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CHAPTER 12(15)Chemical EquilibriumChemical Equilibrium
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CONTENTS
12.1 The Concept of Equilibrium
12.2 The Equilibrium Constant
12.3 Heterogeneous Equilibria
12.4 Calculating Equilibrium Constants
12.5 Application of Equilibrium Constants
12.6 Le Chteliers Principle
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Learning Outcomes Student should be able to use concept of
equilibrium to calculate equilibrium constant and
equilibrium concentrations. Able to apply Le Chaterliers Principle to predict
the direction of a reaction hence apply it incommercial sense in increasing yield, reduce cost
etc. Differentiate homogenous and heterogeneous
equilibria, reaction quotient and Equilibriumconstant
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12.1 The Concept ofEquilibrium
Consider Colorless frozen N2O4.
At room temperature, it decomposes to brownNO2.
N2O4(g) 2NO2(g)
At some time, the color stop changing and wehave a mixture of N2O4 and NO2.
the concentration of all reactants and products
no longer change with time.N2O4 (g) 2 NO2 (g)
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12.1 The Concept ofEquilibrium
N2O4 (g) 2 NO2 (g) Chemical equilibrium is the point at which
the concentrations of all species are constant.(concentrations of reactants and products cease
to change with time)
Chemical equilibrium occurs opposing reactions are proceeding at equal rate.
Rate forward = Rate reverse
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N2O4 (g) - colorless
NO2 (g) - reddish brown
Concentration (color) remains constant at
equilibrium
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12.1 The Concept ofEquilibrium
Consider a simple reaction: A(g) B(g) We can write rate expressions for each reaction:
Forward reaction: A B Rate = k
f[A] k
f= rate constant (forward
reaction)
Reverse reaction: B A
Rate =k
r[B]k
r= rate constant (reversereaction)
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12.1 The Concept ofEquilibrium
For gaseous substances we can use theideal gas equation to convert betweenconcentration and pressure:
PV=nRT so M(Molarity) = (n/V) =(P/RT) For substances A and B:
[A] = (PA
/RT) and [B] = (PB
/RT)
Ratefwd = kf(PA/RT) and Raterev = kr(PB/RT)
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12.1 The Concept ofEquilibrium
Place some pure compound A into a closedcontainer.
As A reacts to form B, the partial pressure of Awill decrease and the partial pressure of B willincrease.
Expect forward reaction rate to slow and reversereaction rate to increase.
Eventually, we get to equilibrium where forward
and reverse rates are equal.
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Haber Process
N2(g) + 3H2(g) 2NH3(g)
The reaction is carried out
under conditions of highpressure and high
temperature. Equilibrium
can be established either by
starting with N2
and H2
or
by starting only with NH3.
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12.1 The Concept ofEquilibrium
At Equilibrium:
Rearrange:
a constant
This mixture is called an equilibrium mixture.
This is an example ofdynamic equilibrium.
RT
Pk
RT
Pk Br
Af =
( )
( )===
r
f
A
B
A
B
k
k
P
P
RTP
RTP
/
/
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12.2 The Equilibrium Constant,
K We can write an expression for the
relationship between the concentration ofthe reactants and products at equilibrium.
This expression is based on the law ofmass action.
For a general reaction:
aA + bB cC + dD Equilibrium expression:[ ] [ ]
[ ] [ ]ba
dc
BA
DCK=
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12.2 The Equilibrium Constant,
K When the reactions and products are all in
gaseous form, equilibrium constant isexpressed in terms of partial pressures of
the gases.
The value of Keq does not depend on the
initial concentrations of reactants andproducts.
( ) ( )
( ) ( )bBa
A
d
D
c
C
PP
PPK=
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12.2 The Equilibrium Constant,
K When the reactants and products are all
in aqueous form, Kc is used for
equilibrium constant.
c indicates the molar concentration(M)
When the reactants and products are all
in gaseous form, Kp is used forequilibrium constant.
p indicates the partial pressures of thegases.
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12.2.1 The Magnitude of
Equilibrium Constants, K K is the ratio of products to reactants.
The larger K the more products are present
at equilibrium. The smaller the K the more reactants are
present at equilibrium.
If K >> 1, then products dominate at
equilibrium and equilibrium lies to the right. K
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The range of equilibrium
constants, K A: reaches equilibrium, little
product.
K= 1/49
B:reaches equilibrium, nearly allproduct
K= 49/1
C: reaches equilibrium with
significant concentrations ofreactant & product.
K= 25/25 =1
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12.2.2 The Direction of the Chemical
Equation and K An equilibrium can be approached from
any direction:
Eg: N2O4 (g) 2 NO2 (g) The equilibrium constant for this reaction
(at 100C) is:
49.6)(
)(
42
2
2
==
ON
NOp
P
PK
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12.2.2 The Direction of the Chemical
Equation and Keq
For the reverse reaction:
2 NO2 (g) N2O4 (g) The equilibrium constant for this reaction (at
100C) is:
The equilibrium constant for a reaction in onedirection is the reciprocal of the equilibriumconstant of the reaction in the opposite direction.
154.0)(
)(2
2
42==
NO
ON
pP
PK
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12.3 Heterogeneous
Equilibria Equilibria in which all reactants and
products are present in the same phase arecalled homogeneous equilibria.
Equilibria in which one or more reactants orproducts are present in a different phase arecalled heterogeneous equilibria.
Consider:
CaCO3(s)
CaO(s) + CO2(g) Experimentally, the amount of CO2 does not
depend on the amounts of CaO and CaCO3.WHY?
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12.3 Heterogeneous
Equilibria The concentration of a pure solid or pure
liquid equals its density divided by itsmolar mass. Neither density nor molar
mass is a variable thus the concentrationsof solids and pure liquids are constant.
[ ]
[ ]3
2
CaCO
PCaOK
CO
=
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12.3 Heterogeneous
Equilibria CaO and CaCO3 are pure solids and have
constant concentrations.
Rearrange:
( )( )2constant
P1constantK 2CO
=
( )( )
( )2
21constant
2constantKK'
COp
CO
PKTherefore
P
=
==
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12.3 Heterogeneous
Equilibria If a pure solid or pure liquid is involved in
the heterogeneous equilibrium, itsconcentration is not included in the
equilibrium constant expression. The amount of CO2 formed (pressure of
CO2) will not depends on the amounts of
CaO and CaCO3 present.
However, they do participate in thereaction and must be present for anequilibrium to be established.
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12.3 Heterogeneous
Equilibria f
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12.4 Calculating EquilibriumConstants, K
Steps: Tabulate initial and equilibrium concentrations (or partial
pressures) for all species in the equilibrium.
If an initial and an equilibrium concentration is given for aspecies, calculate the change in concentration.
Use the coefficients in the balanced chemical equation to
calculate the changes in concentration of all species. Deduce the equilibrium concentrations of all species.
Use these equilibrium concentrations to calculate the value ofthe equilibrium constant.
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Example 1
A mixture of hydrogen and nitrogen in areaction vessel is allowed to attainequilibrium at 472C. The equilibrium
mixture of gases was analysed and foundto contain 7.38 atm H2, 2.46 atm N2, and
0.166 atm NH3. From these data, determine
Kp for
N2(g) + 3H2(g) 2NH3(g)
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Answer 1
N2(g) + 3H2(g) 2NH3(g)( )
( )( )( )
( ) ( )5
3
2
2
1079.2
38.746.2
166.0
322
3
=
=
=
HN
NH
p PP
P
K
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Example 2
Enough ammonia is dissolved in 5.00 L ofwater at 25C to produce a solution that is0.0124 M in ammonia. The solution is thenallowed to come to equilibrium. Analysis of
the equilibrium mixture shows that theconcentration of OH- is 4.64 10-4 M.Calculate Kc at 25C for the reaction:
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
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Answer 2
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)Initial
Change
Eqm
4.64 10-4M
0.0124 M 0 M 0 M
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Answer 2
NH3(aq) + H2O(l)
NH4+
(aq) + OH-
(aq)
Initial
Change
Eqm
4.64 10-4M
0.0124 M 0 M 0 M
+4.64 10-4M +4.64 10-4M-4.64 10-4M
0.0119 M 4.64 10-4M
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Answer 2
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
[ ][ ][ ]
( )( )
5
24
3
4
1081.10119.0
1064.4
+
=
=
=
c
c
K
NH
OHNH
K
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Exercise 1
Methanol is produced commercially by thecatalysed reaction of carbon monoxide and
hydrogen: CO(g) + 2H2(g) CH3OH(g). Anequilibrium mixture in a 2.00 L vessel is found to
contain 0.0406 mol CH3OH, 0.170 mol CO, and
0.302 mol H2 at 500 K. Calculate Kp at this
temperature.
(Answer = 6.22 10-3)
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12.5 Application of EquilibriumConstants, K
The equilibrium constant can be used:
(i) to predict the direction in which areaction mixture will proceed toachieve equilibrium.
(ii) to calculate the concentrations of
reactants and products onceequilibrium has been reached.
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12.5.1 To Predict the Direction ofReaction: Writing Reaction
Quotient Consider: aA + bB pP + qQ
Q is the called reaction quotient.
a,b,p and q are the coefficients in the
balanced chemical equation. The reaction quotient will equal the
equilibrium constant, K, only if the systemis at equilibrium: Q = K at equilibrium.
[ ] [ ]
[ ] [ ]ba
qp
BA
QPQ =
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12.5.1 To Predict the Direction ofReaction: Example
N2(g) + 3H2(g) 2NH3(g) A mixture of 2.00 mol H2; 1.00 mol of N2
and 2.00 mol of NH3 in a 1.00L container
is at 472C. Kp for the reaction at this
temperature is 2.75
10
-5
.
Will N2 and H2 react to form more NH3?
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Answer
N2(g) + 3H2(g) 2NH3(g) First, write the equilibrium-constant expression, Kp (or Q):
To calculate the partial pressure of each gases, use PV=nRT(PH2 = 122 atm, PN2 = 61.2 atm, PNH3 = 122 atm)
( )( )( )32
22
3
HN
NH
PPPQ =
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Answer
Inserting the values into the reaction quotient, Q= 1.34 10-4.
Compare with Kp value, Kp =2.75 10-5.
Therefore the quotient will need to decrease for
the system to achieve equilibrium. Can be achieved if decrease the pressure of NH3 or
increase the pressures of H2 and N2.
Thus the reaction proceeds toward equilibrium by
producing H2 and N2 from NH3, i.e. the reactionproceeds from right to left.
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12.5.1 To Predict the Direction ofReaction
If Q < K,.. moreproduct forms
If Q > K,. more
reactant forms
If Q = K, no net
change
]tan[Re
][Pr
]tan[Re
][Pr
tsac
oductsQ
tsac
oductsKeq
=
=
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12.5.2 Calculating EquilibriumConcentrations
The same step used to calculate theequilibrium constants.
Generally, we do not have a number for the
change in the concentration. Therefore we need to assume that x mol/L of
the species is formed (or used).
The equilibrium concentrations are given asalgebraic expressions.
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Example
A 1.000L flask is filled with 1.00 mol of H2and 2.00 mol of I2 at 448C. Kp for the
reaction at 448C is 50.5.
H2(g) + I2(g) 2HI(g)Question:
What are the partial pressures of H2, I2
and HI in the flask at equilibrium?
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First, calculate the initial partial pressures of H2
and I2 using PV=nRT. (PH2 = 59.17 atm, PI2 =118.4 atm).
Second, construct a table that consists of initial
partial pressures.
Initial 59.17 atm118.4 atm 0 atm
Change
Equilibrium
H2(g) + I2(g) 2HI(g)
Answer
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From the equation, 1 mol of H2
+ 1 mol of I2
willproduce 2 moles of HI.
The partial pressures of H2and I2 will decreaseand the partial pressure of HI will increase.
Initial (atm) 59.17 118.4 0
Change (atm) - x -x +2xEquilibrium(atm)
59.17 x 118.4 x 2x
H2(g) + I2(g) 2HI(g)
Answer
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Substitute the values into equilibrium-constantexpression, Kp (value is 50.5).
Answer
( )
( )( )
( )
( ) ( )5.50
4.11817.59
222
22
=
==
xx
x
PP
PK
IH
HIp
( )
3.556.137
2
4
01054.31097.85.46
1001.76.1775.504
2
532
322
orx
a
acbbxusethen
xx
xxx
=
=
=+
+=
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Substitute x values into the expressions forequilibrium partial pressures.
Using x = 137.6 will lead to negative partialpressures of H2 and I2 which are not chemically
meaningful. So x = 55.3. PH2 = 59.17 x = 3.87 atm, PI2 = 118.4 x = 63.1
atm, PHI = 2x = 110.6 atm.
Check:
Answer
( )
( )( )( )
( ) ( )1.50
1.6387.3
6.11022
22
===
IH
HIp
PP
PK
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12.6 Le Chteliers Principle
If a system at equilibrium is disturbed by a change in
temperature,
Pressure (by changing the volume),
the concentration of one of the components (adding orremoving)
the system will shift its equilibrium position to reducethe effect of the disturbance.
Changes in concentration or pressure cause shifts inequilibrium but K remains constant.
Change in temperature increases or decreases the Kvalue.
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12.6.1 Change in Reactant or
Product ConcentrationsA system at equilibrium:
- if we add a substance (a reactant or a product)
the reaction will shift reestablish equilibriumby consuming part of the added substance.
- removal of a substance reestablish byforming more of the substance.
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N2(g) + 3H2(g)
2NH3(g)
i) Addition of H2 shift to reduce H2 conc.
More NH3 will form N2 will be reduced
ii) Addition of NH3 shift to reduce NH3 conc.
added NH3 will decomposeto form N2 and H2
iii) Removal of NH3 shift from left to right -
formation of more NH3.
Concen
tra
tio
n
Time
InitialEquilibrium
H2
NH3
N2
H2 added at
this time
Equilibriumreestablished
ExampleExample
::
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12.6.2 Change in Pressure
Pressure will change if volume of the containerchanges.
Volume decreases pressure increases
equilibrium will shift towards the less number ofgas molecules
Volume increases pressure decreases equilibrium shift towards the more number ofgas molecules
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12.6.2 Change in Pressure
E.g: PCl5(g) PCl3(g) + Cl2(g) No. of gas molecules is more on RHS
P , equilibrium shift to LHS producing morePCl5 (g)
P , equilibrium shift to RHS producingPCl3(g) + Cl2(g)
No change will occur if we increase the totalpressure by the addition of a gas that is notinvolved in the reaction.
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Changes in concentration or total pressure
cause shifts in equilibrium without changingthe value of Keq .
Applying Le Chteliers principle:
Consider heat as a chemical reagent.
Endothermic reaction (heat as a reactant)
A + B + Heat CExothermic reaction (heat as a product)
A + B C + Heat
12.6.3 Change in Temperature
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When the temperature is increased, the
equilibrium shifts in the direction thatabsorbs heat.
In an endothermic reaction, H>0:
- increasing temperature or adding heat,the equilibrium shifts to the right (product)and K increases.
In an exothermic reaction,
H
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Endothermic
A + B + HeatC
ExothermicA + B C +
Heat
IncreaseIncrease
Decrease
Reaction KTemperature
Decrease
Decrease
Increase
Increase
Decrease
12.6.3 Change in Temperature
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Example
Consider the following equilibrium.
N2O4(g) 2NO2(g) H = +58.0 kJ.Which direction will the equilibrium shift if:
a) N2O4 is added
b) NO2 isremoved
c) N2 is added
d) the volume is increased.
e) the temperature is decreased
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a) N2O4 is added
Concentration of N2O4 increases, reaction
shifts to the right to decrease back N2O4.
b) NO2 isremoved
Concentration of NO2decreases, reactionshifts to the right to increase back NO2.
c) N2 is added
No effect, no shift in equilibrium position
N2O4(g) 2NO2(g) H = +58.0 kJ( )
( )422
ON
2
p P
P
K
NO=
Answer
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d) The volume is increased.
V P , the equilibrium shifts in the direction toincrease P by producing more gas molecules. It
shifts to the right, 1 mol N2O4 moleculesgenerates 2 mol NO2 molecules
e) the temperature is decreased.
The reaction is endothermic, heat is a reactant.Equilibrium shifts to the left, forming more N2O4,decreasing Kp.
N2
O4
(g) 2NO2
(g) H = +58.0
kJ
Answer
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12.6.4 The Effect of Catalyst
A catalyst lowers the activation energybarrier for the reaction.
Therefore, a catalyst will decrease thetime taken to reach equilibrium.
The catalyst has no effect on theequilibrium position.
A catalyst does not effect thecomposition of the equilibrium mixture.
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Haber synthesis of ammoniaAmmonia is used for production of fertilizer,
explosives and polymers.
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N2 (g) + 3H2 (g) 2NH3 (g) H = -91.8 kJ/mol
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Haber Process
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ExampleMethanol (CH
3OH) is manufactured by reaction of carbon
monoxide with hydrogen in the presence of a ZnO/Cr2O3catalyst
CO (g) + 2H2
(g) CH3OH (g) Ho = - 91 kJ
Does the amount of methanol (CH3OH) INCREASE,
DECREASE orREMAIN THE SAME when an equilibriummixture of reactants and product is subjected to the following
changes? The temperature is increased.
CO is added.
The volume of the vessel is decreased.
The catalyst is removed
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Answer
Decrease. Because the reaction isexothermic. Reaction goes to the left(endothermic) to reduce back the T, Kpdecreases, so concentration decreases.
Increase. Because CO is a reactant,
increasing CO, shift equilibrium to right. Increase. When the volume increases,
pressure decreases. The reaction shifts to thesides with fewer molecules i.e. shifts to the
right. Remains the same. Additional or removal of acatalyst does not affect the equilibriumcomposition.
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END of CHAPTER 12
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