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Ch. 13 Equilibrium
Chemical Equilibrium
 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.

(BDVD)
Dynamic Equilibrium
 Reactions continue to take place.
A + B
C + D ( forward)
C + D
A + B (reverse)
 Initially there is only A and B so only the
forward reaction is possible
 As C and D build up, the reverse reaction
speeds up while the forward reaction
slows down.
 Eventually the rates are equal.
Reaction Rate
Forward Reaction
Equilibrium
Reverse reaction
Time
What is equal at Equilibrium?
 Rates are equal.
 Concentrations are not.
 Rates are determined by
concentrations and activation energy.
 The concentrations do not change at
equilibrium.
13.2 Law of Mass Action
 jA + kB
lC + mD
 j, k, l, m are coefficients
 The law of mass action is represented by
the equilibrium expression:
 K = [C]l[D]m
[A]j[B]k
PRODUCTSpower
REACTANTSpower
 K is called the equilibrium constant.

is how we indicate a reversible
reaction. [x] represents concentration.
Playing with K
 If we write the reaction in reverse.
 lC + mD
jA + kB
 Then the new equilibrium constant is
 K’ = [A]j[B]k = 1/K
[C]l[D]m
Playing with K
 If we multiply the equation by a
constant
 njA + nkB
nlC + nmD
 Then the equilibrium constant is
K’ = [A]nj[B]nk
[C]nl[D]nm
n
j[B]k)n
([A]
K
=
=
([C]l[D]m)n
K is CONSTANT
 At any temperature.
 Temperature affects rate.
 The equilibrium concentrations don’t have
to be the same only K.
 Equilibrium position is a set of
concentrations at equilibrium.
 One value at each temperature, but there
are an unlimited number of possibilities.
 Usually written without units.
Equilibrium Expression
4NH3(g) + 7O2(g)
4
4NO2(g) + 6H2O(g)
6
NO2 H 2 O
K
4
7
NH 3 O2
What is the equilibrium expression for this
reaction? #19a
N2(g) + O2(g)
2NO(g)
What is the equilibrium expression for this
reaction? #19b
N2O4 (g)
2NO2(g)
Calculate K
 N2(g) + 3H2(g)
2NH3 (g)
 In the above reaction, at a given
temperature. K = 0.013.
 Calculate the value of K in the
following reactions. #21ab
 1/2N2(g) + 3/2H2(g)
NH3(g)
 2NH3(g)
N2(g) + 3H2(g)
Calculate K
 N2(g) + 3H2(g)
2NH3(g)
 Initial
At Equilibrium
 [N2]0 =1.000 M [N2] = 0.921M
 [H2]0 =1.000 M [H2] = 0.763M
 [NH3]0 =0 M
[NH3] = 0.157M
Calculate K
 N2(g) + 3H2(g)
2NH3(g)
 Initial
At Equilibrium
 [N2]0 = 0 M
[N2] = 0.399 M
 [H2]0 = 0 M
[H2] = 1.197 M
 [NH3]0 = 1.000 M
[NH3] = 0.203M
 K is the same no matter what the
amount of starting materials (at the
same temperature).
13.3 Equilibrium and Pressure
 Some reactions are gaseous
 PV = nRT
 P = (n/V)RT
 P = CRT
 C is a concentration in moles/Liter
 C = P/RT
Equilibrium and Pressure
2SO2(g) + O2 (g)
2SO3 (g)
 In term of partial pressures:

Kp =
(PSO3)2
(PSO2)2 (PO2)
 In terms of concentration:
Kc =
[SO3]2
[SO2]2 [O2]
Practice
 N2(g) + 3H2(g)
2NH3(g)
 In the reaction above, the following
equilibrium pressures were observed.
 PNH3 = 0.89 atm
 PN2 = 0.62 atm
 PH2 = 0.029 atm
 Calculate the value for the
equilibrium constant Kp at this
temperature.
K v. Kp
 For
jA + kB
lC + mD
 Kp = K(RT)n
 n = sum of coefficients of gaseous
products minus sum of coefficients of
gaseous reactants.
Practice
 Kp = K(RT)n
 At 25°C, K = 2.01 x 104 for the reaction
 CO(g) + Cl2(g)
COCl2(g)
 What is the value of Kp at this
temperature?
Practice #32
 For which reaction in #31 is Kp = K?
Homogeneous Equilibria
 So far every example dealt with
reactants and products where all
were in the same phase.
 We can use K in terms of either
concentration or pressure.
 Units depend on reaction.
13.4 Heterogeneous Equilibria
 Are equilibria that involve more than one
phase.
 If the reaction involves pure solids or pure
liquids, the concentration of the solid or
the liquid doesn’t change.
 The position of a heterogeneous
equilibrium does not depend on the
amounts of pure solids or liquids present.
 As long as they are not used up we can
leave them out of the equilibrium
expression.
For Example
 H2(g) + I2(s)
 K = [HI]2
2HI(g)
[H2][I2]
 But the concentration of I2 does not
change, therefore:
K = [HI]2
[H2]
13.5 Applying the Equilibrium
Constant
 Reactions with large K (>>1),
essentially to completion. Large
negative E.
 Reactions with small K (<<1) consist
mostly of reactants.
 Time to reach equilibrium is related to
rate and AE. It is not related to size of
K.
The Reaction Quotient
 Tells you the directing the reaction will go
to reach equilibrium
 Calculated the same as the equilibrium
constant, but for a system not at
equilibrium by using initial
concentrations.
 Q = [Products]coefficient
[Reactants] coefficient
 Compare value to equilibrium constant
What Q tells us
 If Q<K
• Not enough products
• Shift to right
 If Q>K
• Too many products
• Shift to left
 If Q=K system is at equilibrium
Using the Reaction Quotient
 For the reaction
 2NOCl(g)
2NO(g) + Cl2(g)
 K = 1.55 x 10-5 M at 35ºC
 In an experiment 0.10 mol NOCl,
0.0010 mol NO(g) and 0.00010 mol
Cl2 are mixed in 2.0 L flask.
 Which direction will the reaction
proceed to reach equilibrium?
Using the Reaction Quotient
For the synthesis of ammonia at 500°C,
the equilibrium constant is 6.0x10-2.
Predict the direction in which the
system will shift to reach equilibrium in
each of the following cases. Ex. 13.7
a. [NH3]0 = 1.0 x 10-3 M; [N2]0 = 1.0 x 10-5 M;
[H2]0 = 2.0 x 10-3 M
b. [NH3]0 = 1.0 x 10-4 M; [N2]0 = 5.0 M;
[H2]0 = 1.0 x 10-2 M
Problems Involving Pressure
For the reaction N2O4(g)
2NO2(g) KP
= 0.133 atm. At equilibrium, the pressure
of N2O4 was found to be 2.71 atm?
Calculate the equilibrium pressure of
NO2(g) EX.13.8
Problems Involving Pressure
At a certain temperature a 1.00 L flask
initially contained 0.298 mol PCl3(g) and
8.70 x 10-3 mol PCl5(g). After the system
reached equilibrium, 2.00 x 10-3 mol Cl2(g)
was found in the flask. Calculate the
equilibrium concentrations of all species
and the value of K. EX.13.9
PCl5(g)
PCl3(g) + Cl2(g)
No Equilibrium [x] Ex. 13.10
 Carbon monoxide reacts with steam
to produce carbon dioxide and
hydrogen. At 700K the equilibrium
constant is 5.10. Calculate the
equilibrium concentrations of all
species if 1.00 mol of each component
is mixed in a 1.00 L flask.
Intro to ICE
2 SO3(g)
2SO2 (g) + O2(g)
At a certain temperature. 12.0 mol
SO3 is placed into a 3.0 L rigid
container. The SO3 dissociates during
the reaction. At equilibrium, 3.0 mol
SO2 is present. What is the value of K
for this reaction?
(#43)
Intro to ICE
2NH3
N2 + 3H2
At a certain temperature. 4.0 mol
NH3 is introduced into a 2.0 L
container. The NH3 partially
dissociates during the reaction. At
equilibrium, 2.0 mol NH3 remains.
What is the value of K for this
reaction?
(#44) WAssign
What if you’re not given
equilibrium concentration?
H2 + F2
2HF
The Equilibrium constant for the
above reaction is 115 at a certain
temperature. 3.000 mol of each
component was added to a 1.500 L
flask. Calculate the equilibrium
concentrations of all species.
(ex.13.11)
Practical Example
 H2(g) + F2(g)
2HF(g)
 K = 1.15 x 102 at 25ºC
 Calculate the equilibrium
concentrations if a 3.00 L container
initially contains 3.00 mol of H2 and
6.00 mol F2 .
 [H2]0 = 3.00 mol/3.00 L = 1.00 M
 [F2]0 = 6.00 mol/3.00 L = 2.00 M
 [HF]0 = 0
 Q= 0<K so more product will be
formed.
 Assumption since K is large reaction
will go to completion.
 Stoichiometry tells us H2 is LR, it will
be smallest at equilibrium let it be x
 Set up table of initial, change and
equilibrium in concentrations.
H2(g)
1.00 M
Initial
Change
Equilibrium
F2(g)
2.00 M
2HF(g)
0M
 For H2 and F2 the change must be -X
 Using to stoichiometry HF must be +2X
 Equilibrium = initial + change
H2(g)
F2(g)
2HF(g)
Initial
1.00 M 2.00 M
0M
Change -X
-X
+2X
Equili
1.00 -X 2.00-X
2X
 Therefore, ice chart looks like this.
 Change in HF = twice change in H2
H2(g)
F2(g)
2HF(g)
Initial
1.00 M 2.00 M
0M
Change -X
-X
+2X
Equili
1.00 -X 2.00-X
2X
 Now plug these values into the equilibrium
expression
 K = 1.15 x 102 = (2X)2
(1.00-x)(2.00-x)
 Solving this gives us a quadratic equation.
Quadratic Calculator
H2(g)
F2(g)
2HF(g)
Initial
1.00 M 2.00 M
0M
Change -X
-X
+2X
Equili
1.00 -X 2.00-X
2X
 Now plug these values into the equilibrium
expression
 K = 1.5 x 102 = (2X)2
(1.00-x)(2.00-x)
 Solving this gives us a quadratic equation.
 Quadratic gives us 2.14 mol/L and 0.968
mol/L. Only 0.968 is reasonable.
 [H2] = 1.00 M - 0.968 M = 3.2 x 10-2M
 [F2] = 2.00 M - 0.968 M = 1.032 M
 [HF] = 2(0.968 M) = 1.936 M
 If substituted into the equilibrium
expression we get 1.13 x 102 which
very close to given K.
is
Practice
H2O(g) + Cl2O(g)
2HClO(g)
K = 0.090
 In an experiment 1.0 g H2O(g) and 2.0 g
Cl2O are mixed in a 1.00 L flask.
 Calculate the equilibrium concentrations.
(#48)
13.6 Solving Equilibrium Problems
1.
2.
3.
4.
5.
6.
7.
Balance the equation.
Write the equilibrium expression.
List the initial concentrations.
Calculate Q and determine the shift to
equilibrium.
Define equilibrium concentrations.
Substitute equilibrium concentrations
into equilibrium expression and solve.
Check calculated concentrations by
calculating K.
Problems with small K
K< .01
Process is the same
 Set up table of initial, change, and
equilibrium concentrations.
 Choose X to be small.
 For this case it will be a product.
 For a small K the product
concentration is small.
For example
 For the reaction
2NOCl
 K= 1.6 x 10-5
2NO +Cl2
 If 1.00 mol NOCl is put in a 2.0 L
container, what are the equilibrium
concentrations?
 K = [NO]2[Cl2] = 1.6 x 10-5
[NOCl]2
 [NOCl]0= 0.50M, [NO]& [Cl2] =0
2NOCl
Initial
0.50
Change
-2X
Equil
0.50 -2X
2NO
0
+
Cl2
0
+2X
+X
2X
X
K = [NO]2[Cl2] = (2x)2(x) = 1.6 x 10-5
[NOCl]2
(0.50 -2x)2
 Since K is so small, we we can make an approximation
that 0.50-2x = 0.50
 This makes the math much easier. X = 1.0 x 10-2
5% Rule
 Many of the systems we will deal with have very
small equilibrium constants.
 When this is the case, there will be very little shift to
the right to reach equilibrium. Since x is so small, we
will ignore it. However, the final value must be
checked against the initial concentration. If the
difference is less than 5%, then our assumption is
valid.
 In the previous problem X = 1.0 x 10-2
 0.50 -2x = 0.50 - 2(1.0 x 10-2) = 0.48
 Is 1.0 x 10-2 five percent or less than 0.50?
 0.02/0.50 x 100 = 4%
Practice Problem
 For the reaction
N2O4(g)
2NO2(g)
 K = 4.0 x 10-7
 1.0 mol N2O4(g) is placed in a 10.0L
vessel. Calculate the concentrations
of all species at equilibrium.

(#52)
Practice Problem
 For the reaction
COCl2(g)
CO(g) + Cl2(g)
 Kp = 6.8 x 10-9
 If COCl2(g) at an initial pressure of
1.00 atm decomposes, calculate the
equilibrium pressures of all species?

(#54)
Practice Problem
 At 25°C, Kp = 2.9 x 10-3
 NH4OCONH2(s)
2NH3g) + CO2(g)
 In an experiment, a certain amount of
NH4OCONH2(s) is placed in an
evacuated rigid container and
allowed to come to equilibrium.
Calculated the total pressure in the
container at equilibrium.

(#55)
Practice Problem #56WA
 2AsH3(g)
2As(s) + 3H2(g)
 In an experiment pure AsH3 was
placed in a rigid, sealed flask at a
pressure of 392.0 torr. After 48 hours
the pressure was observed to be
constant at 488.0 torr.
 Calculate the Equilibrium pressure of
H2(g).
 Calculate Kp for this reaction. (#56)
Le Chatelier’s Principle
 If a change is applied to a system at
equilibrium, the position of the
equilibrium will shift in a direction
that tends to reduce the change.
 3 Types of change: Concentration
(adding or reducing reactant or
product), Pressure, Temperature.
Change amounts of reactants
and/or products
 Adding product makes Q>K
 Removing reactant makes Q>K
 Adding reactant makes Q<K
 Removing product makes Q<K
 Determine the effect on Q, will tell you the
direction of shift.
 The system will shift away from the added
component.
Change Pressure
 By changing volume.
 When volume is reduced, the system will
move in the direction that has the least moles
of gas.
 COCl2(g)
CO(g) + Cl2(g)
 When volume is increased, the system will
move in the direction that has the greatest
moles of gas.
 Because partial pressures (and conc.) change a
new equilibrium must be reached.
 System tries to minimize the moles of gas.
Change in Pressure
 By adding an inert gas.
 Partial pressures of reactants and
product are not changed.
 No effect on equilibrium position.
Change in Temperature
 Affects the rates of both the forward
and reverse reactions.
 Doesn’t just change the equilibrium
position, changes the equilibrium
constant.
 The direction of the shift depends on
whether it is exo - or endothermic.
Exothermic
 H < 0
 Releases heat.
 Think of heat as a product.
 N2(g) + 3H2(g)
2NH3(g) + Heat
 Raising temperature push toward
reactants.
 Shifts to left.
Endothermic
 H > 0
 Absorbs heat.
 Think of heat as a reactant.
 Heat + COCl2(g)
CO(g) + Cl2(g)
 Raising temperature push toward
products.
 Shifts to right.
Free Energy and Equilibrium
G tells us spontaneity at current conditions.
When will it stop?
It will go to the lowest possible free energy
which may be an equilibrium.
At equilibrium G = 0, Q = K
Gº = -RTlnK from [G = Gº + RTlnK]
Gº
=0
<0
>0
K
=1
>1
<1
Free Energy and Equilibrium
The overall reaction for the corrosion of iron by
oxygen is
4Fe(s) + 3O2(g)
2Fe2O3(s)
Use the following data, calculate the
equilibrium constant for this reaction at 25°C.
Substance
Fe2O3(s)
Fe(s)
O2(g)
H°f (kj/mol)
-826
0
0
Gº = -RTlnK
S°(J/K•mol)
90
27
205
Ex. 16.15
Practice
Calculate K for the following reaction
(at 298 K)
H2(g) + Cl2(g)
2HCl
H°f (kj/mol) = -184 S°(J/K•mol) = 20.
If each gas is placed in a flask such that the
pressure of each gas is 1 atm, in which direction
will the system shift to reach equilibrium at
25°C?
Gº = -RTlnK