Chemical
Equilibrium
Chapter
7
COURSE NAME: CHEMISTRY 101
COURSE CODE: 402101-4
1
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Equilibrium is a state in which there are no observable
changes as time goes by.
Chemical equilibrium is achieved when:
•
the rates of the forward and reverse reactions are equal and
•
the concentrations of the reactants and products remain
constant
Physical equilibrium
H2O (l)
H2O (g)
Chemical equilibrium
NO2
N2O4 (g)
2NO2 (g)
1
N2O4 (g)
2NO2 (g)
equilibrium
equilibrium
Start with N2O4
Start with NO2 & N2O4
equilibrium
Start with NO2
2
N2O4 (g)
K=
2NO2 (g)
[NO2]2
[N2O4]
aA + bB
K=
= 4.63 x 10-3
cC + dD
[C]c[D]d
[A]a[B]b
Law of Mass Action
3
aA + bB
cC + dD
K=
[C]c[D]d
[A]a[B]b
Equilibrium Will
K>1
Lie to the left
Favor products
K<1
Lie to the right
Favor reactants
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Homogenous equilibrium applies to reactions in which all
reacting species are in the same phase.
N2O4 (g)
Kc =
2NO2 (g)
[NO2]2
Kp =
[N2O4]
2
PNO
2
PN2O4
In most cases
Kc  Kp
aA (g) + bB (g)
cC (g) + dD (g)
Kp = Kc(RT)Dn
Dn = moles of gaseous products – moles of gaseous reactants
= (c + d) – (a + b)
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Homogeneous Equilibrium
CH3COOH (aq) + H2O (l)
[CH3COO-][H3O+]
Kc′ =
[CH3COOH][H2O]
CH3COO- (aq) + H3O+ (aq)
[H2O] = constant
[CH3COO-][H3O+]
= Kc′ [H2O]
Kc =
[CH3COOH]
General practice not to include units for the equilibrium constant.
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7
The equilibrium concentrations for the reaction between carbon
monoxide and molecular chlorine to form COCl2 (g) at 740C are
[CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate
the equilibrium constants Kc and Kp.
CO (g) + Cl2 (g)
COCl2 (g)
[COCl2]
0.14
=
= 220
Kc =
[CO][Cl2]
0.012 x 0.054
Kp = Kc(RT)Dn
Dn = 1 – 2 = -1
R = 0.0821
T = 273 + 74 = 347 K
Kp = 220 x (0.0821 x 347)-1 = 7.7
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The equilibrium constant Kp for the reaction
2NO2 (g)
2NO (g) + O2 (g)
is 158 at 1000K. What is the equilibrium pressure of O2 if the
PNO = 0.400 atm and PNO = 0.270 atm?
2
Kp =
2
PNO
PO2
2
PNO
2
PO2 = Kp
2
PNO
2
2
PNO
PO2 = 158 x (0.400)2/(0.270)2 = 347 atm
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Heterogenous equilibrium applies to reactions in which
reactants and products are in different phases.
CaCO3 (s)
[CaO][CO2]
Kc′ =
[CaCO3]
Kc = [CO2] = Kc′ x
[CaCO3]
[CaO]
CaO (s) + CO2 (g)
[CaCO3] = constant
[CaO] = constant
Kp = PCO2
The concentration of solids and pure liquids are not
included in the expression for the equilibrium constant.
PCO 2 = Kp
PCO 2 does not depend on the amount of CaCO3 or CaO
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Consider the following equilibrium at 295 K:
NH4HS (s)
NH3 (g) + H2S (g)
The partial pressure of each gas is 0.265 atm. Calculate Kp
and Kc for the reaction?
Kp = PNH PH S = 0.265 x 0.265 = 0.0702
3
2
Kp = Kc(RT)Dn
Kc = Kp(RT)-Dn
Dn = 2 – 0 = 2
T = 295 K
Kc = 0.0702 x (0.0821 x 295)-2 = 1.20 x 10-4
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A+B
C+D
Kc′
C+D
E+F
Kc′′
A+B
E+F
Kc
[C][D]
Kc′ =
[A][B]
Kc = Kc′ x Kc′′
[E][F]
Kc′′=
[C][D]
[E][F]
Kc =
[A][B]
If a reaction can be expressed as the sum of two or more reactions, the equilibrium
constant for the overall reaction is given by the product of the equilibrium
constants of the individual reactions.
N2O4 (g)
K=
[NO2]2
[N2O4]
2NO2 (g)
= 4.63 x
10-3
2NO2 (g)
N2O4 (g)
[N2O4]
1
= 216
K′ =
=
2
K
[NO2]
When the equation for a reversible reaction is written in the
opposite direction, the equilibrium constant becomes the
reciprocal of the original equilibrium constant.
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Writing Equilibrium Constant Expressions
1. The concentrations of the reacting species in the condensed phase are
expressed in M. In the gaseous phase, the concentrations can be
expressed in M or in atm.
2. The concentrations of pure solids, pure liquids and solvents do not
appear in the equilibrium constant expressions.
3. The equilibrium constant is a dimensionless quantity.
4. In quoting a value for the equilibrium constant, you must specify the
balanced equation and the temperature.
5. If a reaction can be expressed as a sum of two or more reactions, the
equilibrium constant for the overall reaction is given by the product of
the equilibrium constants of the individual reactions.
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Chemical Kinetics and Chemical Equilibrium
A + 2B
kf
kr
ratef = kf [A][B]2
AB2
rater = kr [AB2]
Equilibrium
ratef = rater
kf [A][B]2 = kr [AB2]
kf
[AB2]
= Kc =
kr
[A][B]2
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The reaction quotient (Qc) is calculated by substituting the
initial concentrations of the reactants and products into the
equilibrium constant (Kc) expression.
IF
•
Qc > Kc system proceeds from right to left to reach equilibrium
•
Qc = Kc the system is at equilibrium
•
Qc < Kc system proceeds from left to right to reach equilibrium
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Calculating Equilibrium Concentrations
1. Express the equilibrium concentrations of all species in
terms of the initial concentrations and a single unknown x,
which represents the change in concentration.
2. Write the equilibrium constant expression in terms of the
equilibrium concentrations. Knowing the value of the
equilibrium constant, solve for x.
3. Having solved for x, calculate the equilibrium
concentrations of all species.
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At 1280oC the equilibrium constant (Kc) for the reaction
Br2 (g)
2Br (g)
Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012
M, calculate the concentrations of these species at equilibrium.
Let x be the change in concentration of Br2
Initial (M)
Change (M)
Equilibrium (M)
[Br]2
Kc =
[Br2]
Br2 (g)
2Br (g)
0.063
0.012
-x
+2x
0.063 - x
0.012 + 2x
(0.012 + 2x)2
= 1.1 x 10-3
Kc =
0.063 - x
Solve for x
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(0.012 + 2x)2
= 1.1 x 10-3
Kc =
0.063 - x
4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x
4x2 + 0.0491x + 0.0000747 = 0
2 – 4ac

-b
±
b
x=
ax2 + bx + c =0
2a
x = -0.0105 x = -0.00178
Initial (M)
Change (M)
Equilibrium (M)
Br2 (g)
2Br (g)
0.063
0.012
-x
+2x
0.063 - x
0.012 + 2x
At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M
At equilibrium, [Br2] = 0.062 – x = 0.0648 M
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Le Chatelier’s Principle
If a stress is applied to a system at equilibrium, the system will
change to relieve that stress and re –establish equilibrium. It is
like the “undo” button on your computer!
Factors that Affect Equilibrium
•
•
•
•
Concentration
Temperature
Pressure For gaseous systems only!
The presence of a catalyst
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Concentration Changes
• Add more reactant  Shift to products
• Remove reactants  Shift to reactants
• Add more product  Shift to reactants
• Remove products  Shift to products
Reaction Quotient
The reaction quotient for an equilibrium system is the same as the equilibrium
expression, but the concentrations are NOT at equilibrium!
N2O4(g)  2NO2(g)
Q = [NO2]2 / [N2O4]
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• Changes in concentration are best understood in terms of
what would happen to “Q” if the concentrations were
changed.
• N2O4(g)  2NO2(g)
Q = [NO2]2 / [N2O4]
•Q = Keq at equilibrium
•If Q< K then there are too many reactants, the
reaction will shift in the forward direction (the
products)
•If Q>K then there are too many products, the
reaction will shift to the reactants.
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Temperature Changes
• Consider heat as a product in exothermic reactions
A + B = AB + Heat
– Add heat  Shift to reactants
– Remove heat  Shift to products
• Consider heat as a reactant in endothermic reactions
A + B + heat = AB
– Add heat  Shift to products
– Remove heat  Shift to reactants
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Pressure Changes
• Only affects equilibrium systems with unequal moles of
gaseous reactants and products.
N2(g) + 3H2(g) = 2NH3(g)
• Increase Pressure
* Stress of pressure is reduced by reducing the number
of gas molecules in the container . . . . . .
• There are 4 molecules of reactants vs. 2 molecules of
products.
– Thus, the reaction shifts to the product ammonia.
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PCl5(g) = PCl3(g) + Cl2(g)
• Decrease Pressure
– Stress of decreased pressure is reduced by increasing
the number of gas molecules in the container.
• There are two product gas molecules vs. one reactant
gas molecule.
• Thus, the reaction shifts to the products.
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Presence of a Catalyst
• A Catalyst lowers the activation energy and
increases the reaction rate.
• It will lower the forward and reverse reaction
rates,
• Therefore, a catalyst has NO EFFECT on a system
at equilibrium!
• It just gets you to equilibrium faster!
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Presence of an Inert Substance
• An inert substance is a substance that is notreactive with any species in the equilibrium
system.
• These will not affect the equilibrium system.
• If the substance does react with a species at
equilibrium, then there will be a shift!
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Le Châtelier’s Principle - Summary
Change Equilibrium
Constant
no
Change
Shift Equilibrium
Concentration
yes
Pressure
yes*
no
Volume
yes*
no
Temperature
yes
yes
Catalyst
no
no
*Dependent on relative moles of gaseous reactants and products
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Question 1
The equilibrium constant Keq for a certain reaction
will change if __________ changes.
A) pressure
B) time
C) volume
D) temperature
E) reactant concentrations
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Question 2
What is the relationship of the equilibrium constants for the
following two reactions?
(1) 2 NO2(g) ↔ N2O4(g);
(2) N2O4(g) ↔ 2 NO2(g)
A) Keq of reaction (1) is the reciprocal of Keq of reaction
(2).
B) Keq of reaction (2) is the reciprocal of Keq of reaction
(1)
C) Keq of reaction (1) = Keq of reaction (2)
D) Answers A and B are both correct
E) There is no relationship between the Keqs of these
reactions
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Question 3
What is the correct equilibrium constant expression
for the following reaction? 2 Cu(s) + O2(g) → 2
CuO(s)
A) Keq = 1/[O2]2
B) Keq = [CuO]2/[Cu]2
C) Keq = [CuO]2/[Cu]2[O2]
D) Keq = [O2]
E) Keq = 1/[O2]
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Question 4
Consider the reaction: 2 SO2(g) + O2(g) ↔ 2 SO3(g). If,
at equilibrium at a certain temperature, [SO2] = 1.50 M,
[O2] = 0.120 M, and [SO3] = 1.25 M, what is the value of
the equilibrium constant?
A) 5.79
B) 6.94
C) 8.68
D) 0.14
E) None of the above
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Question 5
PCl5 dissociates according to the reaction: PCl5(g) ↔
PCl3(g) + Cl2(g). One mole of PCl5 was placed in one liter
of solution. When equilibrium was established, 0.5 mole of
PCl5 remained in the mixture. What is the equilibrium
constant for this reaction?
A) 0.25
B) 0.5
C) 1.0
D) 2.5
E) None of the above
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Question 6
At elevated temperatures, solid silicon reacts with
chlorine gas to form gaseous SiCl4. At some
temperature, the equilibrium constant for this
reaction is 0.30. If the reaction is started with 0.10
mol of SiCl4 in a one-liter flask, how much Cl2 will
be present when equilibrium is established?
A) 0.18 mol
B) 0.090 mol
C) 0.030 mol
D) 0.30 mol
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Question 7
Consider the reaction: CO2(g) + H2(g) ↔ CO(g) + H2O(g), for which Kc =
0.64 at 900 K. If the initial CO2 and of H2 are each 0.100 M, what will be
the equilibrium concentrations of each species after the reaction reaches
equilibrium?
A) [CO2] = 0.044 M; [H2] = 0.044M; [CO] = 0.056 M; [H2O] = 0.056 M
B) [CO2] = 0.056 M; [H2] = 0.056 M; [CO] = 0.044 M; [H2O] = 0.044 M
C) [CO2] = 0.020 M; [H2] = 0.020 M; [CO] = 0.080 M; [H2O] = 0.080 M
D) [CO2] = 0.080 M; [H2] = 0.080 M; [CO] = 0.020 M; [H2O] = 0.020 M
E) None of the above
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Question 8
At some temperature, the reaction: 3 ClO- ↔ ClO3- + 2 Cl- has an
equilibrium constant Kc = 3.2 x 103. If the components of this reaction are
mixed such that their initial concentrations are [Cl-] = 0.05 M; [ClO3-] =
0.32; and [ClO-] = 0.74, is the mixture at equilibrium, yes or no? If the
mixture is not at equilibrium in which direction, left to right or right to left,
will reaction occur so that the mixture can reach equilibrium?
A) There is not enough information given to answer this question
B) Yes, the mixture is at equilibrium now
C) No. Left to right
D) No. Right to left
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Question 9
Consider the following endothermic reaction: H2(g) + I2(g)
↔ 2 HI(g). If the temperature is increased,
A) more HI will be produced
B) some HI will decompose, forming H2 and I2
C) the magnitude of the equilibrium constant will decrease
D) the pressure in the container will increase
E) the pressure in the container will decrease
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Question 10
Consider the following reaction at equilibrium: NO2(g) + CO(g)
↔ NO(g) + CO2(g). Suppose the volume of the system is
decreased at constant temperature, what change will this cause
in the system?
A) A shift to produce more NO
B) A shift to produce more CO
C) A shift to produce more NO2
D) No shift will occur
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Question 11
Which of the following statements is incorrect regarding
equilibrium?
A) Chemical equilibrium is a reversible process with no net
change in concentrations of the products and reactants.
B) Physical equilibrium can not exist between phases.
C) A chemical equilibrium with all reactants and products in the
same phase is homogeneous.
D) none of the above
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Question 12
Which of these four factors can change the value of the equilibrium?
A) catalyst
B) pressure
C) concentration
D) temperature
Question 13
Which general rule helps predict the shift in direction of an equilibrium reaction?
A) Le Chatelier's principle
B) Haber process
C) Equilibrium constant
D) Bosch theory
Question 14
There are guideline to help write equilibrium constants.
A) True
B) False
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Question 1
A chemical reaction is at equilibrium when
A) the concentrations of reactants is equal to the concentrations of products
B) the limiting reagent has been completely depleted
C) the rate of the forward reaction equals the rate of the reverse reaction
D) A and B .
Question 2
What is the correct equilibrium constant expression for the following reaction? 2 NO2(g) ↔
2 NO(g) + O2(g )
A) Keq = [O2]2[NO]/[NO2]2
B) Keq = [O2][NO]2/[NO2]2
C) Keq = [NO2]2/[NO]2 [O2 ]
D) Keq = [NO2]2/[NO][O2]2
E) None of the above
Question 3
The equilibrium constant for the reaction 2 NH3(g) → N2(g) + 3 H2(g) is 3 x 10-3 at some
temperature. What is Keq for the reaction 0.5 N2(g) + 1.5 H2(g) ↔ NH3(g) at the same
temperature ?
A) 0.003
B) 0.05
C) 18
D) 20
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Question 4
Consider the reaction that describes the Haber process for the production of
ammonia (NH3): N2(g) + 3 H2(g) ↔ 2 NH3(g) for which Kc at 300 oC is 9.5. Calculate
Kp for this reaction at 300 oC .
A) Kp = 4.3 x 10-3
B) Kp = 9.5
C) Kp = 2.1 x 104
D) Kp = 1.6 x 10-2
Question 5
For the reaction H2(g) + I2(g) ↔ 2 HI(g), Kc = 12.3 at some temperature T. If [H2] =
[I2] = [HI] = 3.21 x 10-3 M at that temperature, which one of the following statements
is true ?
A) The concentration of HI will rise as the system approaches equilibrium
B) The system is at equilibrium, so the concentrations will not change
C) The concentrations of H2 and I2 will increase as the system approaches
equilibrium
D) The concentrations of H2 and HI will decrease as the system approaches
equilibrium
E) Not enough information is given to answer the question
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Question 6
Kc for the reaction 2 NH3(g) ↔ N2(g) + 3 H2(g) is 3 x 10-3 at some temperature. A 1.0 L
mixture containing 1.0 mol of NH3, 0.50 mol of N2, and 0.15 mol of H2 is prepared at this
temperature When equilibrium is reached ,
A) there will be more N2 and H2 present
B) there will be more NH3 present
C) there will be less N2 but more H2 present .
D) there will be less NH3 and less N2 present
E) No shift will occur
Question 7
If the equilibrium constant for the reaction PCl5 ↔ PCl3 + Cl2 is 1.0, how many moles of PCl5
must be placed into one liter of solution in order to obtain 0.50 mol of PCl3 when the system
reaches equilibrium ?
A) 0.25
B) 0.50
C) 0.75
D) 1.00
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Question 8
When the reaction CH3Cl + OH- ↔ CH3OH + Cl- is started with 0.1 mol of CH3Cl and 0.2
mol of OH-, 0.03 mol of CH3OH is present when the system reaches equilibrium. Calculate
the equilibrium constant for the reaction .
A) 0.18
B) 0.08
C) 0.0009
D) 0.30
Question 9
At some temperature, Kc for the reaction PCl5(g) ↔ PCl3(g) + Cl2(g) is 0.020. If 0.10 mol
of PCl5 and 0.20 mol of PCl3 are added to a 1-L flask, what will be the Cl2 concentration
when equilibrium is reached ?
A) 0.020 M
B) 8.7 x 10-3 M
C) 0.0052 M
D) 0.12 M
Question 10
Consider the following reaction: H2(g) + I2(g) ↔ 2 HI(g) for which, at some temperature, Kc
= 4.0. When the reaction is started with equimolar quantities of H2 and I2 and equilibrium is
reached, 0.20 mol of HI is present. How much H2 was used to start the reaction ?
A) 0.10 mol
B) 0.23 mol
C) 0.20 mol
D) 4.0 mol
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