Chemical equilibrium

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TOPIC 4: CHEMICAL EQUILIBRIUM
Lecturer and contacts
Mr. Vincent Madadi
Department of Chemistry, University of Nairobi
P. O. Box 30197-00100,
Nairobi, Kenya
Chemistry Dept. Rm 114
Tel: [phone] ext 2185
Email: [email protected], [email protected]
Website: http://www.uonbi.ac.ke/staff/vmadadi
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Introduction
•Chemical equilibrium is based on the fact that reversible
reactions do not go to completion
•Chemical equilibrium exists when two opposing reactions occur
simultaneously at the same rate.
•A chemical equilibrium is a reversible reaction where the
forward reaction rate is equal to the reverse reaction rate.
•Thus, Molecules are continually reacting, even though the
overall composition of the reaction mixture does not change.
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Introduction cont.
•The RATE of the forward reaction equals the RATE
of the reverse reaction
•Symbolically, this is represented as:
•aA(g) + bB (g) ⇌cC (g) + dD (g)
•The equilibrium constant can be expressed as:
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Introduction cont.
•Graphically, this is a representation of the rates for
the forward and reverse reactions for this general
reaction.
•aA(g) + bB (g) ⇌cC (g) + dD (g)
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Law of mass action
•At constant temperature, the rate of chemical reaction is
proportional to active mass of the reacting substances
•Active mass = is a thermodynamic quantity defined as a=
fC,
•Where a= active mass; f= activity coefficient, C = molar
concentration
•For gaseous systems at low pressure or dilute solutions,
f= 1, thus a = C
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Equilibrium law
•For the reaction , aA(g) + bB (g) → cC (g) + dD (g)
•Rate of forward reaction R
1= k
1[A]
a
[B]
b
•Rate of backward reaction R
2= k
2[C]
c
[D]
d
•At equilibrium,
k
1[A]
a
[B]
b
= k
2[C]
c
[D]
d
Thus,
K
1/k
2= [C]
c
[D]
d
/[A]
a
[B]
b
= Kc
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Introduction cont.
•Therefore , for the general reaction:
a A(g) + bB (g) ⇌c C (g) + d D (g)
•Eqiliburium law can define a constant
Kc is defined for a reversible reaction at a given temperature as the
product of the equilibrium concentrations (in M) of the products,
each raised to a power equal to its stoichiometric coefficient in the
balanced equation, divided by the product of the equilibrium
concentrations (in M) of the reactants, each raised to a power
equal to its stoichiometric coefficient in the balanced equation.
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Introduction cont.
•For gaseous reactions, concentration can be expressed
in terms of partial pressure
•Where, aA + Bb ⇌lL + mM,
•If PA, PB, PL and PM are the partial pressures of the
gaseous species, then at equilibrium
•Kp = (P
L)
i
(P
M)
m
/(P
A)
a
(P
B)
b
•If concentration is in mole fraction (X), the equilibrium
constant can be expressed in form of K
x
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Introduction cont.
•Where
K
x= (X
L)
I
(X
M)
m
/(X
A)
a
(X
B)
b
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Relationship between Kc, Kp and Kx
•For ideal gas, partial pressure of the component in the
mixture is given by,
•Pi = (m/v)RT
•Where m = number of moles of the gas occupying volume
V,
•Thus m/v = Ci = Molar concentration
•Therefore, Pi = CiRT,
•Substituting the value of Pi into the expression for Kp
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Relationship between Kc, Kpand Kx...
•Kp = (CLRT)
l
(CMRT)
m
/(CART)
a
(CBRT)
b
•Kp = (CL)
l
(CM)
m
(RT)
l
(RT)
m(CA)
a
(CB)
b
(RT)
a
(RT)
b
•Kp = (CL)
l
(CM)
m
(RT)
l+m
(CA)
a
(CB)
b
(RT)
a+b
•But
Thus Kp = Kc (RT)
Δn
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Relationship between Kc, Kpand Kx...
•From Kp = Kc (RT)
Δn
•When Δn = +ve, the number of moles of reactants are
less than the number of moles of products , thus Kp >Kc
•When Δn = -ve, the number of moles of reactants are
more than the number of moles of products , thus Kp <Kc
•When Δn = 0, the number of moles of reactants equals
the number of moles of products , thus Kp =Kc
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Relationship between Kc, Kpand Kx...
•For Kp and Kx,
•Note that the mole fraction of a component of a given
gas mixture is given by Xi = Pi/P
•Or Pi = XiP
•Where P = total pressure of the gaseous mixture
•Substituting the value of partial pressure into the
expression for Kp
•Then,
Kp = (P
L)
l
(P
M)
m
/(P
A)
a
(P
B)
b
•Or Kp = (X
LP)
l
(X
MP)
m
/(X
AP)
a
(X
BP)
b
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Relationship between Kc, Kpand Kx...
•This can be expressed as,
Kp = (X
L)
l
(X
M)
m
(P)
l
(P)
m
(X
A)
a
(X
B)
b
(P)
a
(P)
b
•Or Kp = (X
L)
l
(X
M)
m
(P)
(l+m)
(X
A)
a
(X
B)
b
(P)
(a+b)
This can be reduced to,
•Or Kp = Kx (P)
(l+m) =
Kx P
Δn
(P)
(a+b)
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Relationship between Kc, Kpand Kx...
•Therefore,
Kp = Kx P
Δn
= Kc (RT)
Δn
•Hence, Kc = Kx (P/RT)
Δn
•Equivalent to,
Kc = Kx (n/v)
Δn
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Factors That Affect Chemical Equilibrium
1) concentration
2) pressure
3) volume
4) temperature
5) Catalysts- have no effect on position of
equilibrium
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Le Chatelier’s Principle
•If an external stress is applied to a system at
equilibrium, the system adjusts itself in such a way
that the stress is partially offset.
1) Changes in Concentrations
•Increase the yield of product by
–increasing concentration of reactant
–removing the product from the equilibrium
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Le Chatelier’sPrinciple cont.
•Eaxample
N
2 (g) + 3H
2 (g) 2NH
3 (g)
0.683 M 8.80 M 1.05 M
•Increase the concentration of NH
3to 3.65 M, the position of
equilibrium shifts to the left
Q =[NH
3]
2
= ( 3.65 )
2
= 0.0286
[N
2] [H
2]
3
(.683) (8.80)
3
But Kc =[NH
3]
2
= ( 1.05 )
2
=
[N
2] [H
2]
3
(.683) (8.80)
3
•Q >Kc
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Le Chatelier’sPrinciple cont.
BaSO
4(s) Ba
2+(
aq) + SO
4
2-(aq)
K
c= [Ba
2+
] [SO
4
2-]
•By adding Ba
2+
(aq), [SO
4
2-] decreases, but BaSO
4(s)
increases
•By add SO
4
2-
(aq), [Ba
2+
] decreases but BaSO
4(s)
increases
•Add BaSO
4(s) no change
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Changes in Volume and Pressure
•Little effect on reactions in solution
•Effect can be large on reactions in the gas phase
• Increase in pressure shifts the equilibrium to the side with the
fewer moles of gas
•Example:
•How does the position of equilibrium change as the pressure is increased?
five moles of gaseous reactant two moles of gaseous product
•Increase in pressure causes an increase in products at the expense of
reactants
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Changes in Temperature
•Changes the value of the equilibrium constant
–A temperature increase favours an endothermic reaction,
and a temperature decrease favours an exothermic reaction
•Example:
1) Endothermic reactions
•Shifts in the equilibrium position for the reaction:
58 kJ + N
2O
4(g) 2NO
2(g)
•Increase temperature: Favours forward reaction
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The Effect of a Catalyst
•Increases rate, but has no effect on position of
equilibrium
1effect on forward and reverse processes is the same
1a catalyst increases both the rate of both the forward
and reverse reactions
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The Haber Process
•Maximize the yield of ammonia by: carrying out the
reaction at high pressure
ΔH
o
= -92.6 kJ/mol
1.
2.
•Maximize the yield of ammonia by: carrying out the
reaction at low temperatures
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The Haber Process
•In practice the reaction is carried out at 500 ºC
because the rate is too slow at lower temperatures
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Disruption and restoration of equilibrium
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At the left, the concentrations of the three components do not change with time
because the system is at equilibrium.
By adding more hydrogen to the system, disrupts the equilibrium. A net reaction then
ensues that moves the system to a new equilibrium state (right) in which the quantity
of hydrogen iodide has increased; in the process, some of
the I
2and H
2are consumed.
The new equilibrium state contains more hydrogen than did the initial state, but not as
much as was added; as the LeChâtelier principle predicts, the change we
made (addition of H
2) has been partially counteracted by the "shift to the right".
Equilibrium constant calculations
•One litre of equilibrium mixture from the following
system at a high temperature was found to contain 0.172
mole of phosphorus trichloride, 0.086 mole of chlorine,
and 0.028 mole of phosphorus pentachloride. Calculate
Kc for the reaction.
•Ans: Reaction PCl
5 PCl
3+ Cl
2
• Expression for Kc
• Substitute the values
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Equilibrium constant calculations cont.
•The decomposition of PCl
5was studied at another temperature.
One mole of PCl
5was introduced into an evacuated 1.00 litre
container. The system was allowed to reach equilibrium at the
new temperature. At equilibrium 0.60 mole of PCl
3was present in
the container. Calculate the equilibrium constant at this
temperature.
Ans: Reaction PCl
5⇌PCl
3 + Cl
2
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Equilibrium constant calculations cont.
• Kc = [PCl
3][Cl
2]
[PCl
5],
• Kc = (0.60)(0.60) = 0.90
• (0.40)
Significance of Kc
•Large equilibrium constants indicate that most of the reactants
are converted to products.
• Small equilibrium constants indicate that only small amounts
of products are formed.
Kc > 1 mostly products
Kc < 1 mostly reactants
Kc ~ 1 equal amounts of products and reactants
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Equilibrium constant calculations cont.
•The mass action expression or reaction quotient has the
symbol Q.
•Q has the same form as Kc, but the difference between
them is that the concentrations used in Q are not
necessarily equilibrium values.
•The Reaction Quotient
aA + bB ⇌cC + dD
•Q will help us predict how the equilibrium will respond
to an applied stress.
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Equilibrium constant calculations cont.
•To make this prediction we compare Q with Kc.
Q = [C]
c
[D]
d
[A]
a
[B]
b
•Q < K Products are formed
•Q = K System is at equilibrium
•Q > K Reactants are formed
•Example
•The equilibrium constant for the following reaction is
49 at 450°C. If 0.22 mole of I
2, 0.22 mole of H
2, and
0.66 mole of HI were put into an evacuated 1.00-liter
container, would the system be at equilibrium?
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Equilibrium constant calculations cont.
•Ans.
H
2(g) + I
2(g) → 2 HI(g)
0.22 M 0.22 M 0.66 M
•The Reaction Quotient
• Q= [HI]
2
= (0.66)
2
= 9.0
H
2[ ] [I
2] (0.22)(0.22)
•Q = 9.0 but Kc = 49
•Thus Q < Kc
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Equilibrium constant calculations cont.
•The equilibrium constant, Kc, is 3.00 for the following
reaction at a given temperature. If 1.00 mole of SO
2
and 1.00 mole of NO
2are put into an evacuated 2.00 L
container and allowed to reach equilibrium, what will
be the concentration of each compound at
equilibrium?
SO
2(g) + NO
2(g) ⇌SO
3(g) + NO (g)
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Equilibrium constant calculations cont.
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Equilibrium constant calculations cont.
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Equilibrium 0.0800+ x 0.0600+ x 0.790 -2x

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Equilibrium constant calculations cont.
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Types of chemical equlibria
•There are two types of chemical equilibria:
1) Homogeneous equilibrium
•All the reactants and products are in the same phase
2) Heterogeneous equilibrium
•The reactants and products are in two or more
different phases
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Types of chemical equiliburia cont.
•Heterogeneous equilibria have more than one phase
present.
•For example, a gas and a solid or a liquid and a gas.
CaCO
3(s) ⇌CaO(s) + CO
2(g)
•How does the equilibrium constant differ for
heterogeneous equilibria?
•Pure solids and liquids have activities of unity.
•Solvents in very dilute solutions have activities that are
essentially unity.
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Heterogeneous equilibrium
•The thermodynamic equilibrium, Ka is given by:
Ka = (aCaO )(aCO
2) Eq.1
(aCaCO
3)
•Where a = activity of various species
•For pure solids and liquids, activity are taken as unity at
all temperatures
•Thus, = aCaCO
3= 1 = aCaO
•The equation reduces to Ka = aCO
2 Eq. 2
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Heterogeneous equilibrium cont.
•If the gas behaves ideally, activity reduces to pressure. Thus the
equation 2 becomes:
Kp = P CO
2
•The equilibrium constant for heterogeneous reactions is
independent of the amount of a pure solid (liquid) phase
•Thus equilibrium constant involves gaseous constituents but does
not include any term for concentration of either pure solid or
liquid
•Thus Kp of heterogeneous reaction is generally called condensed
equilibrium
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Example- calculations involving Heterogeneous
equilibrium
•At a certain temperature, Kp for the dissociation of
the solid calcium carbonate is 4 x 10
-2
atm and for the
reaction Cs + CO
2⇌2CO is 2 atm respectively.
Calculate the pressure of CO at this temperature
when solid carbon, CaO and CaCO
3are mixed and
allowed to attain equilibrium
•Ans. CaCO
3(s) ⇌CaO(s) + CO
2(g) eq. 1
• Cs + CO
2(g) ⇌2CO(g) eq.2
•Kp for the reaction 1 is:
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Example- calculations involving Heterogeneous
equilibrium cont.
•Kp = aCaO.aCO
2/aCaCO
3= P CO
2
•K’p for reaction 2
K’p = (aCO)
2
/(aC)(aCO
2) = P
2
CO/P CO
2
K’pKp = P CO
2. P
2
CO/P CO
2= P
2
CO
Thus, P CO = (KpK’p)
½
= ((4.0 x 10
-2
)(2.0))
½
= 0.28 atm
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Example- calculations involving Heterogeneous
equilibrium cont.
•A solid NH
4HS is placed in a flask containing 1.50 atmospheric
ammonia. What is the partial pressures of ammonia and H
2S
when equilibrium is attained? (K
p= 0.11)
•Ans: Reaction,
NH
4HS
(S)⇌NH
3+ H
2S
P NH
3= 1.5 + P H
2S
Kp = (P NH
3) (P H
2S)
Thus 0.11 = (1.50 + P H
2S) (P H
2S)
This implies,
P H
2S = 0.06 atm or -1.56 atm
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Example- calculations involving Heterogeneous
equilibrium cont.
•But 0.06 atm is more correct,
•Hence P NH
3= 1.50+0.06 = 1.56 atm
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Example involving Kc, Kp & Kx
•At 21.5
o
C and total pressure of 0.787 atm, N
2O
4is
48.3% converted into NO
2. Calculate the value of Kp
and Kc for the reaction
•Ans:
•Reaction N
2O
4 (g)⇌2NO
2 (g)
•Initial
:a 0
•At t= t, a-x 2x
•Let a= 100, then x = 48.3
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Example involving Kc, Kp & Kx
•Therefore,
(a-x) = 100 - 48.3 = 51.7
Thus 2x = 48.3 x 2 = 96.6
Hence (n
T)aq. = 51.7 + 96.6 = 148.3
PNO
2= (96.6/148.3) X PT = 96.6X 0.0787/148.3
PN
2O
4= 51.7x 0.0787/148.3
Kp = (PNO
2)
2
= 96.6x0.0787/148.3
PN
2O
4 51.7 x0.0787/148.3
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Example involving Kc, Kp & Kx
•Kp = 9.58 x 10
-3
atm
•From th first principles,
•Kp = Kc(RT)
Δn
but Δn = 2-1 = 1
•Thus Kc = Kp/RT (R = 8.31 J/mol K = 0.0821
Latmmol
-1
K
-1
)
= 9.58x10
-3
/0.082 x (273 +21.5)
= 0.00396 molL
-1
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Shifts in chemical equilibrium
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Example: Le Chatelier’s principle in
biological systems
•Carbon monoxide poisoning. Carbon monoxide, a product of
incomplete combustion that is present in automotive exhaust
and cigarette smoke, binds to hemoglobin 200 times more tightly
than does O
2.
•This blocks the uptake and transport of oxygen by setting up a
competing equilibrium O
2-hemoglobin hemoglobin CO-
hemoglobin
•Air that contains as little as 0.1 percent carbon monoxide can tie
up about half of the hemoglobin binding sites, reducing the
amount of O
2reaching the tissues to fatal levels.
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Le Chatelier’s principle in biological
systems cont.
•Carbon monoxide poisoning is treated by administration of
pure O
2which promotes the shift of the above equilibrium to
the left.
•This can be made even more effective by placing the victim in a
hyperbaric chamber in which the pressure of O
2can be made
greater than 1 atm.
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Examples of chemical equilibrium
calculations
•The commercial production of hydrogen is carried out by treating
natural gas with steam at high temperatures and in the presence
of a catalyst (“steam reforming of methane”):
CH
4+ H
2O ⇌CH
3OH + H
2
•Given the following boiling points: CH
4(methane) = –161°C, H
2O
= 100°C, CH
3OH = 65°, H
2= –253°C, predict the effects of an
increase in the total pressure on this equilibrium at 50°, 75° and
120°C.
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Ans:

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Examples of chemical equilibrium
calculations cont.
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Comment: This net reaction describes the dissolution of limestone by acid; it is
responsible for the eroding effect of acid rain on buildings and statues. The first
reaction is “driven” by a second reaction having a large equilibrium constant.
From the standpoint of the LeChâtelier principle, the first reaction is “pulled to the
right” by the removal of carbonate by the hydrogen ion. “Coupled” reactions of this
type are widely encountered in all areas of chemistry, and especially in biochemistry,
in which a dozen or so reactions may be linked in this way.
Examples of chemical equilibrium
calculations cont.
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Examples of chemical equilibrium
calculations cont.
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Examples of chemical equilibrium
calculations cont.
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Examples of chemical equilibrium
calculations cont.
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Examples of chemical equilibrium
calculations cont.
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Examples of chemical equilibrium
calculations cont.
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Examples of chemical equilibrium
calculations cont.
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Examples of chemical equilibrium
calculations cont.
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Examples of chemical equilibrium
calculations cont.
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