chemistry form 6 sem 1 06
TRANSCRIPT
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CHEMISTRY LOWER 6SEMESTER 1
HAPTER 6CHEMICAL EQUILIBRIA
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6.0 Introduction
Among the reactionsdiscussed so far, most of
them are irreversible
reactions. For example, the
dissociation of hydrogenperoxide, H2O2 is an
irreversible process.
Equation :
2 H2O2 (l) 2 H2O (l) + O2 (g)
So, if the graph of
concentration of hydrogen
peroxide and hydrogen gas
against time is plotted
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6.1 Reversible reaction and dynamic equilibria
Actually, the example cited above, is one of a few chemical
reactions proceed in only one direction, Most of the chemicalreaction occur around us are usually reversible.
At the start of a reversible process, the reaction proceeds toward
the formation of products. As soon as some product molecules are
formed, the reverse process begins to take place and reactant
molecules are formed from product molecules.
Chemical equilibrium is achieved when the rates of the
forward and reverse reactions are equal and the concentrations ofthe reactants and products remain constant.
An example of a reversible reaction, is the dissociation of
dinitrogen tetraoxide, N2O4, into nitrogen dioxide, NO2, where
the chemical equation can be written asN2O4 (g) 2 NO2 (g)
Diagram below shows 3 different graphs of changes in the
concentration of N2O
4and NO
2. In each case, equilibrium is
established to the right of the vertical line
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a) Initially only NO2present b) Initially only N2O4presentc) Initially a mixture of NO2and N2O4 is present.
Observation : Observation : Observation :Since there are only NO2, reaction
will move to backward to form
some N2O4. at the same time
concentration of NO2 decrrease
Since there are only N2O4,
reaction will move to forward to
dissociate some N2O4 and form
NO2. At the same time
concentration of N2O4 decrrease
Depending on the situation,
concentration of N2O4 and NO2may varies. From the graph, the
concentration of NO2 decrease as
it form more N2O4.
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6.2 Homogeneous Equilibria and Equilibrium
Constant
The term homogeneous equilibrium applies to reactions in
which all reacting species are in the same phase. Example of
the dissociation of N2O4 above is one of homogeneous gas-
phase equilibrium since both reactant and product are gases.
N2O4 (g) 2 NO2 (g)
As mentioned earlier, a reversible reaction is to say achieved a
chemical equilibrium, when the rate of forward reaction is the
same as rate of backward reaction. Consider the following equilibrium reaction.
aA + bB cC + dD
Since a reversible reaction only take one single steps, hence
the stoichiometry coefficient shall be the order of reaction in
which it can be expressed as the following
rate of forward reaction ;
rate of backward reaction ;
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When reaction achieved equilibrium
rate of forward reaction = rate of backward reaction
ba
dc
BA
DC
k
k
][][
][][
1 =
So, k [A]a[B]b = k1 [C]c[D]d
tconsk
Since tan1
=
1
=kkKcOr
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6.2.1 Equilibrium constant of concentration, Kc
According to equilibrium law (mass law action) states that,
at a given temperature, when the amount of
concentration of reactants in a chemical reaction, is
divided by the amount of concentration of products is a
constant where the concentration of each substance israised to the power equal to coefficient in the balanced
reversible chemical equation
,
pA (aq) + qB (aq) rC (aq) + sD (aq)
whereA and B are reactants ; C and D are products ;p , q ,
r , s are stoichiometry coefficient of reaction
Kc of the reaction above can be written as
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Note that the [X] is used to express concentration for both
products and reactants. Hence, in terms of equilibrium
constant, KC, is known as the equilibrium constant ofconcentration. The c stands for concentration. Since the
unit of concentration for products / reactants is mol dm-3.
So, the unit of KC
depend on the total concentration of
products over the reactants.
Kc can also be used to express substances in gaseous
hase
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Chemical equation Equilibrium constant, K C
a) CH3COOH (aq) + C3H7OH (aq) CH3COOC3H7 (aq) + H2O
b) 2 Fe3+ (aq) + 2 I- (aq) 2 Fe2+ (aq) + I2 (aq)
]OHHC][COOHCH[]OH][HCOOCCH[K
733
2733c=
2232
22
c]I[]Fe[]I[]Fe[K
+
+
=
4
c) 2 N2O5 (g) 4 NO2 (g) + O2 (g)
d) AgNO3 (aq) Ag+
(aq) + NO3-
(aq)
e) H2SO4 (aq) + 8 HI (aq)
4 I2 (g) + H2S (g) + 4 H2O (l)
2
52
22c]ON[K =
]AgNO[
]NO][Ag[
K3
3
c
+
=
8
42
4
2
4
22c
]HI][SOH[
]OH[]I][SH[K =
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Example 1 : Consider the following equation
3 A + 2 B 2 C + D
Given the concentration of A, B, C and D are
1.20 mol dm-3 ; 0.920 mol dm-3 ; 1.10 mol dm-3 ; 1.30 mol dm-3 respectively. Calculate the
KC for the reaction above
Example 2 : Given the KC for following
equation 2 E + F G + 2 H
is 4.30. Given the concentration of E, F and G
are 0.35 mol dm-3 ; 0.46 mol dm-3 ; 0.30 moldm-3 respectively. Calculate the concentration
of H in the reaction above.
Example 3 :When 60.0 g of sulphur dioxide,
SO react with 46.0 of ox en to form 58.7
Example 4 : An equilibrium mixture contains
1.25 mol of h dro en, 2.00 mol of bromine
KC = [C]2[D] / [A]3[B]2
Kc = (1.10)
2
(1.30) / (1.20)
3
(0.920)
2
KC = 1.08 mol-2 dm6
KC = [H]2[G] / [F] [E]2
4.30 = (H)2(0.30) / (0.46)(0.35)2[H] = 0.90 mol dm-3
g of sulphur trioxide in 1 dm3
vessel. Calculatethe KC of system.[RAM S = 32 , O = 16]
and 0.50 hydrogen bromide in a 4.0 dm3
ofvessel. Calculate the KC of the system
H2 + Br2 2 HBr2 SO2 + O2 2 SO3
mol = 60.0 / 64 46.0 / 32 58.7 / 80
= 0.9375 = 1.438 = 0.7338
KC = [SO3]2 / [SO 2]
2[O2]
Kc = (0.7338/1)2 / (0.9375/1)2(1.438/1)
KC = 0.426 mol-1 dm3
KC = [HBr]2 / [H2] [Br2]
Kc = (0.50/4.0)2 / (1.25/4.0) (2.00 /4.0)
KC = 0.100
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6.2.2 Equilibrium constant of partial pressure, Kp
Different from Kc, which can be use to express the
equilibrium constant of both aqueous and gaseous phase,
Kp (equilibrium constant of partial pressure) is specifically
an equilibrium constant used to expressed for gaseous
substance. For example, in the reaction of production of ammonia,
N2 (g) + 3 H2 (g) 2 NH3 (g)
Equilibrium constant of concentration,
Kc
Equilibrium constant of partial pressure,
Kp
We can expressed equilibrium constant as both Kc and Kpas followIn terms of KC, it is written as
3
22
2
3c
]H][N[
]NH[K =
3
HN
2
NH
P)P)(P(
)P(K
22
3=
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a) H2 (g) + I2 (g) 2 HI (g)
b) 2 SO2 (g) + O2 (g) 2 SO3 (g)
c) PCl5 (g) PCl3 (g) + Cl2 (g)
)P)(P(
)P(K
22 IH
2
HIP =
)P()P(
)P(K
22
3
O
2
SO
2
SO
P =
)P)(P(23 ClPCl
d) 2 N2O5 (g) 4 NO2 (g) + O2 (g)
e) N2O4 (g) 2 NO2 (g)
)P( 5PClP
2ON
O
4
NO
P )P(
)P()P(K
52
22=
)P(
)P(K
42
2
ON
2
NO
P=
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Example 5 : The conversion of oxygen to
ozone at high altitude can be represented
by the following equation.3 O2 (g) 2 O3 (g).
In an equilibrium mixture, the pressure of
oxygen gas and ozone are 0.120 atm and
0.200 atm respectively. Calculate the KPof reaction.
Example 6 : A mixture of hydrogen and
nitrogen gas is made up in the ratio of 3 :
1 by volume, and is left to attainequilibrium at 1.01 kPa. The temperature
is kept at 650 K. At constant temperature
and pressure, the equilibrium mixture
contains 20% of ammonia. Calculate the
KP at equilibrium.
3
2
O
PP
)P(K 3= N2 + 3 H2 2 NH3
KP = (0.200)2/ (0.120)3
KP = 23.1 atm-1
2
At eq : 1 : 3 = 80% 20%20% 60%
Since Pa = xa . Ptot(0.20)(1.01) (0.60)(1.01) (0.20)(1.01)
Pgas : 0.202 kPa 0.606 kPa 0.202 kPa
Kp = (PNH3)2 / (PN2) (PH2)
3
KP = (0.202)2/ (0.202) (0.606)3
KP = 0.908 kPa-2
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Example 7 : Chlorine and carbon monoxide react to form phosgene,
COCl2, according to the equation.
Cl2 (g) + CO (g) COCl2 (g).Chlorine gas was placed in a reaction vessel connected to a manometer and
its pressure was found to be 0.69 atm. Carbon monoxide, with an initial
pressure of 0.48 atm was then allowed to react with Cl2 in the vessel. At
equilibrium, the total pressure was found to be 0.82 atm. Calculate KP.
CO (g) + Cl2 (g) COCl2 (g)
Initial : 0.48 atm 0.69 atm 0 atm
At equilibrium : (0.48 x) (0.69 x) xPtotal = 0.82atm
So, (0.48 x) + (0.69 x) + x = 0.82atm
x = 0.35 atm
At equilibrium : 0.13 atm 0.34 atm 0.35 atmKP = (PCOCl2) / ( PCl2) (PCO)
KP = (0.35) / (0.13)(0.34)
KP = 7.9 atm-1
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6.2.3 Relationship between KC and KP.
In general,KC is not equal toKP, because the partial
pressures of reactants and products are not equal to their
concentrations expressed in moles per liter. A simple
relationship betweenKP andKc can be derived as follows.
Let us consider the following equilibrium in the gas phasewA (g) + xB (g) y C (g) + z D (g)
Equilibrium constant of partial Equilibrium constant of
Expressing
Equilibrium
constant
pressure, Kp concentration, Kc
Kp
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Assuming ideal gas behaviour, where PV = nRT, where V is
the volume of container. Rearrange the equation
@ P = [conc]RT
For each partial pressure of the gas
Substituting the partial pressure of each gas into the
equilibrium constant of partial pressure
PA = [A] RT PB = [B] RT PC = [C] RT PD = [D] RT
Factorised RT from equation will make the equation
become :
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The general equation relating KC and KP can be expressed
KP = KC (RT)n
where n = (total mol of gaseous products) - (total mol of
gaseous reactants)
From the equation, there may be a possibility where KP =
KC is when the total amount of mol of product is equal to
the total amount of mol of reactant in a balanced chemicalequation.
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6.3 Heterogeneous equilibria
Heterogeneous equilibrium results from a reversible
reaction involving reactants and products that are indifferent phases
Consider a heterogeneous equilibrium system of calcium
carbonate, when it is heated in a vesselCaCO3 (s) CaO (s) + CO2 (g)
In the heterogeneous system, the equilibirum constant can
be expressed as
*KC' is used in this case to distinguish it from the final form
of equilibrium constant derived shortly
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The concentration of a solid, like its density, is an intensive
property and does not depend on how much of the substance is
present. For example, the molar concentration of calciumcarbonate (density: 2.83 g/cm3) at 20C is the same, whether we
have 1 gram or 1 ton of the metal
[CaCO3] =
For this reason, the terms [CaCO3] and [CaO] are themselves
constants and can be combined with the equilibrium constant.
We can rearran e and sim lif e uation above b writin
where KC, the new equilibrium constant, is conveniently
expressed in terms of a single concentration, that of CO2. Notethat the value of KC does not depend on how much CaCO3 and
CaO are present, as long as some of each is present at
equilibrium
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Another example of heterogeneous equilibria can be
explained using hydrolysis of ester.
CH3COOCH3 (aq) + H2O (l) CH3COOH (aq) + CH3OH (aq)
In the heterogeneous system, the equilibirum constant can
be expressed as
Concentration of a liquid, similar to solid, is another
intensive property and does not depend too, on ow much thesubstance presence. In 1 liter (1 dm3) of pure water, the
molarity can be calculated accordingly
[H2O] =
*Similar to the example above, Kc does not depend on how
much the concentration of water used during hydrolysis as
long as water is present in equilibrium
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From both of example above, in expressing equilibrium
constant of a heterogeneous equilibria, pure solid and pure
liquid can be ignored from the expression of the equilibriumconstant
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Reversible reaction K C KP
1. CO2 (g) + C (s) 2 CO (g) ]CO[
]CO[K
2
2
c=
)P(
)P(K
2CO
2
COP =
2. 3 Fe (s) + 4 H2O (g)
Fe3O4 (s) + 4 H2 (g)
3. Ca(OH)2 (s) + 2 NH4Cl (aq)
CaCl2 (aq) + 2 NH3 (g) + H2O (l)
4
2
42
c]OH[
]H[K =
2
NHP )P(K 3=
2
4
2
32c
]ClNH[]NH][CaCl[K =
4
OH
4
HP
)P()P(K
2
2=
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9.3.1 Dissociation reactions
Dissociation reaction ~ ..
.
For example,
N2O4 (g)
2 N2O5 (g)2 HI (g)
PCl5 (g)
Reaction where large molecule decomposed
to form smaller molecules
2 NO2 (g)
4 NO2 (g) + O2 (g)H2 (g) + I2 (g)
PCl3 (g) + Cl2 (g)
Most dissociation reactions are reversible reaction, meaningthat the dissociation process proceed until a certain extent
before reaching equilibrium.
The extent of dissociation of the compound is measured by
If = 1 it indicates molecules are ... dissociated. If
< 1, molecules are dissociated.
Example, if = 0.20, % of molecules are dissociated.
degree of dissociationcompletely
partially
20
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The following example shows the method how is usually
Kc is calculated when a system with single reactant is
dissociated into its component until it achievedequilibrium.
PCl5 (g) PCl3 (g) + Cl2 (g)
Initial mol 1 mol 0 0
degree of dissociation, - + +
At equilibrium 1 -
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Example 8 : At a temperature of 398oC, hydrogen iodide (HI)has a
degree of dissociation of 25.0%
a) Calculate the KC of the above reaction at 398o
C.a) 2 HI (g) H2 (g) + I2 (g)
Initial : 1.00 0.00 0.00
When 25% dissociate : 0.25 + 0.25 / 2 + 0.25 / 2
At equilibrium 0.75 0.125 0.125
2
22C
]HI[
]I][H[K = 028.0
)(
))((275.0
125.0125.0
==V
VVCK
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Example 9 : At 300 K and 1.0 atm, dinitrogen tetraoxide, N2O4, is 20%
dissociated.
a) Calculate the equilibrium constant of pressure for the reactionb) Calculate the degree of dissociation at 300 K and the total pressure of
0.20 atm.
c) Calculate the pressure that must be applied so that the degree of
dissociation is lowered to 15%a) N2O4 (g) 2 NO2 (g)
Initial 1.00 mol 0 mol
en ssoc a e . + .
At equilibrium 0.80 mol 0.40 mol Total mol = 1.2 mol
atm3
2atm00.1
20.1
80.0P
42ON == atm
3
1atm00.1
20.1
40.0P
2NO ==
)P(
)P(K
42
2
ON
2
NO
P= atmKP 17.0)(
)(
32
2
31
==
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Calculate the degree of dissociation at 300 K and the total pressure of
0.20 atm.
b) N2O4 (g) 2 NO2 (g)Initial 1.00 mol 0 mol
When dissociated + 2
At equilibrium 1.00 + 2 Total mol = 1.00 +
atm200.01
1P
42ON
+
= atm20.0
1
2P
2NO +
=
= 0.416
atm167.0)P(
)P(K
42
2
ON
2NO
P ==
)atm200.01
1(
)atm20.01(K
2
P
+
+
=
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Calculate the pressure that must be applied so that the degree of
dissociation is lowered to 15%
c) N2O4 (g) 2 NO2 (g)Initial 1.00 mol 0 mol
When dissociated 0.15 + 2 (0.15)
At equilibrium 0.85 mol 0.300 mol Total mol = 1.15
atmP15.1
85.0P
42ON = atmP
15.1
30.0P
2NO =
P = 1.8 atm
atmPPK
ON
NO
P 17.0)()(
42
2
2 == atmatmP
atmPKP 17.0
)15.1
85.0(
)15.1
.(
2
=
=
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Example 10 : At 500oC, the KP for the reaction
N2 (g) + 3 H2 (g) 2 NH3 (g)
is 1.5 x 10-5 atm-2. Calculate the pressure that has to be applied to thesystem containing a mixture of N2 and H2 in the molar ratio 1 : 3 so
that half of the reactants are converted to ammonia at equilibrium
N2 (g) + 3 H2 (g) 2 NH3 (g)
Initial 1.00 mol 3.00 mol 0 mol
When dissociated 0.50 3 (0.50) + 2 (0.50)
At equilibrium 0.50 mol 1.50 mol 1.00 mol
Total mol = 0.50 mol + 1.50 mol + 1.00 mol = 3.00 mol
P = 5.96 x 10-3 atm
atmP00.3
50.0P
2N = atmP
00.3
50.1P
2H = atmP
00.3
00.1P
3NH =
25
3
HN
2
NH
P atm10x5.1)P)(P(
)P(
K22
3
== 25
3
21
61
2
3
1
P atm10x5.1)P)(P(
)P(K
==
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Example 11 : Consider the heterogeneous equilibrium
MO (s) + H2 (g) M (g) + H2O (g)
At ToC and pressure of 120 kPa, 1.5 mol of metal oxide, MO and 1.0mol of H2 reacted to produce 0.012 mol of M and 0.012 mol H2O at
equilibrium. Calculate a value for the equilibrium constant, KP at ToC.
MO (s) + H2 (g) M (g) + H2O (g)
Initial : 1.5 mol 1.0 mol 0 mol 0 mol
Mol dissociate : - 0.012 0.012 + 0.012 mol + 0.012 mol
At equilibrium : 1.488 mol 0.988 mol 0.012 mol 0.012 mol
total mol = 0.988 mol + 0.012 mol + 0.012 mol= 1.012 mol
kPa117kPa120012.1
988.0P
2H == kPa42.1kPa120
012.1
012.0PM ==
kPa42.1kPa120012.1012.0P OH2 ==
kPa117
kPa42.1xkPa42.1
)P(
)P)(P(K
2
2
H
MOH
P == KP
= 0.017 kPa
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9.4 Position of Equilibrium
In general, the equilibrium constant helps us to predict the direction
in which a reaction mixture will proceed to achieve equilibrium and tocalculate the concentrations of reactants and products once
equilibrium has been reached. Since
For example, The equilibrium constantKc for the formation of
hydrogen iodide from molecular hydrogen and molecular iodine in the
as hase
]reactatns[
]products[KC
H2 (g) + I2 (g) 2HI(g) KC = 54.3 at 430C. Suppose that in a certain experiment we place 0.243 mole of H2, 0.146
mole of I2, and 1.98 moles of HI all in a 1.00 L container at 430C.
Inserting the starting concentrations in the equilibrium constant
expression, we write
111)146.0)(243.0(
)98.1(
][][
][ 2
0202
20 === cCC QQIH
HIQ
Reaction quotient (Qc) is a value obtained by
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Reaction quotient (Qc) is a value obtained by
substituting the initial concentrations into the equilibrium
constant expression in order to determine position ofequilibrium compare to equilibrium constant (KC) .
When value of Qc is large, we shall have more products
over reactants while, when value of Qc is small, we shall
have more reactants over products.
To determine the direction in which the net reaction will
roceed to achieve e uilibrium we com are the values of
QC and KC.
QC < KC
The ratio of initial concentrations of products to
reactants is too small. To reach equilibrium,
reactants must be converted to products. Thesystem proceeds from left to right (consuming
reactants, forming products) to reach equilibrium.
At this moment, we described it as equilibrium
shift to right
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QC =
KC
The initial concentrations are equilibrium
concentrations. The system is at equilibrium.
At this moment, we described it as
equilibrium remain unchanged
QC >
KC
The ratio of initial concentrations of products
to reactants is too large. To reach equilibrium,
products must be converted to reactants. The
system proceeds from right to left (consumingproducts, forming reactants) to reach
equilibrium. At this moment, we described it as
equilibrium shift to left
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6.6 Factors that Affect the Position of Equilibrium
Once a system has achieved equilibrium, it is possible to change
the chemical composition of equilibrium mixture by changing the
condition of the reaction. In other words, position of equilibrium
can be altered.
2 important factors which must be considered in chemicalequilibria are
the position of equilibrium
e ra e a w c equ r um s ac eve .
The factors which affect the position of equilibrium can be
deduced using Le Chatelier's principle.
Le Chateliers Principle stated that if an external stress is
applied to a system at equilibrium, the system adjusts in such away that the stress is partially offset as the system reaches a new
equilibrium position
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The factors that affected the position of equilibrium
are
the concentration of reactants and products
the temperature of the experiment
t e pressure n react ons nvo v ng gases
9 4 1 The Effect of Concentration on Equilibrium
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9.4.1 The Effect of Concentration on Equilibrium.
If the concentration of the reactants are increased in a reaction at
equilibrium, the equilibrium will .., formingmore By this way, some of the reactants will .....,
and vice versa. The equilibrium constant will remain ..
Changing the concentration of one or more of the reactants or products
of a reaction at equilibrium disturbed the original equilibrium, but the
value of .. remain the same and the system
changes until equilibrium is re-established.
Example, in the reaction of the formation of nitrogen dioxide
shift its position to rightproducts be decreased
the same
equilibrium constant
2 NO (g) + O2 (g) 2 NO2 (g) Equilibrium is established at one point.
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[NO]
[O2]
[NO2]
Then, the equilibrium is disturbed by adding nitrogen monoxide gas
into the gas vessel. When this happened, concentration of NO . According to Le Chateliers Principle, equilibrium will shift to
to decrease the concentration of
At the same time, concentration of NO2 : .
concentration of O2 :
increaserightNO
increasedecrease
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Equation Substance alteredEquilibrium
positionConcentration of other substances
1. H2 + I2 2 HI [H2] increased [I2] [HI]
2. 2 Fe3+ + 2 I- 2 Fe2+ + I2 [Fe2+] decreased [Fe3+] [I2]
Shift to
rightdecrease increase
Shift to
rightdecrease increase
3. N2 + 3 H2 2 NH3 [H2] decrease [N2] [NH3]
4. 2 SO2 + O2 2 SO3 [SO3] increased [SO2] [O2]
5. N2O4 2 NO2 [NO2] decreased [N2O4]
Shift toleft
increase decrease
Shift to
leftincrease increase
Shift to
rightdecrease
9 4 2 The effect of Pressure on a system in equilibrium
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9.4.2 The effect of Pressure on a system in equilibrium
Change in pressure of system will only affect the position
of equilibrium of reaction involving ..
Result in a change in the of both forward and
backward reactions.
Result in a change in the of equilibrium mixture Will not change . or . at constant
temperature.
gas
pressure
positionrate, k Equilibrium, K C
Le Chateliers principle stated that, When the pressure of a gaseous system increased, equilibrium will
shift to the position with .. total amount of gaseous mole
When the pressure of a gaseous system decreased, equilibrium will
shift to the position with .... total amount of gaseous mole Equilibrium constant, KC and KP will remain
less
moreconstant
Consider the reaction ; N O (g) 2 NO (g)
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Consider the reaction ; N2O4 (g) 2 NO2 (g).
The pressure in the vessel increased
According to Le Chateliers principle, the system will react to(i) decrease the pressure of the system
(ii) decrease the concentration of all species involved.
The results are :
Position will shift to the position with the least total number of
moles. From the equation above, equilibrium shift to
Concentration of N O : where are NO :
left
increase decrease
KC and KP remain constant
F t E ilib i Ch f t ti f
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EquationFactor
altered
Equilibriu
m position
Changes of concentration of
substances
2 SO2 (g) + O2 (g)
2 SO3 (g)P
[SO2] [SO3]
4 NH3 (g) + 5 O2 (g)
4 NO (g) + 6 H2O (g)P
[NH3] [NO]
Shift to
right decrease increase
Shift to
right decrease increase
CO (g) + SO3 (g) CO2 (g) + SO2 (g)
P [CO] [SO3]
CaCO3 (s) CaO (s) + CO2(g) P
[CO2]
2 N2O5 (g) 4 NO2 (g) + O2(g) V [N2O5] [NO2]
Nochange No change No change
Shift to
left decrease
Shift to
right decrease increase
Adding a noble gas (inert gas) on an equilibrium.
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Adding a noble gas (inert gas) on an equilibrium.
(i) Adding an inert gas into the system at a constant
pressure and the volume of vessel remain.. but .. the partial pressure
of the gases involved in the system.
(ii) This will cause the equilibrium to shift to the directionwith the .. total number of moles.
Factor Equilibriu
Changes of concentration of
unchanged lowered
more
altered m position substances
PCl5 (g) PCl3 (g) + Cl2 (g) + Ne[PCl5] [PCl3]
2 H2 (g) + O2 (g) 2 H2O (g) + Ar[H2] [H2O]
2 HI (g) H2
(g) + I2
(g) + Kr[I2] [HI]
Shift to
right decrease increase
Shift toleft increase decrease
No
change remain remain
9.4.3 Effect of Temperature on an Equilibrium
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p q
In a reversible reaction, the changes in temperature will affect
. . . .
In kinetic chemistry earlier, we used . equation to
explain the effect on temperature toward the rate of reaction
Rate of forward reactionRate of backward reaction
k, rate constantK, Equilibrium constant
Arrhenius
RT
EA
eAk
= n equ r um, t e e ect o temperature towar an equ r um system
can be explained by using equation
At different temperature, Vant Hoff equation can be derived to become
=
211
2 11lnTTR
H
K
K
cRT
HKln +
=
Vant Hoff
K = equilibrium constant, KCH = enthalpy change
R = gas constant, 8.31 J mol-1 K-1
T = temperature in Kelvin, K
Exothermic reaction Endothermic reaction
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KC
T
KC
T
From the graph, the equilibrium
constant ...... withtemperature.
From the graph, the equilibrium
constant ......withtemperature.
decrease increase
According to Le Chateliers principle, the changes of temperature will affect
the position of equilibrium by
o When temperature increased, equilibrium will shift to the position wherethe system is countered by .. the temperature, which is toward
the position of . process
o When temperature decreased, equilibrium will shift to the position where
the system is countered by .. the temperature, which is toward
the position of . Process
decreaseendothermic
increase
exothermic
Factor Equilibriu Rate Equilibriu
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Equation ProcessFactor
altered
Equilibriu
m position
Rate
constant
Equilibriu
m constant
2 HI (g) H2 (g) + I2 (g) Endothermic T
2 H2O2 (l) 2 H2O (l) + O2 (g) Exothermic T
Shift to
right
increase increase
Shift to
leftincrease decrease
CH4(g) + 2O2(g)
CO2 (g) + 2H2OExothermic T
2 NO2 (g) N2O4 (g) Endothermic T
4 NO2 (g) + O2 (g)
2 N2O5 (g) Exothermic T
Shift to
rightdecrease increase
Shift toleft decrease decrease
Shift to
left
increase decrease
Example 12 At a specific temperature T, a 1.00 dm3 flask was found to
t i ilib i i t f 0 500 l f l h di id 0 100 l f
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contain an equilibrium mixture of 0.500 mol of sulphur dioxide, 0.100 mol of
oxygen and 4.60 mol of sulphur trioxide.
a) Calculate the equilibrium constant, KC of the reaction.
b) Calculate the number of moles of oxygen that must be forced into the mixture
to increase the number of moles of sulphur trioxide at equilibrium to 4.70 mol.
2 SO2 (g) + O2 (g) 2 SO3 (g)
At equilibrium 0.500 mol 0.100 mol 4.60 mol
]O[]SO[
]SO[K
2
2
2
2
3c =
31
C100.02500.0
2
00.160.4
c dmmol846K][][
][K ==
b) 2 SO2 (g) + O2 (g) 2 SO3 (g)
At equilibrium 0.500 mol 0.100 mol 4.60 mol
Amount of mol insert 0.100 mol 0.0500 + x + 0.100
At new equilibrium 0.400 mol 0.0500 + x 4.70 mol
..
]O[]SO[
]SO[K
2
2
2
2
3c=
31
00.1
X0500.02
00.1400.0
2
00.170.4
c dmmol846][][
][K
+ ==
X = 0.113 mol
Example 13 Consider the equilibrium N2O4 (g) 2 NO2 (g).
A 30 C d 1 0 30% f N O i di i d
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At 30oC and 1.0 atm., 30% of N2O4 is dissociated
a) calculate the equilibrium constant, KP, for the reaction
b) calculate the pressure needed to increase the dissociation to 40% at
fixed temperature.
a) N2O4 (g) 2 NO2 (g)
Initial 1.00 mol 0 molWhen dissociated 0.30 + 2 (0.30)
At equilibrium 0.70 mol 0.60 mol Total mol = 1.30
atm538.0atm00.130.1
70.0P 42ON == atm462.0atm00.130.1
60.0P 2NO ==
)P(
)P(K
42
2
ON
2
NO
P = atmKP 40.0
)538.0(
)462.0( 2==
b) calculate the pressure needed to increase the dissociation to 40% at
fi d t t
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fixed temperature
N2O
4(g) 2 NO
2(g)
Initial 1.00 mol 0 mol
When dissociated 0.40 + 2 (0.40)
At equilibrium 0.60 mol 0.80 mol Total mol = 1.40
atmP4286.0atmP40.160.0P
42ON == atmP5714.0atmP
40.180.0P
2NO ==
P 2 2
)P(K
42
2
ON
P = atm396.0)P4286.0(
.KP == P = 0.53 atm
Example 14 Consider the reaction N2 (g) + 3 H2 (g) 2 NH3 (g)
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The KP values at different temperatures are given below.
Plot a graph of ln KP against 1/T. From your graph, determinethe value of H of the reaction
T (oC) 25 127 227 327 427
Kp (atm-2) 6.76 x 105 4.07 x 101 3.55 x 10-2 1.66 x 10-3 7.76 x 10-5
1/T K-1 3.36 x 10-3 2.50 x 10-3 2.00 x 10-3 1.67 x 10-3 1.43 x 10-3
ln Kp (atm-2) 13.4 3.71 3.34 6.40 9.46
Gradient = 3.71 (6.40) / 2.50 x 10-3 1.67 x 10-3
Gradient = H / R = 12180
H = 101 kJ / mol
6.5 The Industrial Application of Chemical Equilibria
i id d b Ch i l I d i i d i
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3 main aspects considered by Chemical Industries in producing
chemical substance
i) ii) iii)
In this Chapter, there are 2 industrial processes which are discussed.
1. Haber process process of synthesizing . in industry.
N2 (g) + 3 H2 (g) 2 NH3 (g) H = 92 kJ / mol
Consider the Haber reaction using Le Chateliers Principle
High yield Short time Low cost
ammonia
In order to produce large amount (high yield) equilibrium must shift
to .
Pressure : to make equilibrium shift to right, the pressure must be
as right side of the equation has total amount of moles
Temperature : to make equilibrium shift to right, the temperature must
be as the forward reaction is an . reaction.
Adding catalyst (Iron) will not change the . but it will
only increase the
right
high lesser
low exothermicyield of product
rate of reaction
2. Contact Process process of synthesizing . in industriessulphuric acid
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2 SO2 (g) + O2 (g) 2 SO3 (g) H = -90 kJ/mol
Consider the Contact process reaction using Le Chateliers Principle
In order to produce large amount (high yield) equilibrium must shift
to
Pressure : to make equilibrium shift to right, the pressure must be
right
. as r g t s e o t e equat on as tota amount o mo es
Temperature : to make equilibrium shift to right, the temperature
must be as the forward reaction is an reaction.
Adding catalyst will not change the . but it will onlyincrease the
low exothermic
yield of productrate of reaction