Chapter 15
Chemical Equilibrium
8–1
John A. Schreifels
Chemistry 212
Chapter 15-1
Overview
• Describing Chemical Equilibrium
– Chemical Equilibrium – A Dynamic Equilibrium (the link to Chemical
Kinetics)
– The Equilibrium Constant.
– Heterogeneous Equilibria; solvents in homogeneous equilibria.
• Using the Equilibrium Constant
– Qualitatively Interpreting the Equilibrium Constant
– Predicting the direction of a Reaction
– Calculating Equilibrium Concentrations
• Changing Reaction Conditions; Le Châtelier’s Principle.
– Removing Products or Adding Reactants
– Changing the Pressure or Temperature
– Effect of a Catalyst.
John A. Schreifels
Chemistry 212
8–2
Chapter 15-2
Chemical Equilibrium – A Dynamic Equilibrium
• Upon addition of reactants and/or products, they react until a
constant amount of reactants and products are present =
equilibrium.
• Equilibrium is dynamic since product is constantly made
(forward reaction), but at the same rate it is consumed (reverse
reaction).
8–3
John A. Schreifels
Chemistry 212
Chapter 15-3
The Equilibrium State
• Not all reactants are completely converted to product.
• Reaction equilibria deal with the extent of reaction.
• Arrows between reactants and products separate them and
qualitatively indicate the extent of reaction.
– Single arrow points to dominant side: H2(g) + O2(g)  H2O(g)
– Double arrow indicates both reactants and products present after
equilibrium obtained: N2O4 (g)  2NO2(g).
• Equilibrium exists when rates of forward and reverse reaction
are the same.
E.g. When rate of N2O4 decomposition equal the rate of
formation of N2O4, reaction at equilibrium
N2O4 (g)  2NO2(g).
• Equilibrium can be obtained from any mixture of reactants and
products.
John A. Schreifels
Chemistry 212
Chapter 15-4
8–4
The Link Between Chemical Equilibrium and
Chemical Kinetics
Reactions with one elementary step:
• At equilibrium: Rf = Rr.
E.g. decomposition of N2O4:
• kf[N2O4] = kr[NO2]2. or
k f [NO 2 ] 2
Kc 

k r [N2O 4 ]
where Kc is called the equilibrium constant.
• At equilibrium the ratio of concentrations equals a constant.
• A generalized form of this expression that describes the
equilibrium condition:
aA + bB + cC + ...  mM + nN + oO ....
Kc 
John A. Schreifels
Chemistry 212
[M]m [N]n [O]o
8–5
[A]a [B]b [C]c
Chapter 15-5
The Link Between Chemical Equilibrium and
Chemical Kinetics II
Reactions with more than one elementary step
• Use the rate limiting step and
• “Steady State Approximation”: at equilibrium the rates of the
forward and reverse reactions are equal.
• Eliminate intermediates.
• E.g.: Determine the equilibrium expression for the two step
decomposition of ozone.
1.
k f ,1
O3 (g)  O2 (g)  O (g)
k r,1
Fast
k f ,2
2.
Sum
O 3 (g)  O(g)  2O 2 (g)
k r,2
Slow
2O3 (g)  3O2.
• Step 2 (rate limiting step)
John A. Schreifels
Chemistry 212
8–6
Chapter 15-6
Equilibrium Constant: Multi-step
• Rate equations lead to:
k f ,2 [O 3 ][O]  k r,2 [O 2 ] 2
[O 2 ] 2

k r,2 [O 3 ][O]
k f ,2
• Remove intermediate, O:
k f ,1[O 3 ]  k r,1[O 2 ][O]
[O] 
k f ,1 [O 3 ]
k r,1 [O 2 ]
• Substitute into the first equation to get: K c


k f ,1 k f ,2
k r,1 k r,2
[O 2 ] 3
[O 3 ] 2
• Conclusion: Form of equilibrium expression is
independent of mechanism.
John A. Schreifels
Chemistry 212
8–7
Chapter 15-7
The Equilibrium Constant Kc
• Equilibrium constant of reverse reaction: K c,r  1
aA + bB  cC + dD K c,f 
cC + dD  aA + bB K c,r 
d
[D] [C]
c
K c, f
[ A ]a [B]b
[ A ]a [B]b
[D] d [C]c
E.g.1 Determine the equilibrium constant for the reaction:
½N2(g) + 3/2 H2(g)  NH3(g)
given N2(g) + 3H2(g)  2NH3(g) Kc = 1.7x102
E.g.2 determine the equilibrium constant for the formation of
HI(g) if the equilibrium concentration of H2, I2 and HI are 0.0060
M, 0.106 M, and 0.189 M, respectively.
H2(g) + I2(g)  2HI(g) Kc = ?
E.g. 3 Determine the equilibrium constant for the reaction below:
½ H2(g) + ½ I2(g)  HI(g) Kc = ?
8–8
John A. Schreifels
Chemistry 212
Chapter 15-8
Equilibrium Constant Kc, Kp
• For gases Kc and Kp are used.
• KP same format as Kc except pressures used instead
of concentrations.
E.g. Write out the equilibrium expression for KP using
the reaction below:
N2(g) + 3H2(g)  2NH3(g) KP = ?
E.g.2 What is equilibrium expression (KP) for the
reaction below?
½ N2(g) + 3/2 H2(g)  NH3(g) KP = ?
E.g.3 Determine the equilibrium constant (KP) for the
formation of one mole of ammonia if at 500K, PNH3 =
0.15 atm, PN2 = 1.2 atm. and PH2 = 0.81 atm.
8–9
John A. Schreifels
Chemistry 212
Chapter 15-9
Relationship between Kc and Kp
•
Relationship between concentration and pressure obtained from the
ideal gas law.
n
– Recall PV = nRT or PA  A RT
V
 [ A ]RT
– Substitute for P in equilibrium expression. Consider the reaction:
aA + bB  cC + dD
KP 

c d
PC
PD
a b
PA
PB

[C]RT c [D]RT d

[ A ]RT a [B]RT b
[C] c [D] d RT c  d
[ A ] a [B]b RT a  b
K P  K c RT (c  d)  (a  b)  K c RT n
– Use this relationship to relate KP and Kc
E.g. Determine Kp for the synthesis of ammonia at 25°C;
N2(g) + 3H2(g)  2NH3(g) Kc = 4.1x108
E.g. 2 Determine Kp for the synthesis of HI at 458°C;
½ H2(g) + ½ I2(g)  HI(g) Kc = 0.142.
John A. Schreifels
Chemistry 212
8–10
Chapter 15-10
Heterogeneous Equilbria
• Do not include a solvent or solid in the equilibrium
expression. Their composition is constant and
included in the equilibrium constant.
– Water concentration is 55.5 M; very high compared with
reactants and products.
– The concentration of a solid such as CaCO3 stays the same
as long as some solid is present.
E.g.: Determine the equilibrium expression for the
reaction:
CaCO3(s) + C(gr)  CaO(s) + 2CO(g).
E.g.2 Determine the equilibrium expression for the
reaction of acetic acid with water.
CH3COOH(aq) + H2O(l)  CH3COO(aq) + H3O+(aq)
John A. Schreifels
Chemistry 212
8–11
Chapter 15-11
Reaction Sequence
• As with H in thermo. Kc can
be determined from a
reaction sequence.
• Consider the reactions to the
right. The third reaction is
the result of the sum of the
other two (called a reaction
sequence..
aA + bB cC + dD
cC + dD eE + fF
aA + bB eE + fF
K1 
K2 
K3 
[D]d [C]c
[ A ]a [B]b
[E] e [F] f
[C]c [D] d
[E]e [F]f
a
b
[ A ] [B]
 K1  K 2

[C]c [D]d
a
[ A ] [B]
E.g.: Determine the Kp for the reaction:
CaCO3(s) + C(gr)  CaO(s) + 2CO(g).
Given:
CaCO3(s)  CaO(g) + CO2(g) , K1 = 0.039 atm. = PCO2, and
C(g) + CO2(g)  2CO(g), K2 = 1.9 atm.
John A. Schreifels
Chemistry 212
b

[E]e [F]f
[C]c [D]d
8–12
Chapter 15-12
Applications of the Equilibrium Constant
• Extent of reaction: The magnitude of the equilibrium constant
allows us to predict the extent of the reaction.
– Very large K (e.g. 1010)  mostly products.
– Very small K (e.g. 1010)  mostly reactants.
– When K is around 1, a significant amount of reactant and
product present in the equilibrium mixture.
E.g. For each of the following decide which species will
predominate at equilibrium.
AgCl(s) Ag+(aq) + Cl(aq) Ksp = 1.8x1010
Ag+(aq) + 2NH3(aq) Ag(NH ) 
3 2
H2CrO4(aq) + H2O(l) HCrO  (aq) +
4
+
H3O (aq)
John A. Schreifels
Chemistry 212
K f  1.7x10 7
Ka1 = 0.15
8–13
Chapter 15-13
Using the Equilibrium Constant
•
•
•
•
•
•
Direction of Reaction: the reaction quotient can be used to determine
the direction of a reaction with certain initial concentrations.
Reaction Quotient: For the general reaction,
aA + bB  cC + dD,.
[D]dt [C]ct
Qc 
[A]at [B]bt
where the t refers to the time that concentrations of the mixture are
measured; not necessarily at equilibrium.
Comparison of Qc with Kc reveals direction of reaction.
When only reactants Qc = 0; leads to
– If Qc < Kc, products will form.
When only products present, Qc  .
– If Qc > Kc, reactants will form.
When Qc = Kc, no net reaction.
E.g. Determine direction of reaction: H2(g) + I2(g)  2HI(g). Assume
that [H2]o = [I2]o = [HI]o = 0.0020M at 490oC for which Kc = 46.
John A. Schreifels
Chemistry 212
8–14
Chapter 15-14
Calculating Equilibrium Concentrations
• Using initial concentrations, stoichiometry and Kc, equilibrium
concentrations of all components can be determined.
E.g. Determine equilibrium amounts of hydrogen and iodine if
you started with exactly 1.00 mol H2(g) and 2.00 mol I2 in a 1.00
L flask; after equilibrium had been attained, [HI] = 1.86 M.
H2(g) + I2(g)  2HI(g)
E.g.2 [H2]o = [I2]o = 24 mM, were mixed and heated to 490oC in
a container. Calculate equil. composition. Given that Kc = 46.
Solution: set up an equilibrium table and solve for unknown
after substitution into equilibrium expression.
H2(g) +I2(g)2HI(g)
H2(g) +
I2  2HI(g)
Init. Conc.
0.024 M
0.024 M
0
+2x
 to Equil.
x
x
8–15
Equil. 0.024 M  x 0.024 M  x
+2x
John A. Schreifels
Chemistry 212
Chapter 15-15
Using the Equilibrium Constant (cont)
•
•
E.g.2 For the reaction: PCl5(g)  PCl3(g) + Cl2(g), Kc = 0.800 M at
340oC. Find the equilibrium amounts of these compounds if they all
start out at 0.120 M.
Solution: Determine Qc so that you know which direction the reaction
will proceed. Qc = 0.120 < Kc  reaction proceeds to products.
Init. Conc.
 to Equil.
Equil.
•
•
.
PCl5(g)
Cl2+
PCl3(g)
0.120M
0.120M
0.120M
+x
+x
x
0.120 M  x 0.120 M + x 0.120 M + x
Equilibrium concentrations of each component are in last row.
Substitute into equilibrium expression and solve for x.
It may be necessary to rearrange so that quadratic equation can be
used.

0.120 M  x 2
Kc 
0.120 M  x
Rearrange to ax2 + bx + c = 0; determine a, b, c and substitute into
quadratic equation:
 b  b 2  4ac
x
2a
John A. Schreifels
Chemistry 212
8–16
Chapter 15-16
Factors that Alter the Composition of an
Equilibrium Mixture
• A change to the system, which is initially at
equilibrium, can cause a change in the equilibrium
composition.
• Le Châtelier’s Principle: “If a stress is applied to a
reaction mixture at equilibrium, reaction occurs in the
direction that relieves the stress.”
• Types of stress on equilibrium:
– Concentration of reactants or products. You can add or
remove one or more components in a reaction mixture.
– With gases changing the pressure or volume is a way of
changing the concentrations of all components in the
mixture.
– Change temperature.
John A. Schreifels
Chemistry 212
8–17
Chapter 15-17
Changes in Concentrations
• Addition or removal of either reactant or product shifts the
equilibrium to reduce the excess compound.
• Removal of product or addition of reactant has the same effect;
they shift the equilibrium to the right.
E.g. Consider the formation of NH3(g); the equilibrium conditions
are listed in the first row; then half of the NH3(g) is removed. We
calculate Qc and compare it with Kc. Qc < Kc  reaction goes to
right to form more NH3(g).
N2(g)+
3 H2
Equil. Conc. 0.50 M 3.00 M
After NH3(g) 0.50 M 3.00 M
2 NH3(g)
1.98 M
0.99 M
Kc = 0.291
Qc = 0.073
E.g.2 Instead lets add N2 so that its concentration increases 10x.
This produces a similar effect. Since hydrogen is a reactant (in
denominator), Qc will become much smaller than Kc. This means
that product must form.
N2(g)+
3 H2 2 NH3(g)
Equil. Conc. 0.50 M 3.00 M
1.98 M Kc = 0.291
1.98 M Qc = 0.0291
After N2(g) 5.0 M 3.00 M
John A. Schreifels
Chemistry 212
8–18
Chapter 15-18
Changes in Concentrations 2
• Removal of reactant or addition of product shifts the equilibrium in
the opposite direction – to the left.
• E.g. Half of N2(g) is removed. We calculate Qc and compare it with
Kc. Notice that it is greater than the equilibrium constant. Therefore
the reaction proceeds to the left to form more N2(g).
N2(g)+
3 H2 2 NH3(g)
Equil. Conc. 0.50 M 3.00 M
1.98 M Kc = 0.291
1.98 M Qc = 0.582
After N2 0.25 M 3.00 M
• E.g.2 Add NH3 so that its concentration increases by a factor of 2.
This produces a similar effect. Since ammonia is a product (in
numerator), Qc > Kc. Reaction proceeds to left since there is too
much product.
N2(g)+
3 H2 2 NH3(g)
Equil. Conc. 0.50 M 3.00 M
1.98 M Kc = 0.291
8–19
3.96 M Qc = 1.162
After NH3 0.50 M 3.00 M
John A. Schreifels
Chemistry 212
Chapter 15-19
Changes in Pressure and Volume
• Equilibria containing gases are sensitive to changes in
pressure (volume).
– Since pressure and volume are inversely proportional, an
increase in pressure will give same result as a decrease in
volume.
– A decrease in the volume increases concentration of each of the
components in the reaction mixture.
– A decrease in V shifts equilibrium to side with least number of
moles of gas.
• E.g. Determine the effect of a 2.00x increase in pressure
(volume decreases by factor of 2;Vf = ½ VI). Concentrations of
reactants and products increase by a factor of 2. Since Qc <
Kc, the reaction proceeds to the right to form more NH3(g).
N2(g)+ 3 H2 2 NH3(g)
Equil. Conc. 0.50 M 3.00 M
1.98 M Kc = 0.291
3.96 M Qc = 0.073
After P 1.00 M 6.00 M
John A. Schreifels
Chemistry 212
8–20
Chapter 15-20
Changes in Pressure and Volume 2
E.g.2 Determine the effect of a factor of 2.00 decrease in the
pressure (volume increases by a factor of 2;Vf = 2 Vi).
N2(g)+
3 H2 2 NH3(g)
Equil. Conc. 0.50 M 3.00 M
1.98 M Kc = 0.291
0.99 M Qc = 1.16
After P 0.25 M 1.50 M
• We need to be vigilant so that we do not include any solids or
solvents in this, since they do not affect the equilibrium.
• E.g. Determine the ratio of Qc/Kc when the pressure increases
by a factor of two and from this discuss the effect on the
equilibrium: C(s) + CO2(g)  2CO(g).
8–21
John A. Schreifels
Chemistry 212
Chapter 15-21
Changes in Temperature
• T  Qc.
• T produces Kc
• Whether Kc increases or decreases depends upon two factors,
T and Hrxn.
– An increase in temperature causes the equilibrium to shift towards
the endothermic side;
– A decrease in temperature causes the equilibrium to shift in the
opposite direction.
E.g. Will an increase or a decrease in temperature increase the
amount of CO produced from the following reaction
2CO2(g)2 CO(g) +O2(g)
H = 566 kJ
The Effect of a Catalyst
• It has no effect on the equilibrium.
• It speeds up attainment of equilibrium.
• Kc related to the H and not Ea
John A. Schreifels
Chemistry 212
8–22
Chapter 15-22