Chapter 13
Chemical Equilibrium:
How can things that are
moving seem to be standing
still?
Chapter 13
Table of Contents
13.1
13.2
13.3
13.4
13.5
13.6
13.7
The Equilibrium Condition
The Equilibrium Constant
Equilibrium Expressions Involving Pressures
Heterogeneous Equilibria
Applications of the Equilibrium Constant
Solving Equilibrium Problems
Le Châtelier’s Principle
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2
Section 13.1
The Equilibrium Condition
Objectives
• To define the Chemical Equilibrium
• To describe the equilibrium condition in terms of
reactant and product concentration
• To describe the equilibrium condition in terms of
forward and reverse reaction rates
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3
Section 13.1
The Equilibrium Condition
A Gaseous Solute
C = kP
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4
Section 13.1
The Equilibrium Condition
Liquid/Vapor Equilibrium
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Section 13.1
The Equilibrium Condition
Chemical Equilibrium
• So far, we have seen chemical equations written as:
• H2O(g) + CO(g)  H2(g) + CO2(g)
• Or stated as such that No Reaction occurs
• This may be an oversimplification, whereas…
• H2O(g) + CO(g)
H2(g) + CO2(g)
• Is a more thorough description of what is happening
in virtually every chemical reaction.
• But, what do these symbols mean?
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6
Section 13.1
The Equilibrium Condition
Chemical Equilibrium
• The double arrow sign,
, represents an
equilibrium condition.
• Equilibrium is: The state where the
concentrations of all reactants and products
remain constant with time.
• On the molecular level, there is frantic activity.
Equilibrium is not static, but is a highly dynamic
situation.
• Equilibrium is :
Macroscopically static
•
Microscopically dynamic
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Section 13.1
The Equilibrium Condition
N2(g) + 3H2(g)
2NH3(g)
Concentrations reach constant levels where the rate of the
forward reaction equals the rate of the reverse reaction.
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Section 13.1
The Equilibrium Condition
The Changes with Time in the Rates of Forward and Reverse Reactions
•
•
Bees wake up and start leaving to collect honey…time passes.
the # of bees leaving = the # of bees entering…equilibrium
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9
Section 13.1
The Equilibrium Condition
Concept Check
Consider an equilibrium mixture in a closed
vessel reacting according to the equation:
H2O(g) + CO(g)
H2(g) + CO2(g)
You add more H2O(g) to the flask. How does
the concentration of each chemical compare to
its original concentration after equilibrium is
reestablished? Justify your answer.
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Section 13.1
The Equilibrium Condition
Objectives Review
• To define the Chemical Equilibrium
• To describe the equilibrium condition in terms of
reactant and product concentration
• To describe the equilibrium condition in terms of
forward and reverse reaction rates
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Section 13.2
Atomic
The
Equilibrium
Masses Constant
Objectives
• To define the equilibrium constant (K)
• To write the equilibrium constant expression for
various chemical reactions
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Section 13.2
Atomic
The
Equilibrium
Masses Constant
Consider the following reaction at equilibrium:
jA + kB
lC + mD
l
m
j
[A]
[B]k
[C] [D]
K=
•
•
•
•
A, B, C, and D = chemical species.
Square brackets = concentrations of species at equilibrium.
j, k, l, and m = coefficients in the balanced equation.
K = equilibrium constant (given without units).
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13
Section 13.2
Atomic
The
Equilibrium
Masses Constant
Write the equilibrium constant expression:
• H2O(g) + CO(g)
• 4NH3(g) + 7O2(g)
H2(g) + CO2(g)
4NO2(g) + 6H2O(g)
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Section 13.2
Atomic
The
Equilibrium
Masses Constant
Conclusions About the Equilibrium Expression
• Equilibrium expression for a reaction is the
reciprocal of that for the reaction written in
reverse.
• When balanced equation for a reaction is
multiplied by a factor of n, the equilibrium
expression for the new reaction is the original
expression raised to the nth power; thus
Knew = (Koriginal)n.
• K values are usually written without units.
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Section 13.2
Atomic
The
Equilibrium
Masses Constant
More Conclusions About the Equilibrium Expression
• K always has the same value at a given
temperature regardless of the amounts of
reactants or products that are present initially.
• For a reaction, at a given temperature, there are
many equilibrium positions but only one value
for K.
 Equilibrium position is a set of equilibrium
concentrations.
 K = [Products]

[Reactants]
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Section 13.2
Atomic
The
Equilibrium
Masses Constant
Objectives Review
• To define the equilibrium constant (K)
• To write the equilibrium constant expression for
various chemical reactions
• Work Session: Pg 613 # 1, 6, 12, 17& 18 a&b,
19
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Section 13.3
The Mole Expressions Involving Pressures
Equilibrium
Objectives
• To write the Equilibrium Expression for gas
reactions (Kp)
• To calculate K from Kp
• To see how the 5-Step Problem Solving Method
can help you calculate without having to
“KNOW” how to do a problem…
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Section 13.3
The Mole Expressions Involving Pressures
Equilibrium
• K involves concentrations.
• Kp involves pressures.
Pr oducts
Coefficient
K =
Reac tan ts
Coefficient
(Pr odPr essure)Coefficient
Kp =
Coefficient
(ReactPr essure)
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Section 13.3
The Mole Expressions Involving Pressures
Equilibrium
Example
N2(g) + 3H2(g)
2NH3(g)
Write the K and Kp expressions:
NH3 
3
N2 H2 
K =
P 

=
P P 
2
2
Kp
NH
3
N
2
3
H
2
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Section 13.3
The Mole Expressions Involving Pressures
Equilibrium
Example
N2(g) + 3H2(g)
2NH3(g)
Calculate Kp given the equilibrium pressures at a
certain temperature:
PNH = 2.9  10 2 atm
3
PN = 8.9  10 1 atm
2
PH = 2.9  10 3 atm
2
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Section 13.3
The Mole Expressions Involving Pressures
Equilibrium
Example
N2(g) + 3H2(g)
2NH3(g)
P 

=
P P 
2
Kp
NH
3
N
2
Kp =
 8.9
H
2
 2.9
 10
Kp = 3.9  104
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 10
1

2 2
 2.9
 10

3 3
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22
Section 13.3
The Mole Expressions Involving Pressures
Equilibrium
The Relationship Between K and Kp
Kp = K(RT)Δn
• Δn = sum of the coefficients of the gaseous
products minus the sum of the coefficients of
the gaseous reactants.
• R = 0.08206 L·atm/mol·K
• T = temperature (in kelvin)
• Using the value of Kp (3.9 × 104) from the
previous example, calculate the value of K at
35°C.
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Section 13.3
The Mole Expressions Involving Pressures
Equilibrium
Example
N2(g) + 3H2(g)
2NH3(g)
Using the value of Kp (3.9 × 104) from the previous
example, calculate the value of K at 35°C.
Kp = K  RT 
n
3.9  10 = K  0.08206 L  atm/mol  K  308K 
4
 2 4 
K = 2.5  107
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Section 13.3
The Mole Expressions Involving Pressures
Equilibrium
The Power of the 5 Step!
• Take a look at the AP reference sheet…
• Even though you may only see an equation once,
or perhaps never, and you may think that you
forget or never learned how to use the equation…..
• The 5 Step is the “HOW” of working with
equations…
• Being more familiar is more comfortable, however,
take comfort in the Power of the 5 Step!
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Section 13.3
The Mole Expressions Involving Pressures
Equilibrium
Objectives Review
• To write the Equilibrium Expression for gas
reactions (Kp)
• To calculate K from Kp
• To see how the 5-Step Problem Solving Method
can help you calculate without having to
“KNOW” how to do a problem…
• Work Session pg 615 # 21, 26
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Section 13.4
Heterogeneous Equilibria
Objectives 13.4 and 13.5
• To define homogeneous and heterogeneous
equilibria
• To understand why pure solids and liquids are
left out of the equilibrium expression
• To evaluate the extent of a reaction based on
the numerical value of K
• To use the Reaction Quotient (Q) to determine
if a system is at equilibrium and what directional
shift will occur to reach EQ
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Section 13.4
Heterogeneous Equilibria
Homogeneous Equilibria
• Homogeneous equilibria – involve the same
phase:
N2(g) + 3H2(g)
2NH3(g)
HCN(aq)
H+(aq) + CN-(aq)
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Section 13.4
Heterogeneous Equilibria
Heterogeneous Equilibria
• Heterogeneous equilibria – involve more than
one phase:
2KClO3(s)
2KCl(s) + 3O2(g)
2H2O(l)
2H2(g) + O2(g)
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Section 13.4
Heterogeneous Equilibria
• The position of a heterogeneous equilibrium
does not depend on the amounts of pure solids
or liquids present.
 The concentrations of pure liquids and solids
are constant. (Think of them as 1)
2KClO3(s)
2KCl(s) + 3O2(g)
K =  O2 
3
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Section 13.5
Applications of the Equilibrium Constant
The Extent of a Reaction
• A value of K much larger than 1 means
that at equilibrium the reaction system will
consist of mostly products – the equilibrium
lies to the right.
 Reaction goes essentially to completion.
 A + B  AB
Pr oducts
Coefficient
K =
Reac tan ts
Coefficient
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Section 13.5
Applications of the Equilibrium Constant
The Extent of a Reaction
• A very small value of K means that the
system at equilibrium will consist of mostly
reactants – the equilibrium position is far to
the left.
 Reaction does not occur to any
significant extent. A + B  No Rx
Pr oducts
Coefficient
K =
Reac tan ts
Coefficient
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Section 13.5
Applications of the Equilibrium Constant
Concept Check
If the equilibrium lies to the right, the value for K
is __________.
large (or >1)
If the equilibrium lies to the left, the value for K
is ___________.
Coefficient
small (or <1)
Pr oducts

K =
Coefficient
Reac tan ts
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Section 13.5
Applications of the Equilibrium Constant
Exercise
Consider the reaction represented by the equation:
Fe3+(aq) + SCN-(aq)
FeSCN2+(aq)
•
Trial #1:
6.00 M Fe3+(aq) and 10.0 M SCN-(aq) are mixed at a
certain temperature and at equilibrium the concentration
of FeSCN2+(aq) is 4.00 M.
2
FeSCN 
K =
Fe3  SCN 
What is the value for the equilibrium constant for this
reaction? Need [at Equilibrium] –Use ICE Table…
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Section 13.5
Applications of the Equilibrium Constant
Set up ICE Table (Initial, Change, Equil’ Conditions)
Fe3+(aq) + SCN–(aq)
Initial
Change
Equilibrium
6.00
- 4.00
2.00
10.00
– 4.00
6.00
FeSCN2+(aq)
0.00
+4.00
4.00
FeSCN2 
4.00 M 

K =
=
3

Fe  SCN   2.00 M  6.00 M 
K = 0.333 (1:1 Stoich)
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Section 13.5
Applications of the Equilibrium Constant
Exercise
Consider the reaction represented by the equation:
Fe3+(aq) + SCN-(aq)
FeSCN2+(aq)
• Trial #2:
Initial: 10.0 M Fe3+(aq) and 8.00 M SCN−(aq) (same
temperature as Trial #1) (Means you know K)
Equilibrium:
? M FeSCN2+(aq) (ICE TABLE)
(NEXT SLIDE)
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Section 13.5
Applications of the Equilibrium Constant
Exercise
Consider the reaction represented by the equation:
Fe3+(aq) + SCN-(aq)
FeSCN2+(aq)
Trial #2:
Initial: 10.0 M Fe3+(aq) and 8.00 M SCN−(aq) (same
temperature as Trial #1) (Means you know K)(ICE
TABLE)
I
C
E
FeSCN2 
K=
Fe3  SCN 
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5.00 M FeSCN2+
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Section 13.5
Applications of the Equilibrium Constant
Exercise
Consider the reaction represented by the equation:
Fe3+(aq) + SCN-(aq)
FeSCN2+(aq)
• Trial #3:
Initial: 6.00 M Fe3+(aq) and 6.00 M SCN−(aq)
Equilibrium:
? M FeSCN2+(aq) (ICE Set Up K)
(Don’t solve)
2
FeSCN 
K =
Fe3  SCN 
3.00 M FeSCN2+
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Section 13.5
Applications of the Equilibrium Constant
Exercise (Start with the ICE, Keq, and Work
Through the Algebra, no Quadratic needed)
Consider the reaction represented by the equation
(assume K = 0.333) :
Fe3+(aq) + SCN-(aq)
FeSCN2+(aq)
Trial #1
Trial #2
Trial #3
Fe3+
9.00 M
3.00 M
2.00 M
SCN5.00 M
2.00 M
9.00 M
FeSCN2+
1.00 M
5.00 M
6.00 M
Find the equilibrium concentrations for all species.
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Section 13.5
Applications of the Equilibrium Constant
Exercise (Answer)
Trial #1: [Fe3+] = 6.00 M; [SCN-] = 2.00 M; [FeSCN2+] = 4.00 M
Trial #2: [Fe3+] = 4.00 M; [SCN-] = 3.00 M; [FeSCN2+] = 4.00 M
Trial #3: [Fe3+] = 2.00 M; [SCN-] = 9.00 M; [FeSCN2+] = 6.00 M
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Section 13.5
Applications of the Equilibrium Constant
Reaction Quotient, Q
• To determine if a system is at EQ, apply
the law of mass action using initial
(instantaneous) concentrations instead of
equilibrium concentrations.
• In other words, calculate the Q expression
and see if the conditions are the same as
the K expression…
Pr oducts
Coefficient
Q=
Reac tan ts
Coefficient
 K ?
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Section 13.5
Applications of the Equilibrium Constant
Reaction Quotient, Q
Pr oducts
Coefficient
Q=
Reac tan ts
Coefficient
 K ?
• Q = K; The system is at equilibrium. No shift will
occur.
• Q > K;
 Consuming products and forming reactants, until
equilibrium is achieved.
 The system shifts to the ……
 ….left.
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Section 13.5
Applications of the Equilibrium Constant
Reaction Quotient, Q
Pr oducts
Coefficient
Q=
Reac tan ts
Coefficient
 K ?
• Q < K;
 Consuming reactants and forming products, to
attain equilibrium.
 The system shifts to the …………
 ………right.
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Section 13.5
Applications of the Equilibrium Constant
For the synthesis of ammonia at 500o C, Keq = 6.0 X 10-2.
Predict if the following system is in equilibrium. If not, indicate
the direction in which the system will shift in order to reach eq’.
a. [NH3] = 1.0 X 10-3M; [N2] = 1.0 X 10 -5M; [H2] = 2.0 X 10 -3 M
a.
1.25 X 107 Left shift
b. [NH3] = 2.0 X 10-4M; [N2] = 1.5 X 10 -5M; [H2] = 3.54 X 10 -1 M
@ EQ’M
c. [NH3] = 1.0 X 10-4M; [N2] = 5.0M; [H2] = 1.0 X 10 -2 M
2.0 X 10 -3 Right Shift
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Section 13.5
http://www.usca.edu/chemistry/genchem/sigfig.htm
Applications of the Equilibrium Constant
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Section 13.5
Applications of the Equilibrium Constant
Objectives Review 13.4 and 13.5
• To define homogeneous and heterogeneous
equilibria
• To understand why pure solids and liquids are
left out of the equilibrium expression
• To evaluate the extent of a reaction based on
the numerical value of K
• To use the Reaction Quotient (Q) to determine
if a system is at equilibrium and what directional
shift will occur to reach EQ
• Work Session: Pg 615 # 29, 31, 33, 37
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Section 13.6
Solving Equilibrium Problems
Objectives
• To practice using the ICE method for solving
Equilibrium Problems
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Section 13.6
Solving Equilibrium Problems
Solving Equilibrium Problems
1) Write the balanced equation for the
reaction.
2) Write the equilibrium expression using the
law of mass action.
3) List the initial concentrations.
4) Calculate Q, and determine the direction of
the shift to equilibrium.
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Section 13.6
Solving Equilibrium Problems
Solving Equilibrium Problems
5) Define the change needed to reach
equilibrium, and define the equilibrium
concentrations by applying the change to
the initial concentrations.
6) Substitute the equilibrium concentrations
into the equilibrium expression, and solve
for the unknown.
7) Check your calculated equilibrium
concentrations by making sure they give
the correct value of K.
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Section 13.6
Solving Equilibrium Problems
Concept Check
A 2.0 mol sample of ammonia is introduced into a
1.00 L container. At a certain temperature, the
ammonia partially dissociates according to the
equation:
NH3(g)
N2(g) + H2(g)
At equilibrium 1.00 mol of ammonia remains.
Calculate the value for K. (Balance)(ICE)
K = 1.69
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Section 13.6
Solving Equilibrium Problems
Concept Check
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Section 13.6
Solving Equilibrium Problems
Concept Check
A 1.00 mol sample of N2O4(g) is placed in a 10.0 L
vessel and allowed to reach equilibrium according to
the equation:
N2O4(g)
2NO2(g)
K = 4.00 x 10-4
Calculate the equilibrium concentrations of: N2O4(g)
and NO2(g). (ICE,
Concentration of N2O4 = 0.097 M
Concentration of NO2 = 6.32 x 10-3 M
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Section 13.6
Solving Equilibrium Problems
Concept Check
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Section 13.6
Solving Equilibrium Problems
Concept Check
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Section 13.6
Solving Equilibrium Problems
Concept Check
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Section 13.6
Solving Equilibrium Problems
Objectives Review
• To practice using the ICE method for solving
Equilibrium Problems
• Work Session: page 616 39, 41, 43*, 45 (setup
only), 47, 51(setup only), 53 (setup only)
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Section 13.7
Le Châtelier’s Principle
Objectives
• To understand Le Chatelier’s Principle
• To predict shifts in equilibrium
• HPhen (aq)
•
CLEAR
H+ (aq) + Phen- (aq)
PINK
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Section 13.7
Le Châtelier’s Principle
• Le Chatelier’s Principle
• If a change is imposed on a system
at equilibrium, the position of the
equilibrium will shift in a direction that
tends to reduce that change.
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Section 13.7
Le Châtelier’s Principle
Effects of Changes on the System
1. Concentration: The system will shift away from
the added component. If a component is
removed, the opposite effect occurs.
2. Temperature: K will change depending upon
the temperature (endothermic – energy is a
reactant; exothermic – energy is a product).
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Section 13.7
Le Châtelier’s Principle
Effects of Changes on the System
3. Pressure:
a) The system will shift away from the added
gaseous component. If a component is
removed, the opposite effect occurs.
b) Addition of inert gas does not affect the
equilibrium position.
c) Decreasing the volume shifts the equilibrium
toward the side with fewer moles of gas.
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Section 13.7
Le Châtelier’s Principle
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Section 13.7
Le Châtelier’s Principle
Equilibrium Decomposition of N2O4
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Section 13.7
Le Châtelier’s Principle
Predicting Shifts in Equilibria
• Which way will the equilibrium shift if: (don’t forget volume…)
N2(g) + 3H2(g)
HCN(aq)
2NH3(g)
H+(aq) + CN-(aq) H = -12 kJ
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Section 13.7
Le Châtelier’s Principle
Objectives
• To understand Le Chatelier’s Principle
• To predict shifts in equilibrium
• Work Session: pg 617 # 57, 59, 63
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Section 13.7
Le Châtelier’s Principle
Objectives
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