AP Chemistry
Mr. Markic
Page 1 of 12
Chapter 14 - Chemical Equilibrium
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
N2O4 (g)  2NO2 (g)
N2O4 (g)  2NO2 (g)
Figure 14.2
Change in the concentrations of NO2 and N2O4 with time, in three situations.
(a) Initially only NO2 is present.
(b) Initially N2O4 is present
(c) Initially NO2 and N2O4 are present.
Note, that even though equilibrium is reached in all cases, the equilibrium concentrations of NO2 and N2O4 are not
the same

N2O4 (g)
K=
[NO2 ]2
[N2 O4]
= 4.63 x 10-3
aA + bB
K=
2NO2 (g)

cC + dD
[C]c [D]d
[A]a [B]b
Law of Mass Action
AP Chemistry
Mr. Markic
Page 2 of 12
Equilibrium Will
K >> 1
K << 1
Lie to the right
Lie to the left
Favor products
Favor reactants
Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.
N2O4 (g)  2NO2 (g)
[NO2 ]2
2 O4]
PNO2 2
Kc = [N
Kp = 𝑃
N2 O4
In most cases Kc  Kp
aA (g) + bB (g)  cC (g) + dD (g)
Kp = Kc(RT)
n
∆n = moles of gaseous products – moles of gaseous reactants
= (c + d) – (a + b)
Homogeneous Equilibrium
CH3COOH (aq) + H2O (l)  CH3COO- (aq) + H3O+ (aq)
Kc ’ =
[CH3 COO− ][H3 O+ ]
[CH3 COOH][H2 O]
[H2O] = constant
Kc =
[CH3 COO− ][H3 O+ ]
[CH3 COOH]
= Kc’ [H2O]
General practice not to include units for the equilibrium constant.
Sample Exercise
Write the expressions for KC, and KP if applicable, for the following reversible reactions at equilibrium:
(a)
HF(aq) + H2O(l)  H3O+(aq) + F-(aq)
(b)
2NO(g) + O2(g)  2NO2(g)
(c)
CH3COOH(aq) + C2H5OH(aq)  CH3COOC2H5(aq) + H2O(l)
AP Chemistry
Mr. Markic
Practice Exercise
Write Kc and Kp for the decomposition of dinitrogen pentoxide.
Page 3 of 12
2NO(g) ↔ 4NO2(g) + O2(g)
Sample Exercise
The following equilibrium process has been studied at 230°C.
2NO(g) + O2(g)  2NO2(g)
In one experiment the concentrations of the reacting species at equilibrium are found to be [NO] = 0.0542 M, [O2] =
0.127 M, and [NO2] = 15.5 M. Calculate the equilibrium constant (Kc) of the reaction at this temperature
Practice Exercise
Carbonyl chloride (COCl2), also called phosgene, was used in World War I as a poisonous gas. The equilibrium
concentrations for the reaction between carbon monoxide and molecular chlorine to form carbonyl chloride
CO(g) + Cl2 ↔ COCl2(g)
At 74°C are [CO] = 1.2 x 10-2 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constant (Kc).
The equilibrium constant KP for the decomposition of phosphorous pentachloride to phosphorous trichloride and
molecular chlorine
PCl5(g)  PCl3(g) + Cl2(g)
Is found to be 1.05 at 250°C. If the equilibrium of the partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm,
respectively, what is the equilibrium partial pressure of Cl2 at 250°C?
The equilibrium constant Kp for the reaction:
2NO2(g) ↔ 2NO(g) + O2(g)
Is 158 at 1000 K. Calculate 𝑃𝑂2 if 𝑃𝑁𝑂2 = 0.400 atm and 𝑃𝑁𝑂 = 0.270 atm.
AP Chemistry
Mr. Markic
Page 4 of 12
Methanol (CH3OH) is manufactured industrially by the reaction: CO(g) + 2H2(g)  CH3OH(g)
The equilibrium constant (Kc) for the reaction is 10.5 at 220°C. What is the value of KP at this temperature?
For the reaction: N2 (g) + 3H2 (g) ↔ 2NH3(g)
KP is 4.3 x 10-4 at 375°C. Calculate Kc for the reaction.
Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.
CaCO3 (s)  CaO (s) + CO2 (g)
Kc ’ =
[CaO][CO2 ]
[CaCO3 ]
Kc = [CO2] = Kc’ x
[CaCO3 ]
[CaO]
Kp = PCO2
[CaCO3] = constant
[CaO] = constant
The concentration of solids and pure liquids are not included in the expression for the equilibrium
constant.
CaCO3 (s)
 CaO (s) + CO2 (g)
PCO2 = Kp
PCO 2 does not depend on the amount of CaCO3 or CaO
Sample Exercise
Write the equilibrium constant expression Kc, and Kp if applicable, for each of the following heterogeneous systems
(a) (NH4)2Se(s)  2NH3(g) + H2Se(g)
(b) AgCl(s)  Ag+(aq) + Cl-(aq)
(c) P4(s) + 6Cl2(g)  4PCl3(l)
Write the equilibrium constant expression for Kc and Kp for the formation of nickel tetracarbonyl, which is used to
separate nickel from other impurities: Ni(s) + 4CO(g) ↔ Ni(CO)4(g)
AP Chemistry
Mr. Markic
Page 5 of 12
Consider the following heterogeneous equilibrium:
CaCO3(s)  CaO(s) + CO2(g)
At 800°C, the pressure of CO2 is 0.236 atm. Calculate (a)Kp and (b) Kc for the reaction at this temperature
Consider the following equilibrium at 395 K: NH4HS(s) ↔ NH3(g) + H2S(g)
The partial pressure of each gas is 0.265 atm. Calculate Kp and Kc for the reaction.
Review of Concepts
For which of the following reactions is Kc equal to Kp?
(a) 4NH3(g) + 5O2(g) ↔ 4NO(g) + 6H2O(g)
(b) 2H2O2(aq) ↔ 2H2O(l) + O2(g)
(c) PCl3(g) +3NH3(g) ↔ 3HCl(g) + P(NH2)3(g)
A + B  C + D Kc ’
[C][D]
[E][F]
Kc’ = [A][B]
Kc’’ = [C][D]
C + D  E + F Kc’’
[E][F]
A + B  E + F Kc
Kc = [A][B]
so
Kc = Kc’ x Kc’’
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)  2NO2 (g)
[NO2 ]2
2 O4]
K = [N
= 4.63 x 10-3
2NO2 (g)  N2O4 (g)
[N2 O4]
K’ = [NO
2
2]
1
=K
= 216
When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant
becomes the reciprocal of the original equilibrium constant.
AP Chemistry
Mr. Markic
Page 6 of 12
Sample Exercise
The reaction for the production of ammonia can be written in a number of ways:
(a) N2(g) + 3H2(g)  2NH3(g)
(b)
1
2
3
N2(g) + 2 H2(g)  NH3(g)
(c)
1
3
N2(g) + H2(g) 
2
3
H3(g)
Write the equilibrium constant expression for each formulation. (Express the concentrations of the reacting species in
mol/L.)
(d) How are the equilibrium constants related to one another?
Practice Exercise
Write the equilibrium expression (Kc) for each of the following reactions and show how they are related to each other:
2
(a) 3O2(g) ↔ 2O3(g)
(b) O2(g) ↔ O3(g)
3
Review of Concepts
From the following equilibrium constant expression, write a balanced chemical equation for the gas-phase reaction.
Kc =
[𝑁𝐻3 ]2 [𝐻2 𝑂]4
[𝑁𝑂2 ]2 [𝐻2 ]7
Writing Equilibrium Constant Expressions
• 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.
•
The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions.
•
The equilibrium constant is a dimensionless quantity.
•
In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature.
•
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.
Chemical Kinetics and Chemical Equilibrium
𝐾𝑓
A + 2B ↔ AB2
Kr
Rate = kf [A][B]2
Rate = kr [AB2]
ratef = rater
𝑘𝑓
𝑘𝑟
[AB ]
2
= Kc = [A][B]
2
kf [A][B]2 = kr [AB2]
Review of Concepts
The equilibrium constant (Kc) for the reaction A ↔ B + C is 4.8 x 10-2 at 80°C. If the forward rate constant is 3.2 x 102 s-1,
calculate the reverse rate constant.
AP Chemistry
Mr. Markic
Page 7 of 12
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
Sample Exercise
At the start of the reaction, there are 0.249 mol N2, 3.21 x 10-2 mol H2, and 6.42 x 10-4 mol NH3 in a 3.50-L reaction vessel
at 375°C. If the equilibrium constant (Kc) for the reaction
N2(g) + 3H2(g)  2NH3(g)
Is 1.2 at this temperature, decide whether the system is at equilibrium. If it is not, predict which way the net reaction
will proceed.
The equilibrium constant (Kc) for the formation of nitrosyl chloride, and orange-yellow compound, from nitric oxide and
molecular chlorine 2NO(g) + Cl2(g)  2NOCl
is 6.5 x 104 at 35C. In a certain experiment, 2.0 x 10-2 mole of NO, 8.3 x 10-3 mole of Cl2. and 6.8 moles of NOCl are
mixed in a 2.0-L flask. In which direction will the system proceed to reach equilibrium?
Review of Concepts
The equilibrium constant (Kc) for the A2 + B2 ↔ 2AB reaction is 3 at a certain temperature. Which of the diagrams
shown here corresponds to the reaction at equilibrium? For those mixtures that are not at equilibrium, will the net
reaction move in the forward or reverse direction to reach equilibrium?
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.
AP Chemistry
Mr. Markic
Page 8 of 12
Sample Exercise
A mixture of 0.500 mol H2 and 0.500 mol I2 was placed in a 1.00-L stainless-steel flask at 430°C. The equilibrium constant
Kc for the reaction H2(g) + I2(g)  2HI(g) is 54.3 at this temperature. Calculate the concentrations of H2, I2, and HI at
equilibrium.
For the same reaction and temperature as in the prior example, suppose that the initial concentrations of H2, I2, and HI
are 0.00623 M, 0.00414 M, and 0.0224 M, respectively. Calculate the concentrations of these species at equilibrium
Le Châtelier’s Principle
If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially
offset as the system reaches a new equilibrium position.
Changes in Concentration
N2 (g) + 3H2 (g) ↔ 2NH3 (g)
Equilibrium
shifts left to
offset stress
Add
NH3
Change
Increase concentration of product(s)
Decrease concentration of product(s)
Increase concentration of reactant(s)
Decrease concentration of reactant(s)
Shifts the Equilibrium
left
right
right
left
AP Chemistry
Mr. Markic
Page 9 of 12
aA (g) + bB (g)  cC (g) + dD (g)
Change
Increase concentration of product(s)
Decrease concentration of product(s)
Increase concentration of reactant(s)
Decrease concentration of reactant(s)
Shifts the Equilibrium
Sample Exercise
At 720°C, the equilibrium constant Kc for the reaction: N2(g) + 3H2(g)  2NH3(g) is 2.37 x 10-3. In a certain experiment
the equilibrium concentrations are [N2] = 0.683 M, [H2] = 8.80 M, and [NH3] = 1.05 M. Suppose some NH3 is added to
mixture so that its concentration is increased to 3.65 M
(a) Use Le Chatelier’s principle to predict the shift in direction of the net reaction to reach a new equilibrium
(b) Confirm your prediction by calculating the reaction quotient QC and comparing its value with Kc
Practice Exercise
At 430°C, the equilibrium constant (KP) for the reaction: 2NO(g) + O2(g)  2NO2(g)
Is 1.5 x 105. In one experiment, the initial pressures of NO, O2, and NO2 are 2.1 x 10-3 atm, 1.1 x 10-2 atm, and 0.14 atm,
respectively. Calculate QP, and predict the direction that the net reaction will shift to reach equilibrium
Changes in Volume and Pressure
A(g) + B(g) ↔ C(g)
Change
Increase pressure
Decrease pressure
Increase volume
Decrease volume
Shifts the Equilibrium
Side with fewest moles of gas
Side with most moles of gas
Side with most moles of gas
Side with fewest moles of gas
AP Chemistry
Mr. Markic
Page 10 of 12
Consider the following equilibrium systems, predict the direction of the net reaction in each case as a result of increasing
the pressure (decreasing the volume) on the system at constant temperature):
(a) 2PbS(s) + 3O2(g)  2PbO(s) + 2SO2(g)
(b) PCl5(g)  PCl3(g) + Cl2(g)
(c) H2(g) + CO2(g)  H2O(g) + CO(g)
Consider the equilibrium reaction involving nitrosyl chloride, nitric oxide, and molecular chlorine
2NOCl(g) ↔ 2NO(g) + Cl2(g)
Predict the direction of the net reaction as a result of decreasing the pressure (increasing the volume) on the system at
constant temperature.
Review of Concepts
The diagram here shows the gaseous reaction 2A ↔ A2 at equilibrium. If the pressure is decreased by increasing the
volume at constant temperature, how would the concentrations of A and A2 change when a new equilibrium is
established?
Changes in Temperature
Change
Increase temperature
Decrease temperature
Exothermic Rx
K decreases
K increases
Endothermic Rx
K increases
K decreases
Adding a Catalyst
• does not change K
• does not shift the position of an equilibrium system
• system will reach equilibrium sooner
Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant
or shift equilibrium.
uncatalyzed
Le Châtelier’s Principle
Change
Concentration
Shift Equilibrium
yes
Change Equilibrium
Constant
no
Pressure
yes*
no
Volume
yes*
no
Temperature
yes
yes
Catalyst
no
no
*Dependent on relative moles of gaseous reactants and products
catalyzed
AP Chemistry
Mr. Markic
Page 11 of 12
Review of Concepts
The diagram shown here represents the reaction X2 + Y2 ↔ 2XY at equilibrium at two temperatures (T2 > T1). Is the
reaction endothermic or exothermic?
Sample Exercise
Consider the following equilibrium process between dinitrogen tetrafluoride (N2F4) and nitrogen difluoride (NF2):
N2F4(g)  2NF2 (g)
H° = 38.5 kJ/mol
Predict the changes in the equilibrium if:
(a) The reacting mixture is heated at constant volume
(b) Some N2F4 gas is removed from the reacting mixture at constant temperature and volume
(c) The pressure of the reacting mixture is decreased at constant temperature
(d) A catalyst is added to the reacting mixture
Practice Exercise
Consider the equilibrium between molecular oxygen and ozone
3O2(g) ↔ 2O3(g)
ΔH° = 284 kJ/mol
What would be the effect of:
(a) Increasing the pressure of the system by decreasing the volume
(b) Adding O2 to the system at constant volume
(c) Decreasing the temperature
(d) Adding a catalyst
AP Chemistry
Mr. Markic
Page 12 of 12
Big Idea 6: Any bond or intermolecular attraction that can be formed can be broken. These two processes are in dynamic
competition, sensitive to initial conditions and external perturbations.
Duration: Early March
Textbook Chapter: 14
Enduring Understanding
Essential Knowledge
6.A: Chemical equilibrium is a
6.A.1: In many classes of reactions, it is important to consider both the forward and
dynamic, reversible state in which
reverse reaction.
rates of opposing processes are
6.A.2: The current state of a system undergoing a reversible reaction can be
equal.
characterized by the extent to which reactants have been converted to products. The
relative quantities of reaction components are quantitatively described by the reaction
quotient, Q.
6.A.3: When a system is at equilibrium, all macroscopic variables, such as concentrations,
partial pressures, and temperature, do not change over time. Equilibrium results from an
equality between the rates of the forward and reverse reactions, at which point Q = K.
6.A.4: The magnitude of the equilibrium constant, K, can be used to determine whether
the equilibrium lies toward the reactant side or product side.
6.B: Systems at equilibrium are
6.B.1: Systems at equilibrium respond to disturbances by partially countering the effect
responsive to external
of the disturbance (Le Chatelier’s principle).
perturbations, with the response
6.B.2: A disturbance to a system at equilibrium causes Q to differ from K, thereby taking
leading to a change in the
the system out of the original equilibrium state. The system responds by bringing Q back
composition of the system.
into agreement with K, thereby establishing a new equilibrium state.
Learning Objective
5.16 The student can use Le Chatelier’s principle to make qualitative predictions for systems in which coupled reactions that
share a common intermediate drive formation of a product.
5.17 The student can make quantitative predictions for systems involving coupled reactions that share a common intermediate,
based on the equilibrium constant for the combined reaction.
5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product
(based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction
can produce large amounts of product for certain sets of initial conditions.
6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental
processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying
chemical reactions or processes.
6.2 The student can, given a manipulation of a chemical reaction or set of reactions (e.g., reversal of reaction or addition of two
reactions), determine the effects of that manipulation on Q or K.
6.3 The student can connect kinetics to equilibrium by using reasoning about equilibrium, such as Le Chatelier’s principle, to
infer the relative rates of the forward and reverse reactions.
6.4 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K, use
the tendency of Q to approach K to predict and justify the prediction as to whether the reaction will proceed toward products
or reactants as equilibrium is approached.
6.5 The student can, given data (tabular, graphical, etc.) from which the state of a system at equilibrium can be obtained,
calculate the equilibrium constant, K.
6.6 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K, use
stoichiometric relationships and the law of mass action (Q equals K at equilibrium) to determine qualitatively and/or
quantitatively the conditions at equilibrium for a system involving a single reversible reaction.
6.7 The student is able, for a reversible reaction that has a large or small K, to determine which chemical species will have very
large versus very small concentrations at equilibrium.
6.8 The student is able to use Le Chatelier’s principle to predict the direction of the shift resulting from various possible stresses
on a system at chemical equilibrium.
6.9 The student is able to use Le Chatelier’s principle to design a set of conditions that will optimize a desired outcome, such as
product yield.
6.10 The student is able to connect Le Chatelier’s principle to the comparison of Q to K by explaining the effects of the stress on
Q and K.