The document discusses the concept of chemical equilibrium, highlighting that it is a dynamic state where the forward and reverse reactions occur at equal rates, resulting in constant concentrations of reactants and products. It includes explanations of reversible reactions, the law of mass action, and the relationship between equilibrium constants and concentrations or partial pressures. Additionally, the document notes that the equilibrium can be established from either the reactant or product side and that catalysts can speed up reactions without shifting the equilibrium position.

1. Written By
Department of Pharmacy
University of Rajshahi
Rajshahi-6205, Bangladesh

2. Fig: Molecules of A and B colliding (lff'fi1 ~>Gm> to give C and 0 and those ofC and D
colliding to give A and B.
The molecules of A and Bin the equilibrium mixture collide with each other 10 form C and 0. Likewise C and D collide
to give hack A and 8. The collisions of molecules in a closed system arc a ceaseless (~) phenomenon. Therefore
collisions of A and B giving C and D (forward reaction) and collisions of C and D giving back A and B (reverse reaction)
continue to occur even at equilibrium. while concentrations remain unchanged. Thus. chemical equilibrium is dynamic
equilibrium.
+
When this reaction attains equilibrium. the concentrations of A and B. as also of C and D remain constant with time.
Apparently 1"51l"'ll'-''1~C"1 it appears that the equilibrium is dead. But it is not so. The equilibrium is dynamic. Actually the
forward and reverse reactions arc laking place at equilibrium hut the concentrations remain unchanged.
The dynamic nature of chemical equilibrium can be easily understood on the basis of the kinetic molecular model.
~' Q
~ 0 + G)
Q G
Question: Explain the term. "Chemical equilibrium is dynamic equilibrium".
Answer: Let us consider the reaction.
A+B¢=C+D
Fig: The equilibrium of a reaction.
Time__..()
-,~····...rForward reaction rate
·,'---
_,,.L,-----~Q1Jilib11um
. /// Reverse reaction rate attained
rRate
Question: What is chemical equilibrium'! Explain its nature.
Chemical equilibrium: The state of a reversible reaction when the two opposing reactions occur at the same rate and the
concentrations of reactants.aud products do not change with the time.
Example/Nature.of dienikal t'quilibrium: Let us consider the reaction.• , ' , T ( ~.
A+8¢=C+D
If we start with A and Bin a closed vessel, the forward reaction proceeds to form C and D. the concentrations of A and B
decrease and those of C andD increase continuously. A~'·a ~Ult the rateof ; ;
forward reaction also decreases 'and the.rate o( reverse reaction increases.
Eventually (~1. the rate of the two opposing reactions equals and the system
attains a state of cquilibrium.[Example .!l'~ ~) .
Furthermore !l.!1~'31. the true equilibrium of a reaction can be
attained from both sides.Thus the equilibrium concentrations of the
reactants and products an: the sane whether (llfii'21 we start with A
and B or C and D. 1 Nature <!l~ ~. ~ &!]!1~'8C!I~graph C'i<li.
Examples: A few common examples or reversible reactions arc listed below:
2N021~> ¢= N20~1~>
Hirn> + 121111 ¢= 2Hl12>
CH.1COOH111 + C2H~0Hm ~ Cll.1COOC2H~111 + ll2<>m
PCI~<•> ¢:: PCl.1"' + Cl21~1
CnC0.11,1 ~ C1101,, + C021~,
Question: What is reversible reaction? Ciive example.
Reversible reaction: A reaction, which can go in the forward and backward direction simultaneously (~:~1 is
called a reversible reaction.
All chemical reactions do not proceed to completion 1'11!1f'31. In most reactions two or more substances react to li11111 .
products which themselves react to give back the original substances. Thus A and U may react to form C and D. which react
together to reform A and B.
A + B --+. C + I>
A + n +-- C + I)
Such a reaction is represented hy writing a pair of arrows between the reactants and products. The arrow pointing right
indicate» the lorward reaction while left shows the reverse reaction.
A+B¢=C+O
Chemical Equilibrium
.•
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3. Question: (jive explanation or CC]uil ibrium constant or equilibrium law.
•
Question: State and explain the law of mass action.
Law of mass action: Two Norwegian chemist. Guldberg and Wage basis on their large experiment, in I 864, gave a law
called the law of mass action, It states that:
The rate of a chemical reaction is proportional to the active masses of the reactants. .
By the term 'active mass' is meant the molar concentration. It is expressed by enclosing the formula of the substance in
square bracket with that is [ ].
Explanation: Let us consider a chemical reaction.
A+B<=C+D
Let. (A]. [13J. [CJ and [OJ represent the molar concentration of A, B. C and D respectively at equilibrium. According to
the law of mass action.
Rate of forward reaction oc [A][B] = K1[A)[B]
Rate of reverse reaction cc [C][D] = !<iC][D]
From the above equations. it stands proved that. the rate of a reaction is proportional to the molar concentration of the
reactants.
v, Value of equilibrium constant docs not depend upon the initial (~1
·concentration or reactants.
vi. At equilibrium the Gibbs free energy {G) is minimum and any change laking place at equilibrium proceeds without
change in free energy that is.1G=O.
f-1a· Th~ v"•·tv.-.1 P·'t'I:'-:.f th!!!'! <':IJt"Vfl!'t 1rid1.-:~t~r. 1·~p1d
inr::r ~.~$~ ,.,, 1 et.e- on ..,ddlt:.ton of •:..,t;aly!;-1 ..
•:°"~ •ly~t. add'!'• I .•tHi-1
.r- t:lrn"'~'"
----··- ·----------- ..
--~ IJn<:-!lf:-!lly-sll!"d l!'quilibtium
- - ..:. - - •..:.::tt.alysf!'l'd eQuil+brium
Question: Explain the term. "a catalyst cannot change the equilibrium point".
iv. Catalyst: When a catalyst is added to a system in equilibrium. ·r
it speeds up the rate of both the forward and the reverse reaction to an
equal extent ('lll~'l1. Therefore a catalyst cannot change the equilibrium point
except that it is achieved earlier. This enhances the rate of the reaction. rc•u:11'11
The rapid increase of the rate of an equilibrium reaction on
the addition ofa catalyst is illustrated in the figure.
111. Equilibrium cannot be attained in an open vessel: The equilibrium can be established only if the reaction vessel
is closed and no part of the reactants or products is allowed to escape out. In an open vessel. the gaseous reactants or products
may escape into the atmosphere leaving (511'11 <ffil1 behind no possibility of attaining equilibrium.
Fh;1: The left Qraph shows how equiltbr·lum Is at:ta,ned for start with hydn:u;1en and Iodine. The' t'i•;Jht .;1raph·shows t:he at.talnn·•..,.n~ t:•f
eQuilibrlum for start. with 2HI. Eq•-iillbr·iurn cor-cer-cr e+tor-rs In both ce se s are the serne ,
Tin1e ,,..Tim~----..,.0
II2 • I 2
.-----
HI
------
I
..... I
l 'onccnuution -e-, .... ~---;
-- -.---··--- _ _._ H_I _Conccn-
trution
1.:on-=•ntr·etton
"::On~tant:
II,. I,
':oni::cnt::r·.otion
~onst:.!Jnt:
I
Question: Write down the characteristics of chemical equilibrium.
The characteristics of chemical equilibrium are given below:
i. Constancy of concentration: When a chemical equilibrium is established in a closed vessel at constant
temperature. concentrations of the various species in the reaction mixture become constant.
The reaction mixture at equilibrium is called equilibrium mixture.
The concentrations at equilibrium arc cal led equilibrium concentrations: The equilibrium concentrations are represented
by square brackets with subscript eq that is ( Lq·
ii. Equilibrium can be initiated from either side: The state of equilibrium of a reversible reaction can be approached
1~ ~-am1 whether we start with reactants or products. For example. the equilibrium is established if we start with 11.' and
I~ or .2 II.
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4. Relation between KP and K.,
Let us consider a general reaction
jA + kB ¢:: IC ~ mD
where all reactants and products are gases. We can write equilibrium constant exprexsionin terms ofpartial pressure as
I m
(p,) (p/)) .
Kl' .0 • 4 (I)
(p_,)'(f711)
[C]''[D]"
K = ----
c [A]" [B]''
Where Kc is the equilibrium constant.
uestion: llow you write the equilibrium constant expression in terms of partial pressure?
nswer: When all the reactants and products arc gases. we can also formulate the equilibrium constant expression in
of partial pressure. (Fite relationship between the partial pressure (p) of any one gas in the equilibrium mixture and the
molar concentration follow!Jfrom the general ideal gas equation
PV = nRT
11
~p=(-)RT
v
The quantity Y,. is the molar concentration. Thus.
P =(molar concentration) x RT
That is. the partial pressure of a gas in the equilibrium mixture is directly proportional to its molar Concentration at a
given temperature. 1
'f1hereforc. we can write the equilibrium constant expression in terms of partial pressure instead of molar concentrations.8
For a general reaction
IL1~1 + mM,~1 ¢:: yY,~, + zZ,~1
The equilibrium law or the equilibrium constant may be written as
K = (Pr)'.(p/)°
" ( >' ( )'"P1. P.f
Herc K" is the equilibrium constant, the subscript p referring to partial pressure.
Equilibrium constant expression for a reaction in general terms
The general reaction may be written as
aA + bB <= cC + dD
where a. b. c and d are numeral quotients of the substance A. B. C and D respectively. The equilibrium constant
expression is
Equilibrium constant/Equilibrium law: Let us consider a general reaction
A+B<=C+D
l.ct, [A). I BJ. IC] and [DJ represent the molar concentration of A. B. C and D respectively al equilibrium point.
According to the law of mass action.
Rate of forward rcaction « [Alll31 ~ Ki[AllB]
Rate of reverse reaction oc [ Cll DI -r: KliCJl DI
Where K1 and K~ arc rate constant for the forward and reverse reactions.
At equilibrium. rate of forward reaction ~' rate of reverse reaction.
Therefore. Ki[ All Bl K-iJC II DI
K, [C][D] .
:=) - ~ (1)
K~ [A][H]
At any spcci fie temperature. ";{: is constant since both K1 and K~ are constants. The· ratio ){.: is called equilibrium
constant and is represented by the symbol Kc or K. The equation (i) may be written as.
[C][D]
=>Kc= ---
[A][B] -~w~.,,...
The equation is known as the J'iU~ij~~nstant expression or equilibrium Law,
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5. The molar concentration ofCl2 = 11
moll-
I a-· r) _
The molar concentration of PCI, '" -1.- moll
The molar concentration or PCI 1 .. 11
moll.'
Calculation of Kc
let us a mol of PCl5 is heated in V liter of vessel and at equilibrium a mol of PCl5 dissociates to x mol of PCI_, and x 11101
of Cl~ gas. Thus (a-x) mols of PCI, remains undissociatc.
At equilibrium,
Application of law of mass action
i. Dissociation Of PCl5
When heated. PCl5 dissociated to PCl1.and Cl2 gas. At equilibrium. the reaction is represented by the following equation
PCl~1,1 PCl.,c•I + Cl2ci1
At primary state: a mol 0- 0
At equilibrium: (a-x) mol x mol x mol
=atm
(atm)2
(atm)
(p,;), )1
Kr=-,.-~~
( /7. .o, )
(ii) When the total number of moles of the reactants and products are unequal
In such reactions K,. will have units (rnol/litre)" and K11 will have units (atm)", where n is equal to the total number of
moles of products minus the totai number of moles of reactants. Thus for the reaction. i , •
N204ci1 ¢::: 2NOic21
K ~' [NOJ~= (mo/ I L)2 = mol/L
c
[N204] (mo/ IL)
(utm)2
----- =No units
(atn'r)( aIm)
([J/11)2
Kr~--=---
(f711)(Pr:)
Units of equilibrium constant
In the equilibrium expression the units of Kc and KP depend on the specific reaction.
(i) Wht'n the total number of moles of reactants and products are equal
In the equilibrium expression of these reactions, the concentration or pressure terms in the numerator 1<ow~c-rn ff<I• and
denominator 1~1 exactly cancel out. Thus Kc or Kr for such a reaction is without units. Taking example of the reaction .
H21e1 + 121~1 ¢::: 2HIM
[Hlf (moil L)2
K "' = · ~ No units
c [H2][12] (mo! I L)(mol IL)
The quantity /{ is the molar concentration. Thus the partial pressure of individual gases A. 13, C and Dare:
P., ~. [A)RT: P11 = [B]RT: Pc ~0 [C]RT and P,, = [D]RT:
Substituting these values in equation (i), we have
[C]' (RT)' [ D]"' (RT)"'
K.,.., [Ar (RT)1[B]1 (RT)1
[C]'[DJ"'. (RT)''"'
=>K .,. x----
r [A]'[JJ]1 (/?T)''1
::::) KP"' Kc X (RT)"""" f1·k1
=> Kr= K x (RT)'" (ii)
where £n"' (I+m) - (i+k) is the difference in the sums of the coefficients for the gaseous products and reactants. This is
the relationship between Kr and K,.
From the expression (ii) it is clear that when ~11 ~- 0, then KP.~ K..-
p
Assuming ·~ m1 that all these gases constituting the equilibrium mixture obey the ideal gas equation. the partial .
pressure (p) ofa gas is ·
II
(-)RT
v
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6. t-
f. Xf
(_;!.~; ~! }~
111: lxl 1: I
I 1111:
Thus. K,
The molar concentration of H2 = ;': moll,"
The molar concentration ofI, = 1'~ moll-
Calculation of Kc
Let us 2a 11101 of HI is heated in V liter of vessel and
at equilibrium 2a mol of HI dissociates to x mol of f-12 and
x mol of 12 gas. Thus (2a - 2x) rnols of HI remains
undissociate.
At equilibrium.
The molar concentration of Ill"
12"1~2r1
moll·
At primary state:
At equilibrium:
When healed. HI dissociated to H2 and 12 gas. At equilibrium. the reaction is represented by the following equation
2Hl(a> H:c•> + l2(KI . '
2amol 0 0
(2a-2x) mol x mol x mol
Calculation of Kr
At equilibrium, the total number of moles" :!a - 2x -1-
b x 1 :!x :?a + b·- x
iii. Dissociation of 111
Ca-xi!(/,-_, .•
,,
I ~11· ~ rl.' I /l- I
---;-:·,--· x ·-;:-
4 I
-,-::
Thus.
ob -· .r
• 2a+l>-x P
P. =o.and the partial pressure of Gi 1s
If p is the total pressure of the mixture, then al
equi Iibri um:
Tl 1· I f~ . P. = .l!!::lLpie pat ra pressure o D"-'1..'s so! La+b=»
. l .;:
The partial pressure of~·' is P.w>, = 2a:~_,P
Calculation of Kc
Let us 2a mol of S01 is heated with b mol of O'.! in Y
liter of vessel and at equilibrium 2x mol of SO, is
produced. Thus (2a-2x) moles of SO! andtb - x ) moles of
o! gas remain undissociate.
At equilibrium. ·~
The molar concentration ofS02 "'
12",~2'1
moll,"
TI I
. .0 11>-r1
11
..
ie mo ar concentrauon ot : •• -1.-. mo
The molar concentration ofCJ2 c• ~· moll,"
1so: 1: x10: 1,
T hus. Kc = ""--""'"-'--'-"""""1so, I: ·
2S0.c~1
0
2x mol
02121
b
(b-x) mol
2S02c21 +
la mol
(2a-2x) mol
At primary state:
Al equilibrium:
ii. Formation of SO.,
When S02 gas reacts with O! gas. then SOJ gas is produced. Al equilibrium. the reaction is represented by the following
equation
The partial prcs~urc of PCI 1 is /~.0, .~ ~ p
,, v
~rThe partial pressure o]' PCI~ is f~n, -;
K" "° "---,.·""p-
'" '!)/'·
:;I: JJ
I·,,' (II+')
--,x--.-111 I ' I~ I (I -- ·' ) ,,
Calculation of K,.
At equilibrium. the total number of moles .: a x 1 x
+ X .. a I X
If p is the total pressure of the mixture. then at
equilibrium:
::::) K, ·- -, ,,-.'-,,-, .
I
11··
and the partial pressure ofCl. is f>c,! = ";' {'11n,1x1n.1
I ts t, I
Thus.
,~ x ,'.
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7. N02 gas. At equilibrium, the reaction 1s represented by the
That is.
Calculation of Kr
As Kr= Kc (RT)'" and here .1n = 2 - 2 = 0
Thus. K,. "° Kc
K -· ~...!
I' -- (a ~ r II h-.r IThat is,
Calculation of K,.
As Kl' " K, (RT)''" and here .1n = 2 · 2 -r: 0
Tints. K1• z: Kc
4(11- I I
_,_
~X 4a
4a 41a · ,,.·I'
/~1! x P,!
(P,,,)2
Thus.
2 N01e1
0
2x mol
2 Hl,~1
0
2x mol
. ca-,.,
The molar concentration ofN2 = -1.- moll."
1·1 I . 'O ,,,_,., Iic mo ar concentration of 2 "' -1.- mo L
~< molar concentration of NO " ~,· moll
~ (; / Thus. K, - / .~:~;;;;, 1
~ Di~sociation of N.o, .
When heated in a closed vessel. N20~ dissociated to
following equation
Calculation of Kc
Let us a mol of N2 is heated with b mol of 02 in V
liter of vessel and at equilibrium 2x mol of NO is
produced. Thus (a-x) moles of N2 and (b- x ) moles of 02
gas remain undissociate.
At equilibrium.
021e1
b
[b-x) mo!
+,N2«e1
a mol
(a-x) mo!
At primary state:
At equilibrium:
Calculation of Kc
Let us a mol of H, is heated with b mol of 12 in V liter
of vessel and at equilibrium 2x mo! of ill is produced.
Thus (a-x) moles of H2 and (f>-l)t) moles of ij.gas remain
undissociatc. : . .
At equilibrium.
lhc molar concentration of H2 = 1'1;·' 1
moll.
Thl' molar concentration of 12 .~ T moll.
The molar concentration of HI -.· ~.' moll."
.. fHll:
Thus. K< = IH1111. I
12 1a1
b
(b-xj mol
H2,a1 +
At primary state; a mol
At equilibrium: (a-x) mol
and the parual pressure of 12 is /~. · 2~,
P
Calculation of Kr
At equilibrium, the total number of moles r: 2a 2x 1
X I X 2a
If p is the total pressure of the mixture. then at
equilibrium:
The partial pressure of HI is P =
2"- ~.r pIll 211
The partial pressure of II! is P - ...LpII.' :!11 ·
__,_
4f t1 ·· r )1
v, Formation of NO
When N2 gas reacts with 02 gas. then NO gas is produced: At equilibrium, the reaction is represented by the following
equation
iv. Formation of HI
When 11.' gas reacts with I! gas. then HI gas is produced. At equilibrium. the reaction is represented by the following
equation
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8. .. .t
AG " "(1" i .,.., t , _ R·1··111 Cle x an x .... .. . .u u A.-H-11 I (11)
a'.; x a,; x .....
where 6G-. is the difference in free energy of the reaction when all reactants and products are in their standard state.
1· ,/
ac X an X ......
If 11
,, is denoted by .I, then the equation (ii) becomes
a, Xa11 x .....
LCj ""8G' t RTln .I (iii)
The equation (iii) is called Van 't Hoff reaction isotherm.
For the reaction at equilibrium. 8G ""0
Then. 8G0 0-., - RTln .I
Thermodynamic dcrlvation ('1llml'I) of chemical equilibrium or Relation between 6G0 and K
I .ct us consider a general reaction
aA 1 bB + ...... <= cC + dD 1 ......
The chemical potential of a substance in a mixture is related to its activity by the expression
µ c µ" + RT In a ..... :(i)
where µn is the chemical potential of the pure substance. R is gas constant and Tis the absolute temperature.
For a mole of the substance A we can write using the equation (i)
aµ,=a(µ~,+RTlna,)
and si milady
b P11 "'· b (µ0R +RT In a11)
c µ< cc c (µ0< I- RT In a1)
d P1> ~. d (µ"" 1 RT In all)
The change in free energy for the reaction ix given by
6(; = (;prndur" - (;n·arl•ut'
On substitution we get. f
~G = (c µ( + d µIJ I- .... )- (a p 1 h µII I .... )
={c {µ0; +RT In ad + d !p''ll 1- RT In all} 1 l : la{µ' 1 RT In a.d i. b {µ"11 ~-RT In au} ' ... j
-: [{ c µ0
< + d µ ·1J t ...... } - { a µ'' . 1 bp "11 + l I
i.nRTlna1'.+RTlna1/11 ...... }-{ RTlna,"-i RTlna11h+ ...... ]
,. "llcXllnX ......
=({cµ·+dµA1>'·····-}-{a~1°,·~hµ''111 ..... }ltRTln ,, 1,
(11Xll11X .....
Calculation of Kr
At equilibrium. the total number of moles -i. a - x 1
2x a 1 x
•
( ~p)~
1t±.r
Thus.
moll.1.lThe molar concentration of N<>: -' 1
The partial pressure of N02 is Pvo. = },', P
(r.; )~
Kr·
P,,o,
Calculationof K,
Let us a mol of N10~ is heated in V liter of vessel and
at equilibrium a mol of N:O~ dissociates to 2x 11101 of
N01. Thus (a-x) mols of PC"I, remains undissociatc.
At equilibrium.
The molar concentration of N!O~ ~ T mol 1.
If p is the total pressure of the mixture. then at
equilibrium:
The partial pressure of N!O~ is P. .s: ~ p;"!(), a I ,.
2N021.1
()
2xmol
N20~ 1e1
11 mot
(a-x) mot
Al primary slate:
At equilibrium:
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9. Van't llofrs equation in term of Kc
We know lhat equilibrium constant in term' ol partial pressure (1.1.) and 111 lc1111s ol l'1llll'l'llllali1111 (11) :11e 1l'lalt'd 111
each other by the relation
K = K x (RT)1"
r '
Taking logarithms we get
In Kr " ln K, + 6n In RT
Differentiating with respect to temperature. we get
_!!._(I K )=..!!....(I K )+- l!!..n n " .n n c I
from equation (i), we get
T{ ,;~ (6G'')lr - Mi" - RT: { ,;~ (In K"llr ......... (ii)
We know that Gibb's Helmholtz equation for a reaction in the standard state can be written as
6G0 == 611" + T{ ,;', (.1G''l}r
=> T{ ,;11 (6G")}r, Mi" - 611' (iii)
Comparing (ii) and (iii). we have
6G'' - Ml'" oG' - RT: { ,;~ (In Kp)}11
~Ml~··· RT~· { ft-(ln K11)l11
·1110 · .t I K . )=> /U7 ·. Ji' ( n · p)p (iv
Equation (iv) is known as Vau't Hoff's equation.
Where. .11 I enthalpy change or till' reaction al constant pressure when all reactants and products arc in their standard
state.
Differentiating the above equation with respect temperature. al constant pressure. p we get.
{ ,;~ (l~G0)}r ~ ,;~ (- RTIn Kr)r
c - RT { ;~(In Kpl}p- R In Kr,;~ (T)
- RT{ .~ (In Kr)},,.. R In x,
Multiplying both sides by T, we have
T{ ,;~ (Mi'')}p-'- - RT: f Ji (In Krllr· RT In Kr
= - RT In Kr- RT2 { ~~ (lnKp)}1,
Temperature dependence of equilibrium constant or Van't Hoff's equation
We know. from Vant Hoff reaction isotherm.
l(i · - RTln K ....... (i)
where ,1.(j" is the change in standard free energy of the reaction.
The sign of6G0 indicates whether tlie forward or reverse reaction is spontaneous
Considering the equation (iv). we can have three possibilities depending on the sign of DG." for the reaction.
1. If 6G'' is negative. log K must be positive and the reaction proceeds spontaneously in the forward reaction.
11. If .1.Ci" is positive. log K must be negative and K is less than one. The reverse reaction is then spontaneous.
iii. If Mi' '0, log k z- 0 and K 0
I. The reaction is at equilibrium.
.1.G~ · - RTln K
=>6G° , - 2.303 RT log K - · - li,;)
The equation (iv) is also called Vau't Hoffreaction isotherm.
have.
~. . "
~G"~-RTln "cxa,,x ......
II t.
a, xa11 x .....
As 6G~ is Ihe free energy of the reaction in the standard state and is constant at a givcn temperature. The gas constant R
' ,/
a,xa,,x .
and T arc also constants, The factor ,, " is a constant and represented hy K. Therefore. from equation (ii'i) 11·l·
a,xa11x .....
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10. Homogeneous equilibrium: Equilibria in which the reactants and products arc in the same phase. are called
homogeneous equilibria. For example. reaction between acetic acid and ethyl alcohol to form ethyl acetate.
CH,COOHrn + C211~0Hrn ~ Cll.,COOC2115;11 + 1120m
or
_I! .. ~. x !_1
I l
If V litre be the total volume of the equilibrium mixture. the concentrations of the ~a1:ious species are:
[CH1COOH Jee¥
[C,H,OHJ ~ !2:::L- . l
rc11,cooc,11,1 "' f
[H,O], f
The equilibrium constant expression may be written as:
f X/K" ----'--
•
Fig Heterogeneous equilibria.
Liquid equllibriurm lhc chemical equilibrium in which all tht: reactants and products are in the liquid phase. arc
referred to as the liquid equilibria. Like the gas-phase equilibria. the liquid equilibria are also called homogeneous equilihria.
For example. reaction between acetic acid and ethyl alcohol to form ethyl acetate. ·
. CH.1COOHm + C2H~Ollrn ¢:: CH.,COOC2H~(ll + H20m
Equilibrium constant expression: Let us start with 'a' moles of acetic acid and 'b· moles of alcohol. If x moles of
acetic acid react with x moles ethyl alcohol. x moles of ester and x moles of water arc produced when the-equilibrium is
established. Now, the moles present at equilibrium arc:
CH1COOH :c (a-x) moles
C1H~qH = (b-x) moles
CH,COOC1!1, -- x moles
Or. in terms of partial pressures.
1l
---+--Cu
1.04 atm
Heterogeneous equilihrium: The equilibria in which the reactants and products are not all in same phase. arc 'called
heterogeneous equilibria. The decomposition of calcium carbonate upon heating to form calcium oxide and co~ IS an
example of heterogeneous equilibrium.
CaCO.,c•i ~ Ca<>csi + C<>2c21
. ..;>
Equilibrium constant expression: According 1Jkir mass action. the
equilibrium constant expression of decomposition of CaC0.1
K = jCO. 11<..<>1
I 01< o, I
But CaC01 and CaO are pure solids. The concentration of a pure solid
is fixed and cannot vary. Thus the concentrations of pun: solids arc not
included in the equilibrium constant expression.
Ignoring the concentrations ofCaC01 and Cao, the equilibrium
. constant expression for the decomposition of CaC'O, may be written as,
Kc= [C02l
Question: What arc heterogeneous and liquid equilibrium and homogeneous equilibrium'? Write the equilibrium
constant expression of them.
=>L(lnK Jc- ...JL
,// ( «rThus.
But we know. .:'.II L'L 1 ~nl{T
Where •is the heat of the reaction at constant volume.
~JE .
j lron: Vant llofrs equation we know. !(':' c ,;~.(In Kp)]
'1 (I K ) ..,Ji n I' -1
su •t
Iii - T
.II .~nR I
Ill
=--? ,;~ (In K<)
=-=>fi-(lnK1)
=> ,;~ (In K1)
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"../.·.·,,· ·,/•.
... . ..
;
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11. I
l-quilibruun shrfl "" mu-case of I'
A+B<=CA+B<=C
Moles increaseMoles decrease
t quilihrium shili lquilibruuu shili
Following the same line of argument. a decrease in the concentration of A by its removal from the equilibrium mixture.
will he undone by shift of the equilibrium position to the left This reduces the concentration of the product C.
Effect of a change in pressure
When pressure is increased on a gaseous equilibrium reaction. the equilibrium will shirt in a direction. which tends to
decrease the pressure.
Let us consider a general reaction.
A+B<=C
. The combination of A and B produces a decrease of number of molecules while the decomposition of C into A and R
results in the increase of molecules. Therefore, by theincrease of pressure on the equilibrium it will shift to right and give
more C. A decrease in pressure will cause the opposite effect. The equilibrium will shift to the left when C will decompose to
form more ofA and 8.
Effect of a change in concentration
When concentration of any of the reactants or products is changed. the equilibrium shirts in a direction so as to reduce
the change in concentration that was made.
I.et us consider a general reaction.
A+R<=C
When a reactant. say. A is added at equilibrium, its concentration is increased. The forward reaction alone occurs
momentarily. According to Le Chatelicr's principle, ;i new equilibrium will be established so as lo reduce the concentration
pf A. 'thus the addition of A causes the equilibrium to shiti to right. This increases the concentration ofthe product C.
Add111on lfrmoval .
~A+B<=C 'A+B<=C
Le Chateliers principle
In 1884. the French chemist Henry Le Chatclier proposed a general principle, which applies all systems in equilibrium.
This important principle called the L<: Chatclicr'x principle. It may be stated as: '
When a stress is applied on a system in equilibrium, the system tends to adjust itself so as to reduce the stress.
There arc three ways in which the stressISl<>fJ can he caused on a chemical equilibrium:
1. Changing the concentration of a reactant or product.
11. Changing the pressure or volume of the system.
111. Changing the temperature.
Thus when applied to a chemical reaction in equilibrium, l.e Chatcliers principle can he stated as:
If a change in concentration, pressure or temperature is caused to a chemical reaction in equilibrium, the
equiltbrium will shift to the right or left so as to minimize the change.
.'!.:..-.~~ x !.~.--v
I I
1)1/• I)
or LI
K .r-
I 11,oJ i:
Th1: equilibrium constant expression may be written as:
fXf
v
I
Equilibrium constant expression: Let us start with 'u ' moles of acetic acid and ·h· moles of alcohol. If x moles of
acetic acid react with x moles ethyl alcohol. x moles or ester and x moles of water arc produced when the equilibrium is
established. Now, the moles present at equilibrium are:
CH,COOll = (a-x ) moles
C2H50H c (b-x) moles
CH,COOC,11, ·· x moles
II 20 ·· x 11101 cs
If V litre he the total volume of the equilibrium mixture. the concentrations ofthe various species arc:
[CH,COOH )~ T
[C,11,0H] = T
1c1-1,cooc,11,1
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12. Manufacture of Sulphuric acid (Contact process)
The chief reaction used in the process is
2S0'11t, + Oi1c1 ¢:::: 2S0_,1~1 + 42 Kcal
Following information is revealed ·~ <1>"lf!1 by the above equation:
(a) the reaction is exothermic.
(b) the reaction proceeds with a decrease in number of moles.
The favourable conditions for the maximum yield of S0_1 arc:
i. Low temperature: Since the forward reaction is exothermic. the equilibrium will shift on the right at low
temperature. An optimum temperature between 400 - 500°C is required for the maximum yield of S0_1.
ii. High pressure: Since the number of moles is decreased in the forward reaction, increase of pressure will shifi the ·
equilibrium to the right. Thus for maximum yield of SO,. 2 to 3 atm pressure is used. -
lquilibriurn shift
Jn an endothermic reaction
X + Y +Heat ¢:::: Z
The increase of temperature will shift the equilibrium to the right as it involves the absorption of heat. This increase the
concentration of the prod11:ctZ. I cmpcrafurc increase, ·
/ I lcat added
X + Y + Ueat ¢= l. •
lquilibrium shili
Effect of change in temperature
When temperature of a reaction is increased. the equilibrium shifts in a direction in which heal is absorbed.
lei us consider an exothermic reaction.
A + B (== C + Heat
When the temperature of the system is increased, heat is supplied to it from outside. According to Le Chatcliers
principle, the equilibrium will shift to the left. which involves the absorption of heal. This would result in the increase of the
reactants A and B. Icmpcraturc increase.
/ lkaladdcd
A + B ¢:::: C + Heat
H21~1+ l!1e1 = 21111e1
The reaction in which the number of product molecules is unequal to the number of reactant molecules. are affected by
pressure changes. In such a case change in pressure can increase or decrease the rate of reaction, An example of such reaction
is:
The reactions in which the number of product molecule is equal to the number of reactant molecules, are unaffected by
pressure changes. In such a case the system is unable to undo the increase or decrease pf pressure. An example of such
reaction is:
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1
' .
•
Application of the Le Chatelier 's principle
Synthesis of Ammonia (Haber process)
The manufacture of ammonia by Haber-process is represented by the equation
N21~1 + 3112111 (== 2NH~1~1 + 22 Kcal
A look at the equation provides the following information:
(a) the reaction is exothermic.
(h) the reaction proceeds with a decrease in the number of moles.
The favourable conditions for the maximum yield of NH, are:
i. Low temperanire: By applying le Chateliers principle. low temperature will shift 'the equilibrium to the right.
This gives greater yield of ammonia. In actual practice a temperatures of about 450°C is used when the percentage in the
equilibrium mixture is 15.
ii. High pressure: High pressure on the reaction at equilibrium favors the shifi of the equilibrium to the right. This is
because. the forward reaction proceeds with a decrease in the number of moles. A pressure of about 200 aim is applied in
practice.
iii. Catalyst: Low temperature is necessary for higher yield of ammonia. But at relatively low temperature, the rate of
reaction is slow and the equilibrium is attained in a long time. To increase the rate of reaction and thus quicken the attainment
of equilibrium. a catalyst is used. Finf!Jy divided iron containing molybdenum is employed in actual practice. Molybdenum
acts as a promoter that increases the Ii fe and efficiency of the catalyst.
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13. . "'
Math: Bahl and Bahl.
111. Catalyst: At the low temperature used in the reaction. the rate of reaction is slow and the equilibrium is attained
slowly. A catalyst is used to speed up the establishment of the equilibrium. Vanadium pentoxide is commonly used.
Manufacture of Nitric acid (Birkeland-Eyde process)
Nitric acid is prepared on a large scale by making use of the reaction:
N2c11 + 02110 <= 2NOca1 - 43.2 Kcal
The equation tells us that (a) the reaction proceeds with no change in the number of moles.
(b) the reaction is endothermic and proceeds by absorption of heat.
The favourable conditions for the maximum yield of NO are:
i. High temperature: Since the forward reaction is endothermic, increase of temperature will favour it. Thus a high
temperature of the order of 3000"(' is employed to get high yield 1~ ~ ~ct"r:l ~<11111 ofnitric acid.
ii. No effect of pressure: Since the forward rcactiou involves 111 change· '11 the number ofm<;les a change in pressure
has no effect 011 the equilibrium.
iii. High concentration: The formation of nitric oxide is favoured by using high conceutraticn ofthereactants.
j
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