Chapter 15
Chemical
Equilibrium
CHE1102, Chapter 14
Learn, 1
Chemical Equilibrium
– 2 opposing reactions occur simultaneously
at the same rate ⇌
D
E
E
D
When the rate D
E is equal to rate E
the condition of equilibrium has been
established
D ⇌ E
D,
CHE1102, Chapter 14
Learn, 2
CHE1102, Chapter 14
Learn, 3
Dynamic Equilibrium
2 NO (g) + O2 (g)
2 NO2 (g)
2 NO2 (g)
2 NO (g) + O2 (g)
2 NO (g) + O2 (g)
⇌
2 NO2 (g)
CHE1102, Chapter 14
Learn, 4
Dynamic Equilibrium
The same equilibrium is reached whether we
start with only reactants (N2 and H2) or with
product, NH3.
N2(g) + 3 H2(g) ⇌ 2 NH3(g)
CHE1102, Chapter 14
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The Equilibrium Constant
When a process is at equilibrium, it has
been found experimentally, for all
processes, that the ratio of products to
reactants is constant !
Equilibrium constant, K
[Products]
K = [Reactants]
Kc = for [ ] in molar, M
Kp = for [ ] in partial pressure
in atm for gases
CHE1102, Chapter 14
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Mass Action Expression
2 NO (g) + O2 (g)
⇌
2 NO2 (g)
[NO2]2
K = [NO]2 [O ]
2
CHE1102, Chapter 14
Learn, 7
The Equilibrium Constant
• Consider the generalized reaction
aA + bB
cC + dD
• The equilibrium expression for this reaction
would be
Kc =
[C]c[D]d
[A]a[B]b
CHE1102, Chapter 14
Learn, 8
Derivation of Mass Action Expression
Consider a system at equilibrium in
which both the forward and reverse
reactions are being carried out; we write
its equation with a double arrow:
N2O4(g) ⇌ 2 NO2(g)
CHE1102, Chapter 14
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Comparing Rates
• For the forward reaction
N2O4(g) → 2 NO2(g)
• The rate law is
Rate = kf [N2O4]
• For the reverse reaction
2 NO2(g) → N2O4(g)
• The rate law is
Rate = kr [NO2]2
CHE1102, Chapter 14
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The Meaning of Equilibrium
• At equilibrium
Ratef = Rater
kf[N2O4] = kr[NO2]2
• Rewriting this, it becomes the mass
action expression for the equilibrium
constant, Keq.
kf
Keq = k
r
=
[NO2]2
[N2O4]
CHE1102, Chapter 14
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N2O4 (g)
Kc =
⇌
2 NO2 (g)
[NO2]2
[N2O4]
When equilibrium is established, the concentration of reactants
and products vary….but the RATIO of P/R remains CONSTANT
!
CHE1102, Chapter 14
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2 NO (g) + O2 (g)
⇌
2 NO2 (g)
[NO2]2
Kc = [NO]2 [O ]
2
At equilibrium, the following [ ]’s are
found:
[NO2] = 0.896 M
[NO] = 0.0126 M
[O2] = 0.00413 M
Determine the equilibrium constant, Kc
CHE1102, Chapter 14
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By convention, equilibrium constants,
Kc are always UNITLESS !!
CHE1102, Chapter 14
Learn, 14
Determine the equilibrium constant, Kc
for the reverse process
2 NO2 (g)
[NO]2[O2]
Kc = [NO ]2
2
⇌
2 NO (g) + O2 (g)
This equilibrium expression
is the reciprocal of the
previous example
1
Kc = 1.22 x 106
= 8.17 x 10-7
The value of the equilibrium constant, K, for a reaction in one direction is
the reciprocal of the equilibrium reaction written in the reverse direction
CHE1102, Chapter 14
Learn, 15
Manipulating Equilibrium Constants
2 NO (g) + O2 (g)
⇌ 2 NO2 (g)
reverse reaction
2 NO2 (g)
⇌
2 NO (g) + O2 (g)
Kc = 1.22 x 106
reciprocal
Kc = 8.17 x 10-7
CHE1102, Chapter 14
Learn, 16
Manipulating Equilibrium Constants
The equilibrium constant of a reaction that
has been multiplied by a number is the
equilibrium constant raised to a power that
is equal to that number:
N2O4(g)
2N2O4(g)
2NO2(g)
[NO2]2
Kc =
= 0.212 at 100 C
[N2O4]
4
[NO
]
2
2 at 100 C
4NO2(g) Kc =
=
(0.212)
[N2O4]2
CHE1102, Chapter 14
Learn, 17
Manipulating Equilibrium Constants
The equilibrium constant for a net reaction made
up of two or more steps is the product of the
equilibrium constants for the individual steps.
2 NO2(g)
NO3(g) + NO(g)
NO3(g) + CO(g)
NO2(g) + CO(g)
[NO][NO 3 ]
[NO 2 ]2
NO2(g) + CO2(g)
NO(g) + CO2(g)
Kc =
1
Kc =
2
Kc =
3
[NO][NO3 ]
[NO2 ]2
[NO2 ][CO2 ]
[NO3 ][CO]
[NO][CO2 ]
[NO2 ][CO]
[NO][CO2 ]
[NO 2 ][CO 2 ]


[NO 2 ][CO]
[NO 3 ][CO]
Therefore
CHE1102, Chapter 14
Learn, 18
when K >> 1
(bigger than 1000)
[Products] > [Reactants]
P
K=
R
“Equilibrium lies to the right”
“Equilibrium lies to the left”
[reactants] > [products]
when K << 1
(smaller than 0.001)
P
K= R
CHE1102, Chapter 14
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If K  1
(0.001 – 1000)
The equilibrium mixture will have similar (or
comparable) amounts of reactants and products
K is constant and does NOT vary with [ ]
K does vary with temperature
CHE1102, Chapter 14
Learn, 20
Kc and Kp are Related
Kp = Kc (RT) Δn
R = 0.08206 L·atm/mole·K
T = temperature in Kelvin
n = (sum of coefficients of gaseous products)
− (sum of coefficients of gaseous reactants)
CHE1102, Chapter 14
Learn, 21
The production of ammonia from hydrogen and
nitrogen has a Kp of 41. If the reaction is run at
400 K what would Kc be?
N2 (g) + 3 H2 (g)
A. 41
B. 2.3 x 10-5
C. 24
D. 4.42 x 104
E. None of these
2 NH3(g)
Δn = (2 – 4) = -2
Kc = Kp/(RT)Δn
Kc= 41/(0.082057 × 400)-2
Kc = 4.42 x 104
CHE1102, Chapter 14
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An Equilibrium Problem
A closed system initially containing
1.000 x 103 M H2 and 2.000 x 103 M
I2 at 448 C is allowed to reach equilibrium.
Analysis of the equilibrium mixture shows that the
concentration of HI is 1.87 x 103 M. Calculate Kc
at 448 C for the reaction taking place, which is
H2(g) + I2(g)
2HI(g)
CHE1102, Chapter 14
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What Do We Know?
Initially
[H2], M
[I2], M
[HI], M
1.000 x 103
2.000 x 103
0
Change
At equilibrium
1.87 x 103
CHE1102, Chapter 14
Learn, 24
[HI] Increases by 1.87 x 10−3 M
Initially
Change
At equilibrium
[H2], M
[I2], M
[HI], M
1.000 x 103
2.000 x 103
0
+1.87 x 10−3
1.87 x 103
CHE1102, Chapter 14
Learn, 25
Stoichiometry tells us [H2] and [I2]
decrease by half as much.
[H2], M
[I2], M
[HI], M
Initially
1.000 x 103
2.000 x 103
0
Change
−9.35 x 10−4
−9.35 x 10−4
+1.87 x 103
At equilibrium
H2(g) + I2(g)
1.87 x 103
2HI(g)
CHE1102, Chapter 14
Learn, 26
We can now calculate the equilibrium
concentrations of all three compounds
[H2], M
[I2], M
[HI], M
Initially
1.000 x 103
2.000 x 103
0
Change
9.35 x 104
9.35 x 104
+1.87 x 103
6.5 x 10−5
1.065 x 10−3
1.87 x 103
At equilibrium
CHE1102, Chapter 14
Learn, 27
Therefore, the equilibrium constant is:
[HI]2
Kc = [H ] [I ]
2
2
(1.87 x 103)2
= (6.5 x 105)(1.065 x 103)
= 51
CHE1102, Chapter 14
Learn, 28
A container is initially charged with
2.00 M phosgene, COCl2 (g) at 395 °C. An
equilibrium with carbon monoxide and
chlorine gas is established. The
equilibrium concentration of chlorine gas
was found to be 0.0398 M. What is Kc for
this reaction ?
1. Write the balanced chemical reaction
2. Build a chart under the reaction
3. Fill in the appropriate information (this is hard !!!)
4. Solve the problem that is presented
CHE1102, Chapter 14
Learn, 29
A container is initially charged with
0.260 atm Cl2 (g) and 0.520 atm Br2 (g)
at 75 °C. The reactants combine to
produce BrCl (g). Kp = 56.9 at this
temperature. What are the partial
pressures of all species at equilibrium ?
CHE1102, Chapter 14
Learn, 30
A container is initially charged with
0.18 M CH4 (g) and 0.18 M CCl4 (g) at 455 °C.
The reactants combine to produce CH2Cl2 (g).
Kc = 0.559 at this temperature. What are the
molarities of all species at equilibrium ?
CHE1102, Chapter 14
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Homogeneous vs. Heterogeneous
• Homogeneous equilibria occur when
all reactants and products are in the
same phase.
• Heterogeneous equilibria occur
when something in the equilibrium is in
a different phase.
• The value used for the concentration
of a pure substance is always 1.
CHE1102, Chapter 14
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Reaction Quotient, Q
– an equilibrium expression for a reaction
NOT necessarily at equilibrium
When Q < K, the reaction will proceed to the right
(toward products) to establish equilibrium
When Q > K, the reaction will proceed to the left
(toward reactants) to establish
equilibrium
When Q = K, the reaction is at equilibrium
CHE1102, Chapter 14
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CHE1102, Chapter 14
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Homogeneous equilibrium – all species are
in the same
phase
Heterogeneous equilibrium – all species are
NOT in the same
phase
* ALL pure liquids and solids are left
out of the equilibrium expression
Aqueous solutions, ex. NaCl (aq), are
always included in the equilibrium
expression
CHE1102, Chapter 14
Learn, 35
The Decomposition of CaCO3—
A Heterogeneous Equilibrium
• The equation for the reaction is
CaCO3(s) ⇌ CaO(s) + CO2(g)
• This results in
Kc = [CO2]
and
Kp = PCO2
CHE1102, Chapter 14
Learn, 36
Heterogeneous Equilibria
Write equilibrium laws for the following:
Ag+(aq) + Cl–(aq)
AgCl(s)
1
Kc =
+
[Ag ][Cl ]
H3O+(aq) + H2PO4–(aq)
H3PO4(aq) + H2O(l)
+
-
[H3O ][H2PO4 ]
Kc =
[H3PO4]
CHE1102, Chapter 14
Learn, 37
Fact: When a system is at equilibrium, it will
remain at equilibrium forever, unless disturbed
by some outside force.
outside force:
- change in concentration
- change in temperature
- change in pressure (gas phase)
Le Châtelier’s Principle – when a “stress” is applied to
a system at equilibrium, the
reaction “shifts” to relieve
the stress, and re-establish
the condition of equilibrium
Henry Le Chatelier 1850 – 1936
CHE1102, Chapter 14
Learn, 38
N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g)
CHE1102, Chapter 14
Learn, 39
A Stress Applied to System
CHE1102, Chapter 14
Learn, 40
Le Châtelier states, “if a system is at equilibrium,
and you…
1. add reactant, the reaction shifts to the Right,
to consume the excess reactant and restore equilibrium”
2. add product, the reaction shifts to the Left,
to consume the excess product and restore equilibrium”
3. remove reactant, the reaction shifts to the Left,
to replace the lost reactant and restore equilibrium”
4. remove product, the reaction shifts to the Right,
to replace the lost product and restore equilibrium”
CHE1102, Chapter 14
Learn, 41
Changes in applied pressure affects
gas phase reactions
N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g)
Increase in pressure results in the reaction shifting
toward the side with the fewest number of gas particles
Decrease in pressure results in the reaction shifting
toward the side with the largest number of gas particles
CHE1102, Chapter 14
Learn, 42
N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g)
CHE1102, Chapter 14
Learn, 43
Changes in concentration or pressure result in reactions
shifting but do NOT affect the numerical value of the
equilibrium constant, K
Changes in temperature also result in reactions shifting
and DOES affect the numerical value of the equilibrium
constant, K
Changes in Temperature Affects Equilibrium
predicting how temperature affects equilibria requires
thermodynamic knowledge of the equilibrium reaction
heat is a reactant for an endothermic reaction and
heat is a product for an exothermic reaction
CHE1102, Chapter 14
Learn, 44
Changes in Temperature
CHE1102, Chapter 14
Learn, 45
Catalysts
• Catalysts increase the rate of both the forward
and reverse reactions.
• Equilibrium is achieved faster, but the equilibrium
composition remains unaltered.
• Activation energy is lowered, allowing equilibrium
to be established at lower temperatures.
CHE1102, Chapter 14
Learn, 46