Download - TOPIC : EQUILIBRIUM
Page | 1
CHEMISTRY: CHEMICAL EQUILIBRIUM
COURSE : B.Sc I
PAPER : I
TOPIC : EQUILIBRIUM
prepared by : Dr. KAMAL KISHORE SINGH
Chief coordinator
Department of chemistry, Nalanda
Open university , patna , Bihar.
Page | 2
CHEMISTRY: CHEMICAL EQUILIBRIUM
CHEMICAL EQUILIBRIUM
Chemical reactions can be divided into two categories
Reversible reaction
Those reactions in which products can react to form reactants back are called reversible
reactions. They proceed in both the directions and do not reach completion e.g.
3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4H2(g)
N2(g) + 3H2(g) 2 NH3(g)
2 SO2(g) + O2(g) 2 SO3
Irreversible reactions
Those reactions which proceed only in one direction and it is not easily possible to
convert the products into reactants are known as Irreversible reactions
Page | 3
CHEMISTRY: CHEMICAL EQUILIBRIUM
They reach completion e.g.
BaCl2 + Na2SO4 BaSO4 + 2NaCl
AgNO3 + NaCl AgCl + NaNO3
Equilibrium
The stage at which the rate of forward reaction becomes equal to the rate of backward
reaction is called equilibrium. It is the condition in which the properties of system do not
change with time.
Physical Equilibrium
When there is equilibrium between two physical states of same substances it is called
physical equilibrium e.g.
H2O(s) H2O (1)
NH4Cl(s) + aq NH4Cl(aq)
Chemical Equilibrium
When there is equilibrium between two processes involving change of chemical species,
it is called chemical equilibrium.
Characteristics of Chemical Equilibrium
1. When chemical equilibrium is attained, there is no further change in pressure,
concentration, density or colour at a fixed temperature.
2. It is dynamic in nature i.e. reaction does not stop after the equilibrium is reached rather
it appears to stop because there is no further change in concentration, pressure or density
because rate of forward reaction is equal to rate of backward reaction.
3. It can be approached from either direction e.g.
H2 (g) + I2 (g) 2 HI (g)
Page | 4
CHEMISTRY: CHEMICAL EQUILIBRIUM
2HI (g) H2 (g) + I2 (g)
4. A catalyst can hasten the approach of equilibrium but does not alter the equilibrium.
Forward reaction
The reaction which proceeds from reactants to products is called forward reaction. The
rate of forward reaction is denoted by rf.
Backward reaction
The reaction which proceeds from products to reactants is called backward reaction. The
rate of backward reaction is denoted by rb.
Active Mass
The concentration of a reactant in moles per litre or partial pressure of a reacting gas in
atmosphere is known as its active mass. It is expressed by enclosing the symbol or
formula of the substance in square brackets e.g. [A]. These days active mass is replaced
by molar concentration.
Law of mass action
The rate at which a substance reacts at any instant is proportional to its active mass
(molar concentration) and the rate of chemical reaction is proportional to product of
active mass (molar concentration) of reactants e.g.
A B + C
Rate [A]
or, Rate = k [A] where ‘k’ is rate constant or velocity constant
A + B C + D
Rate [A] [B]
Rate = k[A] [B]
2A + B C + D
Page | 5
CHEMISTRY: CHEMICAL EQUILIBRIUM
Rate [A]2 [B]
Rate = k [A]2 [B]
Thus, the concentration term is raised to the power equal to Stoichiometeric coefficient in
the chemical equation.
Equilibrium Constant
It is defined as ratio of product of molar concentration of products to the product of molar
concentration of reactants raised to the power equal to their Stoichiometric coefficient in
a chemical reaction at particular temperature. The equilibrium constant measures the
extent to which a reaction takes place before equilibrium is reached.
If equilibrium constant (K) is large, the equilibrium mixture will have high proportion of
products. If it is small, the equilibrium mixture will contain high proportion of reactants.
aA + bB cC + dD
a b c df bk [A] [B] r [C] [D]
a b c df f b br k [A] [B] r k [C] [D]
Where kf and kb are rate constants for forward and backward reactions respectively.
At equilibrium fr = br
a b c df bk [A] [B] k [C] [D]
c d
fC a b
b
k [C] [D]K
k [A] [B]
Where Kc is equilibrium constant in terms of molar concentration of reactants and
products.
In case of gaseous reactions, it is more convenient to use partial pressures instead of
molar concentrations. The equilibrium constant in terms of partial pressures of reactants
and products is called Kp.
Page | 6
CHEMISTRY: CHEMICAL EQUILIBRIUM
aA bB cC dD
c d
p a b
(pC) (pD)K
(pA) (pB) Reaction coefficient
p q
c m m
[C] [D]Q
[A] [B]
Note 1 : If Q < K it means equilibrium has not reached the and system is moving
towards right.
At equilibrium p q
c m m
(pC) (pD)Q
(pA) (pB)
Note 2 : Q > K equilibrium has not reached. The reaction must proceed backwards.
pv = nRT
n
p RT CRTv
where ‘c’ is molar concentration in mol L-1
c d c d c d
(c d) (a b)p ca b a b a b
{[C]RT} {[D]RT} [C] [D] (RT)K K (RT)
{[A]RT} {[B]RT} [A] [B] (RT)
np cK K (RT) where n = (c + d) − (a + b)
Where n is no. of moles of gaseous products − no. of moles of gaseous reactants.
n may be zero, positive or negative e.g.
If n = 0, Kp and Kc are equal and have no units. In other case i.e. for n = +ve or −ve,
Kp is expressed as (Unit of pressure n) e.g. (atm)n, Kc is expressed as (Unit of
concentration)n e.g. (mol L−1)n
Relationship between degree of dissociation and Vapour density
In all cases, where number of moles are being changed from reactants to products, so
Vapour density will also change. If we consider, say dissociation of PCl5, the initial and
equilibrium states are represented as follows
5 3 2PCl (g) PCl (g) Cl (g)
Page | 7
CHEMISTRY: CHEMICAL EQUILIBRIUM
Initial 1 0 0
Moles at Equilibrium 1 − x x x
Total number of moles = 1 − x + x + x = 1 + x
Here we start with 1 mole of PCl5 and ‘x’ is degree of dissociation. If 1 mole at
temperature T and pressure P occupies a volume ‘v’ litres, the volume of resulting
mixture will be (1 + x) litres, where (1 + x) represents the number of moles at
equilibrium.
M
DV
1
Dvolume
(since mass is constant)
If ‘D’ is density of gas before dissociation and ‘d’ is density of equilibrium mixture then
in case of dissociation of PCl5
1
DV
....(i)
1
d(1 x)V
....(ii)
From (i) and (ii) D (1 x)V
d V
D
1 xd
D D d
x 1d d
D d
xd
For a molecule of gas which on dissociation gives ‘n’ molecules of products:
Page | 8
CHEMISTRY: CHEMICAL EQUILIBRIUM
D d
xd(n 1)
This relationship can be used for calculation of degree of dissociation from “Vapor
Density Measurement”
Illustration 1
pK for the reaction 5 3 2PCl PCl Cl at 250°C is 0.82. Calculate the degree of
dissociation at given temperature under a total pressure of 5 atm. What will be the degree
of dissociation if the equilibrium pressure is 10 atm. at same temperature.
Solution
Let 1 mole of 5PCl be taken initially. If `x’ moles of 5PCl dissociate at
equilibrium, its degree of dissociation = x.
Moles 5PCl 3PCl 2Cl
Initially 1 0 0
At equilibrium 1 – x x X
Total moles 1 – x + x + x = 1 + x
P = 5 atm. pK 0.82
PCl PCl Cl5 3 2
1– x xP xPP , P , P
1 x 1 x 1 x
PCl Cl5 2
p
PCl5
(P )(P )K
(P )
Page | 9
CHEMISTRY: CHEMICAL EQUILIBRIUM
2
p 2
xK P 0.82
1– x
2
2
x (5)0.82
1– x
2 0.82x
5.82
x = 0.375 (or 37.5%)
Now the new pressure P = 10 atm.
Let y be the new degree fo dissociation. As the temperature is same (250°C), the
value of pK will remain same.
2 2
p 2 2
(y) P (y) 10K 0.82
1– (y) 1– (y)
0.82
y10.82
or y = 0.275 (or 27.5%)
Note : By increasing pressure, degree of dissociation has decreased, as
depicted in illustration 1
xP
i.e., the system shifts to reverse direction. We can also
compare the result by applying Le Chatelier’s principle.
Illustration 2
Vapour density of the equilibrium mixture of NO2 and N2O4 is found to be 40 for the
equilibrium
N2O4 2NO2
Calculate (i) abnormal molecular weight
ii) degree of dissociation
iii) percentage of NO2 in the mixture
Solution
Page | 10
CHEMISTRY: CHEMICAL EQUILIBRIUM
i) N2O4 2NO2
Observed value of vapour density (d) = 40
Hence, abnormal molecular weight = 40 × 2 = 80
ii) D × 2 = theoretical molecular weight = 92
92
D 462
D d 46 40
x 0.15d 40
iii) N2O4 2NO2
Initial mol 1 0
At equilibrium (1 – x) 2x
0.85 0.30
Total moles at equilibrium = (1 + x) = 1 + 0.15 = 1.15
Percentage of NO2 = 2x 0.30
100 1001 x 1.15
= 26.08%
Heterogeneous equilibrium
When reactants and products are in different physical states, it is called heterogeneous
equilibrium e.g.
CaCO3 (s) CaO(s) + CO2(g) Kp = PCO2
3Fe (s) + 4H2O(g) Fe3O4(s) + 4H2(g) 4
2P 4
2
(pH )K
(pH O)
NH4 HS(s) NH3(g) + H2S(g) Kp = (pH2S) (pNH3)
Page | 11
CHEMISTRY: CHEMICAL EQUILIBRIUM
The molar conc. of solid and liquid reactant or product in equilibrium involving gases is
taken as unity. Kp is written for heterogeneous reactions and analysis is done in terms of
partial pressure.
Le-Chartelier’s Principle
When a system in equilibrium is subjected to change in temperature, pressure or
concentration, the equilibrium shifts in that direction so as to counterbalance the effect of
the change.
This principle is extremely useful in predicting the effect of changes in temperature,
pressure or concentration upon a system at equilibrium.
Effect of temperature
In case of exothermic reaction, increase in temperature favours backward reaction where
as decrease in temperature favours forward reaction e.g.
N2(g) + 3H2(g) 2NH3 (g) + heat
Increase in temperature, generally favours an increase in rate but extent of increase will
be different for backward and forward reactions owing to the differences in their energies
of activation. So the numerical value of ‘k’ (equilibrium constant) will change with
change in temperature.
The variation is expressed by the relation:
2 2 1
1 1 2
K T TH2.303log
K R T T
Where K1 and K2 are equilibrium constant at temperature T1 and T2 respectively
H = Standard enthalpy of reaction
Note 1 : For endothermic reaction
H = +ve and if T2 > T1 then 2
1
Klog ve
K
log K2 > log K1 K2 > K1
Page | 12
CHEMISTRY: CHEMICAL EQUILIBRIUM
i.e. K increases with increase in temperature for endothermic reactions.
Note 2 : For exothermic reactions
H = − ve and T2 > T1 2
1
Klog ve
K
log K2 < log K1 K2 < K1
i.e. K decreases with increase in temperature for exothermic reactions.
Note 3 : The new equilibrium state has a new value of equilibrium constant ‘k’ on
changing the temperature
Effect of pressure
If there is an overall increase in number of moles of products from reactants, low pressure
is favourable for forward reactions e.g. in dissociation of PCl5(g) low pressure is
favourable
PCl5 PCl3 (g) + Cl2 (g) n = 1
If there is decrease in number of moles from reactants to products, high pressure is
favourable for forward reaction e.g. high pressure is favourable for manufacture of NH3.
N2 (g) + 3H2(g) 2 NH3(g) n = −2
If the number of moles of reactants and products are equal is no effect of pressure e.g.
N2(g) + O2 (g) 2NO (g) n = 0
Effect of Concentration
If at equilibrium some quantity of reactant or product is added, the equilibrium will shift
in the direction in which added substance is consumed e.g. in N2(g) + 3H2 2NH3(g).
If at equilibrium, we increase the concentration of reactant (N2 or H2 or both), the rate of
forward reaction will increase. But if we increase the concentration of NH3, the reaction
proceeds towards backward direction. If we increase the conc. of NH3, the reaction
proceeds towards backward direction. If we decrease the conc. of NH3, rate of forward
reaction will increase. Therefore, product should be removed as soon as it is formed.
Page | 13
CHEMISTRY: CHEMICAL EQUILIBRIUM
Effect of Catalyst
Catalyst increases the rate of forward as well as backward reaction equally but
equilibrium is reached faster in presence of catalyst. It does not effect value of ‘k’.
Effect of addition of inert gas
At constant volume, the various concentration terms are not changed. Hence the
equilibrium will not be affected. However at constant pressure, the addition of inert gas
will increase the volume of the system and thus decrease the concentration and partial
pressures of the species involved. Hence the equilibrium will shift in the direction in
which concentration of species increase i.e. degree of dissociation should increase i.e.
formation of products should be favoured if n = +ve. If n = −ve, addition of inert gas
will favour backward reaction.
Note 1 : If K1 is equilibrium constant for
A + B C + D then 1
[C] [D]K
[A] [B]
K2 for C + D A + B is equal to 1
1
K
2
1
[A][B] 1K
[C] [D] K
K3 for nA + nB nC + nD is , n1(K ) , ‘n’ can be in fraction also.
n n
n3 13 n
[C] [D]K (K )
[A] [B]
K4 for nC + nD nA + nB is n
1
1
(K )
n n
4 n n n1
[A] [B] 1K
[C] [D] (K )
Page | 14
CHEMISTRY: CHEMICAL EQUILIBRIUM
Note 2 : If K1 be equilibrium constant for P Q and K1 = Q
[P]
If K2 be equilibrium constant for R S then K2 =
S
R
K for P + R Q + S in equal K1 × K2
1 2
[Q] [S]K K K
[P] [R]
Note 3 : If K1, be equilibrium constant or A B 1
[B]K
[A]
K2 be equilibrium constant for C D 2
[D]K
[C]
‘K’ for A + D B + C
1
2
K[B] [C]K
[A] [D] K
Illustration 3
Consider the reaction 2 2 2 2SO Cl (g) SO (g) Cl (g) ; at 37 C the value of
equilibrium constant for the reaction is 0.0032. It was observed that concentration of the
three species is 0.050 mol/lit. each at a certain instant. Discuss what will happen in the
reaction vessel ?
Solution
In this question, concentration of three species i.e. 2 2SO Cl , 2SO and 2Cl (g) each is
given, but is is not mentioned that whether the system is at equilibrium or not. So first
check it. Find reaction coefficient for given equation.
2 2
2 2
[SO ][Cl ] (0.05)(0.05)Q 0.05
[SO Cl ] (0.05)
Page | 15
CHEMISTRY: CHEMICAL EQUILIBRIUM
Q K , so system is not at equilibrium state.
As Q > K, the concentrations must adjust till Q = K for equilibrium. This can
happen only if reaction shifts backwards, and products recombine to give back reactants.
Hence in the reaction vessel, the system will move backward so that it can achieve
equilibrium state.
Relationship between KP and KC
Consider a gaseous reaction
1 2 1 2m A m B n C n D
n n1 2c D
p m m1 2A B
P PK
P P
…(i)
n n1 2
c m m1 2
[C] [D]K
[A] [B] …(ii)
Applying ideal gas equation
PV = nRT
n
P RTV
or P = CRT
Where C is molarity
A A B BP C RT ; P C RT
C C D DP C RT ; P C RT
Substittuing this in equation (i)
Page | 16
CHEMISTRY: CHEMICAL EQUILIBRIUM
n n1 2C D
m m1 2A B
C RT C RT
C RT C RT
=
n n1 2n n1 2C D
m m m m1 2 1 2A B
RTC C
C C RT
= (n n ) (m m )1 2 1 2cK (RT)
np cK K (RT)
n = number of moles of products in gaseous state – number of moles of reactants in
gaseous state.
If n 0
p cK K
Gaseous equilibria are of two types
Type I : In which n 0
p cK K
e.g. 2(g) 2(g) (g)H I 2HI
2(g) 2(g) (g)N O 2NO
Type II : In which n 0
So p cK K
e.g. 5(g) 3(g) 2(g)PCl PCl Cl ; n>0
2(g) 2(g) 3(g)N 3H 2NH ; n 0
Illustration 4
Page | 17
CHEMISTRY: CHEMICAL EQUILIBRIUM
For a homogenous gaseous reaction X 2Y Z, at 473 K, the value of cK 0.35
concentration units. When 2 moles of Y are mixed with 1 mole of X, at what pressure
60% of X is converted to Z?
Solution
Since pressure is to be calculated, so first find pK using the relation between cK and
pK ,
c n 1– 3 –2K 0.35, R 0.0821, T 473,
n 2 –4p cK K (RT) 0.35 (0.0821 473) 2.32 10
The expression for pK is : Zp 2
X Y
pK
p (p )
Moles X Y Z
Initially 1 2 0
At equilibrium 1 – x 2 – 2x X
total moles Tn 3 – 2x
Let P = equilibrium pressure
X Y
1– x 2 – 2xP P, P P,
3 – 2x 3 – 2x z
xP P
3 – 2x
2
p 2 2 2
xP
x(3 – 2x)3 – 2xKP (1– x)(2 – 2x)1– x 2 – 2x
P P3 – 2x 3 – 2x
Page | 18
CHEMISTRY: CHEMICAL EQUILIBRIUM
x = 0.6 (given)
2
–4p 2 2 2
0.6(3 –1.2)K 2.32 10
P (1– 0.6) –1.2)
2 2 2P (1.8 10 )
P = 180 atm
HOMOGENOUS GASEOUS REACTIONS
Here, we will discuss some important reversible reactions and the calculations for Kc and
Kp.
Illustration 5
The value of cK for the reaction : 2 2H (g) I (g) 2HI(g) is 45.9 at 773 K. If one
mole of 2H , two mole of 2I and three moles of HI are taken in a 1.0 L flask, find the
concentrations of 2I and HI at equilibrium at 773 K.
Solution
Reaction quotient = 2 2
2 2
[HI] 3 9
[H ][I ] 1 2 2
Note : When n 0 , not only p cK K , but volume terms always cancels in the
expression of K.
Q K (=45.9). Hence the reaction proceeds to forward direction to achieve
equilibrium. Let x mole of 2H and 2I combine to produce 2x mole of HI.
Moles 2H 2I 2HI
Initially 1 2 3
At equilibrium 1 – x 2 – x 3 + 2x
Page | 19
CHEMISTRY: CHEMICAL EQUILIBRIUM
2
C
2 2
[HI]K 45.9
[H ][I ]
concentration of species at equilibrium are :
2 2[H ] (1– x) /1, [I ] (2 – x) /1,[AB] (3 2x) /1
2
2
C
3 2x
(3 2x)1K 45.9
1– x 2 – x (1– x)(2 – x)
1 1
x = 0.68
2[I ] 2 – x 2 – 0.68 1.32M
(Note that volume = 1.0 L)
[HI] = 3 + 2x = 3 + 1.36 = 4.36 M
Illustration 6
0.15 mol of CO taken in a 2.5 L flask is maintained at 750 K along with a catalyst so that
the following reaction can take place; CO(g) + 2 3H (g) CH OH(g) . Hydrogen is
introduced until the total pressure of the system is 8.5 atm. At equilibrium 0.08 mol of
methanol is formed. calculate:
i) pK and cK
ii) The final pressure if the same amount of CO and 2H as before are used but no catalyst
so that the reaction does not take place.
Solution
2 3CO(g) 2H (g) CH OH(g)
Page | 20
CHEMISTRY: CHEMICAL EQUILIBRIUM
3C 2
2
[CH OH]K
[CO][H ]
Let y moles of 2H were initially.
Moles CO(g) 22H (g) 3CH OH
Initially 0.15 y 0
At equilibrium 0.15 – x y – 2x X
x = 0.08 (given)
moles of CO = 0.15 – 0.08 = 0.07,
moles of 2H y – 0.16, moles of 3CH OH 0.08
total moles = Tn 0.07 (y – 0.16) 0.08
Equilibrium pressure (P) = 8.5 atm
Volume of vessel (V) = 2.5 L, T = 750 K
Using the Gas equation, PV = nRT, we have :
T
PV 8.5 2.5n 0.345
RT 0.0821 750
0.345 = 0.07 + (y – 0.16) + 0.08
y = 0.355 moles
moles of 2H at equilibrium = y – 0.16 = 0.355 – 0.16 = 0.195
Page | 21
CHEMISTRY: CHEMICAL EQUILIBRIUM
Now find cK as follows :
3C 2 2
2
0.08[CH OH] 2.5K
[CO][H ] 0.07 0.195
2.5 2.5
= 187.84 concentration units .
Now find pK using the relation : np cK K (RT)
n 1–3 –2
–2pK 187.84 (0.0821 750) = 0.0495 atm units.
ii) when no reaction takes place, then the total pressure is simply due to 2H and CO
present initially.
T 2n n(CO) n(H ) 0.15 0.355 0.505
nRT 0.505 0.0821 750
P 12.438 atmV 2.5
FORMATION OF NH3
N2(g) + 3H2(g) 2NH3(g)
Suppose a moles of N2 are heated with b moles of H2 in a V litre container. Let x moles
of N2 be reacted at equilibrium.
N2(g) + 3H2(g) 2NH3(g)
Initial moles conc. a
V
b
V
2x
V
Conc. at eq. a x
V
b 3x
V
2x
V
Page | 22
CHEMISTRY: CHEMICAL EQUILIBRIUM
2
2 23
c 3 32 2
4x
[NH ] VK[N ][H ] a x b 3x
V V
If a = 1 and b = 3 then
2 2 2 2
c 3 4
4x V 4x VK
(1 x)(3 3x) 27(1 x)
If x 1
2 2
c
4x VK constant
27
2
2
1x
V or
1x
V but
1P
V So, x P
Thus fraction of N2 converted into NH3 is proportional to total pressure.
Illustration 7
Ammonia under a pressure of 15 atm. At 27°C is heated to 347°C is a closed vessel in the
presence of catalyst. Under these conditions, NH3 partially decomposes to H2 and N2. The
vessel is such that the volume remains effectively constant, whereas the pressure
increases to 50 atm. Calculate the % age of NH3 actually decomposed.
Solution
Ammonia decomposes to 2N and 2H as follows :
3 2 22NH N 3H
3 3NH NH
at 27°C at 347°C
(15 atm.) (P = ?) V remains constant
Page | 23
CHEMISTRY: CHEMICAL EQUILIBRIUM
First, let us find initial pressure of 3NH at 347°C.
P T (V is constant)
1 2 1 22
1 2 1
P P P T 15 620p 31atm.
T T T 300
Partial pressure 2NH3 3H2 N2
Initial 31 0 0
At equilibrium 31 – x 3x
2
x/2
Now final equilibrium pressure = 50 atm.
3 x
50 31– x x2 2
x = 19 atm.
% 3NH decomposed 19
100 61.2%31
Alternative method:
3 2 22NH 3H N
Let x be the degree of dissociation
Moles 32NH 23H 2N
At equilibrium 1 – x 3/2x x/2
Page | 24
CHEMISTRY: CHEMICAL EQUILIBRIUM
total moles = 1 + x
initial moles initial pressure
final moles final pressure
1 31
1 x 50
19x
31 % dissociation
19100 61.2%
31
Illustration 8
Solid NH4HS(s) (Ammonium hydrogen sulphide) dissociates to give NH3(g) and H2S(g) and
is allowed to attain equilibrium at 100°C. If the value of Kp for its dissociation is found to
be 0.34. find the total pressure at equilibrium and partial pressure of each component.
Solution
NH4HS(s) NH3(g) + H2S(g); since NH4HS is a solid, hence, NH HS4a 1 and its
undissociated amount will not effect the total pressure (due to gaseous NH3 and H2S
only). Let ‘x’ be its moles decomposed at equilibrium and P be the equilibrium pressure.
Moles NH4HS NH3 H2S
Initial 1 0 0
at equilibrium 1 – x x X
Total moles at equilibrium = moles of NH3 and H2S = 2x
P = ? Kp = 0.34
H S NH2 3
x P x PP P & P P
2x 2 2x 2
(for equimolar proportions, partial pressures are equal)
Page | 25
CHEMISTRY: CHEMICAL EQUILIBRIUM
P H S NH NH HS2 3 4K P P (a 1)
P P
0.342 2
2P
0.34 P 4 0.344
= 1.17
NH3
P 1.17P 0.585 atm
2 2 H S2
P 1.17P 0.585 atm
2 2
HETROGENEOUS SIMULTANEOUS EQUILIBRIUM
Illustration 9
Two solid compounds A and C dissociates into gaseous product at temperature T as
follows:
i) A(s) B(g) + D(g)
ii) C(s) E(g) + D(g)
At 20C pressure over excess solid A is 50 atm and that over excess solid C is 68 atm.
Find the total pressure of the gases over the solid mixture.
Solution
When A(s) dissociates alone in a vessel, the pressure over excess solid A is 50 atm. The
product gases B and D are present in the molar ratio of 1:1, this means that the pressure
of each gas will be half of total pressure.
P1K = 25 25 = 625 atm2
P2K = 34 34 = 1156 atm2
But when A(s) and C(s) are placed together in the container, the degree of dissociation of
each will be suppressed by the presence of other. (Le-Chatelier Principle)
A(s) B(g) + D(g)
Page | 26
CHEMISTRY: CHEMICAL EQUILIBRIUM
x x (x < 25)
C(s) E(g) + D(g)
y y (y < 34)
we have to deal with two equilibriums simultaneously. Write separate dissociation for
each equilibrium and then make the concentration or pressure of common species as
same.
A(s) B(g) + D(g)
X (x + y)
C(s) E(g) + D(g)
Y (x + y)
625 = x (x +y) …(1)
1156 = y(x +y) …(2)
From eq. (1) and eq. (2) we get
x = 14.8 atm, y = 27.37 atm.
Total pressure of the gases over the solid mixture=2(x +y)=84.34 atm.
EFFLORESCENT AND DELIQUESCENT
The moisture content of a gas in often expressed in terms of the dew point which is the
temperature to which a gas must be cooled before it becomes saturated with water
vapours. At this temperature, water or ice (depending on the temperature) will be
deposited on a solid surface. Dew point of H2O is 43°C at which vapour pressure of ice
formed is 0.07 Torr.
CaCl2·2H2O(s) CaCl2(s) + 2H2O(g)
CaCl2 (s) has its desiccating properties due to the formation of CaCl2·2H2O(s)
Page | 27
CHEMISTRY: CHEMICAL EQUILIBRIUM
Kp =
2
2H O2
0.07p atm
760
A substance is the most effective drying agent, if it has the lowest vapour pressure of
water in equilibrium with it.
A hydrated salt will lose water (a property called efflorescent) below the vapor pressure
of water at that temperature.
Na2SO4· 10H2O(s) Na2SO4 (s) + 10H2O(g)
At 0°C, Kp = 4.08 × 10-25, Kp = 10H O2
p
H O2p = (4.08 × 10-25)1/10
= 3.64 × 10-3
Hence, NaSO4 · 10H2O will effloresce at a vapour pressure below 3.64 × 10-3 atm at 0°C,
or above relative humidity.
Relative humidity = 2
2
partial pressure of H O100
vapour pressure of H O
An anhydrous salt will absorbs water below relative humidity.
Illustration 10
Equilibrium constants are given (in atm) for the following reactions at 0°C
a) SrCl26H2O(s) SrCl22H2O(s) + 4H2O(g), Kp = 6.89 × 10–12
b) Na2HPO412H2O(s) Na2HPO47H2O(s) + 5H2O(g), Kp = 5.25 × 10–13
c) Na2SO410H2O(s) Na2SO4(s) + 10H2O(g) Kp = 4.08 × 10–25
The vapour pressure of water at 0°C is 4.58 torr.
Page | 28
CHEMISTRY: CHEMICAL EQUILIBRIUM
i) Calculate the pressure of water vapour in equilibrium at 0°C with each of (a), (b) and
(c).
ii) Which is the most effective drying agent at 0°C?
iii) At what relative humidities with Na2SO410H2O be efflorescent when exposed to air
at 0°C?
iv) At what relative humidities will Na2SO4 be deliquescent (i.e,. absorb moisture) when
exposed to air at 0°C?
Solution
From the values of Kp, partial pressure of H2O in each case is calculated. Smaller the
vapour pressure, better the drying the agent.
i) For a) 4
12 3p H O H O2 2
K P 6.89 10 , P 1.62 10 atm
b) 5
13 3p H O H O2 2
K P 5.25 10 , P 3.53 10 atm
c) 10
25 3p H O H O2 2
K P 4.08 10 , P 3.64 10 atm
ii) SrCl22H2O is the most effective drying agent since it has the lowest vapour pressure
of water in equilibrium with it.
iii) The hydrate, Na2SO410H2O will lose water (efflorescence) below 3.64 × 10–3 atm
(2.77 torr) and relative humidity (R.H.) at 0°C will be
2.77
(R.H.) 100 60.5%4.58
iv) The dehydrated salt will absorb water above 60.5% R.H.
Illustration 11
For the reaction : 2 2 2CO (g) H (g) CO(g) H O(g), K is 0.63 at 700°C and 1.66 at
1000°C.
Page | 29
CHEMISTRY: CHEMICAL EQUILIBRIUM
a) What is the average H for the temperature range considered ?
b) What is the value of K at 800°C ?
Solution
a) 1T 700 273 973 K
2T 1000 273 1273 K
1 2K 0.63, K 1.66
using the Vant’ Hoff equation :
2 2 1
1 1 2
K T – THlog
K 2.303R T T
1.66 H 1273 – 973
log0.63 2.303 2 1273 973
3H 8.0 10 cal 8.0 kcal
Note : The units of R and H must be same.
b) Let 2K be the equilibrium constant at 2T 1073 K
1T 973 K and then 1K 0.63
3
2K 8.0 10 1073 – 973log
0.63 2.303(1.99) 1073 973
2K 0.93
Le Chatelier’s Principle and Physical Equilibrium
Le Chatelier’s principle, as already stated, is applicable to all types of equilibria
involving not only chemical but physical changes as well. A few examples of its
application to physical equilbria are discussed below.
1. Vapour pressure of a liquid: Consider the equilibrium
Page | 30
CHEMISTRY: CHEMICAL EQUILIBRIUM
Liquid Vapour
It is well known that the change of a liquid into its vapour is accompanied by absorption
of heat whereas the conversion of vapour into liquid state is accompanied by evolution of
heat. According to Le Chatelier’s principle, therefore, addition of heat to such a system
will shift the equilibrium towards the right. On raising the temperature of the system,
liquid will evaporate. This will raise the vapour pressure of the system. Thus, the vapour
pressure of a liquid increases with rise in temperature.
2. Effect of pressure on the boiling point of a liquid: The conversion of liquid into
vapour, as represented by the above equilibrium, is accompanied by increase of pressure
(vapour pressure). Therefore, if pressure on the system is increased, some of the vapours
will change into liquid so as to lower the pressure. Thus, the application of pressure on
the system tends to condense the vapour into liquid state at a given temperature. In order
to counteract it, a higher temperature is needed. This explains the rise of boiling point of
a liquid on the application of pressure.
3. Effect of pressure on the freezing point of a liquid (or melting point of a solid): At
the melting point, solid and liquid are in equilibrium:
Solid Liquid
Now, when a solid melts, there is usually a change, either increase or decrease, of
volume. For example, when ice melts, there is decrease in volume, or at constant volume,
there is decrease in pressure. Thus, increase of pressure on ice water system at a
constant temperature will cause the equilibrium to shift towards the right, i.e., it will
cause the ice to melt. Hence, in order to retain ice in equilibrium with water at the higher
pressure it will be necessary to lower the temperature. Thus, the application of pressure
will lower the melting point of ice.
When sulphur melts, there is increase in volume or at constant volume, there is increase
in pressure. From similar considerations, it follows that if the pressure on the system,
sulphur (solid) sulphur (liquid) is increased, the melting point is raised.
4. Effect of temperature on solubility: In most cases, when a solute passes into
solution, heat is absorbed, i.e., cooling results. Therefore according to Le Chatelier’s
principle, when heat is applied to a saturated solution in contact with solute, the change
will take place in that direction which absorbed heat (i.e., which tends to produce
cooling). Therefore, some more of the solute will dissolve. In other words, the solubility
of the substance increases with rise in temperature.
Page | 31
CHEMISTRY: CHEMICAL EQUILIBRIUM
Dissociation of a few salts (e.g., calcium salts of organic acids) is accompanied by
evolution of heat. In such cases, evidently, the solubility decreases with rise in
temperature.
Illustration 12
2H and 2I are mixed at 400°C in a 1.0 L container and when equilibrium is established,
the following concentration are present : [HI] = 0.49 M, 2[H ] 0.08 M and 2[I ] 0.06
M. If now an additional 0.3 mol of HI are added, what are the new equilibrium
concentrations, when the new equilibrium 2 2H I 2HI is re-established ?
Solution
First determine the equilibrium constant CK for 2 2H I 2HI .
2 2
C
2 2
[HI] (0.49)K 50
[H ][I ] (0.08)(0.06)
When 0.3 mole of HI are added, equilibrium is disturbed. At instant,
[HI] = 0.49 + 0.3 = 0.79 M
CQ K since 2
C
(0.79)K 130
(0.08)(0.06)
backward reaction dominate and the equilibrium shifts to the left.
Let 2x = concentration of HI consumed (while going left) then concentration of each of
2 2H & I formed = x
[HI] = 0.79 – 2x, 2[H ] 0.08 x , 2[I ] 0.06 x and CK 50
2
C
(0.79 – 2x)K 50
(0.08 x)(0.06 x)
246x 10.2x – 0.35 0
Page | 32
CHEMISTRY: CHEMICAL EQUILIBRIUM
x = 0.033 or –0.25 (neglecting the –ve value)
Finally, the equilibrium concentrations are :
[HI] = 0.79 – 2x = 0.79 – 0.033 2 = 0.724 M
2[H ] 0.08 x 0.08 0.033 0.113 M
2[I ] 0.06 x 0.06 0.033 0.093
RELATION BETWEEN STANDARD FREE ENERGY CHANGE (G0) AND
EQUILIBRIUM CONSTANT
The Gibbs free energy function is a true measure of chemical affinity under conditions of
constant temperature and pressure. The free energy change in a chemical reaction can be
defined as
G = G(products) – G(reactants)
When G = 0, there is no net work obtainable. The system is in a state of equilibrium.
When G is positive, net work must be put into the system to effect the reaction,
otherwise it cannot take place. When G is negative, the reaction can proceed
spontaneously with accomplishment of the net work. The larger the amount of this work
that can be accomplished, the farther away is the reaction from equilibrium. For this
reason –G has often been called the driving force of the reaction. From the statement of
the equilibrium law, it is evident that the driving force depends on the concentration of
the reactants and products. It also depends upon the temperature and pressure which
determine the molar free energies of the reactants and products.
The reaction conducted at constant temperature (i.e., in a thermostat)
–G = – H + TS
The driving force is made up of two parts, –H term and TS term. The –H term is the
heat of reaction at constant pressure and TS is heat involved when the process is carried
out reversibly. The difference is the amount of heat of reaction which can be converted
into net work (–G), i.e., total heat minus unavailable heat.
Page | 33
CHEMISTRY: CHEMICAL EQUILIBRIUM
If the reaction is carried out at constant volume, the decrease in Helmholtz function –G
= –E + TS would be the proper measure of affinity of the reactant or the driving force
of the reaction.
Now we can see why Berthellot and Thompson were wrong in assuming that driving
force of the reaction was the heat of reaction. They neglected the TS term. The reasons
for the apparent validity of their principle was that for many reactions, H term far
outweighs the TS term. This is especially true at low temperature, since at higher
temperature, TS term increases.
The fact that driving force for a reaction is large (G is large negative quantity) does not
mean that the reaction will necessarily occur under any given conditions.
For example, the reaction
2 2 2
1H O H O; G 228.6kJ
2
does not occur at the laboratory temperature. The reaction mixture may be kept for years
without any detectable formation of water. Here H factor favours, but S factor
disfavours the reaction.
Similarly, the reaction
2Mg + O2 → 2MgO; G = – 570.6kJ
is not favoured. However, the thermite reaction
2Al + 3
2O2 → Al2O3
with large value of – G proceeds favourably.
Standard Free Energy and Equilibrium Constant
The change in free energy for a reaction taking place between gaseous reactants and
products represented by the general equation.
aA bB cC dD
Page | 34
CHEMISTRY: CHEMICAL EQUILIBRIUM
According to Van’t Hoff reaction isotherm
c d0 C D
a bA B
p pG G RTln
p p
= G0 + RTlnQp
the condition for a system to be at equilibrium is that
G = 0
Thus at equilibrium
c d0 0 0C D
pa bA B
p p0 G RTln G RTln K
p p
Whence G0 = – RTlnK0p
Hence 0
0p
Gln K
RT
Note
1. In the reaction, where all gaseous reactants and products; K represents Kp
2. In the reaction, where all solution reactants and products; K represents Kc
3. A mixture of solution and gaseous reactants; Kx represents the thermodynamic
equilibrium constant and we do not make the distinction between Kp and Kc.
we may conclude that for standard reactions, i.e., at 1 M or 1 atm.
When G0 = –ve or K 1: forward reaction is feasible
G0 = +ve or K 1: reverse reaction is feasible
G0 = 0 or K = 1: reaction is at equilibrium (very rare)
Illustration 13
Page | 35
CHEMISTRY: CHEMICAL EQUILIBRIUM
NO and Br2 at initial pressures of 98.4 and 41.3 torr respectively were allowed to react at
300K. At equilibrium the total pressure was 110.5 torr. Calculate the value of equilibrium
constant , Kp and the standard free energy change at 300K for the reaction
(g) 2(g) (g)2NO Br 2NOBr
Solution
From the initial and final pressures of the reaction species, partial pressures of each are
determined, hence Kp and G0 are calculated
2NO()g) + Br2(g) 2NOBr(g)
Initial 98.4 41.3 0
Used up 2x x ––
Produced –– –– 2x
Equilibrium (98.4 – 2x) (41.3 – x) 2x
Total pressures at equilibrium = NO Br NOBr2P P P
= (98.4 2x) (41.3 x) 2x 139.7 x 110.5
This gives, x = 29.2
Hence, NOP (98.4 2 29.2) 40 Torr
Br2P (41.3 29.2) 12.1 Torr
NOBrP 2 29.2 58.4 Torr
2 2
-1 -1NOBrp 2 2
NO Br2
P (58.4)K 0.1762 Torr 133.88 atm
P P (40) (12.1)
0G 2.303RT log K 2.303 8.314 300 log133.88
Page | 36
CHEMISTRY: CHEMICAL EQUILIBRIUM
= 12216.26J = 12.22 kJ mol–1
SOLVED EXAMPLES
Example 1
The equilibrium constant for the reaction 2 2 32SO O 2SO at 1000K is 3.5. What
would partial pressure of oxygen gas have to be, to give equal moles of 2SO and 3SO ?
(a) 0.29 atm (b) 3.5 atm
(c) 0.53 atm (d) 1.87 atm.
Solution
2SO3
p 2OSO O 22 2
(p ) 1K
p(p ) p
Here, partial pressure of 3SO and 2SO same because moles of 2SO and 3SO are equal
at equilibrium.
O2p
1p 0.285
K
Ans. (a)
Example 2
Page | 37
CHEMISTRY: CHEMICAL EQUILIBRIUM
5PCl is 50% dissociated into 3PCl and 2Cl at 1 atmosphere. It will be 40% dissociated
at :
(a) 1.75 atm (b) 1.84 atm
(c) 2.00 atm (d) 1.25 atm.
Solution
5 3 2PCl PCl Cl
Let x = degree fo dissociation of 5PCl at pressure P
y = degree of dissociation of 5PCl at pressure p
2 2
p 2 2
px p yK
(1– x ) (1– y )
given, x = 0.50, p = 1 atm
y = 0.40, p ?
1 0.25 p 0.16
0.75 0.84
p 1.75atm
Ans. (a)
Example 3
For the reaction
2 4(s) 3(g) 2(g)NH COONH 2NH CO
The equilibrium constant –5 3pK 2.9 10 atm . The total pressure of gases at equilibrium
when 1.0 mole of reactant was heated will be :
Page | 38
CHEMISTRY: CHEMICAL EQUILIBRIUM
(a) 0.0194 atm (b) 0.0388 atm
(c) 0.0580 atm (d) 0.0667 atm.
Solution
2 4(s) 3(g) 2(g)NH COONH 2NH CO
Initially 1 0 0
At equi. (1–x) 2x x
Total moles of gaseous substances at equilibrium = 3x
Equilibrium pressure = P
NH CO3 2
2 pp p, p
3 3
2p NH CO3 2
K (p ) p
–5 342.9 10 p
27
p = 0.0580 atm
Ans. (c)
Example 4
One mole of 2 4(g)N O at 300 K is kept in a closed container under one atmosphere. It is
heated to 600 K when 20% of 2 4(g)N O decomposes to 2(g)NO . The resultant pressure is
(a) 1.2 atm(b) 2.4 atm
(c) 2.0 atm(d) 1.0 atm.
Solution
Page | 39
CHEMISTRY: CHEMICAL EQUILIBRIUM
2 4 2N O 2NO
Initially 1 0
At equi. (1–x) 2x
Total mole at equilibrium = 1 + x = 1 .2 x 0.2
1 2 2
1 1 2 2
P P P1or
n T n T 1 300 1.2 600
or 2P 2.4atm
Ans. (b)
Example 5
For the reversible reaction, 2(g) 2(g) 3(g)N 3H 2NH at 500°C, the value of pK is
1.44 –510 when partial pressure is measured in atmospheres. The corresponding value
of cK with concentration in mole –1litre , is
(a) 1.44 –5 –210 / (0.082 500) (b)
–5 –21.44 10 / (8.314 773)
(c) 1.44 –5 210 / (0.082 773) (d)
–5 –21.44 10 / (0.082 773) .
Solution
– ng
c pK K (RT)
Since gn –2
–5 2cK 1.44 10 (0.082 773)
Ans. (d)
Example 6
Page | 40
CHEMISTRY: CHEMICAL EQUILIBRIUM
One mole of N2O(g) is kept in a closed container along with gold catalyst at 450 K under
one atmosphere. It is heated to 900 K when it dissociates to N2(g) and O2(g) giving an
equilibrium pressure of 2.4 atm. The degree of dissociation of N2O(g) is
(a) 20% (b) 40%
(c) 50% (d) 60%.
Solution
122 (g) 2(g) 2(g)N O N O
0P (1– ) 0P 0P2
When temperature of 2N O is doubled, its pressure also doubles.
Pressure of 2N O at 900K, 0P 2atm
Total pressure at equilibrium = 2.4 atm
0P 1 2.42
0.4
Degree of dissociation of 2 (g)N O 40%
Ans. (b)
Example 7
For the reaction : (g) 2(g) 2(g)2HI H I ; the degree of dissociation ( ) of HI(g) is
related to equilibrium constant pK by the expression
(a) p1 2 K
2
(b)
p1 2K
2
Page | 41
CHEMISTRY: CHEMICAL EQUILIBRIUM
(c) p
p
2K
1 2K (d)
p
p
2K
1 2 K.
Solution
(g) 2(g) 2(g)2HI H I
1– 2
2
2
p 2 2
P2
K(1– ) P
p2 K1–
p
p
2
1 2 K
Ans. (d)
Example 8
For which of the following reactions, the degree of dissociation cannot be calculated from
the vapour density data
I (g) 2(g) 2(g)2HI H I II 3(g) 2(g) 2(g)2NH N 3H
III (g) 2(g) 2(g)2NO N O IV 5(g) 3(g) 2(g)PCl PCl Cl
(a) I and III (b) II and IV
(c) I and II (d) III and IV.
Solution
Page | 42
CHEMISTRY: CHEMICAL EQUILIBRIUM
The degree of dissociation cannot be calculated from the vapour density data if the
number of moles remains unchanged before and after reaching equilibrium.
Ans. (a)
Example 9
At a certain temperature the following equilibrium is established.
(g) 2(g) 2(g) (g)CO NO CO NO
One mole of each of the four gases is mixed in one litre container and the reaction is
allowed to reach equilibrium state. When excess of baryta water is added to the
equilibrium mixture, the weight of white precipitate obtained is 236.4 g. The equilibrium
constant cK of the reaction is
(a) 1.2 (b) 2.25
(c) 2.1 (d) 3.6.
Solution
(g) 2(g) 2(g) (g)CO NO CO NO
Initially 1 1 1 1
At equi. 1–x 1–x 1 + x 1 + x
2 2 3 2CO Ba(OH) BaCO H O
Moles of 3
236.4BaCO 1.2
197
Moles of 2CO at equilibrium = 1.2
or, 1 + x = 1.2; x = 0.2
Page | 43
CHEMISTRY: CHEMICAL EQUILIBRIUM
2 2
c
1 x 1.2K 2.25
1– x 0.8
Ans. (b)
Example 10
In a reversible reaction, study of its mechanism says that both the forward and backward
reactions follow first order kinetics. If the half life of forward reaction 1/2 f(t ) is 400 sec.
and that of backward reaction 1/2 b(t ) is 250 sec. The equilibrium constant of the reaction
is
(g) 2(g) 2(g) (g)CO NO CO NO
(a) 1.6 (b) 0.433
(c) 0.625 (d) 1.109.
Solution
–1f
0.693k sec
400 ; –1
b
0.693k sec
250
f
b
k 250K 0.625
k 400
Ans. (c)
Example 11
At 27°C NO and 2Cl gases are introduced in a 10 litre flask such that their initial partial
pressures are 20 and 16 atm respectively. The flask already contains 24 g of magnesium.
After some time, the amount of magnesium left was 0.2 moles due to the establishment of
following two equilibrium
(g) 2(g) (g)2NO Cl 2NOCl
Page | 44
CHEMISTRY: CHEMICAL EQUILIBRIUM
–12(g) (s) 2(s) pCl Mg MgCl ; K 0.2 atm
The final pressure of NOCl would be
(a) 7.84 atm (b) 18.06 atm
(c) 129.6 atm (d) 64.8 atm.
Solution
p
Cl2
1K 0.2
P Cl2
P at equilibrium = 5 atm
(g) 2(g) (g)2NO Cl 2NOCl
Initially 20 16 0
At equi. 20–2x 16–x–y 2x
2(g) (s) 2(s)Cl Mg MgCl
Initially 16 1 0
At equi. 16–x–y 1– y y
= 0.2
y= 0.8
(since y is the moles while y is pressure in atm)
0.8 0.082 300
y 1.9710
We know that,
16 – x – y 5
Page | 45
CHEMISTRY: CHEMICAL EQUILIBRIUM
or, x = 9.03; NOClP 18.06 atm
Ans. (b)
Example 12
The 3CaCO is heated in a closed vessel of volume 1 litre at 600 K to form CaO and
2CO . The minimum weight of 3CaCO required to establish the equilibrium
3(s) (s) 2(g)CaCO CaO CO is p(K 2.25 atm)
(a) 2g (b) 4.57 g
(c) 10g (d) 100 g.
Solution
p CO2K P 2.25 atm
Number of moles of 2
2.25 1CO
0.0821 600
The minimum moles of 3CaCO required = 0.0457
The minimum weight 3CaCO required = 0.0457 100 = 4.57 g
Ans. (b)
Example 13
An acid reacts with glycerine to form complex and equilibrium is established. If the heat
of reaction at constant volume for above reaction is 1200 cal more than at constant
pressure and the temperature is 300 K, then which of the following expression is true ?
(a) p cK K (b) c pK K
(c) p cK K (d) none of these.
Solution
Page | 46
CHEMISTRY: CHEMICAL EQUILIBRIUM
Given : E – H 1200 cal
H E nRT
n –2
np cK K (RT)
p –3
c
K1.648 10
K
p cK K
Ans. (a)
Example 14
When pure NH3 is maintained at 480°C and a pressure of 1 atm, it dissociates to give a
gaseous mixture containing 20% NH3 by volume. The degree of dissociation of NH3 is
(a) 2/3 (b) 3/2
(c) 5/2 (d) 2/5.
Solution
3(g) 2(g) 2(g)2NH N 3H
Initial a 0 0
At equi. a(1– ) a /2 3a /2
% of 3NH by volume = a(1– ) 1–
100 100a 3a 1
a(1– )2 2
By condition,
Page | 47
CHEMISTRY: CHEMICAL EQUILIBRIUM
1–
10 201
2
3
Ans. (a)
Example 15
Steam at pressure of 1 atm is passed through a furnace at 2000 K wherein the reaction –51
22 (g) 2(g) 2(g) pH O H O ; K 6.4 10 occurs. The percentage of oxygen in the
exit steam would be
(a) 0.32 (b) 0.08
(c) 0.04 (d) 0.16
Solution
122 (g) 2(g) 2(g)H O H O
Initially 2 atm 0 0
At equi. 2 – x x x/2
1/2
–5x(x / 2)6.4 10
2 – x
3/2
–5x6.4 10
2(2 – x)
Dissociation of 2H O is so small that we can assume H O2P 2 atm
i.e. 2 – x 2
3/2 –5x 2 2 6.4 10 x or
3/2 3/2 –6x (2 16) 10
Page | 48
CHEMISTRY: CHEMICAL EQUILIBRIUM
x = 2 –416 10 0.0032
% of 2O in the exit steam 0.0032 / 2
100 0.08%2
Ans. (b)
Example 16
Two moles of an equimolar mixture of two alcohols, 1R — OH and 2R — OH are
esterified with one mole of acetic aicd. If 80% of the acid is consumed and the quantities
of ester formed under equilibrium are in the ratio of 3 : 2, the value of the equilibrium
constant for the esterification of 1R — OH with acetic acid is
(a) 3.3 (b) 3.7
(c) 3.5 (d) 3.9 .
Solution
1 3 2 3 2 1 2R OH CH CO H CH CO R H O
(1– x) (1– x – y) x (x + y)
2 3 2 3 2 2 2R OH CH CO H CH CO R H O
(1– y) (1– x – y) y (x + y)
Since, the amount of acid consumed is 80% in complete esterification process, therefore
x y 0.8
Also, x 3
y 2
x = 0.48 and y = 0.32
Page | 49
CHEMISTRY: CHEMICAL EQUILIBRIUM
Thus, equilibrium constant for first esterification process x(x y)
(1– x)(1– x – y)
C1
0.48 0.8K
0.52 0.2
C1K 3.7
Ans. (b)
Example 17
Solubility of a substance which dissolves with a decrease in volume and absorption of
heat will be favoured by
(a) high P and high T (b) low P and low T
(c) high P and low T (d) low P and high T.
Solution
The substance dissolves with a decrease in volume and absorption of heat. Therefore
according to Le Chalelier’s principle the process of dissolution will be favoured by high
pressure and high temperature.
Ans. (a)
Example 18
X2 + X– X3– (x = iodine)
This reaction is set up in aqueous medium. We start with 1 mol of X2 and 0.5 mol of X–
in 1L flask. After equilibrium is reached, excess of AgNO3 gave 0.25 mol of yellow ppt.
equilibrium constant is
(a) 1.33 (b) 2.66
(c) 2.00 (d) 3.00
Solution
Page | 50
CHEMISTRY: CHEMICAL EQUILIBRIUM
X2 + X– 3X
1 0.5 0
(1 – x) (0.5 – x) x
(0.5 – x) = unreacted X–
X– Ag + 0.25
x = 0.25
3c
2
[X ] 0.25K 1.33
0.75 0.25[X ][X ]
Ans. (a)
Example 19
At temperature T, a compound AB2(g) dissociates according to the reaction 2AB2(g)
2AB(g) + B2(g) with degree of dissociation , which is small compared with
unity. The expression for Kp, in terms of and the total pressure, PT is
(a) 3
TP
2
(b)
2TP
3
(c) 3
TP
3
(d)
2TP
2
.
Solution
For the given equilibrium, the equilibrium concentrations are
2AB2(g) 2AB(g) + B2(g)
Equilib. conc. c(1 – ) c c
2
Page | 51
CHEMISTRY: CHEMICAL EQUILIBRIUM
22
TB AB2
P 22AB2
c(c ) P(P )(P ) 2K
(P ) [c(1– )] [c(1 )]2
; 3
TP
2
PK
2(1– ) 12
Since, is small compared to unity, so 1 – 1 and 1 + 12
.
3
TP
PK
2
Ans. (a)
Example 20
2.0 mol of PCl5 were introduced in a vessel of 5.0 L capacity at a particular temperature.
At equilibrium, PCl5 was found to be 35% dissociated into PCl3 and Cl2. The value of Kc
for the reaction is
(a) 1.89 (b) 0.377
(c) 0.75 (d) 0.075.
Solution
Moles of PCl5 dissociated = 2 35
100
= 0.7
Moles of PCl5 left undissociated = 2 – 0.7 = 1.3 mol
[PCl5] = 1.3
5M, [PCl3] =
0.7
5M, [Cl2] =
0.7
5M
3 2
5
0.7 0.7
[PCl ][Cl ] 5 5K 0.75
1.3[PCl ]
5
Ans. (c)
Page | 52
CHEMISTRY: CHEMICAL EQUILIBRIUM
SOLVED SUBJECTIVE EXAMPLES
Example 1
The equilibrium constant for the reaction H2 + I2 2HI; is found to be 64 at 450°C. If
6 mole of hydrogen are mixed with 3 mole of iodine in a litre vessel at this temperature;
what will be the concentration of each of the three components, when equilibrium is
attained ?
It the volume of reaction vessel is reduced to half; then what will be the effect on
equilibrium ?
Solution
The given equilibrium reaction is :
H2 + I2 2HI
At t = 0 6 3 0
At equilibrium 6 – x 3 – x 2x
concentration at
equilibriu
m
6 x
1
3 x
1
2x
1
2
C
(2x)K
(6 x)(3 x)
2
2
4x64
x 18 9x
Page | 53
CHEMISTRY: CHEMICAL EQUILIBRIUM
2 216(x 9x 18) x
On solving x = 2.84
[HI] = 2x 2 2.84
5.681 1
[H2] = 6 x 6 2.84
3.161 1
[I2] = 3 x 3 2.84
0.161 1
In the present reaction n = 0 hence, volume change will not affect the equilibrium.
Example 2
5 gm of PCl5 were completely vaporized at 250°C in a vessel of 1.9 litre capacity. The
mixture at equilibrium exerted a pressure of one atmosphere. Calculate the degree of
dissociation, KC and Kp for this reaction.
Solution
No. of moles of PCl5
= Weight
Molecularweight =
50.024
208.5
The equilibrium for dissociation of PCl5 maybe represented as
PCl5 PCl3 + Cl2
At t = 0 0.024 0 0
At equilibrium (0.024 - x) X x
Page | 54
CHEMISTRY: CHEMICAL EQUILIBRIUM
Total moles of gas components
= 0.024 – x + x + x = (0.024 + x)
We know, PV = nRT
1 1.9 = (0.024 + x) 0.0821 × 523
(0.024 + ) = 1.9
0.0821 523
x = 0.0202
Degree of dissociation = 0.0202
0.024 = 0.843
[PCl5] = 0.024 x 0.024 0.0202 0.0038
1.9 1.9 1.9
[PCl3] = [Cl2] = x 0.0202
V 1.9
Kc =
2
43 2
5
0.0202
[PCl ][Cl ] 4.08 101.9
0.0038[PCl ] 1.9 0.0038
1.9
= 0.0565 mole/litre
Kp = Kc (RT) n
Where Kc = 0.0565
n = 2 – 1 = 1
Kp = 0.0565 × 0.0821 × 523 = 2.43 atm
Example 3
16 moles of H2 and 4 moles of N2 are sealed in a one litre vessel. The vessel is heated at a
constant temperature until the equilibrium is established, when it is found that the
pressure in the vessel has fallen to 9/10 of its original value. Calculate KC for the reaction
Page | 55
CHEMISTRY: CHEMICAL EQUILIBRIUM
N2(g) + 3H2(g) 2NH3(g)
Solution
The given equilibrium is
N2(g) + 3H2(g) 2NH3(g)
At t = 0 4 16 0
At equilibrium 4 – x 16 – 3x 2x
Total gaseous moles at equilibrium
= 4 – x + 16 – 3x + 2x = (20 – 2x)
Since, pressure has fallen to 9/10 of its original value, hence no. of mole will also fall up
to the same extent.
(20 – 2x) = 9
20 1810
x = 1
[N2] = 4 x 4 1
31 1
mole/litre
[H2] = 16 3x
1
= 13 mle/litre
[NH3] = 2x
1 = 2 mole/litre
KC = 2 2
3
3 32 2
[NH ] 2
[N ][H ] (3)(13) = 6.07 10–4
Page | 56
CHEMISTRY: CHEMICAL EQUILIBRIUM
Example 5
The value of K for the reaction
O3(g) + OH(g) H(g) + 2O2(g)
Changed from 0.096 at 298K to 1.4 at 373K. Above what temperature will the reaction
become thermodynamically spontaneous in the forward direction assuming that H0 and
S0 values for the reaction do not change with change in temperature? Given that, S2980
=10.296JK–1.
Solution
We have,
0
2 2 1
1 1 2
K (T T )Hlog
K 2.303R T T
01.4 H 373 298
log0.096 2.303 8.314 373 298
H0= 33025J
Now the temperature above which the forward reaction will be spontaneous is actually
the temperature at which the reaction attains equilibrium, that is, when
K = 1 or log K = 0
G0 = – 2.303 RT log K = – 2.303 RT log 1.0 = 0
From thermodynamics, we get
G0 = H0 – TS0
0 = 33025 – T × 10.296
or T = 320.75 K
Example 6
Page | 57
CHEMISTRY: CHEMICAL EQUILIBRIUM
A mixture of SO3, SO2 and O2 gases is maintained in a 10.0 litre flask at a temperature at
which equilibrium constant Kc for the reaction
2SO2(g) + O2(g) 2SO3(g) is 100.
i) If the number of mole of SO2 and SO3 in the flask are equal; how many mole of O2 are
present ?
ii) If the number of mole of SO3 in the flask is twice the number of mole of SO2,how
many mole of O2 are present ?
Solution
The given equilibrium is
2SO2(g) + O2(g) 2SO3(g)
2
3C 2
2 2
[SO ]K
[SO ] [O ] … (i)
i) If [SO3] = [SO2]
Then C
2
1K
[O ]
2
C
1 1[O ] 0.01
K 100 mole/litre
Total moles of O2 present = 0.01 10 = 0.1 mole
Volume of vessel is 10 litre
ii) When [SO3] = 2[SO2]
Then 2
3C 2
2 2
[SO ]K
[SO ] [O ] from Eq. (i)
2
4100
[O ]
Page | 58
CHEMISTRY: CHEMICAL EQUILIBRIUM
[O2] = 4/100 = 0.04 mole/litre
Total moles of O2 in vessel at equilibrium = 0.04 10 = 0.4 mole
Example 7
A saturated solution of iodine in water contains 0.33g of I2 per litre of solution. More
than this can dissolve in KI solution because of the following equilibrium.
I2(aq) + I– I3–
A 0.1 M KI solution actually dissolves 12.5g of I2/litre, most of which is converted to I3–.
Assuming that the concentration of I2 in all saturated solutions is the same, calculate the
equilibrium constant for the above reaction.
Solution
I2 + I– I3–
Initially 12.5
M254
0.1M 0
At equb. 0.33
M254
12.17
0.1 M254
12.17M
254
Thus,
3
2
12.17[I ] 254K 708
0.33 13.23[I ][I ]
254 254
Example 8
For the reaction Ag(CN)2– Ag+ + 2CN–, the KC at 25°C is 4 10–19. Calculate
[Ag+] in solution which was originally 0.1 M in KCN and 0.03 M in AgNO3.
Solution
Page | 59
CHEMISTRY: CHEMICAL EQUILIBRIUM
2KCN + AgNO3 Ag(CN 2) + KNO3 + K+
0.1 0.03 0 0 0
(0.1 – 0.06) 0 0.03 0.03 0.03
[Ag(CN 2) ] = 0.03 M
Now use Ag(CN 2) Ag+ + 2CN–
0.03 0 0.04
(0.03 – a) a 0.04 + a = 0.04
Since CK is too small and dissociation of Ag(CN 2) is very less and thus,
0.04 + a 0.04 and 0.03 – a 0.03
[Ag(CN 2) ] = 0.03; [Ag+] = a; [CN–] = 0.04
Now 2 2
C
2
[Ag ][CN ] a (0.04)K
0.03[Ag(CN) ]
a = 7.5 10–18
Example 9
When baking soda is heated in a sealed tube, following equilibrium exists:
2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g)
If the equilibrium pressure is 1.04 atm at 398 K, calculate the equilibrium constant for the
reaction at 398 K.
Solution
Since, there are only two gaseous species in the above equilibrium
Page | 60
CHEMISTRY: CHEMICAL EQUILIBRIUM
CO H O2 2
1p p
2 total pressure
1
1.04 0.52atm2
2p CO H O2 2
K p p [0.52]
Example 10
For the reaction
NH3(g) 1
2N2(g) +
3
2H2(g)
Show that degree of dissociation of NH3 is given as :
1/2
p
3 3 P1
4 K
where ‘P’ is equilibrium pressure. If Kp of the above reaction is 78.1 atm at 400°C,
calculate KC.
Solution
NH3(g) 1
2N2(g) +
3
2H2(g). Total moles
Initial moles t = 0 1 0 0 1
Moles equilibrium 1 – /2 3 /2 1 +
Partial pressure 1
p1
p2(1 )
3p
2(1 )
1/2 3/2
2 2p
3
[N ] [H ]K
[NH ]
Page | 61
CHEMISTRY: CHEMICAL EQUILIBRIUM
1/2 3/23
p p2(1 ) 2(1 )
1p
1
2
2
p 27
4(1 )
Solving for ` ’ we get :
1/2
p
3 3 p1
4 K
We know,
np CK K (RT)
78.1 = CK (0.0821 673)1
KC = 1.413 moles litre–1
Example 11
Kp for the reaction N2O4 (g) 2NO2 (g) is 0.66 at 46C. Calculate the percent
dissociation of N2O4 at 46C and a total pressure of 0.5 atm. Also calculate the partial
pressure of N2O4 and NO2 at equilibrium.
Solution
This Example can be solved by two methods.
Method 1: Let the number of moles of N2O4 initially be 1 and is the degree of
dissociation of N2O4.
N2O4 2NO2
Initial moles 1 0
Moles at equilibrium 1– 2
Total moles at equilibrium = 1– + 2 = 1+
Page | 62
CHEMISTRY: CHEMICAL EQUILIBRIUM
N O T2 4
1p P
1
NO T2
2p P
1
Kp =
2 2 2NO2 T
2N O2 4
p 4 P 4 0.5
p 1 1 1
= 0.5 i.e., 50% dissociation
Hence, partial pressure of N2O4 = 0.167 atm.
and partial pressure of NO2 = 0.333 atm.
Method 2 : Let the partial pressure of NO2 at equilibrium be p atm, than the partial
pressure of N2O4 at equilibrium will be (0.5–p)atm.
Kp =
2p0.66
0.5 p
p2 + 0.66 p–0.33 = 0
On solving p = 0.333 atm.
NO2p = 0.333 atm and N O2 4
p = 0.167 atm.
Example 12
At 1000 K, the pressure of iodine gas is found to be 0.112 atm due to partial dissociation
of I2(g) into I. Had there been no dissociation, the pressure would have been 0.074 atm.
Calculate the value of Kp for the reaction: I2(g) 2I(g).
Solution
Analysing in terms of pressure directly
Page | 63
CHEMISTRY: CHEMICAL EQUILIBRIUM
Partial pressure I2 2I
Initial 0.074 0
At equilibrium 0.074- p 2p
total pressure at equilibrium
= (0.074 – p) + 2p = 0.112 (given) p = 0.038 atm
Kp =
2 2
I
I2
P 2p
P 0.074 p
Substituting value of p Kp = 0.16 atm
Example 13
Determine Kc for the reaction ½N2(g) + ½O2(g) + ½Br2(g) NOBr(g) from the
following information (at 298oK)
Kc = 2.4 1030 for 2NO(g) N2(g) + O2(g) ; '''CK = 1.4 for NO(g) + ½Br2(g) NOBr(g)
Solution
N2(g) + O2(g) 2NO (g)
Kc/ =
30c
1 1
K 2.4 10
i) ½N2(g) + ½O2(g ) NO(g)
Kc// =
/ 30 15c
1K 10 0.6455 10
2.4
ii) NO(g) + ½Br2(g) NOBr(g) '''CK = 1.4.
(i) + (ii) gives the net reaction:
Page | 64
CHEMISTRY: CHEMICAL EQUILIBRIUM
½N2(g) + ½O2(g) + ½Br2(g) NOBr(g)
K =
1 1 1 1 1 1
2 2 2 2 2 22 2 2 2 2 2
NO NOBr NOBr
N O NO Br N O Br
= ''CK '''
CK = 0.6455 10-15 1.4 = 9.037 10-16
Example 14
In a mixture of N2 and H2 initially in a mole ratio of 1:3 at 30 atm and 300oC , the
percentage of ammonia by volume under the equilibrium is 17.8. Calculate the
equilibrium constant (KP) of the mixture, for the reaction
N2(g) + 3H2(g) 2NH3(g)
Solution
Let the initial moles of N2 and H2 be 1 and 3 respectively (this assumption is valid as KP
will not depend on the exact no. of moles of N2 and H2. One can even start with x and
3x).
Alternatively
N2(g) + 3H2(g) 2NH3
Initial 1 3 0
At eqb 1x 33x 2x
Since % by volume of a gas is same as % by mole,
2x
4 2x=0.178
4 0.178
x 0.302(2 2 0.178)
Mole fraction of H2 at equilibrium = 3 3x
0.61654 2x
Mole fraction of N2 at equilibrium = 1 0.6165 0.178 = 0.2055
KP =
2
NH T3
3
N T H T2 2
X P
X P X P
2
3
0.178 30
(0.2055 30) (0.6165 30)
= 7.31 10-4 atm–2
Example 15
Page | 65
CHEMISTRY: CHEMICAL EQUILIBRIUM
The density of an equilibrium mixture of N2O4 and NO2 at 1 atm. and 348 K is 1.84 g
dm-3. Calculate the equilibrium constant of the reaction
N2O4(g) 2NO2(g).
Solution
Let us assume that we start with C moles of N2O4(g) initially.
N2O4(g) 2NO2(g)
Initial 0 0
At equilibrium C(1) 2C
Where is the degree of dissociation of N2O4(g)
Since =Total moles at equilibrium Vapour density initial
Total moles initially Vapour density at equilibrium
Initial vapour density = 92
462
C(1 ) 46
C d
Since vapour density and actual density are related by the equation,
V.D. = RT
2P
=
1.84 0.082 348
2
= 26.25
1 + = 46
1.75226.25
= 0.752
Kp =
2 2
T
T
2 C 2 0.75P 1
C(1 ) 1.752
C(1 ) 0.248P 1
C(1 ) 1.752
= 5.2 atm.
Example 16
In an evacuated vessel of capacity 110 litres, 4 moles of Argon and 5 moles of PCl5 were
introduced and equilibrated at a temperature of 250oC. At equilibrium, the total pressure
of the mixture was found to be 4.678 atm. Calculate the degree of dissociation, of PCl5
and KP for the reaction PCl5 PCl3 + Cl2 at this temperature.
Solution
Page | 66
CHEMISTRY: CHEMICAL EQUILIBRIUM
PCl5(g) PCl3(g) + Cl2(g)
Initial moles 5 0 0
At equilibrium 5x x x
Total moles = 5 + x + 4 (including moles of argon) = 9 + x
Since total moles = PV 4.67 110
RT 0.082 523
= 12
x = 3
=3
0.65 ; KP =
23
4.6712
24.67
12
= 1.75
Example 17
256 g of HI was heated in a sealed bulb at 444°C till the equilibrium was attained. The
acid was found to be 22% dissociated at equilibrium. Calculate equilibrium constants for
synthesis and dissociation of HI?
Solution
2HI (g) H2 (g) + I2 (g)
Initial moles 256
128 0 0
Moles at equilibrium 2–2
(given = 0.22)
Equilibrium constant for dissociation
KC=
2
22 2
=
2
0.22 0.22
2 0.44
= 0.0199
Equilibrium constant for synthesis = C
1 1
K 0.0199 = 50.25
(H2 + I2 2HI )
Page | 67
CHEMISTRY: CHEMICAL EQUILIBRIUM
[ Note : The equilibrium reaction considered for dissociation is 2HI H2 + I2 and not
HI ½ H2 + ½ I2 because for synthesis, the reaction is with 1 mole each of H2 and I2
i.e., H2 + I2 2HI ]
Example 18
Variation of equilibrium constant K with temperature T is given by Van’t Hoff equation
log K = log A –0H
2.303RT
log K
T–1
O
P
A graph between log K and T–1 was a straight line as shown and having OP = 10 and tan
= 0.5. Calculate
i) equilibrium constant at 298 K, and
ii) and equilibrium constant at 798 K, assuming H to be independent of temperature.
Solution
To calculate equilibrium constant, we need to know A and H, which are calculated as –
The given equation represents a straight line of slope = oH
2.303 R
= –tan = –0.5
H = 2.303 8.314 0.5 = 9.574 J/mol
Intercept = log A = OP = 10
log K = log A–oH 9.574
102.303 RT 2.303 8.314 298
K = 9.96 109
Page | 68
CHEMISTRY: CHEMICAL EQUILIBRIUM
Now, to calculate equilibrium constant at some other temperature, we will use the
expression
2
1
Klog
K=
o
1 2
H 1 1
2.303 R T T
log 2
9
K
9.96 10 =
9.574
2.303 8.314
500
798 298
K2 (equilibrium constant at 798 = 9.98 109
Example 19
The equilibrium constant KP of the reaction
2SO2(g) + O2(g) 2SO3(g)
is 900 atm–1 at 800C. A mixture containing SO3 and O2 having initial partial pressures of
1 atm and 2 atm respectively is heated at constant volume to equilibrate. Calculate the
partial pressure of each gas at 800C.
Solution
It can be seen that as SO2 is not present initially, so equilibrium cannot be established in
the forward direction. Therefore it is established from reverse direction. Let n be the
increase in partial pressure of O2. Then at equilibrium the partial pressures of SO2, O2 at
SO3 are (0+2n), (2+n) and (1–2n) in atm respectively.
2SO2(g) + O2(g) 2SO3(g)
(0+2n) (2+n) (1–2n)
Also,
22SO3
P 2 2SO O2 2
p 1 2nK
p p 0 2n 2 n
As n is small (because equilibrium constant for the reverse reaction is very small i.e.,
1/900), it can be neglected in comparison to 2 and also
1–2n can be taken approximately to 1.
Page | 69
CHEMISTRY: CHEMICAL EQUILIBRIUM
900 = 2
1
4n 2
Solving for n, we get n = 0.0118
Hence, O2p = 2+n = 2.0118 atm
SO2p = 2n = 0.0236 atm
SO3p = 1–2n = 1–0.0236 = 0.9764 atm
Example 20
In an experiment 5 moles of HI were enclosed in a 5 litre container. At 717 K equilibrium
constant for the gaseous reaction 2HI (g) H2 (g) + I2 (g) is 0.025. Calculate the
equilibrium concentrations of HI, H2 and I2. What is the fraction of HI that decomposes.
Solution
Let 2n be the number of moles of HI which is decomposed, the number of moles of H2
and I2 produced will be n moles each. Then molar concentrations of various species at
equilibrium are –
[HI] = 5 2n
5
mol/l, [H2] =
n
5 mol/l, [I2] =
n
5 mol/l
Also, KC =
2 2
2
H I
HI =
2
n n
5 5
5 2n
5
0.025 =
2
2
n
5 2n
Solving for n, we get n = 0.6
[HI] = 5 2 0.6
5
=
3.8
5= 0.76 mol/l
Page | 70
CHEMISTRY: CHEMICAL EQUILIBRIUM
[H2] = 0.6
5 = 0.12 mol /l
[I2] = 0.6
5 = 0.12 mol /l
Fraction of HI decomposed = 2 0.6
5
= 0.24 or 24%