Assignment chemical equilibrium_jh_sir-4168

Topic Page No.
Theory 01 - 05
Exercise - 1 06 - 20
Answer Key 42 - 43
Contents
Chemical Equlibrium
Syllabus
Chemical Equlibrium :
Types of Equlibrium, Relation between Kp
and Kc
, Homogeneous and Hetrogeneous Equlibrium,
Relation Between Degree of Dissociation and Vapour Density, Reaction Quotient, FactorsAffecting
Equlibrium Constant, Le-Chateliers Principle, Simultaneous Equlibrium.
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CHEMICAL EQUILIBRIUM_ADVANCED# 1
CHEMICAL EQUILIBRIUM
Types of chemical reactions
Types of chemical reactions
Irreversible reaction Reversible reaction
1 The reaction which proceeds in one direction (forward
direction) only.
1 The reaction which proceed in both the direction under the same
set of experimental conditions.
2 Reactants are almost completely converted into products.
Products do not react to form reactants again.
2 Reactants form products and products also react to form reactants
in backward direction. These are possible in closed vessels .
3 Do not attain equilibrium state. 3 Attain the equilibrium state and never go to completion.
4 Such reactions are represented by single arrow {} 4 Represented by double arrow ( ) or
5 Examples – 5 Examples :–
(a) Precipitation reactions e.g.
NaCl(aq) + AgNO3(aq)  NaNO3(aq) + AgCl 
(a) Homogeneous reactions- only one phase is present
(b) Neutralization reactions e.g
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O
(c) 2KClO3 (s) 2KCl(s) + 3O2(g)
(d) Reactions in open vessel :–
Even a reversible reaction will become irreversible if it is
carried out in open vessel. Ex.
CaCO3(s) CaO(s) + CO2(g)
NH4HS(s) NH3(g) + H2S(g) (b)


( )
(i) Gaseous phase–
H2(g) + I2(g) 2HI(g)
N2(g) + O2(g) 2NO(g) [Birkland eyde process (HNO3)]
N2(g) + 3H2(g) 2NH3(g) (Haber’s process)
(ii) Liquid phase
CH3 COOH(l) + C2H5OH(l) CH3COOC2H5(l)+ H2O(l)
Heterogeneous reactions– More than one phases are present
CaCO3(s) CaO(s) + CO2(g)
NH4HS(s) NH3(g) + H2S(g) Closed
vessel
Open
vessel
STATE OF CHEMICAL EQUILIBRIUM :
At equilibrium :
(i) Rate of forward reaction (rf) = rate of backward reaction (rb) (dynamic nature)
(ii) All measurable parameters become constant with respect to time.

Types of equilibria on the basis of process
Types of equilibria on basis of physical state

Homogeneous equilibrium Heterogeneous equilibrium
When all reactants and products When more than one phase are
are in same phase present
H2(g) + Cl2(g) 2HCl(g) 3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)
SO2(g) + NO2(g) SO3(g) + NO(g) 2Na2O2(s) + 2H2O() 4NaOH + O2(g)
CHARACTERISTICS OF CHEMICAL EQUILIBRIUM :
The nature and the properties of the equilibrium state are the same regardless of the direction from which it
is achieved. It can be achieved in both directions.
Equilibrium is dynamic in nature.
It means that at microscopic level reaction has not stopped. It appears that no change is occuring but But
both the opposing reactions are proceeding at the same rate. So there is no net change.Thus equilibrium is
not static in nature.
A catalyst can alter the rate of approach of equilibrium but does not change the state of equilibrium. By using
catalyst, the equilibrium can be achieved in different (more/less) time, but the relative concentrations of
reactants and products are same irrespective of the presence or absence of a catalyst.
Equilibrium can be observed by constancy of some observable properties like colour, pressure, concentration,
density, temperature, refractive index etc.which may be suitable in a given reaction.
At equilibrium, free energy changeG = 0
Equilibrium state can be affected by altering factors like pressure, volume, concentration and temperature
etc.(Le chateliers Principle).
System moves toward an equilibrium state spontaneously even if it is disturbed. It will return to original state.
LAW OF MASS ACTION :
To illustrate the law of mass action, consider the following general reaction at constant temperature,
aA(g) + bB(g)  cC(g) + dD(g)
or Rf = kf[A]a[B]b …(i)
Where kf = rate constant of forward reaction
or Rr = kr[C]c [D]d …(ii)
where kr = rate constant of reverse reaction.

At equilibrium,
Rate of forward reaction = Rate of reverse reaction
i.e., Rf = Rr
So, from equations (i) and (ii), we get
kf[A]a [B]b = kr[C]c[D]d
or,
c d
f
a b
r
k [C] [D]
k [A] [B]

or
c d
f
c a b
r
k [C] [D]
K
k [A] [B]
 
Where, Kc is the equilibrium constant in terms of molar concentration.
Equilibrium Constant (Kp) in terms of Partial Pressures:
c d
C D
p a b
A B
P P
K
P P



…(2)
RELATIONSHIP BETWEEN KP AND KC :
n
p cK K (RT)
 
Where, n = (c + d) – (a + b)
Applications of Equilibrium constant :
Predicting the direction of the reaction
Reaction Quotient (Q)
At each point in a reaction, we can write a ratio of concentration terms having the same form as the
equilibrium constant expression. This ratio is called the reaction quotient denoted by symbol Q.
It helps in predicting the direction of a reaction.
The expression Q = ba
dc
]B[]A[
]D[]C[
at any time during reaction is called reaction quotient.The concentrations
[C], [D] , [A], [B] are not necessarily at equilibrium.
Predicting the extent of the reaction :
K =
eq
eq
]ttanac[Re
]oduct[Pr
Case-I If K is large (K > 103) then product concentration is very very larger than the reactant ([Product] >>[Reactant])
Hence concentration of reactant can be neglected with respect to the product. In this case, the reaction is
product favourable and equilibrium will be more in forward direction than in backward direction.
Case-II If K is very small (K < 10–3)
[Product] << [Reactant]
Hence concentration of Product can be neglected as compared to the reactant.
In this case, the reaction is reactant favourable.
Equilibrium constant is dependent only on the temperature.
It means Kp and Kc will remain constant at constant temperature no matter how much changes are made in
pressure, concentration, volume or catalyst.

However if temperature is changed,
log
1
2
k
k
= R303.2
H







21 T
1
T
1
; H = Enthalpy of reaction
If T2 > T1 then K2 > K1 provided H = +ve (endothermic reaction)
K2 < K1 if H = –ve (exothermic reaction)
In the above equation, the unit of R and H/T should be same.
Relation between equilibrium constant & standard free energy change.
Gº = – 2.303 RT log K
Where G° = standard free energy change
T = Absolute temperature,
R = universal gas constant.
Homogeneous liquid system : Formation of ethyl acetate :
The reaction between alcohol and acid to form ester is an example of homogeneous equilibrium in liquid
system.
CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)
KC = ]OHHC][COOHCH[
]OH][HCOOCCH[
523
2523
Observed molecular weight and Observed Vapour Density of the mixture
Observed molecular weight of An(g) = mequilibriuatmolesof.nototal
)g(Aofweightmolecular n
=
))1n(1(a
M.a th

 Mobs =
])1n(1[
Mth

where Mth = theoritical molecular weight (n = atomicity)
Mmixture =
])1n(1[
M nA

, nAM = Molar mass of gas AAn
Vapour density (V.D). : Density of the gas divided by density of hydrogen under same temp &
pressure is called vapour density.
 D = vapour density without dissociation =
2
M nA
d = vapour density of mixture = observed v.d. =
2
Mmix
d
D
= 1 + (n – 1)

0
oT
M)1n(
MM
d)1n(
dD






where MT = Theoritical molecular wt. M0 = observed molecular wt. or molecular wt. of the mixture at equilibrium.
EXTERNAL FACTORS AFFECTING EQUILIBRIUM :
LE-CHATELIER’S PRINCIPLE
When a system (reaction) in equilibrium is subjected to change in any of the factors that determine the
equilibrium condition, the equilibrium shifts in such a manner so as to reduce or counteract the effect
of that change.
Consider the following reaction in equilibrium
2NH3(g) N2(g) + 3H2(g)

(I) Effect of change of concentration :
Increase in concentration of reactant propel the reaction in forward direction and increase in concentration
of product propel the reaction in backward direction.
(II) Effect of change of pressure :
(a) When n = 0 ; no effect
(b) When n  0 ;
If pressure increases the reaction shifts in the direction of lesser number of moles.
If pressure decreases the reaction shifts in the direction of greater number of moles.
(III) Addition of inert gas :
(a) At constant volume – no effect.
(b) At constant pressure –
When n = 0 ;no effect
When n  0 ;Reaction shifts in the direction of greater number of moles.
(IV) Effect of temperature :
(a) If H = +ve (endothermic) an increase in temperature, shifts the reaction in forward direction and vice–
versa.
(b) If H = –ve (exothermic) an increase in temperature, shifts the reaction in backward direction and vice–
versa.
Application of Le-Chatelier Principle to Physical Equilibrium :
(1) Effect of Pressure on Melting Points :
(a) When volume of solid decrease on melting i.e., in liquid state volume is lesser, then by increasing pressure
on such solids the melting point will go down, because at high pressure melting is facilitated eg. Ice,
diamond, Carborundum (SiC) etc.
Solid Liquid
(b) When volume increase on melting then by increasing pressure, the above equilibrium will shift in the reverse
direction, hence process of melting will be lowered and more heat is required, thus melting point will rise
eq. Iron, Copper, NH4Cl, NaCl etc.
(2) Effect of Pressure on Boiling point :
When pressure is raised then condensation of vapour into liquid take place, thus vapour pressure will
decrease. Now more heat is required to equate vapour pressure with atmospheric pressure, hence boiling
point will increase.
FREE ENERGY AND CHEMICAL EQUILIBRIUM :
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
According to Van’t Hoff reaction isotherm
c d
0 C D
a b
A B
p p
G G RTln
p p

   

= G0 + RTlnQp
the condition for a system to be at equilibrium is that
G = 0
Thus at equilibrium
c d
0 0 0C D
pa b
A B
p p
0 G RTln G RTln K
p p
 
      
 
Whence G0 = – RTlnK0
p
Hence
0
0
p
G
ln K
RT



PART - I : OBJECTIVE QUESTIONS
* Marked Questions are having more than one correct option.
Section (A) : Equilibrium, Equilibrium constant ,law of mass action, Reaction Quotient
A-1. A chemical reaction, A B, is said to be in equilibrium when :
(A) rate of forward reaction is equal to rate of backward reaction
(B) conversion of A to B is only 50% complete
(C) complete conversion of A to B has taken place
(D) only 25% conversion of A to B taken place
A-2. The equilibrium concentration of x, y and yx2
are 4, 2 and 2 respectively for the equilibrium
2x + y yx2
. The value of equilibrium constant, KC
is
(A) 0.625 (B) 6.25 (C) 0.0625 (D) 62.5
A-3. 4 mole of A are mixed with 4 mole of B and 2 mole of C are formed at equilibrium, according to the reaction,
A + B C + D
the equilibrium constant is :
(A) 2 (B) 2 (C) 1 (D) 4
A-4. For the following reaction at 250°C, PCl3
(g) + Cl2
(g) PCl5
(g) the value of KC
is 26 then the value of Kp
at same temperature will be
(A) 0.57 (B) 0.61 (C) 0.83 (D) 0.91
A-5. For a reversible reaction, the rate constants for the forward and backward reactions are
2.38 ×10–4
and 8.15 × 10–5
respectively. The equilibrium constant for the reaction is –
(A) 0.342 (B) 2.92 (C) 0.292 (D) 3.42
A-6. If different quantities of ethanol and acetic acid were used in the following reversible reaction,
CH3
COOH() + C2
H5
OH() CH3
COOC2
H5
() + H2
O()
the equilibrium constant will have values which will be ?
(A) different in all cases
(B) same in all cases
(C) higher in cases where higher concentration of ethanol is used
(D) higher in case where higher concentration of acetic acid is used
A-7. Chemical equilibrium is dynamic in nature because –
(A) The equilibrium in maintained quickly
(B) Conc. of reactants and products become same at equilibrium
(C) Conc. of reactants and products are constant but different
(D) Both forward and backward reactions occur at all times with same speed
A-8. Which of the following statements is false in case of equilibrium state –
(A) There is no apparent change in properties with time
(B) It is dynamic in nature
(C) It can be attained from either side of the reaction
(D) It can be attained from the side of the reactants only
A-9. Starting with the reactants , At any moment before a reversible reaction attains equilibrium it is found
that –
(A) The rate of the forward reaction is increasing and that of backward reaction is decreasing
(B) The rate of the forward reaction is decreasing and that of backward reaction is increasing
(C) The rate of both forward and backward reactions is increasing
(D) The rate of both forward and backward reactions is decreasing

A-10. A chemical reaction A B is said to be in equilibrium when -
(A) Complete conversion of A to B has taken place
(B) Conversion of A to B is only 50% complete
(C) Only 10% conversion of A to B has taken place
(D) The rate of transformation of A to B is just equal to rate of transformation of B to A in the system
A-11. In a chemical reaction Kp
is greater than Kc
when
(A) the number of molecules entering into the chemical reaction is more than the number of molecules
produced.
(B) the number of molecules entering into the chemical reaction is the same as the number of molecules
produced.
(C) the number of molecules entering into the chemical reaction is less than the number of molecules
produced.
(D) the total number of moles of reactants is less than the number of moles of products.
A-12. Which of the following statements is not correct about the equilibrium constant ?
(A) Its value does not depend upon the initial conc. of the reactants
(B) Its value does not depend upon the initial conc. of the products
(C) Its value does not depend upon temperature.
(D) Its value does not depend upon presence of catalyst.
A-13. In a chemical equilibrium, the equilibrium constant is found to be 2.5. If the rate constant of backward
reaction is 3.2 × 10–2, the rate constant of forward reaction is -
(A) 8.0 × 10–2 (B) 4.0 × 10–2 (C) 3.5 × 10–2 (D) 7.6 × 10–3
A-14. K1 and K2 are the rate constants of forward and backward reactions. The equilibrium constant K of the
reaction is -
(A) K1 × K2 (B) K1 – K2 (C)
K
K
1
2
(D)
K K
K K
1 2
1 2

–
A-15. The value of KP for the reaction H2(g) + I2(g) 2HI(g) is 50. What is the value of KC
(A) 30 (B) 40 (C) 50 (D) 70
A-16. In which of the following reaction, the value of KP will be equal to KC –
(A) N2(g) + O2(g) 2NO (g) (B) PCl5 (g) PCl3 (g) + Cl2(g)
(C) 2NH3 (g) N2(g) + 3H2(g) (D) 2SO2 (g) + O2(g) 2SO3 (g)
A-17. Select the correct expression regarding the relation between KP and KC for the reaction
aX(g) + bY(g) bZ(g) + aW(g) -
(A) KP = KC(RT)a+b (B) KP =
K
a b
C
( ) 2
(C) KP = KC RT (D) KP = KC
A-18. The equilibrium constant KC for the decomposition of PCl5 is 0.625 mole / lit at 300ºC. Then the value
of KP is -
(A) 2.936 atm (B) 0.0625 atm (C) 6.25 atm (D) 0.00625 atm
A-19. The reaction A(g) + B(g) C(g) + D(g) proceeds to right hand side upto 99.9% when starting bwith
equal moles of A and B. The equilibrium constant K for the reaction will be -
(A) 104 (B) 105 (C) 106 (D) 108
A-20. For the reaction, 2NO2 (g) 2NO (g) + O2(g), KC = 1.8 × 10–6 at 185ºC. At 185ºC, the
value of KC for the reaction -
NO(g) +
1
2
O2 (g) NO2(g) is -
(A) 0.9 × 106 (B) 7.5 × 102 (C) 1.95 × 10–3 (D) 1.95 × 103

A-21. If in the reaction N2O4(g) 2NO2(g),  is the part of N2O4 which dissociates per mole, then the
number of moles at equilibrium will be -
(A) 3 (B) 1 (C) (1–)2 (D) (1 + )
A-22. The equilibrium concentration of X, Y and YX2 are 4, 2 and 2 moles respectively for the equilibrium
2X(g) + Y(g) YX2(g).The value of KC is -
(A) 0.625 (B) 0.0625 (C) 6.25 (D) 0.00625
A-23. At 444º C, the equilibrium constant K for the reaction 2AB(g) AA2(g) + B2(g) is
1
64
. The
degree of dissociation of AB will be -
(A) 10% (B) 20 % (C) 30% (D) 50%
A-24. For the reaction A(g) + B(g) C(g) + D(g), the degree of dissociation  would be –
(A)
1K
K

(B) K +1 (C) K  1 (D) K – 1
A-25. For the reaction : N2O3(g) NO(g) + NO2(g) ; total pressure = P, degree of dissociation = 50%.
Then Kp would be –
(A) 3P (B) 2P (C)
P
3
(D)
P
2
A-26. For N2(g) + 3H2(g) 2NH3(g)
(A) KP = KC (B) KP = KC RT (C) KP = KC (R T)–2 (D) KP = KC (R T)–1
A-27. In the reaction
C2H4(g) + H2(g) C2H6(g), the equilibrium constant can be expressed in units of -
(A) litre–1 mol–1 (B) mol2 litre–2 (C) litre mol–1 (D) mol litre–1
A-28. Equilibrium concentration of HI, I2 and H2 is 0.7, 0.1 and 0.1 moles/litre. Calculate the equilibrium
constant for the reaction :
I2(g) + H2(g) 2HI(g) –
(A) 0.36 (B) 36 (C) 49 (D) 0.49
A-29. The equilibrium constant for the reaction Zn (s)+CO2(g) ZnO (s) + CO (g) is -
(A)
P
P
co
co2
(B)
[ ]
[ ]
ZnO
Zn
(C)
P P
P P
ZnO CO
Zn CO2
(D)
P P
P P
Zn CO
ZnO CO
2
2
A-30. For the reaction
C (s) + CO2 (g) 2CO (g)
the partial pressure of CO2 and CO are 2.0 and 4.0 atm respectively at equilibrium. The Kp for the
reaction is –
(A) 0.5 (B) 4.0 (C) 8.0 (D) 32.0
A-31. The unit of equilibrium constant for the reaction H2 + I2 2HI is –
(A) Mole–1 litre (B) Mole–2 litre (C) Mole litre (D) None
A-32.
K
K
P
C
for the gaseous reaction –
(a) 2 A + 3 B 2C
(b) 2 A 4B
(c) A + B + 2C 4D
would be respectively -
(A) (RT)–3 , (RT)2, (RT)º (B) (RT)–3 , (RT)–2, (RT)–1
(C) (RT)–3 , (RT)2, (RT) (D) None of the above

A-33. In Which of the following equilibria, the value of KP is less than KC -
(A) H2 + I2 2Hl (B) N2 + 3H2 2NH3
(C) N2 + O2 2NO (D) CO + H2O CO2 + H2
A-34. In a reversible reaction A B, the initial concentration of A and B are a and b in moles per litre and the
equilibrium concentrations are ( a  x) and (b + x) respectively; express x in terms of k1
, k2
, a and b.
(A)
21
21
kk
bkak


(B)
21
21
kk
bkak


(C)
21
21
kk
bkak 
(D)
21
21
kk
bkak


A-35. The equilibrium constant of the reaction SO2(g) + ½O2(g) SO3(g) is 4 × 10–3 atm–1/2. The equilibrium
constant of the reaction 2SO3(g) 2SO2(g) + O2(g) would be :
(A) 250 atm (B) 4 × 103 atm (C) 0.25 × 104 atm (D) 6.25 × 104 atm
Section (B) : Degree of Dissociation, Vapour Density ,Average Molar Mass
B-1. For which of the following reactions, the degree of dissociation cannot be calculated from the vapour density
data
I 2H(g)  H2(g)
+ I2(g)
II 2NH3(g)
 N2(g)
+ 3H2(g)
III 2NO(g)  N2(g)
+ O2(g)
IV PCl5(g)
 PCl3(g)
+ Cl2(g)
(A) I and III (B) II and IV (C) I and II (D) III and IV.
B-2. The degree of dissociation of SO3
is  at equilibrium pressure P0
.
Kp
for 2SO3
(g) 2SO2
(g) + O2
(g) is
(A) [(P0
3
)/2(1 – )3
] (B) [(P0
3
)/(2+)(1 – )2
]
(C) [(P0
2
)/2(1 – )2
] (D) None of these
B-3. In a container equilibrium N2
O4
(g) 2NO2
(g)
is attained at 25°C. The total equilibrium pressure in container is 380 torr. If equilibrium constant of above
equilibrium is 0.667 atm, then degree of dissociation of N2
O4
at this temperature will be :
(A)
3
1
(B)
2
1
(C)
3
2
(D)
4
1
B-4. In the dissociation of N2O4 into NO2, (1 + ) values with the vapour densities ratio 





d
D
is as given by :
[-degree of dissociation, D-vapour density before dissociation, d-vapour density after dissociation]
(A) (B) (C) (D)
B-5. In the above question,  varies with
d
D
according to :
(A) (B) (C) (D)

B-6. Before equilibrium is set-up for the chemical reaction N2O4(g) 2NO2(g), vapour densityd of thegaseous
mixture was measured. If D is the theoretical value of vapour density, variation of  with D/d is given by the
graph. What is value D/d at point A?
(A) 0 (B) 0.5 (C) 1 (D) 1.5
B-7. The degree of dissociation of PCl5
() obeying the equilibrium, PCl5
PCl3
+ Cl2
, is approximately
related to the presure at equilibrium by (given  << 1) :
(A)  P (B) 
P
1
(C) 2
P
1
(D)  4
P
1
B-8. For the reaction N2O4 (g) 2NO2 (g), if percentage dissociation of N2O4 are 20%, 45%, 65% & 80%,
then the sequence of observed vapour densities will be :
(A) d20 > d45 > d65 > d80 (B) d80 > d65 > d45 > d20
(C) d20 = d45 = d65 = d80 (D) (d20 = d45) > ( d65 = d80)
B-9. An unknown compound A dissociates at 500ºC to give products as follows -
A(g) B(g) + C(g) + D(g)
Vapour density of the equilibrium mixture is 50 when it dissociates to the extent to 10%. What will
be the molecular weight of Compound A –
(A) 120 (B) 130 (C) 134 (D) 140
B-10. N2O4 dissociates as N2O4(g) 2NO2(g) at 273 K and 2 atm pressure. The equilibrium mixture has
a density of 41. What will be the degree of dissociation -
(A) 14.2% (B) 16.2% (C) 12.2% (D) None
B-11. At 250ºC and 1 atmospheric pressure, the vapour density of PCl5 is 57.9 . What will be the dissociation
of PCl5 –
(A) 1.00 (B) 0.90 (C) 0.80 (D) 0.65
Section (C) : Le Chateliers Principle
C-1. Consider the reaction, CaCO3(s) CaO(s) + CO2(g) ; in closed container at equilibrium. What
would be the effect of addition of CaCO3 on the equilibrium concentration of CO2 -
(A) Increases (B) Decreases
(C) Remains unaffected (D) Data is not sufficient to predict it
C-2. In the melting of ice, which one of the conditions will be more favourable –
(A) High temperature and high pressure (B) Low temperature and low pressure
(C) Low temperature and high pressure (D) High temperature and Low pressure
C-3. In the reaction, 2SO2 (g) + O2 (g) 2SO3 (g) + X cals, most favourable condition of temperature
and pressure for greater yield of SO3 are -
(A) Low temperature and low pressure (B) High temperature and low pressure
(C) High temperature and high pressure (D) Low temperature and high pressure
C-4. On adding inert gas to the equilibrium PCI5(g) PCI3(g) + CI2(g) at constant pressure. The degree
of dissociation will remain –
(A) Unchanged (B) Decreased (C) Increased (D) None of these

C-5. Dissociation of phosphorus pentachloride is favoured by –
(A) High temperature and high pressure (B) High temperature and low pressure
(C) Low temperature and low pressure (D) Low temperature and high pressure
C-6. Adding inert gas to system N2(g) + 3H2(g) 2NH3(g) at equilibrium at constant volume will lead to :
(A) N2 and H2 are formed in abundance
(B) N2, H2 and NH3 will have the same molar concentration
(C) The production of ammonia increases
(D) No change in the equilibrium
C-7. In the reaction N2(g) + 3H2(g) 2NH3(g), the forward reaction is exothermic and the backward
reaction is endothemic. In order to produce more heat it is necessary –
(A) To add ammonia
(B) To add N2 and H2
(C) Increasing the concentration of N2,H2 and NH3 equally
(D) None of the above
C-8. The reaction in which the yield of the products can not be increased by the application of high pressure
is –
(A) PCl3 (g) + Cl2 (g) PCl5 (g) (B) N2 (g) + 3H2 (g) 2NH3 (g)
(C) N2 (g) + O2 (g) 2NO (g) (D) 2SO2 (g) + O2 (g) 2SO3 (g)
C-9. Factors affecting KC is/are -
(A) Increasing concentration of the reactant
(B) Presence of catalyst
(C) Method of writing balanced equation (or stoichiometry of reaction)
(D) Time taken by the chemical reaction
C-10. In the reaction A (g) + B (g) C (g), the backward reaction is favoured by -
(A) Increase in pressure (B) Decrease in pressure
(C) Neither increase nor decrease in pressure (D) Data unpredictable
C-11. Which among the following conditions, increase the yield of the product in the equilibrium,
3 A(g) + B(g) 4C(g) + heat
(A) Increase in pressure (B) Increase in volume
(C) Increase in temperature (D) Decrease in temperature
C-12. When H2 is added to an equilibrium mixture 2HI(g) H2(g) + I2(g), at constant temperature, the -
(A) Value of Kp decreases
(B) Value of Kp increases
(C) The degree of dissociation of HI decreases
(D) Degree of dissociation of HI increases
C-13. For the reaction PCl5 (g) PCl3 (g) + Cl2 (g), the forward reaction at constant temperature is
favoured by -
(A) Increasing the volume of container
(B) Introducing an inert gas at constant pressure
(C) Introducing PCl5 at constant volume
(D) All of these
C-14. Which of the following reaction will be favoured at low pressure -
(A) H2(g) + I2(g) 2HI(g) (B) N2(g) + 3H2(g) 2NH3(g)
(C) PCl5(g) PCl3(g) + Cl2(g) (D) N2(g) + O2(g) 2NO(g)
C-15. In what manner will increase of pressure affect the equation C (s) + H2O (g) CO(g) + H2(g) -
(A) Shift in the forward direction (B) Shift in the reverse direction
(C) Increase in the yield of H2 (D) No effect

C-16. Which of the following will shift the reaction PCl3(g) + Cl2(g) PCl5(g) to the left side-
(A) Addition of PCl5 (B) Increase in pressure
(C) Decrease in temperature (D) Catalyst
C-17. Which of the following equilibrium is not affected by pressure -
(A) N2 (g) + O2 (g) 2NO(g) (B) 2SO2 (g) + O2 (g) 2SO3(g)
(C) 2O3 (g) 3O2(g) (D) 2NO2 (g) N2O4(g)
C-18. According to Le Chatelier principle, an increase in the temperature of the following reaction will
N2 + O2 2NO – 43200 cal
(A) Increase the yield of NO (B) Decrease the yield of NO
(C) Not effect on the yield of NO (D) Not help the reaction to proceed
C-19. When a reversible reaction has reached the state of equilibrium -
(A) The forward reaction stops
(B) The backward reaction stops
(C) The whole reaction stops
(D) The forward and backward reaction proceed with same speed
C-20. During thermal dissociation of gas, the vapour density -
(A) Remains same (B) Will be increased
(C) Will be decreased (D) Some times increases some times decreases
C-21. For the reaction CO(g) + H2
O(g) CO2
(g) + H2
(g) at a given temperature the equilibrium amount of CO2
(g)
can be increased by :
(A) adding a suitable catalyst (B) adding an inert gas
(C) decreasing the volume of container (D) increasing the amount of CO(g)
C-22.* For the reaction : PCl5
(g) PCl3
(g) + Cl2
(g)
The forward reaction at constant temperature is favoured by
(A) introducing chlorine gas at constant volume
(B) introducing an inert gas at constant pressure
(C) increasing the volume of the container
(D) introducing PCl5
at constant volume
C-23. Given the following reaction at equilibrium N2(g) + 3H2(g) 2NH3(g). Some inert gas at constant pressure
is added to the system. Predict which of the following facts will be affected.
(A) More NH3(g) is produced (B) Less NH3(g) is produced
(C) No affect on the equilibrium (D) Kp of the reaction is decreased
C-24. For an equilibrium H2O(s) H2O() which of the following statements is true.
(A) The pressure changes do not affect the equilibrium
(B) More of ice melts if pressure on the system is increased
(C) More of liquid freezes if pressure on the system is increased
(D) The pressure changes may increase or decrease the degree of advancement of the reaction depending
upon the temperature of the system
C-25. When a bottle of cold drink is opened, the gas comes out with a fizz due to :
(A) Decrease in temperature
(B) Increase in pressure
(C) Decrease in pressure suddenly which results in decrease of solubility of CO2
gas in water
(D) None
C-26. The equilibrium, SO2Cl2(g) SO2(g) + Cl2(g) is attained at 25°C in a closed container ,if an inert gas,
helium, is introduced at constant volume. Which of the following statement(s) is/are correct.
(A) Concentrations of SO2, Cl2 and SO2Cl2 are changed
(B) No effect on equilibrium
(C) Concentration of SO2 is reduced
(D) Kp of reaction is increasing

C-27. An equilibrium mixture in a vessel of capacity 100 litre contain 1 mol N2
, 2 mol O2
and 3 mol NO. Number of
moles of O2
to be added so that at new equilibrium the conc. of NO is found to be 0.04 mol/lit.:
(A) (101/18) (B) (101/9) (C) (202/9) (D) None of these.
C-28.* 2CaSO4(s) 2CaO(s) + 2SO2(g) + O2(g), H > 0
Above equilibrium is established by taking some amount of CaSO4(s) in a closed container at 1600 K. Then
which of the following may be correct option.
(A) moles of CaO(s) will increase with the increase in temperature
(B) If the volume of the container is doubled at equilibrium then partial pressure of SO2(g) will change at
new equilibrium.
(C) If the volume of the container is halved partial pressure of O2(g) at new equilibrium will remain same
(D) If two moles of the He gas is added at constant pressure then the moles of CaO(s) will increase.
C-29. The yield of product in the reaction, A2
(g) + 2B(g) C(g) + Q kJ
would be higher at :
(A) low temperature and high pressure (B) high temperature and high pressure
(B) low temperature and low pressure (D) high temperature and low pressure
C-30. Manufacture of ammonia from the elements is represented by
N2
(g) + 3H2
(g) 2NH3
(g) + 22.4 kcal
the maximum yield of ammonia will be obtained when the process is made to take place –
(A) at low pressure and high temperature (B) at low pressure and low temperature
(C) at high pressure and high temperature (D) at high pressure and low temperature
C-31. In the reaction, 2SO2
(g) + O2
(g) 2SO3
(g) + X cal, most favourable conditions of temperature and
pressure for greater yield of SO3
are :
(A) low temperature and low pressure (B) high temperature and low pressure
(C) high temperature and high pressure (D) low temperature and high pressure
C-32. For the reaction : 2A(g) + B(g) 3C(g) + D(g)
two mole each of A and B were taken into a flask. The following must always be true when the system
attained equilibrium :
(A) [A] = [B] (B) [A] < [B] (C) [B] = [C] (D) [A] > [B]
C-33. In a vessel containing SO2
, SO3
and O2
at equilibrium, some helium gas is introduced so that total pressure
increases while temperature and volume remain the same. According to the
Le Chatelier’s principle, the dissociation of SO3
:
(A) increases (B) decreases (C) remains unaltered (D) change unpredictably
C-34. The equilibrium SO2
Cl2
(g) SO2
(g) + Cl2
(g) is attained at 25°C in a closed container and an inert gas,
helium, is introduced. Which of the following statements is correct ?
(A) concentrations of SO2
Cl2
, SO2
and Cl2
do not change
(B) more Cl2
is formed
(C) concentration of SO2
is reduced (D) more SO2
Cl2
is formed
Section (D) : Simultaneous Equilibrium
D-1. The two equilibria, AB(aq) AA+(aq) + B(aq) and AB(aq) + B(aq) AB2
(aq) are simultaneously
maintained in a solution with equilibrium constants, K1 and K2 respectively. The ratio of concentration of A+
to AB2
 in the solution is :
(A) directly proportional to the concentration of B– (aq.).
(B) inversely proportional to the concentration of B– (aq.).
(C) directly proportional to the square of the concentration of B– (aq.).
(D) inversely proportional to the square of the concentration of B– (aq.).
D-2. In the preceeding problem, if [A+] and [AB2
] are y and x respectively, under equilibrium produced by adding
the substance AB to the solvents, then K1/K2 is equal to
(A)
2
)xy(
x
y
 (B)
x
)yx(y2

(C)
x
)yx(y2

(D) )yx(
x
y


D-3.* The two equilibrium, AB AA+ + B and AB + B AB2
 are simultaneously maintained in a solution
with equilibrium constants, K1 and K2 respectively. If [A+] and [AB2
] are y and x respectively, under equilibrium
produced by adding the substance AB(s) to the solvents, then
(A) k1/k2 =
2
)xy(
x
y
 (B) k1/k2 = )yx(
x
y
 (C) [B¯]eq. = y – x (D) None of these
Section (E) : Thermodynamics of Equilibrium
E-1. Which is/are correct relation (s) for thermodynamic equilibrium constant
(A) Gº = –2.303 RT log K (B) G = Gº + 2.303 RT log K
(C) Eºcell =
n
0591.0
log K (D) E = Eº –
n
0591.0
logK
E-2. Which is/are correct relation (s) for equilibrium constant K?
(A) Gº = – 2.303 RT log K (B) Ecell
º =
nF
KlogRT303.2
(C) K =
Ó [Pr oduct]
Ó [Reac tant]
(D) log K = log A –
RT2.303
H
E-3. For the reaction H2(g) + I2(g) 2HI(g)
Kc = 66.9 at 350°C and Kc = 50.0 at 448°C. The reaction has
(A) H = + ve (B) H = – ve
(C) H = zero (D) H sign can not be determined
E-4. Variation of log10 K with
T
1
is shown by the following graph in which straight line is at 45°, hence H° is :
(A) + 4.606 cal (B) – 4.606 cal (C) 2 cal (D) – 2cal
E-5.* For a reversible reaction aA + bB cC + dD ; the variation of K with temperature is given by
log
1
2
K
K
=
R303.2
Hº








12 T
1
T
1
then,
(A) K2
> K1
if T2
> T1
for an endothermic change
(B) K2
< K1
if T2
> T1
for an endothermic change
(C) K2
> K1
if T2
> T1
for an exothermic change
(D) K2
< K1
if T2
> T1
for an exothermic change
E-6. Which one of the following oxides is most stable? The equilibrium constants are given at the same temperature:
(A) 2N2
O5
(g) 2N2
(g) + 5O2
(g) ; K = 1.2 × 1034
(B) 2N2
O(g) 2N2
(g) + O2
(g) ; K = 3.5 × 1035
(C) 2NO(g) N2
(g) + O2
(g) ; K = 2.2 × 1030
(D) 2NO2
(g) N2
(g) + 2O2
(g) ; K = 6.71 × 1016
E-7. The equilibrium constant for a reaction A + B C + D is 1 × 10–2
at 298 K and is 2 at 273 K. The
chemical process resulting in the formation of C and D is :
(A) exothermic (B) endothermic
(C) unpredictable (D) there is no relationship between H and K

PART - II : MISCELLANEOUS QUESTIONS
COMPREHENSIONS TYPE
Comprehension # 1
Le chatelier's principle
If a system at equilibrium is subjected to a change of any one of the factors such as concentration, pressure
or temperature, the system adjusts itself in such a way as to Nulify the effect of that change.
Change of pressure : If a system in equilibrium consists of gases, then the concentrations of all the
components can be altered by changing the pressure. To increase the pressure on the system, the volume
has to be decreased proportionately. The total number of moles per unit volume will now be more and the
equilibirum will shift in the direction in which there is decrease in number of moles i.e., towards the direction
in which there can be decrease in pressure.
Effect of pressure on melting point : There are two types of solids :
(a) Solids whose volume decreases on melting, e.g., ice, diamond, carborundum, magnesium nitride and
quartz.
Solid (higher volume) Liquid (lower volume)
The process of melting is facilitated at high pressure, thus melting point is lowered.
(b) Solids whose volume increase on melting, e.g., Fe, Cu, Ag, Au, etc.
Solid (lower volume) Liquid (higher volume)
In this case the process of melting become difficult at high pressure; thus melting point becomes high.
(c) Solubilityof substances : When solid substance are dissolved in water, either heat is evolved (exothermic)
or heat is absorbed (endothermic).
KCl + aq KCl(aq) – heat
In such cases, solubility increase with increase in temperature. Consider the case of KOH; when this is
dissolved, heat is evolved.
KOH + aq KOH(aq) + heat
In such cases, solubility decrease with increase in temperature.
(d) Solubility of gases in liquids : When a gas dissolves in liquid, there is decrease in volume. Thus,
increase of pressure will favour the dissolution of gas in liquid.
1. A gas 'X' when dissolved in water heat is evolved. Then solublity of 'X' will increase :
(A) Low pressure, high temperature (B) Low pressure, low temperature
(C) high pressure, high temperature (D) high pressure, low temperature
2. Au(s) Au()
Above equilibrium is favoured at :
(A) High pressure low temperature (B) High pressure high temperature
(C) Low pressure, high temperature (D) Low pressure, low temperature
3. For the reaction,
2
1
N2(g) +
2
1
O2(g) NO(g)
If pressure is increased by reducing the volume of the container then :
(A) Total pressure at equilibrium will change.
(B) Concentration of all the component at equilibrium will change.
(C) Concentration of all the component at equilibrium will remain same
(D) Equilibrium will shift in the forward direction

Comprehension # 2
Effect of temperature on the equilibrium process is analysed by using the thermodynamics
From the thermodynamics relation
Gº = – 2.30 RT logk .......... (1) Gº : Standard free energy change
Gº = Hº – TSº .......... (2) Hº : Standard heat of the reaction.
From (1) & (2)
– 2.3 RT logk = Hº – TSº Sº : Standard entropy change
 logK = R3.2
ºS
RT3.2
ºH 


 .......... (3)
Clearly if a plot of log k vs 1/T is made then it is a straight line having slope =
R3.2
ºH
and Y intercept = R3.2
S
If at temp. T1 equilibrium constant be k1 and at temperature T2 equilibrium constant be k2 then :
The above equation reduces to:
 log K1 =
R3.2
ºS
TR3.2
ºH
1



 .......... (4)
 log K2 =
R3.2
ºS
TR3.2
ºH
2



 .......... (5)
Substracting (4) from (5) we get
 








211
2
T
1
T
1
R30.2
ºH
K
K
log
From the above relation we can conclude that the value of equilibrium constant increases with increase in
temperature for endothermic reaction but value of equilibrium constant decreases with the increase in
temperature for exothermic reaction.
4. If standard heat of dissociation of PCl5 is 230 cal then slope of the graph of logk vs
T
1
is :
(A) +50 (B) – 50 (C) 10 (D) None
5. For exothermic reaction if S0 < 0 then the sketch of logk vs
T
1
may be :
(A) (B) (C)
1/T
logk (D)
6. If for a particular reversible reaction at :
KC = 57 at 3550C and KC = 69 at 4500 then :
(A) H < 0 (B) H > 0 (C) H = 0 (D) H whose sign can’t be determined
Comprehensions # 3
Equilibrium constants are given (in atm) for the following reactions at 0° C:
SrCl2 6H2O(s)  SrCl2  2H2O (s) + 4H2O(g) Kp = 5 × 1012
Na2HPO4  12 H2O(s)  Na2HPO4  7 H2O (s) + 5H2 O(g) Kp = 2.43 × 1013
Na2SO4 10 H2O(s)  Na2SO4 (s) + 10 H2O (g) Kp = 1.024 × 1027
The vapor pressure of water at 0°C is 4.56 torr.

7. Which is the most effective drying agent at 0°C?
(A) SrCl2  2H2O (B) Na2HPO47 H2O (C) Na2SO4 (D) all equally
8. At what relative humidities will Na2SO4 10 H2O be efflorescent (release moisture) when exposed to air at
0°C?
(A) above 33.33% (B) below 33.33 % (C) above 66.66% (D) below 66.66%
9. At what relative humidities will Na2SO4 be deliquescent (i.e. absorb moisture) when exposed to the air at
0°C?
(A) above 33.33% (B) below 33.33 % (C) above 66.66% (D) below 66.66%
Comprehensions # 4
If we know the equilibrium constant for a particular reaction, we can calculate the concentrations in the
equilibrium
mixture from the initial concentrations. Commonly only the initial concentration of reactants are given.
10. In a study of equilibrium
H2(g) + I2(g)  2HI (g)
1 mol of H2 and 3 mol of I2 gave rise at equilibrium to x mol of HI.
Addition of a further 2 mol of H2 gave an additional x mol of HI. What is x?
(A) 0.5 (B) 1 (C) 1.5 (D) None of these
11. In above problem, what is Kp at the temperature of the experiment.
(A) 1 (B) 2 (C) 4 (D) None of these
12. In a study of equilibrium
2SO2(g) + O2(g) 2SO3(g).
Starting with 2 mole SO2 and 1.5 mole O2 in 5 litre flask. Equilibrium mixture required 0.4 mole KMnO4 in
acidic medium. Hence KC is :
(A) 0.2 (B) 5.0 (C) 675.0 (D) None of these
MATCH THE COLUMN
13. Match the following
List I List II
(a) N2(g)
+ 3H2(g)
2NH3
(P) KP
= KC
(b) H2(g)
+ S(s)
H2
S(g)
(Q) KP
> KC
(c) H2(g)
+ I2(g)
2HI(g)
(R) KP
< KC
(d) 2NaHCO3(s)
Na2
CO3(s)
+ CO2(g)
+ H2
O(g)
(A) a-Q, b-P, c-P, d-R (B) a-P, b-P, c-Q, d-R (C) a-Q, b-R, c-P, d-R (D) a-R, b-P, c-P, d-Q
List I List II
(a) Keq
< 1 (P) Affected by temperature
(b) Degree of dissociation (Q) Affected by pressure
(c) Equilibrium constant (R) Kf
< Kb
(d) Melting of ice (S) Kf
> Kb
(A) a-R, b-P, c-S, d-Q (B) a-P,R, b-P,Q, c-P, d-P,Q
(C) a-R, b-S c-P, d-Q (D) a-Q, b-R, c-S, d-P
List I (Reaction) List II (units)
(a) N2
+ O2
2NO (P) Kc
= m/l
(b) N2
+ 3H2
2NH3
(Q) no unit
(c) PCl5
PCl3
+ Cl2
(R) kp
= atm–2
(d) A(l)
B(g)
(S) Kp
= atm
(A) a-Q, b-R,c-P,S, d-P,S (B) a-Q, b-R,c-P, d-S
(C) a-S, b-Q,c-P,R, d-P,S (D) a-R, b-Q,c-P,S, d-S

16. Match the following ; D - Vapour Density before Dissociation
d - Vapour Density after Dissociation
List I (Reaction) List II (degree of dissociation)
(a) 2NH3
N2
+ 3H2
(P)
d2
dD 
(b) N2
O4
2NO2
(Q)
d
dD 
(c) 4PH3
P4
+ 6H2
(R)
d3
dD 
(d) 5A 6B + 9C (S)
d3
)dD(4 
(A) a-Q, b-P, c-R, d-S (B) a-P, b-Q, c-R, d-S
(C) a-Q, b-Q, c-S, d-P (D) a-S, b-R, c-Q, d-P
17. Match List I with List II
(P is partial pressure of prdouct having stiochiometric coefficient unity)
List I (Reaction) List II (expression of Kp
)
(a) NH4
Cl(s) NH3
(g) + HCl(g) (P) 6P6
(b) NH2
COONH4
(s) 2NH3
(g)+ 2CO2
(g) (Q) P2
(c) A(s) 2B(g) + C(g) + 3D(g) (R) 4P3
(d) A(s) B(g) + C(g) (S) 108P6
(A) a-Q, b-R, c-S, d-Q (B) a-P, b-Q, c-R, d-S
(C) a-S, b-R, c-Q, d-P (D) a-S, b-P, c-R, d-Q
18. Column I Column II
(A) KP < KC (P) N2 + 3H2  2NH3
(B) Introduction of inert gas at (Q) PCl5 (g)  PCl3 (g) + Cl2 (g)
constant pressure will decrease
the concentration of reactants
(C) Kp is dimensionless (R) 2NO2 (g)  N2O4 (g)
(D) Temperature increase will shift (S) NH3 (g) + HI (g)  NH4I (s)
the reaction on product side.
Column Column 
(A) N2(g) + 3H2(g) 2NH3(g) (t = 300ºC) (p) ng > 0
(B) PCl5(g) PCl3(g) + Cl2(g) (t = 50ºC) (q) Kp < Kc
(C) C(s) + H2O(g) CO(g) + H2(g) (r) Kp not defined
(D) CH3COOH() + C2H3OH () CH3COOC2H5() + H2O() (s) Pinitial > Peq.
20. Match the following :
Column I (Assume only reactant were present initially) Column II
(A) For the equilibrium NH4
(s) NH3
(g) + H(g), (p) Forward shift
if pressure is increased at equilibrium
(B) For the equilibrium N2
(g)+ 3H2
(g) 2NH3
(g) (q) No change
volume is increased at equilibrium
(C) For the equilibrium H2
O(g) + CO(g) H2
(g) + CO2
(g) (r) Backward shift
inert gas is added at constant pressure at equilibrium
(D) For the equilibrium PCl5
PCl3
+ Cl2
(s) Final pressure is more than
Cl2
is removed at equilibrium. initial pressure

ASSERTION / REASONING
DIRECTION :
Each question has 5 choices (A), (B), (C), (D) and (E) out of which ONLY ONE is correct.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(C) Statement-1 is True, Statement-2 is False.
(D) Statement-1 is False, Statement-2 is True.
(E) Statement-1 and Statement-2 both are False.
21. Statement - 1. For the reaction H2(g) + 2(g) 2H(g), Kp = Kc.
Statement - 2. Kp of all gaseous reactions is equal to Kc.
22. Statement - 2. Kp is related to Kc by the relation, Kp = Kc (RT)n
Statement - 2. Kp has same units as Kc.
23. Statement - 1. Reaction quotient Q is equal to Keq when the reaction is in equilibrium.
Statement - 2. If a catalyst is added to the reaction at equilibrium, the value of Q remains no
longer equal to Keq.
24. Statement - 1. In the presence of catalyst, the value of equilibrium constant K increases.
Statement - 2. Catalyst increases the rate of forward and backward reaction to same extent.
25. Statement - 1. For the reaction, N2 + O2 2NO, increase in pressure at equilibrium has no effect
on the reaction.
Statement - 2. The reaction is not accompanied by any change in number of moles of gaseous species.
26. Statement-1 : In dilute aqueous solution water is present in such large excess that its concentration remains
essentially constant during any reaction involving water.
Statement-2 : The term [H2O] does not appear in any equilibrium constant expression for a reaction taking
place in dilute aqueous solution.
27. Statement-1 : A net reaction can occur only if a system is not at equilibrium.
Statement-2 : All reversible reactions occur to reach a state of equilibrium.
28. Statement-1 : No term in the concentration of a pure solid or a pure liquid appears in an equilibrium constant
expression.
Statement-2 : Each pure solid or pure liquid is in a phase by itself, and has a constant concentration at
constant temperature.
29. Statement-1 : The reaction quotient , Q has the same form as the equilibrium constant Keq, and is evaluated
using any given concentrations of the species involved in the reaction, and not necessarily equilibrium
concentrations.
Statement-2 : If the numerical value of Q is not the same as the value of equilibrium constant, a reaction will occur.
30. Statement-1 : If the equation for a reaction is reversed, the equilibrium constant is inverted and if the
equation is multiplied by 2, the equilibrium constant is squared.
Statement-2 : The numerical value of an equilibrium constant depends on the way the equation for the
reaction is written.
31. Statement-1 : The dissociation of CaCO3 can be represented as, CaCO3(s) CaO(s) + CO2(g). Some
solid CaCO3 is placed in an evacuated vessel enclosed by a piston and heated so that a portion of its
decomposes. If the piston is moved so that the volume of the vessel is doubled, while the temperature is held
constant, the number of moles of CO2 in the vessel increase.
Statement-2 : The pressure of CO2 in the vessel will remain the same.
32. Statement-1 : A catalyst does not influences the values of equilibrium constant.
Statement-2 : Catalysts influence the rate of both forward and backward reactions equally.
33. Statement-1 : For . If more Cl2
is added the equilibrium will shift in backward
direction hence equilibrium constant will decrease.
Statement-2 : Addition of inert gas to the equilibrium mixture at constant volume, does not alter the equilibrium.

TRUE / FALSE
34. A catalyst does not alter the equilibrium point.
35. For the equilibrium, H2 + I2 2HI, the value of equilibrium constant increases with increase in
concentration of H2.
36. For any reaction greater the value of equilibrium constant greater is the extent of completion of reaction.
37. For the reaction, N2O4 2NO2, Kc = Kp/RT..
38. For the reaction PCl5 PCl3 + Cl2, the degree of dissociation of PCl5 increases with increase in
pressure.
39. The value of equilibrium constant does not depend upon pressure.
40. When a liquid and its vapour are at equilibrium and the pressure is suddenly decreased, cooling occurs.
41. If equilibrium constant for the reaction, A2
+ B2 2AB, is K, then for the backward reaction
AB
2
1
A2 +
2
1
B2, the equilibrium constant is
K
1
.
42. Catalyst makes a reaction more exothermic.
43. For the reaction, CaCO3
(s) CaO(s) + CO2
(g), Kp
= PCO
2
.
44. A catalyst increases the value of the equilibrium constant for a reaction.
45. In case of endothermic reactions, the equilibrium shifts in backward direaction on increasing the temperature.
46. The value of K increases with increases in pressure.
47. For the reaction, H2
+ I2
2HI, the equilibrium constant, K is dimenstionless.
PART - I : MIXED OBJECTIVE
Single Choice Correct :
1. A cylinder provided with a piston has some PCl5
which is in equilibrium with PCl3
and Cl2
. The system is
compressed with the help of piston. Indicate the correct statement :
(A) some more PCl5
will decompose (B) the system remains unaffected
(C) PCl3
and Cl2
will combine to form PCl5
(D) explosion occurs
2. XY2
dissociates as :
XY2
(g) XY(g) + Y(g)
Initial pressure of XY2
is 600 mm Hg. The total pressure at equilibrium is 800 mm Hg. Assuming volume of
system to remain constant, the value of Kp
is :
(A) 50 (B) 100 (C) 200 (D) 400
3. At temperature T, a compound AB2
(g) dissociates according to the reaction
2AB2
(g) 2AB(g) + B2
(g)
with a degree of dissociation ‘x’ which is very small as compared to unity. The expression for Kp
, in terms of
‘x’ and total pressure ‘P’ is
(A)
2
Px3
(B)
3
Px2
(C)
3
Px3
(D)
2
Px2

4. In the equilibrium SO2
Cl2
SO2
+ Cl2
at 2000 K and 10 atm pressure, % Cl2
= % SO2
= 40 by volume.
Then
(A) Kp
= 2 atm (B)
)SO(n
)ClSO(n
2
22
=
4
1
at equilibrium
(C) Kp
= 8 atm (D) n(SOCl2
) = n(SO2
) = n(Cl2
)
5. The KC for the reaction A + B C + D is 9. If one mole of each of A and B are mixed and there
is no change in volume, the number of moles of C formed is–
(A) 0.50 (B) 0.75 (C) 0.90 (D) 1.5
6. ‘a’ moles of PCl5 undergo thermal dissociation as –
PCl5 PCl3 +Cl2, the mole fraction of PCl3 at equilibrium is 0.5. The total pressure is 2.0
atmosphere. The partial pressure of Cl2 at equilibrium is –
(A) 2.5 (B) 1.0 (C) 0.5 (D) None
7. The equilibrium constant of the reaction A + B C + D is 10. If rate constant of forward reaction
is 203, the rate constant of backward reaction is –
(A) 20.3 (B) 10.3 (C) 2.03 (D) 203
8. The equilibrium constants for the reaction X2 2X at 300K and 600K are 10–8 and
10–3 respectively. The reaction is –
(A) Exothermic (B) Endothermic (C) Thermo neutral (D) Slow
9. The reversible reaction
[Cu(NH3)4]2+ + SO3
2– [Cu (NH3)3SO3] + NH3
is at equilibrium. What would not happen if ammonia is added –
(A) [SO3
2– ] would increase
(B) [Cu (NH3)3SO3] would increase
(C) The value of equilibrium constant would not change
(D) [Cu(NH3)4]2+ would increase
10. Two systems
PCl5
(g) PCl3
(g) + Cl2
(g)
and COCl2
(g) CO(g) + Cl2
(g)
are simultaneously in equilibrium in a vessel at constant volume. If some CO(g) is introduced in the vessel at
constant volume, then at new equilibrium the concentration of :
(A) PCl5
is greater (B) PCl3
remains unchanged
(C) PCl5
is less (D) Cl2
is greater
11. To the system, LaCl3
(s) + H2
O(g) LaClO(s) + 2HCl(g) – heat
already at equilibrium, more water vapour is added without altering T or V of the system. When equilibrium is
re-established, the pressure of water vapour is doubled. The pressure of HCl present in the system increases
by a factor of –
(A) 2 (B) 21/2
(C) 3 (D) 4
12. The equilibrium constant for the reaction 2 X (g) + Y (g) 2Z (g) is 2.25. What would be
the concentration of Y at equilibrium with 2.0 moles of X and 3.0 moles of Z in one litre vessel at
equilibrium–
(A) 1.0 moles (B) 2.25 moles (C) 2.0 moles (D) 4.0 moles

13. The relation between KP and KC for the reaction A + B C + 2D is –
(A) KP = KC [RT]–1 (B) KP.KC
–1 = RT (C) KC KP
–1 = RT (D) KP = KC [RT]3
14. In the following reactions –
(a) 2N2O5 (g) 4NO2 (g) + O2 (g) and
(b) 4NO2 (g) + O2 (g) 2N2O5 (g)
Choose the correct fact –
(A) (a) KP > KC (b) KP = KC (B) (a) KP > KC (b) KP < KC
(C) (a) KP = KC (b) KP < KC (D) (a) KP < KC (b) KP > KC
15. In a 10 litre box 2.5 mole iodic acid is taken. After equilibrium
2HI H2 + I2
the concentration of HI is found to be 0.1 mol l –1 The concentration of [H2] at equilibrium in mol –1, is –
(A) 2.4 (B) 0.15 (C) 1.5 (D) 7.5 × 10–2
16. The statement not applicable to an irreversible reaction is –
(A) It goes to completion in the forward direction
(B) On increasing the concentration of the reactants, the rate of the reaction increases
(C)The removal of the reaction products makes the reaction faster
(D) The addition of the reaction products does not influence the rate of reaction
17. At 27°C NO and Cl2
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 equilibria
(g) 2(g) (g)2NO Cl 2NOCl (g) 2(g) (g)2NO Cl 2NOCl  
2(g) (s) 2(s) pCl Mg MgCl ; K 0.2 atm 
–1
2(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.
18. 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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.
19. One mole of N2
O(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 N2
O(g)
is
(A) 20% (B) 40% (C) 50% (D) 60%.
20. KP will be equal to KC under which of the following conditions for the reaction–
aA + bB  cC + dD
(A) (a + b) > ( c+ d ) (B) (a + b) – ( c+d ) = 0
(C) (c + d) > ( a+ b ) (D) (a + c) = ( b+d )
21. For the equilibrium reaction :
2HCl (g) H2 (g) + Cl2 (g) the equilibrium constant is 1.0 × 10–5 what is the concentration of
HCl if the equilibrium concentrations of H2 and Cl2 are 1.2 ×10–3 M and 1.2 × 10–4 M respectively –
(A) 12 × 10–2 M (B) 12 × 10–4 M (C) 12 × 10–3 M (D) 12 × 104 M
22. The value of ng for the reaction : 2Hg(l) + Cl2 (g) Hg2Cl2 (s) is –
(A) – 1 (B) + 2 (C) – 2 (D) 0

23. For the reaction H2 (g) + I2 (g) 2HI (g) at 721 K, the value of equilibrium constant KC is 50.
When the equilibrium concentration of both is 0.5 M, value of KP under the same conditions will be–
(A) 0.02 (B) 0.2 (C) 50.0 (D) 50/RT
24. For an exothermic reaction in which the number of moles of reactants are more than the number of
moles of products. In order to displace the reaction in the reverse direction, what are the favourable
conditions .
(A) High pressure, low temp. and high conc. of the product
(B) Low pressure, high temp. and high conc. of the products
(C) Low pressure, low temp. and high conc. of the products
(D) High pressure, high temp. and low conc. of the products
25. What is wrong about equilibrium state –
(A) G (equi) = 0
(B) The reaction ceases at equilibrium
(C) Equilibrium constant is independent of initial concentrations of reactants
(D) Catalyst has no effect on equilibrium state
26. In A + B + Heat C, then reaction is favoured at –
(A) High temperature and high pressure
(B) High temperature and low pressure
(C) Low temperature and high pressure
(D) Low pressure and low temperature
27. The ratio of KP / KC for the reaction :
CO (g) + 1/2 O2 (g) CO2 (g) is –
(A) 1.0 (B) RT (C) 1/ RT (D) (RT)1/2
28. A 2B, Kp
; C D + E;
'
pK . If degrees of dissociation of A and C are same and Kp
= 2
'
pK , then
the ratio of total pressure p/ 'p = ?
(A) 1/2 (B) 1/3 (C) 1/4 (D) 2
29. For the reaction 2NO2
N2
O4
, if degrees of association of N2
O4
are 25%, 50% 75% and 100%, the
gradation of observed vapour densities is :
(A) d1
> d2
> d3
> d4
(B) d4
> d3
> d2
>d1
(C) d1
= d2
= d3
= d4
(D) None of these
30. N2
O4
 2NO2
, Kc
= 4. This reversible reaction is studied graphically as shown in figure. Select the
correct statements out of I, II and III.
Ti me
Concentration
A D E
F G
C
B
I. Reaction quotient has maximum value at point A
II. Reaction proceeds left to right at a point when 2 4 2[N O ] [NO ] 0.1M 
III. CK Q when point D or F is reached :
(A) I, II (B) II, III (C) I, III (D) I, II, III.

31. When 20g of CaCO3
were put into 10 litre flask and heated to 800°C, 35% CaCO3
remained unreacted at
equilibrium, Kp
for decomposition of CaCO3
is
32. When the equilibrium :
2 NH3 N2 + 3H2
has been established, NH3 is found to be 20% dissociated. The ratio of total number of moles at
equilibrium to the moles of NH3 at equilibrium is –
(A) 3/2 (B) 2/3 (C) 3/1 (D) 1/3
33. For dissociation of PCl5,mole fraction of Cl2 at equilibrium is 0.3. The total pressure is 1.0 atmosphere.
The partial pressure of PCl3 at equilibrium is –
(A) 1.3 (B) 0.3 (C) 0.7 (D) 0.4
34. The value of KP for the reaction 2H2 S (g) 2H2 (g) + S2 (g) is 1.2 × 10–2 at 1065º C. The value
of KC for this reaction is–
(A) 1.2 × 10–2 (B) < 1.2 × 10–2 (C) 83 (D) > 1.2 × 10–2
35. An equilibrium mixture for the reaction
2H2S (g) 2H2 (g) + S2 (g)
had 1 mole of H2S, 0.2 mole of H2 and 0.8 mole of S2 in a 2 litre flask. The value of KC in mol 1–
1 is–
(A) 0.004 (B) 0.080 (C) 0.016 (D) 0.160
36. If KP for the reaction A (g) + 2B (g) 3 C (g) + D (g) is 0.05 atm. at 1000 K. Its KC in
terms of R will be –
(A) 20000 R (B) 0.02 R (C) 5 × 10–5 R (D)
5 10 5
 
R
37. The equilibrium constant for a reaction N2 (g) + O2 (g) 2NO (g) is 4 × 10–4 at 2000 K. In the
presence of catalyst, the equilibrium is attained 10 times faster. The equilibrium constant in the
presence of catalyst, at 2000 K is –
(A) 40 × 10–4 (B) 4 × 10–4
(C) 4 × 10–2 (D) Difficult to compute without more data
38. At 250º C, the vapour density of PCl5 is Y (at equilibrium) and molar mass is Q (Initially). Its degree
of dissociation is then equal to -
(A)
Q Y
Y
–
(B)
Y Q
Q
–
(C)
Y Q
Q
– 2
2
(D)
Q Y
Y
– 2
2
39. 2SO2 + O2 2SO3 + 47 Kcal.
If large excess of oxygen is added to the above system in equilibrium this will -
(A) Shift the equilibrium to the right (B) Shift the equilibrium to the left
(C) Cause no change in the equilibrium (D) Produce only chormous amount of heat
2SO2 + O2 2SO3 + 47 Kcal.
40. For the reaction at equilibrium –
CO2 (g) + H2 (g) CO (g) + H2O (g)
[P – total pressure and x = degree of dissociation]
(A) PCO2 and PH2 is equal to
1
2
– x
P
L
NM O
QP (B) PCO and PH2 is equal to
1
2
– x
P
L
NM O
QP
(C) PH2 and PH O2 is equal to
x
P
2
L
NM O
QP (D) PCO2 and PH O2 is equal to
x
P
2
L
NM O
QP

41. The reaction in which the yield of the products cannot be increased by the application of high pressure
is –
(A) PCl3 (g) + Cl2 (g) PCl5 (g) (B) N2 (g) + O2 (g) 2NO (g)
(C) N2 (g) + 3H2 (g) 2NH3 (g) (D) 2SO2(g) + O2 (g) 2SO3 (g)
42. In the study of the reaction –
Cl2 + PCl3 PCl5
partial pressures of Cl2,PCl3 and PCl5 at equilibrium are 0.1, 0.1 and 0.2 atm respectively at 250º C.
At the same temperature, in another experiment on the same reaction, at equilibrium the partial
pressures of PCl3 and and Cl2 are half those in the first experiment. The partial pressure of the PCl5
at equilibrium in the second experiment is –
(A) One -fourth of the first (B) Half of the first
(C) One - eight of the first (D) One - third of the first
43. In the formation of nitric acid, N2 and O2 are made to combine. Thus
N2 + O2 2NO – Heat
Which of the following condition will favour the formation of NO –
(A) Low temperature (B) High temperature (C) Freezing point (D) All are favourable
44. The oxidation of SO2 to SO3 by oxygen is an exothermic reaction. The yield of SO3 will be maximum
if –
(A) Temperature is increased and pressure is kept constant
(B) Temperature is reduced and pressure is increased
(C) Both temperature and pressure are increased
(D) Both temperature and pressure are decreased
45. Which of the following conditions will be favourable for the formation of HX according to the gaseous
equilibrium :
H2 X2 + heat 2HX –
(A) High temperature and high pressure (B) High temperature and low pressure
(C) Low temperature and low pressure (D) High pressure and low temperature
46. A (l) + B (l) C (l) + D ( l)
One mole of A and one mole of B are mixed in a volume of one litre. If 0.9 mole per litre of C is found
at equilibrium, the equilibrium constant K is –
(A) 1 (B) 81 (C) 10 (D) 10–1
47. X2(g) + Y2(g) 2XY(g) reaction was studied at a certain temperature. In the beginning 1 mole of X2 was
taken in a one litre flask and 2 moles of Y2 was taken in another 2 litre flask and both these containers are
connected so equilibrium can be established. What is the equilibrium concentration of X2 and Y2? Given
Equilibrium concentration of [XY] = 0.6 moles/litre.
(A) 





 3.0
3
1
, 





 3.0
3
2
(B) 





 6.0
3
1
, 





 6.0
3
2
(C) (1 – 0.3) , (2 – 0.3) (D) (1 – 0.6) , (2 – 0.6)
48. Ammonia dissociates into N2
and H2
such that degree of dissociation  is very less than 1 and equilibrium
pressure is P0
then the value of  is [if Kp
for 2NH3
(g) N2
(g) + 3H2
(g) is 27 × 10–8
P0
2
:
(A) 10–4
(B) 4 × 10–4
(C) 0.02 (D) can’t be calculated.
49. At a temperature T, a compoundAB4
(g) dissociates as 2AB4
(g) AA2
(g) + 4B2
(g) with a degree of dissociation
x, which is small compared with unity. The expression of KP
in terms of x and total pressure P is ;
(A) 8P3
x5
(B) 256P3
x5
(C) 4Px2
(D) None of these

50. At 727°C and 1.23 atm of total equilibrium pressure, SO3
is partially dissociated into SO2
and O2
according to
SO3
(g) SO2
(g) + 1/2O2
(g). The density of equilibrium mixture is 0.9 gm/litre. The degree of dissociation is :
(A) 1/3 (B)2/3 (C) l/4 (D) 1/5.
51. A 10 litre box contains O3 and O2 at equilibrium at 2000 K. KP = 4 × 1014 atm for 2O3 (g) 3O2(g).
Assume that 32 OO PP  and if total pressure is 8 atm, then partial pressure of O3 will be :
(A) 8 × 10–5 atm (B) 11.3 × 10–7 atm (C) 9.71 × 10–6 atm (D) 9.71 × 10–2 atm
52. For the reaction CaCO3(s) CaO(s) + CO2(g), the pressure of CO2(g) depends on :
(A) the mass of CaCO3(s) (B) the mass of CaO(s)
(C) the masses of both CaCO3(s) and CaO(s) (D) temperature of the system
53. Ammonia gas at 15 atm is introduced in a rigid vessel at 300 K.At equilibrium the total pressure of the vessel
is found to be 40.11 atm at 300ºC. The degree of dissociation of NH3
will be:
(A) 0.6 (B) 0.4 (C) Unpredictable (D) None of these
54. For the equilibrium CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O(g)
Kp = 2.25 × 10–4 atm2 and vapour pressure of water is 22.8 Torr at 298 K.
CuSO4.5H2O(s) is efflorescent (i.e., loses water) when relative humidity is :
(A) less than 33.3% (B) less than 50 % (C) less than 66.6% (D) above 66.6%
55. Equilibrium constant for the following equilibrium is given at 0ºC.
Na2
HPO4
. 12H2
O (s) Na2
HPO4
. 7H2
O (s) + 5H2
O(g) KP
= 31.25 × 10–13
At equilibrium what will be partial pressure of water vapour :
(A)
5
1
×10–3
atm (B) 0.5 × 10–3
atm (C) 5 × 10–2
atm (D) 5 × 10–3
atm.
56. The equation  =
d)1n(
dD


is correctly matched for :
(A) A(g) nB/2(g) + nC/3(g) (B)A(g) nB/3(g) + (2n/3)C(g)
(C)A(g) (n/2)B(g) + (n/4)C(g) (D)A(g) (n/2)B(g) + C(g)
57. What is the equilibrium constant for the reaction
P4(s) + 5O2(g) P4O10(s) :
(A) Kc =
 5
2O
1
(B) Kc =
 
5
24
104
]O[]P[5
OP
(C) Kc = [O2]5 (D) Kc =
 
5
24
104
]O[]P[
OP
58. For the following three reactions 1, 2 and 3, equilibrium constants are given :
(1) CO(g) + H2O(g)  CO2(g) + H2(g) ; K1
(2) CH4(g) + H2O(g)  CO(g) + 3H2(g) ; K2
(3) CH4(g) + 2H2O(g)  CO2(g) + 4H2(g) ; K3
Which of the following relations is correct ?
(A) K1 2K = K3 (B) K2K3 = K1 (C) K3 = K1K2 (D) K3 · K2
3K1
2
59. For the reaction 3 A (g) + B (g)  2 C (g) at a given temperature , Kc = 9.0 . What must be the volume
of the flask, if a mixture of 2.0 mol each of A , B and C exist in equilibrium?
(A) 6L (B) 9L (C) 36 L (D) None of these
60. 1 mole N2 and 3 mol H2 are placed in a closed container at a pressure of 4 atm. The pressure falls to 3 atm
at the same temperature when the following equilibrium is attained.
N2(g) + 3H2(g)  2NH3(g). The equilibrium constant KP for dissociation of NH3 is:
(A)
5.0
1
× (1.5)3 atm–2 (B) 0.5 ×(1.5)3 atm2 (C)
33
)5.1(5.0 3


atm2 (D) 3
)5.1(5.0
33


atm–2

61. One mole of N2O4 (g) at 300 K is left in a closed container under one atm . It is heated to 600 K when
20 % by mass of N2O4 (g) decomposes to NO2 (g) . The resultant pressure is :
62. For the following gases equilibrium. N2O4 (g)  2NO2 (g)
Kp is found to be equal to Kc. This is attained when temperature is
(A) 0°C (B) 273 K (C) 1 K (D) 12.19 K
63. For the reaction; 2NO2(g)  2NO(g) + O2(g)
Kc = 1.8 × 10–6 at 184° C and R = 0.083 JK–1 mol–1. When Kp and Kc are compared at 184°C, it is found that:
(A) Kp > Kc
(B) Kp < Kc
(C) Kp = Kc
(D) Kp  Kc depends upon pressure of gases
64. PCl5 dissociation a closed container as :
PCl5(g)  PCl3(g) + Cl2(g)
If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is , the partial
pressure of PCl3 will be :
PCl5(g)  PCl3(g) + Cl2(g)
(A) P · 







1
(B) P · 







1
2
(C) P · 







1
(D) P · 







1
65. For the reaction A(g) + 2B(g)  C(g) + D(g) ; Kc = 1012 .
If the initial moles of A,B,C and D are 0.5, 1, 0.5 and 3.5 moles respectively in a one litre vessel. What is the
equilibrium concentration of B?
(A) 10–4 (B) 2 × 10–4 (C) 4 ×10–4 (D) 8 × 10–4
66. A 20.0 litre vessel initially contains 0.50 mole each of H2 and I2 gases. These substances react and finally
reach an equilibrium condition. Calculate the equilibrium concentration of HI if Keq = 49 for the reaction H2 +
I2  2HI.
(A) 0.78 M (B) 0.039 M (C) 0.033 M (D) 0.021 M
67. At 87°C, the following equilibrium is established
H2(g) + S(s)  H2S (g) Kp = 7 × 10–2
If 0.50 mole of hydrogen and 1.0 mole of sulfur are heated to 87°C in 1.0 L vessel, what will be the partial
pressure of H2S at equilibrium?
68. At 675 K, H2(g) and CO2 (g) react to form CO(g) and H2O (g), Kp for the reaction is 0.16.
If a mixture of 0.25 mole of H2(g) and 0.25 mol of CO2 is heated at 675 K, mole % of CO(g) in equilibrium
mixture is :
(A) 7.14 (B) 14.28 (C) 28.57 (D) 33.33
69. In which of the following reactions, increase in the pressure at constant temperature does not affect the
moles at equliibrium :
(A) 2NH3(g)  N2(g) + 3H2(g) (B) C(g) +
2
1
O2(g)  CO(g)
(C) H2(g) +
2
1
O2(g)  H2O(g) (D) H2(g) + I2(g)  2HI(g)
70. The exothermic formation of ClF3 is represented by the equation :
Cl2(g) + 3F2(g)  2ClF3(g) H = – 329 kJ
Which of the following will increase the quantity of ClF3 in an equilibrium mixture of Cl2, F2 and ClF3 :
(A) Increasing the temperature (B) Removing Cl2
(C) Increasing the volume of container (D)Adding F2

71. Densities of diamond and graphite are 3.5 and 2.3 gm/mL.
C (diamond) C (graphite) rH = –1.9 kJ/mole
favourable conditions for formation of diamond are
(A) high pressure and low temperature (B) low pressure and high temperature
(C) high pressure and high temperature (D) low pressure and low temperature
72. The equilibrium SO2Cl2(g)  SO2(g) + Cl2(g) is attained at 25°C in a closed rigid container and an inert gas,
helium is introduced. Which of the following statements is/are correct.
(A) concentrations of SO2, Cl2 and SO2Cl2 do not change
(B) more chlorine is formed
(C) concentration of SO2 is reduced
(D) more SO2Cl2 is formed
73. Following two equilibrium is simultaneously established in a container
PCl5(g) PCl3(g) + Cl2(g)
CO(g) + Cl2(g) COCl2(g)
If some Ni(s) is introduced in the container forming Ni (CO)4 (g) then at new equilibrium
(A) PCl3 concentration will increase (B) PCl3 concentration will decrease
(C) Cl2 concentration will remain same (D) CO concentration will remain same
74. The yield of product in the reaction
2A(g) + B(g) 2C(g) + Q kJ
would be lower at :
(A) low temperature and low pressure (B) high temperature & high pressure
(C) low temperature and to high pressure (D) high temperature & low pressure
75. What is the effect of the reduction of the volume of the system for the equilibrium 2C(s) + O2(g)  2CO (g)?
(A) The equilibrium will be shifted to the left by the increased pressure caused by the reduction in volume.
(B) The equilibrium will be shifted to the right by the decreased pressure caused by the reduction in volume.
(C) The equilibrium will be shifted to the left by the increased pressure caused by the increase in volume.
(D) The equilibrium will be shifted to the right by the increased pressure caused by the reduction in volume.
76. The vapour density of N2O4 at a certain temperature is 30. What is the % dissociation of N2O4 at this
temperature?
(A) 53.3% (B) 106.6% (C) 26.7% (D) None of these
77. The equilibrium constant KP (in atm) for the reaction is 9 at 7 atm and 300 K.
A2 (g)  B2(g) + C2 (g)
Calculate the average molar mass (in gm/mol) of an equilibrium mixture.
Given : Molar mass of A2, B2 and C2 are 70, 49 & 21 gm/mol respectively.
(A) 50 (B) 45 (C) 40 (D) 37.5
78. Vapour density of the equilibrium mixture of the reaction
2NH3 (g)  N2 (g) + 3H2 (g) is 6.0
Percent dissociation of ammonia gas is :
(A) 13.88 (B) 58.82 (C) 41.66 (D) None of these
79. The equilibrium constants
1PK and
2PK for the reactions
X  2Y and Z  P + Q, respectively are in the ratio of 1 : 9. If the degree of dissociation of X and Z be
equal then the ratio of total pressures at these equilibria is :
(A) 1 : 36 (B) 1 : 1 (C) 1 : 3 (D) 1 : 9
80. When N2O5 is heated at temp. T, it dissociates as 52ON  232 OON  , Kc = 2.5. At the same time N2O3
also decomposes as : N2O3  N2O + O2. If initially 4.0 moles of N2O5 are taken in 1.0 litre flask and allowed
to attain equilibrium, concentration of O2 was formed to be 2.5 M. Equilibrium concentration of N2O is
(A) 1.0 (B) 1.5 (C) 2.166 (D) 0.334

81. An exothermic reaction is represented by the graph :
(A) (B) (C) (D)
Multiple Choice Correct :
82. In which of the following reactions is Kp
< Kc
?
(A) CO(g) + Cl2
(g) COCl2
(g) (B) CO(g) + 3H2
(g) CH4
(g) + H2
O(g)
(C) 2BrCl(g) Cl2
(g) + Br2
(g) (D) I2
(g) 2I(g)
83. The dissociation of phosgene, which occurs according to the reaction
COCl2
(g) CO(g) + Cl2
(g)
is an endothermic process. Which of the following will increase the degree of dissociation of COCl2
?
(A) Adding Cl2
to the system (B) Adding helium to the system at constant pressure
(C) Decreasing the temperature of the system (D) Reducing the total temperature of the system
84. The equilibrium of which of the following reactions will not be disturbed by the addition of an inert gas at
constant volume ?
(A) H2
(g) + I2
(g) 2HI(g) (B) N2
O4
(g) 2NO2
(g)
(C) CO2
(g) + 2H2
(g) CH3
OH(g) (D) C(s) + H2
O(g) CO(g) + H2
(g)
85. A box contains CO(g), Cl2
(g) and COCl2
(g) in equilibrium at 1000 K. The removal of CO(g) will
(A) decrease the concentration of COCl2
(B) increase the concentration of Cl2
(C) increase the concentration of COCl2
(D) reduce the concentration of CO as well as Cl2
86. An industrial fuel, ‘water gas’, which consists of a mixture of H2
and CO can be made by passing steam over
red-hot carbon. The reaction is
C(s) + H2
O(g) CO(g) + H2
(g) , H = +131 kJ
The yield of CO and H2
at equilibrium would be shifted to the product side by
(A) raising the relative pressure of the steam (B) adding hot carbon
(C) raising the temperature (D) reducing the volume of the system
87. For the equilibrium 2SO2
(g) + O2
(g) 2SO3
(g), H = -198 kJ, the equilibrium concentration of SO3
will
be affected by
(A) doubling the volume of the reaction vessel
(B) increasing the temperature at constant volume
(C) adding more oxygen to the reaction vessel
(D) adding helium to the reaction vessel at constant volume
88. The following reaction attains equilibrium at high temperature
N2
(g) + 2H2
O(g) + heat 2NO(g) + 2H2
(g)
The yield of NO is affected by
(A) increasing the nitrogen concentration (B) decreasing the hydrogen concentration
(C) compressing the reaction mixture (D) none of these
89. N2
(g) + 3H2
(g)  2NH3
+ heat
In this reaction, the direction of equilibrium will be shifted to the right by
(A) increasing the concentration of nitrogen (B) compressing the reaciton mixture
(C) removing the catalyst (D) decreasing the concentration of ammonia
N2
(g) + 3H2
(g)  2NH3
+ heat
90. The dissociation of ammonia carbonate may be represented by the equation
NH4
CO2
NH2
(s)  2NH3
(g) + CO2
(g)
The equilibrium will shift from right to left if there is
(A) a decrease in pressure
(B) an increase in temperature
(C) an increase in the concentration of ammonia
(D) an increase in the concentration of carbon dioxide

91. Which of the following statements about the reaction quotient Q are correct?
(A) the reaction quotient, Q and the equilibrium constant always have the same numerical value.
(B) Q may be >, < , = Keq
.
(C) Q(numerical value) varies as reaction proceeds
(D) Q = 1 at equilibrium
92. Which of the following factors will increase solubility of NH3
(g) in H2
O?
NH3
(g) + H2
O(aq) NH4
OH(aq)
(A) increase in pressure (B) addition of water.
(C) liquefaction of NH3
(D) decrease in pressure
93. Which of the following factors will affect solubility of CaO in H2
O?
(A) pressure (B) temperature (C) addition of water (D)volume
94. For the following equilibrium
NH4
HS(s) NH3
(g) + H2
S(g)
partial pressure of NH3
will increase
(A) if NH3
is added after equilibrium is established
(B) if H2
S is added after equilibrium is established
(C) temperature is increased
(D) volume of the flask is decreased
95. For the gas phase reaction carried out in a vessel,
C2
H6
C2
H4
+ H2
the equilibrium concentration of C2
H4
can be increased by -
(A) increasing the temperature (B) decreasing the pressure
(C) removing some H2
(D) adding some C2
H6
96. For the reaction
2A(g) + B(g) 2C(g) ; H = +13.6 kJ
which of the following will increase the extent of the reaction at equilibrium
(A) increasing the temperature (B) increasing the pressure
(C) addition of catalyst (D) removing C
97. The reactions in which the yield of the products cannot be increased by the application of high pressure
(A) 2SO2
(g) + O2
(g) 2SO3
(g) (B) NH4
HS(s) NH3
(g) + H2
S(g)
(C) N2
O4
(g) 2NO2
(g) (D) N2
(g) + 3H2
(g) 2NH3
(g)
98. (i) N2
(g) + O2
(g) 2NO(g), K1
(ii) 





2
1 N2
(g) + 





2
1
O2
(g) NO(g) ; K2
(iii) 2NO(g) N2
(g) + O2
(g) ; K3
(iv) NO(g) 





2
1
N2
(g) + 





2
1
O2
(g) ; K4
Correct relation between K1
, K2
and K4
is/are :
(A) K1
× K3
= 1 (B) 41 KK  = 1 (C) 23 KK  = 1 (D) None
99. When NaNO3 is heated in a closed vessel, oxygen is liberated and NaNO2 is left behind. At equilibrium,
NaNO3 (s) NaNO2 (s) +
2
1
O2 (g)
(A) addition of NaNO2 favours reverse reaction
(B) addition of NaNO3 favours forwards reaction
(C) increasing temperature favours forward reaction
(D) increasing pressure favours reverse reaction

100. The dissociation of ammonium carbamate may be represented by the equation
NH4
CO2
NH2
(s) 2NH3
(g) + CO2
(g)
H0
for the forward reaction is negative. The equilibrium will shift from right to left if there is
(A) a decrease in pressure
(B) an increase in temperature
(C) an increase in the concentration of ammonia
(D) an increase in the concentration of carbon dioxide
101. For the reaction PCl5(g)  PCl3(g) + Cl2(g), the forward reaction at constant temperature is favoured by
(A) introducing an inert gas at constant volume
(B) introducing chlorine gas at constant volume
(C) introducing an inert gas at constant pressure
(D) introducing PCl5 at constant volume.
102. When NaNO3 is heated in a closed vessel, oxygen is liberated and NaNO2 is left behind. At equilibrium
(A) addition of NaNO2 favours reverse reaction
(B) addition of NaNO3 favours forward reaction
(C) increasing temperature favours forward reaction
(D) increasing pressure favours reverse reaction
103. For the gas phase reaction, C2H4 + H2  C2H6 (H = – 32.7 kcal), carried out in a closed vessel, the
equilibrium moles of C2H4 can be increased by
(C) removing some H2 (D) adding some C2H6
104. For the gas phase exothermic reaction, A2 + B2  C2, carried out in a closed vessel, the equilibrium moles
of A2 can be increased by
(C) adding inert gas at constant pressure (D) removing some C2
105. Consider the equilibrium HgO(s) + 4I– (aq) + H2O (l) HgI4
2– (aq) + 2OH– (aq), which changes will decrease
the equilibrium concentration of HgI4
2–
(A) Addition of 0.1 M HI (aq) (B)Addition of HgO (s)
(C)Addition of H2O (l) (D)Addition of KOH (aq)
106. Decrease in the pressure for the following equilibria : H2O (s)  H2O(l) result in the :
(A) formation of more H2O (s) (B) formation of more H2O(l)
(C) increase in melting point of H2O(s) (D) decrease in melting point of H2O(s)
PART - II : SUBJECTIVE QUESTIONS
1. For the reaction NOBr (g) NO(g) +
2
1
Br2 (g)
KP = 0.15 atm at 90°C. If NOBr, NO and Br2 are mixed at this temperature having partial pressures 0.5 atm,
0.4 atm and 0.2 atm respectively, will Br2 be consumed or formed ?
2. The KP values for three reactions are 10–5 , 20 and 300 then what will be the correct order of the percentage
composition of the products.
3. 1 mole of N2 and 3 moles of H2 are placed in 1L vessel. Find the concentration of NH3 at equilibrium, if
equilibrium pressure is 1 atm and the equilibrium constant at 400K is
27
4
4. The value of Kc for the reaction, N2(g) + 2O2(g) 2NO2(g) at a certain temperature is 400. calcualte the
value of equilibrium constant for.
(i) 2NO2(g) N2(g) + 2 O2(g) ; (ii) 1/2 N2(g) + O2(g) NO2(g)

5. From the following data :
(i) H2(g) + CO2(g) H2O(g) + CO(g)
K2000K = 4.4
(ii) 2H2O(g) 2H2(g) + O2(g)
K2000K = 5.31 x 10–10
(iii) 2CO(g) + O2(g) 2CO2(g)
K1000K = 2.24 x 1022
State whether the reaction (iii) is exothermic or endothermic?
6. In an experiment starting with 1 mole of ethyl alcohol, 1 mole of acetic acid and 1 mole of water at T0C, the
equilibrium mixture on analysis shows that 54.3% of the acid is esterfied. Calculate the equilibrium constant
of this reaction.
7. Derive an expression for Kc and Kp for the reaction
N2(g) + 3H2(g)  2NH3(g)
Assuming that in a container of volume V, initially 1 mole of N2 and 3 moles of H2 were taken and
at equilibrium 2x moles of NH3 is formed.
8. The degree of dissociation at a certain or given temperature of PCl5 at 2 atm is found to be 0.4. At
what pressure, the degree of dissociation of PCl5 will be 0.6 at the same temperature? Also calculate
the equilibrium constant for the reverse reaction.
9. (i) Calculate the percentage dissociation of H2S(g), if 0.1 mole of H2S is kept in 0.4 litre vessel at
1000K. For the reaction, 2H2S(g)  2H2(g) + S2(g), the value of Kc = 1.6 × 10–6
(ii) A sample of HI was found to be 22% dissociated when equilibrium was reached. What will be the
degree of dissociation if hydrogen is added in the proportion of 1 mole for every mole of HI originally
present, the temperature and volume of the system being kept constant?
10. Calculate the degree of dissociation and Kp for the following reaction.
PCl5(g) PCl3(g) + Cl2(g)
t = 0 a 0 0
t = t a –x x x
Since for a mole, x moles are dissociated
11. The vapour density of PCl5 at 250ºC is found to be 57.9. Calculate percentage dissociation at this
temperature.
12. The vapour density of a mixture containing NO2 and N2O4 is 38.3 at 33°C calculate the no. of moles of
NO2 if 100g of N2O4 were taken initially.
13. What is the effect on the following equilibrium if each of the indicated stresses is applied?
2(g) 2(g)
1
N +O
2 2(g)
NO + heat
(a) Increase in N2 concentration
(b) Decrease in temperature
(c) Increase in volume of vessel
(d) addition of a catalyst

14. Kc for the reaction N2O4  2NO2 in chloroform at 291 K is 1.14. Calculate the free energy change
of the reaction when the concentration of the two gases are 0.5 mol dm–3 each at the same
temperature. (R = 0.082 lit atm K–1 mol–1.)
15. Calculate the pressure of CO2 gas at 700K in the heterogeneous equilibrium reaction
CaCO3(s)  CaO(s) + CO2(g) if G0 for this reaction is 130.2 kJ mol–1.
16. For the equilibrium NiO(s) + CO(g)  Ni(s) + CO2(g), G0 (J mol–1) = – 20,700 – 11.97 T. Calculate
the temperature at which the product gases at equilibrium at 1 atm will contain 400 ppm (parts per
million) of carbon monoxide.
17. (i) Calculate the partial pressure of HCl gas above solid a sample of NH4Cl(s) as a result of its
decomposition according to the reaction:
NH4Cl(s)  NH3(g) + HCl(g) Kp = 1.04 × 10–16
(ii) Calculate the equilibrium constant of a reaction at 300 K if G0 at this temperature for the
reaction is 29.29 kJ mol–1.
18. For the formation of ammonia the equilibrium constant data at 673K and 773K respectively are
1.64 × 10–4 and 1.44 × 10–5 respectively. Calculate heat of the reaction. Given R=8.314 JK–1mol–1.
19. The equilibrium constant KP for the reaction N2(g) + 3H2(g) 2NH3(g) is 1.6  10-4 atm at
400oC. What will be the equilibrium constant at 500oC if heat of the reaction in this temperature
range is 25.14 k cal?
20. 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 ? If the volume of reaction vessel is reduced to half;
then what will be the effect on equilibrium?
21. 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.
22. If a mixture of N2 and H2 in the ratio 1 : 3 at 50 atmosphere and 650°C is allowed to react till equilibrium is
reached. Ammonia present at equilibrium was at 25 atm pressure. Calculate the equilibrium constant for
the reaction. N2(g) + 3H2(g) 2NH3(g)
23. 0.96 g of HI were heated to attain equilibrium 2HI(g) H2 (g) + I2(g). The reaction mixture on titration
requires 15.7 mL of N/10 hypo solution. Calculate degree of dissociation of HI.
24. Would 1% CO2 in air be sufficient to prevent any loss in weight when M2CO3 is heated at 120oC ?
M2CO3(s) M2O(s) + CO2(s)
Kp = 0.0095 atm at 120oC. How long would the partial pressure of CO2 have to be to promote this reaction
at 120oC ?
25. In a container of constant volume at a particular temparature N2 and H2 are mixed in the molar ratio of 9:13.
The following two equilibria are found to be coexisting in the container
N2 (g) + 3H2 (g) 2NH3 (g)
N2 (g) + 2H2 (g) N2H4 (g)
The total equilibrium pressure is found to be 3.5 atm while partial pressure of NH3 (g) and H2(g) are 0.5 atm
and 1 atm respectively. Calculate of equilibrium constants of the two reactions given above.

26. 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.
27. 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.
28. When baking soda is heated in a sealed tube, following equilibrium exits:
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.
29. For the reaction
NH3(g)
2
1
N2(g) +
2
3
H2(g)
Show that degree of dissociation of NH3 is given as :
2/1
pK
P
4
33
1










where ‘P’ is equilibrium pressure. If Kp of the above reaction is 78.1 atm at 400°C, calculate KC.
30. (i) The vapour density of a mixture consisting of NO2 and N2O4 is 38.3 at 26.7°C. Calculate the
number of moles of NO2 in 100 gm mixture.
(ii) Establish a relationship between Kc and Kp for the following reactions.
(a) N2(g) + O2(g)  2NO(g)
(b) 4 2 3(s) 3(g) 2(g) 2 (g)(NH ) CO 2NH +CO +H O4 2 3(s) 3(g) 2(g) 2 (g)(NH ) CO 2NH +CO +H O


31. Calculate the Kc and Kp for the following reactions and also deduce the relationship between Kc
and Kp
(i) 2SO2(g) + O2(g)  2SO3(g) (ii) 2(g) 2(g)
1 3
N + H
2 2
 NH3(g)
32. At 400K for the gaseous reaction
2A + 3B  3C + 4D
the value of Kp is 0.05. Calculate the value of Kc (R = 0.082 dm3 atm K–1 mol–1)

PART - I : IIT-JEE PROBLEMS (PREVIOUS YEARS)
* Marked Questions are having more than one correct option.
1. When 3.06 g of solid NH4
HS is introduced into a two litre evacuated flask at 27ºC,30% of the solid decomposes
into gaseous ammonia and hydrogen sulphide. [JEE-1999, 7/120]
(i) Calculate kC
& kP
for the reaction at 27ºC.
(ii) What would happen to the equilibrium when more solid NH4
HS is introduced into the flask.
2. For a chemical reaction 3X(g) + Y(g) X3
Y(g), the amount of X3
Y at equilibrium is affected by
[JEE-1999, 2/80]
(A) temperature and pressure (B) temperature only
(C) pressure only (D) temperature, pressure and catalyst
3. For the reversible reaction, N2
(g) + 3H2
(g) 2NH3
at 500°C, the value of KP
is 1.44 × 10–5
when
partial pressure is measured in atmospheres. The corresponding value of KC
, with concentration in
mole litre–1
, is [JEE 2000, 1/35]
(A) 2
5
)500082.0(
1044.1




(B) 2
5
)773314.8(
1044.1




(C) 2
5
)773082.0(
1044.1

 
(D) 2
5
)773082.0(
1044.1




4. When two reactants, A & B are mixed to give products C & D, the reaction quotient Q, at the initial stages of
the reaction. [JEE-2000, 1/35]
(A) is zero (B) decrease with time
(C) is independent of time (D) increases with time
5. At constant temperature, the equilibrium constant (KP
) for the decomposition reaction N2
O4
2NO2
is
expressed by KP
=
)x1(
)Px4(
2
2

, where P = pressure, x = extent of decomposition. Which one of the following
statements is true ? [JEE 2001, 1/35]
(A) KP
is increases witn increase of P (B) KP
is increases witn increase of x
(C) KP
is increases witn decrease of x (D) KP
remains constant with change in P and x
6. Consider the following equilibrium in a closed container [JEE 2002, 3/90]
N2
O4
(g) 2NO2
(g)
At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following
statements holds true regarding the equilibrium constant (KP
) and degree of dissociation ()?
(A) neither KP
nor  changes (B) both KP
and  change
(C) KP
changes, but  does not change (D) KP
does not change but  changes
7. N2
+ 3H2
2 NH3
K = 4 × 106
at 298
K = 41 at 400 K
Which statements is correct? [JEE 2006, 3/184]
(A) If N2
is added at equlibrium condition, the equilibrium will shift to the forward direction because according
to IInd
law of thermodynamics the entropy must increases in the direction of spontaneous reaction.
(B) The condition for equilibrium is 3NHG2 = 2HG3 + 2NG where G is Gibbs free energy per mole of the
gaseous species measured at that partial pressure.
(C) Addition of catalyst does not change Kp
but changes H.
(D) At 400 K addition of catalyst will increase forward reaction by 2 times while reverse reaction rate will be
changed by 1.7 times.

8. The value of log10
K for a reaction A B is :
(Given : 
 K298rH = –54.07 kJ mol–1
, 
 K298rS = 10 JK–1
mol–1
and R = 8.314 JK–1
mol–1
; 2.303 x 8.314 x 298
= 5705) [JEE 2007, 3/162]
(A) 5 (B) 10 (C) 95 (D) 100
9. STATEMENT-1 :For every chemical reaction at equilibrium, standard Gibbs energy of reaction is zero.
STATEMENT-2 : At constant temperature and pressure, chemical reactions are spontaneous in the direction
of decreasing Gibbs energy. [JEE 2008, 3/163]
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(C) Statement-1 is True, Statement-2 is False.
(D) Statement-1 is False, Statement-2 is True.
(E) Statement-1 and Statement-2 both are False.
10.* The thermal dissociation equilibrium of CaCO3
(s) is studied under different conditions :
CaCO3
(s) CaO(s) + CO2
(g) [JEE_2013]
For this equilibrium, the correct statement(s) is(are)
(A) H is dependent on T
(B) K is independent of the initial amount of CaCO3
(C) K is dependent on the pressure of CO2
at a given T
(D) H is independent of the catalyst, if any
11. The Ksp
of Ag2
CrO4
is 1.1 × 10–12
at 298 K. The solubility (in mol/L) of Ag2
CrO4
in a 0.1 M AgNO3
solution is:
[JEE_2013]
(A) 1.1 × 10–11
(B) 1.1 × 10–10
(C) 1.1 × 10–12
(D) 1.1 × 10–9
PART - II : AIEEE PROBLEMS (PREVIOUS YEARS)
1. Change in volume of the system does not alter the number of moles in which of the following equilibriums :
[AIEEE 2002]
(1) N2
(g) + O2
(g) 2NO(g) (2) PCl5
(g) PCl3
(g) + Cl2
(g)
(3) N2
(g) + 3H2
(g) 2NH3
(g) (4) SO2
Cl2
(g) SO2
(g) + Cl2
(g)
2. In which of the following reactions, increase in the volume at constant temperature don’t effect the number of
moles of at equilibrium : [AIEEE 2002]
(1) 2NH3
N2
+ 3H2
(2) C (g) + (1/2) O2
(g) CO (g)
(3) H2
(g) + O2
(g) H2
O2
(g) (4) none of these
3. Consider the reaction equilibrium
2SO2
(g) + O2
(g) 2SO3
(g) ; H° = – 198 kJ.
On the basis of Le Chatelier’s principle, the condition favourable for the forward reaction is [AIEEE 2003]
(1) lowering of temperature as well as pressure
(2) increasing temperature as well as pressure
(3) lowering the temperature and increasing the pressure
(4) any value of temperature and pressure.
4. For the reaction equilibrium, N2
O4
(g) 2NO2
(g) the concentrations of N2
O4
and NO2
at equilibrium are
4.8 × 10–2
and 1.2 × 10–2
mol L–1
respectively. The value of Kc
for the reaction is [AIEEE 2003]
(1) 3.3 × 102
mol L–1
(2) 3 × 10–1
mol L–1
(3) 3 × 10–3
mol L–1
(4) 3 × 103
mol L–1

5. What is the equilibrium constant expression for thereaction P4
(s)+ 5O2
(g) P4
O10
(s)? [AIEEE 2004]
P4
(s) + 5O2
(g) P4
O10
(s) ?
(1) KC
= [P4
O10
]/[P4
] [O2
]5
(2) KC
= 1/[O2
]5
(3) KC
= [O2
]5
(4) KC
= [P4
O10
] / 5[P4
] [O2
]
6. For the reaction, CO(g) + Cl2
(g) COCl2
(g) then Kp
/Kc
is equal to : [AIEEE 2004]
(1) 1/RT (2) 1.0 (3) RT (4)RT
7. The equilibrium constant for the reaction, N2
(g) + O2
(g) 2NO(g) at temperature T is 4 × 10–4
. The value
of Kc
for the reaction, NO(g)
2
1
N2
(g) +
2
1
O2
(g) at the same temperature is [AIEEE 2004]
NO(g)
2
1
N2
(g) +
2
1
O2
(g)
(1) 2.5 × 102
(2) 0.02 (3) 4 × 10–4
(4) 50
8. For the reaction, 2NO2
(g) 2 NO(g) + O2
(g), [AIEEE 2005]
(KC
= 1.8 × 10–6
at 184°C)
(R = 0.0831 kJ/(mol.K))
When Kp
and Kc
are compared at 184°C it is found that
(1) Whether Kp
is greater than, less than or equal to Kc
depends upon the total gas pressure
(2) Kp
= Kc
(3) Kp
is less than Kc
(4) Kp
is greater than Kc
9. The exothermic formation of ClF3
is represented by the equation : [AIEEE 2005]
Cl2
(g) + 3F2
(g) 2ClF3
(g) ; rH = – 329 J
which of the following will increase the quantity of ClF3
in an equilibrium mixture of Cl2
, F2
and ClF3
.
(1)Adding F2
(2) Increasing the volume of container
(3) Removing Cl2
(4) Increasing the temperature
10. An amount of solid NH4
HS is placed in a flask already containing ammonia gas at a certain temperature at
0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3
and H2
S gases in the flask.
When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm? The
equilibrium constant for NH4
HS decomposition at this temperature is : [AIEEE 2005]
(1) 0.11 (2) 0.17 (3) 0.18 (4) 0.30
11. Phosphorus pentachloride dissociates as follows in a closed reaction vessel.
PCl5
(g) PCl3
(g) + Cl2
(g)
If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5
is x, the partial
pressure of PCl3
will be : [AIEEE 2006]
(1) 





1x
x
P (2) 





x–1
x2
P (3) 





1x
x
P (4) 





x–1
x
P
12. The equilibrium constant for the reaction, SO3
(g) SO2
(g) +
2
1
O2
(g)
is KC
= 4.9 × 10–2
. The value of KC
for the reaction 2SO2
(g) + O2
(g) 2SO3
(g)will be [AIEEE 2006]
(1) 416 (2) 2.40 × 10–3
(3) 9.8 × 10–2
(4) 4.9 × 10–2
13. For the following three reactions a, b and c, equilibrium constants are given:
(A) CO(g) + H2
O(g) CO2
(g) + H2
(g); K1
(B) CH4
(g) + H2
O(g) CO(g) + 3H2
(g); K2
(C) CH4
(g) + 2H2
O(g) CO2
(g) + 4H2
(g); K3
Which of the following relations is correct ? [AIEEE 2008, 3/105]
(1) K2
K3
= K1
(2) K3
= K1
K2
(3) K3
K2
3
= K1
2
(4) 321 KKK 

14. The equilibrium constants
1
pK and
2
pK for the reactions X 2Y and Z P + Q, respectively are in the
ratio of 1 : 9. If the degree of dissociation of X and Z be equal then the ratio of total pressures at these
equilibria is [AIEEE 2008, 3/105]
(1) 1 : 1 (2) 1 : 3 (3) 1 : 9 (4) 1 : 36
15. If 10–4
dm3
of water is introduced into a 1.0 dm3
flask at 300 K, how many moles of water are in the vapour
phase when equilibrium is established ? [AIEEE 2010, 4/144]
(Given : Vapour pressure of H2
O at 300 K is 3170 Pa ; R= 8.314 J K–1
mol–1
)
(1) 5.56 × 10–3
mol (2) 1.53 × 10–2
mol (3) 4.46 × 10–2
mol (4) 1.27 × 10–3
mol
16. A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the
addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is : [AIEEE 2010, 4/144]
(1) 1.8 atm (2) 3 atm (3) 0.3 atm (4) 0.18 atm
NCERT QUESTIONS
1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the
container is suddenly increased.
(a) What is the initial effect of the change on vapour pressure ?
(b) How do rates of evaporation and condensation change initially ?
(c) What happens when equilibrium is restored finally and what will be the final vapour pressure ?
2. What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2] =
0.60M, [O2] = 0.82M and [SO3] = 1.90M ?
2SO2(g) + O2(g) 2SO3(g)
3. At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of I atoms
I2 (g) 2I (g)
Calculate Kp for the equilibrium.
4. Write the expression for the equilibrium constant, Kc for each of the following reactions:
(i) 2NOCl (g) 2NO (g) + Cl2 (g)
(ii) 2Cu(NO3)2 (s) 2CuO (s) + 4NO2 (g) + O2 (g)
(iii) CH3COOC2H5 (aq) + H2O(l) CH3COOH (aq) + C2H5OH (aq)
(iv) Fe3+ (aq) + 3OH– (aq) Fe(OH)3(s)
(v) I2 (s) + 5F2 2IF5
5. Find out the value of Kc for each of the following equilibria from the value of Kp :
(i) 2NOCl (g) 2NO (g) + Cl2 (g); Kp = 1.8 × 10–2 at 500 K
(ii) CaCO3 (s) CaO(s) + CO2(g); Kp = 167 at 1073 K
6. For the following equilibrium, Kc = 6.3 × 1014 at 1000 K
NO (g) + O3 (g) NO2 (g) + O2 (g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc,
for the reverse reaction ?
7. Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
8. Reaction between N2 and O2– takes place as follows :
2N2 (g) + O2 (g) 2N2O (g)
If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O
at a temperature for which Kc = 2.0 10–37, determine the composition of equilibrium mixture.

9. Nitric oxide reacts with Br2 and give nitrosyl bromide as per reaction given below :
2NO (g) + Br2 (g) 2NOBr (g)
When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature,
0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2.
10. At 450K, Kp = 2.0 × 1010 bar for the given reaction at equilibrium.
2SO2(g) + O2(g) 2SO3 (g)
What is Kc at this temperature ?
11. Asample of HI(g) is placed in flask at a pressure of 0.2 atm.At equilibrium the partial pressure of HI(g) is 0.04
atm. What is Kp for the given equilibrium ?
2HI (g) H2 (g) + I2 (g)
12. A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at
500 K. At this temperature, the equilibrium constant, Kc for the reaction N2 (g) + 3H2 (g) 2NH3 (g) is 1.7
102. Is the reaction mixture at equilibrium ? If not, what is the direction of the net reaction ?
13. The equilibrium constant expression for a gas reaction is,
4 5
3 2
c 4 6
2
[NH ] [O ]
k
[NO] [H O]

Write the balanced chemical equation corresponding to this expression.
14. One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of
water (by mass) reacts with CO according to the equation,
H2O (g) + CO (g) H2 (g) + CO2 (g)
Calculate the equilibrium constant for the reaction.
15. At 700 K, equilibrium constant for the reaction:
H2 (g) + I2 (g) 2HI (g)
is 54.8. If 0.5 mol L–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g)
assuming that we initially started with H2(g) HI(g) allowed it to reach equilibrium at 700K ?
16. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration
of ICl was 0.78 M ?
2ICl (g) I2 (g) + Cl2 (g); Kc = 0.14
17. Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when
it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium ?
C2H6 (g) C2H4 (g) + H2 (g)
18. Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as :
CH3COOH (l) + C2H5OH (l) CH3COOC2H5 (l) + H2O (l)
(i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and
is not a solvent in this reaction)
(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl
acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl
acetate is found after sometime. Has equilibrium been reached ?
19. A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained,
concentration of PCl5 was found to be 0.5 × 10–1 mol L–1. If value of Kc is 8.3 × 10–3, what are the concentrations
of PCl3 and Cl2 at equilibrium ?
PCl5 (g) PCl3 (g) + Cl2 (g)

20. One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon
monoxide to give iron metal and CO2.
FeO (s) + CO (g) Fe (s) + CO2 (g); Kp = 0.265 atm at 1050K
What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are :
pCO= 1.4 atm and pCO2
= 0.80 atm ?
21. Equilibrium constant, Kc for the reaction
N2 (g) + 3H2 (g) 2NH3 (g) at 500 K is 0.061
At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L–1 N2, 2.0 mol
L–1 H2 and 0.5 mol L–1 NH3. Is the reaction at equilibrium ?
If not in which direction does the reaction tend to proceed to reach equilibrium ?
22. Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium :
2BrCl (g) Br2 (g) + Cl2 (g)
for which Kc = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10–3 mol L–1, what is its
molar concentration in the mixture at equilibrium ?
23. At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with soild carbon has
90.55% CO by mass
C (s) + CO2 (g) 2CO (g)
Calculate Kc for this reaction at the above temperature.
24. Calculate a) f G
 and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298K
NO (g) + 1/2O2 (g) NO2 (g)
f G
 (NO2) = 52.0 kJ/mol
f G
 (NO) = 87.0 kJ/mol
f G
 (O2) = 0 kJ/mol
25. Does the number of moles of reaction products increase, decrease or remain same when each of the
following equilibria is subjected to a decrease in pressure by increasing the volume ?
(a) PCl5 (g) PCl3 (g) + Cl2 (g)
(b) CaO (s) + CO2 (g) CaCO3 (s)
(c) 3Fe (s) + 4H2O (g) Fe3O4 (s) + 4H2 (g)
26. Which of the following reactions will get affected by increasing the pressure ?Also, mention whether change
will cause the reaction to go into forward or backward direction.
(i) COCl2 (g) CO (g) + Cl2 (g)
(ii) CH4 (g) + 2S2 (g) CS2 (g) + 2H2S (g)
(iii) CO2 (g) + C (s) 2CO (g)
(iv) 2H2 (g) + CO (g) CH3OH (g)
(v) CaCO3 (s) CaO (s) + CO2 (g)
(vi) 4 NH3 (g) + 5O2 (g) 4NO (g) + 6H2O(g)
27. The equilibrium constant for the following reaction is 1.6 × 105 at 1024K
H2(g) + Br2(g) 2HBr(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.
28. Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic
reaction :
CH4 (g) + H2 O (g) CO (g) + 3H2 (g)
(a) Write as expression for Kp for the above reaction.
(b) How will the values of Kp and composition of equilibrium mixture be affected by
(i) increasing the pressure
(ii) increasing the temperature
(iii) using a catalyst ?

29. Describe the effect of :
(a) addition of H2
(b) addition of CH3OH
(c) removal of CO
(d) removal of CH3OH
on the equilibrium of the reaction:
2H2(g) + CO (g) CH3OH (g)
30. At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3 × 10-3. If
decomposition is depicted as,
PCl5 (g) PCl3 (g) + Cl2 (g) rH
 = 124.0 kJ mol–1
(a) write an expression for Kc for the reaction.
(b) what is the value of Kc for the reverse reaction at the same temperature ?
(c) what would be the effect on Kc if (i) more PCl5 is added (ii) pressure is increased (iii) the temperature is
increased ?
31. Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high
temperature steam. The first stage of two stage reaction involves the formation of CO and H2. In second
stage, CO formed in first stage is reacted with more steam in water gas shift reaction,
CO (g) + H2O (g) CO2 (g) + H2 (g)
If a reaction vessel at 400 °C is charged with an equimolar mixture of CO and steam such that pCO = pH2O =
4.0 bar, what will be the partial pressure of H2 at equilibrium ? Kp = 10.1 at 400°C
32. Predict which of the following reaction will have appreciable concentration of reactants and products:
(a) Cl2 (g) 2Cl (g) Kc = 5 × 10–39
(b) Cl2 (g) + 2NO (g) 2NOCl (g) Kc = 3.7 × 108
(c) Cl2 (g) + 2NO2 (g) 2NO2Cl (g) Kc = 1.8
33. The value of Kc for the reaction 3O2 (g) 2O3 (g) is 2.0 × 10–50 at 25°C. If the equilibrium concentration of
O2 in air at 25°C is 1.6 × 10–2, what is the concentration of O3 ?
34. The reaction, CO (g) + 3H2 (g) CH4 (g) + H2O (g)
is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O
and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium
constant, Kc for the reaction at the given temperature is 3.90.

EXERCISE # 1
PART # I
A-1. (A) A-2. (C) A-3. (C) A-4. (B) A-5. (B) A-6. (B) A-7. (D)
A-8. (D) A-9. (B) A-10. (D) A-11. (C) A-12. (C) A-13. (A) A-14. (C)
A-15. (C) A-16. (A) A-17. (D) A-18. (A) A-19. (C) A-20. (B) A-21. (D)
A-22. (B) A-23. (B) A-24. (A) A-25. (C) A-26. (C) A-27. (C) A-28. (C)
A-29. (A) A-30. (C) A-31. (D) A-32. (A) A-33. (B) A-34. (A) A-35. (D)
B-1. (A) B-2. (B) B-3. (B) B-4. (A) B-5. (B) B-6. (C) B-7. (B)
B-8. (A) B-9. (A) B-10. (C) B-11. (C) C-1. (C) C-2. (A) C-3. (D)
C-4. (C) C-5. (B) C-6. (D) C-7. (B) C-8. (C) C-9. (C) C-10. (B)
C-11. (C) C-12. (C) C-13. (D) C-14. (C) C-15. (B) C-16. (A) C-17. (A)
C-18. (A) C-19. (D) C-20. (C) C-21. (D) C-22.* (BCD) C-23. (B) C-24. (B)
C-25. (C) C-26. (B) C-27. (A) C-28.* (ACD) C-29. (A) C-30. (D) C-31. (D)
C-32. (B) C-33. (C) C-34. (A) D-1. (D) D-2. (A) D-3.* (A) E-1. (A)
E-2. (ABD) E-3. (B) E-4. (B) E-5.* (AD) E-6. (D) E-7. (A)
PART # II
1. (D) 2. (C) 3. (AB) 4. (B) 5. (B) 6. (B) 7. (A)
8. (B) 9. (A) 10. (C) 11. (C) 12. (B) 13. (D) 14. (B)
15. (A) 16. (C) 17. (A) 18. (A) P,R,S (B) P,Q,R,S (C) P,Q,R,S (D) Q
19. (A) q, s (B) p (C) p (D) r 20. (A – r) ; (B – r) ; (C – q) ; (D – p) 21. C 22. C
23. C 24. D 25. A 26. B 27. A 28. A 29. B
30. A 31. A 32. A 33. D 34. T 35. F 36. T
37. T 38. F 39. T 40. T 41. F 42. F 43. T
44. F 45. F 46. F 47.
EXERCISE # 2
1. (C) 2. (B) 3. (A) 4. (C) 5. (B) 6. (B) 7. (A)
8. (B) 9. (B) 10. (C) 11. (B) 12. (A) 13. (B) 14. (B)
15. (D) 16. (C) 17. (B) 18. (B) 19. (B) 20. (B) 21. (A)
22. (A) 23. (C) 24. (B) 25. (B) 26. (A) 27. (C) 28. (A)
29. (B) 30. (B) 31. (A) 32. (A) 33. (B) 34. (B) 35. (C)
36. (D) 37. (B) 38. (D) 39. (A) 40. (A) 41. (B) 42. (A)
43. (B) 44. (B) 45. (B) 46. (B) 47. (A) 48. (C) 49. (A)
50. (B) 51. (B) 52. (D) 53. (B) 54. (B) 55. (D) 56. (B)
57. (A) 58. (C) 59. (A) 60. (B) 61. (B) 62. (D) 63. (A)
64. (A) 65. (B) 66. (B) 67. (A) 68. (B) 69. (D) 70. (D)
71. (C) 72. (A) 73. (B) 74. (D) 75. (A) 76. (A) 77. (C)
78. (C) 79. (A) 80. (D) 81. (A) 82. (AB) 83. (BD)
84. (ABCD) 85. (AB) 86. (AC) 87. (ABC) 88. (ABC) 89. (ABD) 90. (BCD)
91. (BC) 92. (AB) 93. (B*C) 94. (ACD) 95. (ABCD) 96. (BCD) 97. (BC)
98. (ABC) 99. (CD) 100. (BCD) 101. (CD) 102. (CD) 103. (ABCD) 104. (ABC)
105. (CD) 106. (AC)

SUBJECTIVE
1. 2. Kp = 300. 3. 0.76 4. (i) 0.0025 mole litre–1
(ii) 20 litre–1/2
mole–1/2
5. reaction is exothermic. 6. 4.0 7. 24
22
P)x1(27
)x2(x16


8. 0.677 atm
9. (i) 2.34% (ii) 0.29 10. KP
= 2
2
1
P.


11. 80% 12. 0.435
13. (a) Right (b) Right (c) Left (d) No shift will occur
14. – 19.67 lit atm 15. 1.94 × 10–10 atm 16. T = 399K.
17. (i) 1.02 × 10–8
atm (ii) 7.91 × 10–6
18. – 105.216kJ
19. 1.43 × 10–5
20. 0.16 21. 1 22. 1.677  10–3
23. 0.209 24. – 0.0005. 25. 0.6 atm–2 26. 708
27. 7.5  10–18 28. 0.2704 atm 29. 1.413 moles litre–1
30. (i) 0.435 (ii) (a) Kp
= Kc
(b) Kp
= Kc
(RT)4
31. (i)
c
p
K
K
RT
 (ii)
c
p
K
K
RT
 32. 4.648 × 10–5
EXERCISE # 3
PART # I
1. (i) 8.1  105
mol/, 0.049 atm2
(ii) No change. 2. (A) 3. (D) 4. (D)
5. (D) 6. (D) 7. (B) 8. (B) 9. (D) 10.* (ABD)
11. (B)
PART # II
1. (1) 2. (4) 3. (3) 4. (3) 5. (2) 6. (1) 7. (4)
8. (4) 9. (1) 10. (1) 11. (1) 12. (1) 13. (2) 14. (4)
15. (4) 16. (1)
EXERCISE # 4
2. 12.229 3. 2.67 × 104 5. (i) 4.33 × 10–4 (ii) 1.90 6. 1.59 × 10–15
8. [N2] = 0.0482 mol L–1, [O2] = 0.0933 mol L–1, [N2O] = 6.6 × 10–21 mol L–1
9. 0.0352 mol of NO and 0.0178 mol of Br2 10. 7.47 × 1011 M–1 11. 4.0
12. Qc = 2.379 × 103. No, reaction is not at equilibrium. 14. 0.44
15. 0.068 mol L–1 eact of H2 and I2 16. [I2] = [Cl2] = 0.167 M, [ICI] = 0.446 M
17. [C2H6]eq = 3.62 atm
18. (i) [CH3COOC2H5] [H2O] / [CH3COOH] [C2H6OH] (ii) 3.92
(iii) Value of Qc is less than Kc therefore equilibrium is not attained.
19. 0.02 mol L–1 for both. 20. [PCO] = 1.739 atm, [PCO2] = 0.461 atm
21. No, the reaction proceeds to form more products. 22. 3 × 10–4 mol L–1 23. 0.149
24. (a) – 35.0( kJ, (b) 1.365 × 106 27. [PH2
]eq = [PH2
]eq = 2.5 × 10–2 bar, [PHBr] = 10.0 bar
30. (b) 120.48 31. [H2]eq = 0.96 bar 33. 2.86 × 10–28 M 34. 5.85 × 10–2

Assignment chemical equilibrium_jh_sir-4168

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Assignment chemical equilibrium_jh_sir-4168