2/9/2016
The Equilibrium State
Chapter 13 - Chemical
Equilibrium
• Not all chemical reactions go to completion;
instead they attain a state of equilibrium.
• When you hear “equilibrium”, what do you
think of?
• Example: weather patterns: ocean water
evaporates at the same rate that it rains.
They are in equilibrium.
The Equilibrium State
Equilibrium is Dynamic!
• Equilibrium: a state in which the rates
of the forward and reverse reactions
are equal and the concentrations of
reactants and products remain constant
over time.
• Most reactions are reversible.
The conversions of reactants to
products and products to reactants
are still going on, although there is
no net change in the number of
reactant and product molecules.
A + B C+ D
 A and B react to make C and D
 But C and D can also react to make A and B
N2O4 (g)  2NO2 (g)
 In the forward reaction each molecule of N2O4
breaks down to form two molecules of NO2.
 In the reverse reaction two molecules of NO2
combine to form N2O4.
 Equilibrium occurs when the rate at which an
N2O4 molecule breaks apart in the forward
reaction is equal to the rate at which it is
formed by the reverse reaction.
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Exp data for N2O4 (g)  2NO2 (g)
• Figure 13.1 What’s the same/different?
The Equilibrium Constant
The Equilibrium Constant, Kc, tells us which side
of the reaction is favored.
For a reaction: aA + bB  cC + dD
Kc is the equilibrium constant, a value
The fraction is the equilibrium constant expression
Example. Write the equilibrium constant,
Kc, for N2O4(g)  2NO2(g)
2
Kc =
[NO2 ]
[N2O4 ]
Reaction: N2O4 (g)  2NO2 (g)
Kc =
[NO2 ]2
[N2O4 ]
Experiment 1
Kc =
Law of Mass Action
The value of the equilibrium constant,
Kc, is constant at a given temperature
for a reaction at equilibrium.
 The equilibrium concentrations of
reactants and products may be different,
but the value for Kc remains the same.
Kc Characteristics:
1) Equilibrium can be approached from
either direction.
2) Kc does not depend on the initial
concentrations of reactants and
products.
3) Kc does depend on temperature.
4) Kc values are listed without units
 don't include units when calculating Kc.
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The Equilibrium Constant, Kc
• Write the equilibrium constant expressions
(Kc) for the following equations. (Note:
Expressions don’t include solids or liquids!)
a) CO(g) + 3H2(g)  CH4(g) + H2O(g)
b) 2NH3(g)  N2(g) + 3H2(g)
a)
Calculating Kc
2NH3(g)  N2(g) + 3H2(g)
At 500 K, the following concentrations were
measured: [N2] = 3.0 x 10-2 M, [H2] = 3.7 x 10-2
M, [NH3] = 1.6 x 10-2 M. What is Kc?
b)
What is Kc if we reverse this reaction?
Kc versus Kp
When does Kc = Kp?
• Can measure gas pressures instead of molarity.
Pressure is directly proportional to concentration.
• PV = nRT  P = (n/V) RT  P = MRT where n/V = M
• 2NH3(g)  N2(g) + 3H2(g)
Kc 
N2 H2 3
NH3 2
 Kp 
PN PH 3
PNH 2
2
2
They are equal when there are the same #
of gas molecules on both sides of equation.
Example. Does Kc = Kp?
a) H2(g) + F2(g)  2HF(g)
b) 2SO2(g) + O2(g)  2SO3(g)
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• Kp = Kc(RT)n
n = # product gas molecules - # reactant gas molecules
R = 0.08206 L·atm / mol·K
T = temperature in Kelvin
Example Kp
CH4(g) + 2H2S(g)  CS2(g) + 4H2(g)
• At 1000 K, the equilibrium pressure are:
CH4 = 0.20 atm, H2S = 0.25 atm, CS2 =
0.52 atm, and H2 = 0.10 atm.
• What is Kp?
Heterogenous Equilibria
• Homogeneous equil.: all substances are in
one phase (all gas, all solid, etc.)
• Heterogeneous equil.: substances in 2 or
more different phases
• CaCO3(s)  CaO(s) + CO2(g)
• Solids and liquids have constant
concentrations. If you increase the amount
of CaCO3, you also increase its volume.
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Heterogeneous Equilibria
Answers Kc & Kp Expressions
Write Kc and Kp expressions for the
following equations:
a) CaCO3(s)  CaO(s) + CO2(g)
b) 2Cu2S(s) + 3O2(g)  2Cu2O(s) + 2SO2(g)
c) Hg(l) + Hg2+(aq)  Hg22+(aq)
a) CaCO3(s)  CaO(s) + CO2(g)
Exclude pure solids and liquids from Kc & Kp expressions!
•Use Kc or Kp for gas phase reactions
•Use Kc for aq solutions (or both aq and g present)
c) Hg(l) + Hg2+(aq)  Hg22+(aq)
The Equilibrium Constant, Kc
• Usually one side of the equation is favored
(reactants or products).
For Example, 2 H2(g) + O2(g)  2 H2O(g)
• If Kc = 2.4 x 1047 (e.g., a large number)
what does this tell you about the
equilibrium of the reaction?
A. There are more reactants than products
B. There are more products than reactants
C. More information is needed.
Interpreting the Equil. Constant
• If Kc > 103, products are favored over reactants;
reaction goes nearly to completion.
• If Kc < 10-3, reactants are favored over products;
reaction hardly proceeds at all.
• If Kc is in the range 10-3 - 103, appreciable
concentration of reactants and products are present.
Kc = [CO2]
Kp = PCO
2
b) 2Cu2S(s) + 3O2(g)  2Cu2O(s) + 2SO2(g)
The Equilibrium Constant, Kc
2 HBr (g)  H2 (g) + Br2 (g)
• Kc = 2 x 10-19 (e.g., a small number), what
does this tell you about the equilibrium of
the reaction?
A. There are more reactants than products
B. There are more products than reactants
C. More information is needed.
Interpreting the Equil. Constant
• Which system below (all at equilibrium) has
the largest equilibrium constant? Why?
• Problem 13.8
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Using the Equil. Constant
3H2(g) + N2(g)  2NH3(g) Calculate Kc at 510 K.
[H2]eq = 0.104 M, [N2] eq = 0.554 M, [NH3] eq = 0.418 M
• 2H2(g) + O2(g)  2H2O(g) Kc = 2.4x1047 at 500K
 What is the concentration of H2 at equilibrium if
[H2O]eq = 1.0 M and [O2]eq = 1.0 x 10-16 M?
Using the Kc found, calculate PNH3. Given: PH2 = 1.24
atm, PN2 = 2.17 atm
Kp = Kc(RT)n
Reaction Quotient, Q
• Q is set up like K but the reaction mix may
not be at equilibrium. Calculating Q tells us
which direction a reaction must go to reach
equilibrium.
• H2(g) + I2(g)  2HI(g) Kc = 57.0 at 700 K
• If [H2] = 0.010 M, [I2] = 0.20 M, and [HI] =
0.40 M, is this system at equilibrium?
• What needs to happen to reach
equilibrium?
Reaction Quotient, Q
Using Q to predict shift
• Kc = 57.0
[HI]2
(0.40)2
Q

 80
[H2 ][I2 ] (0.010)(0.20)
• Q is larger than Kc (Kc = 57) so there
are too many products.
Rxn must shift left to attain eq
(Prod and React until Q matchs K)
Reaction Quotient, Q
• If Qc < Kc, reaction shifts right (small Q
means not enough products so shifts right
to make more products)
• If Qc > Kc, reaction shifts left (big Q means
too many products so shifts left to convert
products into reactants)
• If Qc = Kc, reaction is at equilibrium (no
shifting)
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Finding Q & Direction of Shift
• Example. For the reaction, B  2A, Kc = 2.
Suppose 3.0 moles of A and 3.0 moles of B are
introduced into a 2.00 L flask.
a) Is this system at equilibrium?
b) In which direction will the reaction proceed to
reach equilibrium?
c) Does the concentration of B increase,
decrease or remain the same as the system
moves towards equilibrium?
ICE method – use to find eq [ ]’s,
given Kc and initial [ ] ‘s
1. I = initial concentration: Initial concentration of
reactants are usually given; initial [Product]'s are
assumed to be 0 unless otherwise specified.
2. C = change in concentration: Assign change as
the variable x; use the stoichiometry of the reaction
to assign changes for all species.
3. E = equilibrium concentration: E = I + C
Note, values in ICE tables can be in terms of moles or
Molarity (or atm for Kp), but values used in the Kc
expression must be in terms of Molarity (or atm for Kp).
Equil. Conc. - Perfect Squares
Calculate the equilibrium concentrations of all species.
H2 (g) + I2 (g)  2HI (g)
Kc = 50.5
1.00 M 1.00 M
I
H2 (g) +
I2 (g) 
2HI (g)
1.00
1.00
0
C
-x
-x
+2x
E
1.00 - x
1.00 - x
+2x
Kc 
2
[HI]
[H2 ][I2 ]
B  2A
Calculating Q for
• [A] = 3.0 mol/2.00 L = 1.5 M;
• [B] = 3.0 mol/2.00 L = 1.5 M
Q
A2  1.52
B 1.5
 1.5 so not at eq (Kc = 2)
• Q too small, so not enough products
• Eq shifts right to attain eq
• B  and A
Solving for Equil. Conc.
• 3 methods for finding equilibrium concentrations:
 Use perfect squares (e.g. H2 + I2  2HI)
 50.5 =
(2x)2
(1.00  x)2
 Assume x is much smaller than initial
concentration; General rule: if K < 10-3, assume
x is small
 Omit –x terms
 Use quadratic if Kc is too big to make
assumption.
 get two values of x; one will give
negative eq concentration(s)
x=
-b  b 2  4ac
2a
Approximation Method- Assume small x
Calculate the equilibrium concentrations for
N2(g) + O2(g)  2 NO(g) Kc = 1.0 x 10-5
0.80 M 0.20 M
2
Kc 
NO 
N2 O2 
Check assumption: (x /initial concentration) * 100% < 5%
Check math by plugging concentrations into Kc
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Quadratic Method
Assumption that x is small isn’t always valid!
What to do?
Quadratic formula!!!
-b  b 2  4ac
x=
2a
•Gives two values of x:
If obtain +x and –x, pick +x!
If both x’s are +, pick smaller x!
Can eq concentrations (M) be – values?
Note: only one x should give you all + eq concs!
Factors Affecting Equil.
Now we look at what factors can cause shifts in
equilibrium. These can alter expenses in industrial
settings where a shift toward creation of more
products is often more economical.
 Concentrations of reactants or products can be
changed.
 Pressure and volume of the system can be
changed (only important for gas phase reactions).
 Temperature can be changed.
 Addition of a catalyst.
 Addition of an inert gas (e.g. Noble gas)
Concentration Changes
• Adding a reactant or product, the equilibria
shifts away from the increase in order to
consume part of the added substance.
• Removing a reactant or product, the
equilibria shifts toward the decrease to
replace part of the removed species.
Quadratic Example
H2(g) + F2(g)  2HF(g), Kc = 1.15x102
3.000 moles of H2 and 6.000 moles of F2 are
placed in a 3.000 L container
Calculate the equilibrium concentrations of all
species.
Le Chatelier’s Principle
• If a stress is applied to a reaction mixture at
equilibrium, net reaction occurs in the
direction that relieves the stress.
• “Stress” means a change in one of the
factors mentioned on the previous slide.
Changes in Concentration
AB K=4
• At equilibrium [A] = 5 M and [B] = 20 M
• What happens if B is removed & [B] drops to
15 M? Is this still at equilibrium? If not, what
is Q? Which way does this system need to
shift?
• What happens if A is added and [A] increases
to 10 M? What is Q? Which direction does it
shift?
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Changes in P and V
• Increase in volume: more space = more gases
allowed. Shift to side with more moles of gas.
• Increase in pressure (same as decrease in
volume): less volume = less gas allowed. Shift
to side with fewer moles of gas.
• If same number of moles on both sides, P and V
don’t affect equilibrium.
• Adding an inert gas (e.g. a noble gas) also
doesn’t cause an equilibrium shift or affect Kc.
Changes in P and V
• N2 (g) + 3 H2 (g)
2 NH3 (g)
• There are 4 moles of reactant and 2 moles
of product.
• What will happen if we increase pressure?
• What will happen if we increase volume?
Changes in Temperature
Changes in Temperature
• Changes in Conc., P, and V just shift
equilibrium to maintain a constant K.
• Changes in Temperature will affect the
value of K.
• We can look at heat exchange (enthalpy) of
a reaction to predict shift.
• Endothermic: heat is a “reactant”
• Exothermic: heat is a “product”
N2O4 (g)
2NO2 (g)
Ho = +57.2 kJ
Endothermic, heat is a reactant, so adding
heat helps the reaction proceed forward.
Heat + reactants  products
• heat, eq shifts 
• heat, eq shifts 
• Endo likes it hot!
Low T
Exothermic Reactions
Exo: Heat is product!
Reactants  Products + heat
• heat, eq shifts 
• heat, eq shifts 
• Opposite shifts for endothermic reactions!
High T
How T affects Kc
Both Kc and the position of the equilibrium system
will vary with temperature:
•Kc is larger when the reaction shifts right.
•Kc is smaller when the reaction shifts left.
Example. The temperature is decreased for the
reaction: 2CO2  2CO + O2, ΔH = 566 kJ.
a) Will the equilibrium shift left or right? b) Does
Kc become larger or smaller?
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Adding a Catalyst
• A catalyst speeds up a reaction so it attains
equilibrium faster.
• If a catalyst is added to a system that is already
at equilibrium, it won’t affect the system at all!
• The catalyst does not affect the equilibrium
concentrations of reactants and products;
thus, the Kc value does not change.
Le Chatelier’s Principle
• Determine how the equilibrium will shift if the
following changes are made:
• 2Cl2(g) + 2H2O(g) 4HCl(g) + O2(g) Ho = +113 kJ
a) Temperature is increased Right
b) Volume is increased
Right
c) Pressure is decreased
Right
d) HCl is added
Left
e) Ne (g) is added
No change
f) Cl2 is added
Right
g) H2O is removed
left
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