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
Δn
Kp  Kc(RT)
Kc 
Kp
(RT)
Kc = 11.7
n

Jespersen/Brady/Hyslop
0.480
(0.0821  298K )
Chemistry: The Molecular Nature of Matter, 6E
1
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Your Turn!
Consider the reaction A(g) + 2B(g)
4C(g)
If the Kc for the reaction is 0.99 at 25ºC, what
would be the Kp?
A. 0.99
Δn=(4 – 3)=1
B. 2.0
C. 24.
Kp = Kc(RT)Δn
Kp= 0.99*(0.082057*298.15)1
D. 2400
Kp = 24
E. None of these
Jespersen/Brady/Hyslop
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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
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
29
Heterogeneous Equilibria
2NaHCO3(s)
Na2CO3(s) + H2O(g) + CO2(g)
 Equilibrium Law =
[Na 2CO 3 (s )][H 2O ( g )][CO 2 ( g )]
K 
[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
 Ex. 1 mol NaHCO3 occupies 38.9 cm3
2 mol NaHCO3 occupies 77.8 cm3
M 
1 mol NaHCO3
M 
 25.7M
0.0389
L
Jespersen/Brady/Hyslop
2 mol NaHCO3
 25.7M
0.0778 L
Chemistry: The Molecular Nature of Matter, 6E
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Heterogeneous Equilibria
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
[Na2 CO3 (s )]
Kc  K
 [H2 O( g )][CO2 ( g )]
2
[NaHCO3 (s )]
 Equilibrium law for heterogeneous system written
without concentrations of pure solids or liquids
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
31
Interpreting KC
 Large K (K>>1)
 Means product rich mixture
 Reaction goes far toward
completion
Ex.
2SO2(g) + O2(g)
2SO3(g)
Kc = 7.0  1025 at 25 ° C
Kc
[SO 3 ]
2
7.0  10


2
1
[SO 2 ] [O 2 ]
Jespersen/Brady/Hyslop
25
Chemistry: The Molecular Nature of Matter, 6E
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Interpreting KC
 Small K (K<<1)
 Means reactant rich
mixture
 Only very small amounts of
product formed
Ex.
H2(g) + Br2(g)
2HBr(g)
Kc = 1.4  10–21 at 25 °C
2
21
[HBr]
1.4  10
Kc 

[H2 ][Br2 ]
1
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
33
Interpreting KC
 K  1
 Means product and
reactant concentrations
close to equal
 Reaction goes only ~
halfway
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
34
 Size of K gives measure of how
reaction proceeds
 K >> 1
 K=1
 K << 1
[products] >> [reactants]
[products] = [reactants]
[products] << [reactants]
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
35
Learning Check
Consider the reaction of 2NO2(g)
N2O4(g)
If Kp = 0.480 at 25°C, does the reaction favor
product or reactant?
K is small (K < 1)
Reaction favors reactant
Since K is close to 1, significant amounts of
both reactant and product are present
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Equilibrium Positions and “Shifts”
 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)
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
37
Relationship Between Q and K
 Q=K
 Q<K
reaction at equilibrium
reactants  products
 Too many reactants
 Must convert some reactant to product to move
reaction toward equilibrium
 Q>K
reactants  products
 Too many products
 Must convert some product to reactant to move
reaction toward equilibrium
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
38
Examples of Le Châtelier’s Principle
Let’s see how this works with changes in
1. Concentration
2. Pressure and volume
3. Temperature
4. Catalysts
5. Adding inert gas to system at constant
volume
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
39
Effect of Change in Concentration
 2SO2(g) + O2(g) → 2SO3(g)
 Kc = 2.4 x 10-3 at 700 oC
 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
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
40
Effect of 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
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
41
Effect of Pressure and Volume Changes
 Consider gaseous system at constant T and n
2
3H2(g) + N2(g)
2NH3(g)
PNH
3
K

P
 If reduce volume (V)
PN2 PH3






2
Expect Pressure to increase (P)
To reduce pressure, look at each side of reaction
Which has less moles of gas
Reactants = 3 + 1 = 4 moles gas
Products = 2 moles gas
Reaction favors products (shifts to right)
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
42
Effect of P and V Changes
Consider gaseous system at constant T and n
Ex.
H2(g) + I2(g)
2 HI(g)
2
PHI
KP 
PH2 PI2
 If pressure is increased, what is the effect on
equilibrium?
 nreactant = 1 + 1 = 2
 nproduct = 2
 Predict no change or shift in equilibrium
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
43
Effect of P and V Changes
2NaHSO3(s)
NaSO3(s) + H2O(g) + SO2(g)
K P  PH2OPSO2
 If you decrease volume of reaction, what is
the effect on equilibrium?
 Reactants: no moles gas = all solids
 Products: 2 moles gas
 V, causes P
 Reaction shifts to left (reactants), as this has
fewer moles of gas
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
44
Effect of P and V Changes
 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
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
45
Effect of Temperature Changes
H2O(s)
H° =+6 kJ (at 0 °C)
H2O(ℓ)
 Energy + H2O(s)
H2O(ℓ)
 Energy is reactant
 Add heat, shift reaction right
3H2(g) + N2(g)
 3 H2(g) + N2(g)
2NH3(g) Hf°= –47.19 kJ
2 NH3(g) + energy
 Energy is product
 Add heat, shift reaction left
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
46
Effect of Temperature Changes
  T shifts reaction in direction that produces
endothermic (heat absorbing) change
  T shifts reaction in direction that produces
exothermic (heat releasing) change
 Changes in T change value of mass action
expression at equilibrium, so K changed
 K depends on T
 T of exothermic reaction makes K smaller
 More heat (product) forces equilibrium to reactants
 T of endothermic reaction makes K larger
 More heat (reactant) forces equilibrium to products
Jespersen/Brady/Hyslop
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47
Catalysts And Equilibrium
 Catalyst lowers Ea
for both forward
and reverse
reaction
 Change in Ea
affects rates kr
and kf equally
 Catalysts have no
effect on
equilibrium
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
48
Effect of Adding Inert Gas
Inert gas
 One that does not react with components of
reaction
Ex. Argon, Helium, Neon, usually N2
 Adding inert gas to reaction at fixed V (n and
T),  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
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
49
Your Turn!
The following are equilibrium constants for the
reaction of acids in water, Ka. Which is the
most efficient reaction?
A. Ka = 2.2×10–3
B. Ka = 1.8×10–5
C. Ka = 4.0×10–10
D. Ka = 6.3×10–3
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
50
Equilibrium Calculations



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
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Calculating KC from 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
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Calculating KC Given Equilibrium
Concentrations
Ex. 1 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
[NO 2 ]
Kc 
[N 2O 4 ]
2
[0.0116]
Kc 
[0.0292]
KC = 4.61  10–3
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Chemistry: The Molecular Nature of Matter, 6E
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Your Turn!
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
[C]3
Kc 
2
B. 0.15
[A] [B]
C. 1.5
[3.0]3
D. 6.75
Kc 
2
[2.0] [1.0]
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Calculating KC Given Initial Concentrations
and One Final Concentration
Ex. 2 2SO2(g) + O2(g)
2SO3(g)
 1.000 mol SO2 and 1.000 mol O2 are placed in
a 1.000 L flask at 1000 K. At equilibrium
0.925 mol SO3 has formed. Calculate KC for
this reaction.
 1st calculate concentrations of each
 Initial
1.00mol
[SO 2 ]  [O 2 ] 
 1.00M
1.00L
 Equilibrium
0.925mol
[SO 3 ] 
 0.925M
1.00L
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How to Solve:
 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
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Next Set up Concentration Table
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 = Amt of SO3 formed
= [SO3] at equilibrium = 0.925 M
[O2] consumed = ½ amt 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
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Chemistry: The Molecular Nature of Matter, 6E
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Ex. 2
 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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Chemistry: The Molecular Nature of Matter, 6E
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Summary of Concentration Table
 Used for most equilibrium calculations (Ch
15, 17, and 18)
1. Equilibrium concentrations are only values
used in mass action expression

Values in last row of table
2. Initial value in table must be in units of
mol/L (M)


[X]initial = those present when reaction
prepared
No reaction occurs until everything is
mixed
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Chemistry: The Molecular Nature of Matter, 6E
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Summary of Concentration Table
3. Changes in concentrations always occur in
same ratio as coefficients in balanced
equation
4. 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 + ()
because [reactant]final  negative
If [reactants] , all entries for reactants in change
row should have minus sign and all entries for
products should be positive
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
60
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. 3 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] = 0.800M and [CO] =0.300M.
What is [H2O] at equilibrium ?
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Calculate [X]equilibrium from Kc and [X]initial
Ex. 3 CH4(g) + H2O(g)
CO(g) + 3H2(g) Kc = 5.67
[CH4] = 0.400 M; [H2] = 0.800M; [CO] =0.300M
 What is [H2O] at equilibrium?
 First, set up equilibrium
3
[CO][H2 ]3
[H2O] 
[CH4 ]K c
[CO][H2 ]
Kc 
[CH4 ][H2O]
 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
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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
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
63
Calculate [X]equilibrium from [X]initial and KC
Ex. 4 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
2
[HI ]
Kc 
 55.64
[H 2 ][I 2 ]
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Ex. 4 Step 2. Concentration Table
Conc (M)
H2(g) +
I2(g)
2HI (g)
Initial
Change
Equil’m
2.00
2.00
0.000
–x
–x
+2x
+2x
2.00 – x
2.00 – x
 Initial [H2] = [I2] = 1.00 mol/0.500L =2.00M
 Amt of H2 consumed = Amt of I2 consumed = x
 Amt of HI formed = 2x
(2x ) 2
(2x ) 2
55.64 

(2.00  x )(2.00  x ) (2.00  x ) 2
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Ex. 4 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 
(2.00  x )
2
2x
7.459 
(2.00  x )
7.459(2.00  x )  2x
14.918  7.459x  2x
14.918  9.459x
14.918
x 
 1.58
9.459
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Chemistry: The Molecular Nature of Matter, 6E
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Ex. 4 Step 4. Equilibrium Concentrations
Conc (M)
H2(g) +
Initial
2.00
Change – 1.58
Equil’m
0.42
I2(g)
2.00
– 1.58
0.42
2HI (g)
0.00
+3.16
+3.16
 [H2]equil = [I2]equil = 2.00 – 1.58 = 0.42 M
 [HI]equil = 2x = 2(1.58) = 3.16
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
67
Calculate [X]equilibrium from [X]initial and KC
Ex. 5 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
[H 2 ][I 2 ]
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Chemistry: The Molecular Nature of Matter, 6E
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Ex. 5 Step 2. Concentration Table
Conc (M)
H2(g) +
I2(g)
2HI (g)
Initial
Change
Equil’m
2.00
2.00
2.00
–x
–x
2.00 – x
2.00 – x
+2x
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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Ex. 5 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
12.918  9.459 x
12.918
x 
 1.37
9.459
Jespersen/Brady/Hyslop
 [H2]equil = [I2]equil = 2.00 – x =
2.00 – 1.37 = 0.63 M
 [HI]equil = 2.00 + 2x
= 2.00 + 2(1.37)
= 2.00 + 2.74
= 4.74 M
Chemistry: The Molecular Nature of Matter, 6E
70
Your Turn!




N2(g) + O2(g) → 2NO(g)
Kc = 0.0123 at 3900 oC
If 0.25 moles of N2 and O2 are placed in a
250 mL container, what are the equilibrium
concentrations of all species ?
A. 0.0526 M, 0.947 M, 0.105 M
B. 0.947 M, 0.947 M, 0.105 M
C. 0.947 M 0.105 M, 0.0526 M
D. 0.105 M, 0.105 M, 0.947 M
Jespersen/Brady/Hyslop
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Your Turn! - Solution




Conc (M)
Initial
Change
Equil
N2(g) +O2(g)
1.00
1.00
–x
–x
1.00 – x 1.00 – x
2NO (g)
0.00
+ 2x
+ 2x
0.250 mol
[N2 ]  [O2 ] 
 1.00M
0.250 L
(2x )2
2x
0.0123 
0.0123 
2
1x
(1  x )
x  0.0526M
Jespersen/Brady/Hyslop
[NO] = 2x = 0.105M
Chemistry: The Molecular Nature of Matter, 6E
72
Calculate [X]equilibrium from [X]initial and KC
Ex. 6
CH3CO2H(aq) + C2H5OH(aq)
acetic acid
ethanol
CH3CO2C2H5(aq) +
ethyl acetate H2O(l)
KC = 0.11
 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?
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Chemistry: The Molecular Nature of Matter, 6E
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Calculate [X]equilibrium from [X]initial and KC
 When KC is very small
Ex. 7 2H2O(g)
2H2(g) + O2(g)
 At 1000 °C, KC = 7.3  10–18
 If the initial H2O concentration is 0.100M,
what will the H2 concentration be at
equilibrium?
 Step 1. Write Equilibrium Law
Kc 
[H2 ]2 [O 2 ]
2
[H2 O]
Jespersen/Brady/Hyslop
 7.3  10
18
Chemistry: The Molecular Nature of Matter, 6E
74
Ex. 7 Step 2. Concentration Table
Conc (M)
2H2O(g)
Initial
0.100
Change
– 2x
Equil’m 0.100 – 2x
7.3  10
18

2H2(g) + O2(g)
0.00
0.00
+2x
+x
+x
+2x
(2x )2 x
2
(0.100  2x )

4x 3
(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
 must prove valid later
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Chemistry: The Molecular Nature of Matter, 6E
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Ex. 7 Step 3. Solve for x
 Assume (0.100 – 2x)  0.100
Conc (M)
2H2O (g)
2H2 (g) +
O2 (g)
Initial
Change
Equil’m
0.100
0.00
0.00
– 2x
+2x
0.100
+2x
+x
+x
 Now our equilibrium expression simplifies to
2
3
(2x ) x
4x
18
7.3  10


(0.100)2 0.010
4x 3  0.010(7.3  10 18 ) = 7.3 × 10–20
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Chemistry: The Molecular Nature of Matter, 6E
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Ex. 7 Step 3. Solve for x
20
7
.
3

10
x3 
 1.8  10 20
4
 Now take cubic 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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Simplifications: When Can You
Ignore x in Binomial (Ci – x)?
 If equilibrium law gives very complicated
mathematical problems
 And if K is small
 Change to reach equilibrium (x term) is also small
 Compare initial concentration Ci in binomial to
value of K
Ci
 400
K
 Use proof to show that dropped x term was
sufficiently small
dropped x term ?
0.05

Ci
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
78
Learning Check
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?
PN2O4
Q
2A ↔
B
2
PNO
I
C
E
0.2
0 atm
x
16
–2x
+x
3.5  10

2
(
0
.
2
)
0.2 – 2x ≈0.2 x
–17
0
x
=
1.4×10
Q
0
2
(0.2)
[B]= 1.4×10–17 M
Q  K shifts right proof: 1.4×10–17/0.2<0.05
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Your Turn!
In the reaction shown, K = 1.8 × 10–5
HC2H3O2(aq) + H2O(l) ↔ H3O+(aq) + C2H3O2–(aq)
If we start with 0.3M HC2H3O2, what will be the
equilibrium concentration of C2H3O2 –?
A. 0.3 M
B. 0.002 M
C. 0.04 M
D. 0.5 M
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Calculating KC Given Initial Concentrations
and One Final Concentration
Ex. 2a
H2(g) + I2(g)  2HI(g) @ 450 °C
 Initially H2 and I2 concentrations are 0.200
mol each in 2.00L (= 0.100M); no HI is
present
 At equilibrium, HI concentration is 0.160 M
 Calculate KC
 To do this we need to know 3 sets of
concentrations: initial, change and
equilibrium
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