Solution of Chemical Equilibrium
Cool Classroom Work
[𝑁𝑂]2
Soln. 1(b)
In the expression for equilibrium constant species in
solid state are not written (that is their mar
concentration are taken as 1)
P4(s) + 5O2(g) ⇌ P4O10(s)
So, Kc = 1/[O2]5
Kc =
Soln. 2(a)
First option is incorrect as the value of KP given is
wrong. It should have been
Soln. 9(a)
𝑃 𝐶𝑂 2
KP =
𝑃 𝐶𝐻 4 × 𝑃 𝑂 2
𝑁2 [𝑂2 ]
NO ⇌ 1 2 N2(g) + 1 2 O2(g)
Kc =
[𝑁2 ]1/2 [𝑂2 ]1/2
[𝑁𝑂]
1/4 × 10−4 = 50
= (1/𝐾𝑐 ) =
K c 1 = 4.9 × 10 – 2 =
Kc 2 =
2
= 4 × 10 – 4
𝑆𝑂2 𝑂2 1/2
[𝑆𝑂3 ]
[SO 3 ]2
[SO 2 ]2 [O 2 ]
K c 2 = (1/Kc 1 )2 = (1 / 4.9 × 10 – 2)2
= 416.5
Soln. 3(a)
Kp = Kc(RT)∆ng
∆ng = 1 – 1.5 = – 0.5
Soln. 10(c)
As equation (c) = equation (a) + equation (b)
So K3 = K1K2
𝐾
Kp = Kc(RT) – ½ = (𝑅𝑇)𝑐1/2
𝐾
So, 𝐾𝑐 = (RT)1/2..s
𝑝
Soln. 11(d)
As 2nd eq is reverse and half of 1st one so
Soln. 4(c)
(Kc) = (𝑅𝑇)∆𝑛 = (8.134 ×700)1
= 3.1 × 10 – 7
Soln. 5(b)
(Kp) = Kc (RT)∆n ……….(i)
(where ∆n = 1 – 2 = – 1)
Substituting this value of ∆n in equation (i),
Kp = KC (RT)
–1
Soln. 6(a)
Kp = Kc(RT)∆n
∆n = 3 – 2 = 1
Kp = Kc (0.0831 × 457)1.
1
K c =
1.8 × 10 −3
𝐾𝑝
𝐾𝑝
1
or 𝐾 = 𝑅𝑇
𝑐
Kc
=
1
1
4 × 10 −4
= 2 × 10 −2 = 50
Soln. 12(a)
For reaction (1)
𝑁𝑂 2
K1 =
𝑁2 𝑂2
and for reaction (2)
K2 =
𝑁2
1
2 𝑂2
1
2
𝑁𝑂
1
therefore K1 = 𝐾 2
2
Soln. 13(d)
Given,
N2 + 3H2 ⇌ 2NH3; K1
N2 + O2 ⇌ 2NO; K2
…(i)
…(ii)
1
Soln. 7(d)
2C(s) + O2(g) ⇌ 2 CO2(g)
∆n = 2 – 1 = + 1
 Kc and Kp are not equal.
We have to calculate
4NH3 + 5O2 → 4NH + 6H2O; K = ?
or 2NH3 + 5O2 → 2 NO + 3H2O
For this equation, K =
Soln. 8(d)
N2(g) + O2(g) ⇌ 2NO(g)
…(iii)
H2 + 2 O2 ⇌ H2O; K3
but K1 =
𝑁𝐻3 2
𝑁2 𝐻2
3
𝑁𝑂 2 𝐻2 𝑂 3
𝑁𝐻3 2 𝑂2 5/2
, 𝐾2 =
𝑁𝑂 2
𝑁2 𝑂2
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Solution of Chemical Equilibrium
& K3 =
𝐻2 𝑂
𝐻2 𝑂2
Now operate,
=
=
𝑁𝑂 2
𝑁2 𝑂2
×
𝑁𝐻3 2 𝑂5 5/2
𝐻2 𝑂 3
𝐻2 3 𝑂2 3/2
𝐾2 .𝐾33
𝐾1
𝐻2 𝑂 3
.
𝐻2 3 𝑂2 3/2
𝑁𝑂 2 𝐻2 𝑂 3
𝐾 =
𝑜𝑟 𝐾3 =
1 2
𝑁2 𝐻2 3
𝑁𝐻3 2
=K
𝐾2 .𝐾33
𝐾1
Soln. 14(d)
N2O4(g) ⇌ 2NO2 (g) (∆n = 1)
Kp > Kc
Since the temperature is kept constant hence both Kp
and Kc remain unchanged. However, when volume is
halved, pressure doubles up (Boyle’s law). So in
order to keep Kp unchanged  must decrease.
N2O4 (g) ⇌ 2NO2
Initial conc.
a
0
 = Degree of

2
dissociation
a–
2
conc. left
Total no. of moles at equilibrium = a –  + 2
= (a + )
Let the pressure at equilibrium = P by (a + ) mole
2
Partial pressure of NO2 = PNO 2 = a+  P
a− 
Partial pressure of PN 2 O 4 = a+  P
Kp =
(𝑃 𝑁𝑂 2 )2
𝑃𝑁 2 𝑂 4
=
44 𝑃
𝑎 2 −2
Since Kp is unchanged

1
Soln. 17(c)
∆ng = – ve → takes place with decrease in number of
moles or pressure, so increase in pressure shifts
equilibrium in forward side.
∆H = – ve → takes place with evolution of heat or
increase in temperature, so decrease in temperature
shifts this equilibrium in forward side.
Soln. 18(d)
Cl2(g) + 3F2(g) ⇌ 2 ClF3(g); ∆H = – 329 kJ.
Favourable conditions
(i) Decrease in temperature
(ii) Addition of reactants
(iii) Increase in pressure, that is, decrease in volume
Soln. 19(d)
Solid ⇌ Liquid
It is an endothermic process. So when temperature is
raised, more liquid is formed. Hence adding heat will
shift the equilibrium in the forward direction.
Soln. 20(c)
For the reaction
BaO2(s) ⇌ BaO(s) + O2(g); ∆H = + ve.
At equilibrium kp = PO 2
[For solid and liquid concentration term is taken as
unity]
Hence, the value of equilibrium constant depends
only partial pressure of O2. Further on increasing
temperature formation of O2 increasing as this is an
endothermic reaction.
𝑃
As P increases  decreases or changes.
Soln. 15(d)
Kp is not based upon pressure and concentration.
Soln. 16(c)
The given reaction, in the forward direction, is
endothermic and proceeds with an increase in number
of moles gaseous constituents. As such high
temperature and low pressure will favour the forward
process. An increase in concentration of I or decrease
in concentration of I2 will shift the equilibrium in the
backward direction.
Soln. 21(b)
N2 + 3H2 ⇌ 2NH3
At equilibrium, addition of a reactant or any catalyst
has no effect. However, the catalyst brings the
equilibrium faster by increasing both forward and
backward reactions.
Soln. 22(d)
For reaction to proceed from right to left
Q
>
Kc
(backward rate)
(forward rate) i.e. the
reation will be fast in backwared direction i.e r b > rf.
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Solution of Chemical Equilibrium
Soln. 23(c)
[𝑁𝑂2 ]2
Kc = [𝑁
2 𝑂4 ]
= 3 × 10
=
−3
Soln. 28(d)
CO2(g) + C(s) ⇌ 2CO(g)
0.5 – x
2x
0.5 – x + 2x = 0.8
x = 0.3
(1.2 × 10 −2 )2
4.8 × 10 −2
mol L – 1
Soln. 24(d)
NH4HS(s) ⇌ NH3(g) + H2S(g)
0
0.5
0
Initial
pressure
At equilibrium 0
0.5 + x
Total pressure = 0.5 + 2x = 0.84
So, x = 0.17 atm
Kp = PNH 3 × PH 2 S = 0.11 atm2
x
Soln. 25(a)
PCl5 ⇌ PCl3 + Cl2
1
0
0
1–x
x
x
PPCl 5 = Total pressure × mole fraction of PCl3
=P
𝑥
0.6 ×0.6
K=
Soln. 29(b)
For the reaction
2AB2(g) ⇌ 2AB(g) + B2(g)
At equi. 2(1 – x)
2x
x
Kc =
𝐴𝐵 2 𝐵2
or Kc =
𝐴𝐵2 2
2(1−x)
(2+x)
× P where P is the total pressure.
2𝑥
(2𝑥)2
KP1 = (1− 𝑥) (P1 / 1 + x)1
Z⇌P+Q
1
0 0
(1 – x) x x
KP2 = (1−𝑥) (P2 / 1 + x)
4 × 𝑃1
𝑃2
1
1
=
𝑃 𝐴𝐵 2 𝑃 𝐵 2
𝑃 𝐴𝐵 2
= 1.27 × 10 – 3
2
𝑥
𝑥 2 × 𝑃 2 .𝑃×
2
1−𝑥 2 ×𝑃 2
𝑥 3 .𝑃 3
=
Soln. 27(d)
The volume occupied by water molecules in vapour
phase is (1 – 10 – 4)dm3, that is approximately 1 dm3.
Pvap V = nH 2 O RT
3170 (Pa) × 1 × 10 – 3 (m3) =
nH 2 O (mol) × 8.314 (J K – 1 mol – 1) × 300 (K)
8.314 ×300
𝑥
PB 2 = 2 × 𝑃
[ 1 – x ≃1]
= 2 ×1 × 𝑃 2
On solving, we get
P1: P2 = 1: 36
3170 × 10 −3
Since x is very small so can be neglected in
denominator
Thus, we get
PAB 2 = (1 – x) × P
PAB = x × P
Now, KP =
=9
nH 2 O =
𝑥
PAB = (2+𝑥) × 𝑃, 𝑃𝐵2 = (2+𝑥) × 𝑃
Soln. 26(a)
X ⇌ 2Y
1
0
(1 – x) 2x
(𝑥)2
2𝑥 2 ×𝑥
2 1−𝑥 2
= x3[(1 – x) can be neglected in denominator (1 – x)
≃ 1]
The partial pressures at equilibrium are calculated on
the basis of total number of moles at equilibrium.
Total number of moles
= 2 (1 – x) + 2x + x = (2 + x)
 PAB 2 =
1+𝑥
= 1.8 atm
0.2
𝑥 3 .𝑃
2
1
𝑜𝑟 𝑥 3 =
2.𝐾𝑃
𝑃
𝑜𝑟 𝑥 =
2𝐾𝑃 3
𝑃
Soln. 30(c)
2A(g) + B(g) ⇌ 3C(g) + D(g)
Mole ratio
2
1
3
1
Molar conc. at 1
1
0
0
t=0
Equilibrium
0.50
0.75 0.75 0.25
Molar concentration
Kc =
𝐶 3 𝐷
𝐴2 𝐵
=
0.75 3 0.25
0.50 2 0.75
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Solution of Chemical Equilibrium
Soln. 31(c)

SO−2
4
Na2SO4 ⇌ 2Na +
0.004
Let x be the degree of dissociation
x
2x
x
conc left 0.004 – x
2x
x
No. of species left in soln. = 0.004 – x + x + 2x
= 0.004 + 2x
The conc. of isotonic
c1 = c2
0.004 + 2x = 0.01
2x = 0.006
X = 0.003 = 3 × 10 – 3 M
% of dissociation =
3 × 10 −3
4 × 10 −3
0.003
× 100 = 0.004 × 100 =
75%
Soln. 32(b)
N2(g) + 3H2(g) ⇌ 2NH3 (g)
∆n = 2 – (1 + 3) = – 2
R = 0.082
T = 500 + 273 K
= 773 K
Kp = Kc × (RT) – 2
1.44 × 10 −5
𝐾
Kc = (𝑅𝑇)𝑝−2 = (0.082 ×773)−2
Cool MCQ Level 1
Soln. 8(c)
𝐾𝑃 2
∆𝐻
1
𝑅
𝑇1
1
Soln. 1(c)
In region (I), the concentration of reactants and
products are constant hence this region shows
equilibrium.
ln 𝐾
Soln. 2(a)
Soln. 9(d)
A(g) + 2B(g) ⇌ 3C(g) + D(g)
∆ng = (3 + 1) – (1 + 2) = 1
Kp = Kc (𝑅𝑇)∆ng
Soln. 3(d)
Use the van’t Hoff equation.
𝑃1
= ln
=
10 −4
10 −6
−𝑇
2
∆𝐻
= 8.314
25
298 ×323
∆rH = 147.41 kJ / mol
0.05
⇒ 𝑅 x 1000 = Kc
Soln. 4(a)
Use the van’t Hoff equation
𝑑 ln 𝐾𝑝
𝑑𝑇
∆𝐻°
= 𝑅𝑇 2 or, ln Kp = 
∴
∆𝐻 1
𝑅 𝑇
+ C.
Soln. 5(a)
For endothermic reactions, the equilibrium constant
increases with increase in temperature.
Soln. 6(a)
5 x 10 −2 x 10 −3
or, Kc =
∆H
∵
logKc = log(constant) –
∴
logKc∝ T [slope = log(constant)]
1
2.303RT
= Kc
5 x 10 −5
𝑅
Soln. 10(d)
∵ Kp = KC(RT)∆ng ; Kp = 1.44 x 10-5
for reaction N2(g) + 3H2(g) ⇌ 2NH3(g)
∆ng = 2 – (1 + 3) = -2; T = 500°C = 773 K
∴ KC =
Soln. 7(b)
The effect of temperature on Kc is given by VantHoff’s equation
𝑅
KP
(RT )∆n g
1.44 x 10 −5
= (0.082 x 773)−2
Soln. 11(c
For the reaction,
N2(g) + 3H2(g) ⇌ 2NH3(g), ∆n = -2.
Kp = Kc(RT)∆n = Kc(RT)-2
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Solution of Chemical Equilibrium
𝐾
Soln. 21(a)
Stability of
1
Stability
∝
∝ equilibrium constant (for
Complex
constant
⇒ Kc = (𝑅𝑇)𝑝−2
dissociation )
Soln. 12(c)
1
3
NH3 (g) ⇌ 2 𝑁2 𝑔 + 2 𝐻2 𝑔
Kc = 4
 Kc for the following reaction:
N2 (g) + 3H2 (g) ⇌ 2NH3 (g) ng = 2 – 4 =  2
1
will be equal to 4 2
Kp = Kc RT
Soln. 22(d)
(i) XeF6 (g) + H2O (g) ⇌ XeOF4 (g) + 2HF (g); K1
(ii) XeO4 (g) + XeF6 (g) ⇌ XeOF4 (g) + XeO3 F2 ; K2
Required reaction: XeO4 (g) + 2HF (g) ⇌ XeO3 F2
(g) + H2O (g)
It is obtained by subtracting (i) from (ii), hence
equilibrium constant of the required equation will be:
∆n g
1
= 4 2 × 800 × 𝑅
−2
=
1
2
4×800 𝑅
Soln. 13(d)
In the reaction 2C(s) + O2 (g) ⇌ 2CO(g)
ng 0 hence Kp and Kc will not be equal.
K2
K1
Soln. 14(b)
Equilibrium constant at a given temperature is always
constant.
Soln. 16(d)
𝐾2
Soln. 19(c)
1
1
𝑃 + 2 𝑄2 + 2 𝑅2
2 2
1
1
𝑃 + 2 𝑄2 ⇌ 𝑃𝑄
2 2
⇌ 𝑃𝑄𝑅
1
× K2 =
and K2 =
[𝑋𝑒𝐹6 ][𝐻2 𝑂]
[𝑋𝑒𝑂𝐹4 ][𝑋𝑒𝑂 3 𝐹2 ]
[𝑋𝑒𝑂 4 ][𝑋𝑒𝐹6 ]
[𝑋𝑒𝑂 3 𝐹2 ][𝐻2 𝑂]
[𝑋𝑒𝑂𝐹4 ][𝐻𝐹]2
1
K1
=
1
2.5×10 5
K2 = 5  10 – 3
1
2.5×10 5
𝐾1
=
[𝑋𝑒𝑂 3 𝐹2 ][𝐻2 𝑂]
[𝑋𝑒𝑂𝐹4 ][𝐻𝐹]2
= K3
Soln. 24(d)
Equilibrium constant Kp is independent of pressure
and concentration, it depends only on temperature.
K
K′ =
PQ + 2 R2⇌ PQR
K1
[𝑋𝑒𝑂𝐹4 ][𝐻𝐹]2
Dividing K2 by K1, we have
Soln. 18(c)
Kp does not depend on the equilibrium pressure.
1
K1 =
K3 =
Soln. 17(b)
K=
Soln. 23(d)
Suppose equilibrium constant for reaction
XeO4(g) + 2HF(g) ⇌ XeO3F2(g) + H2O(g) is K3.
Then
Soln. 15(c)
1
Thus A will be more stable.
Option D is also incorrect because stability of
complex is directly proportional to stability constant.
× 5 × 10−3
= 1  10 – 5
Soln. 20(b)
On heating the colour deepens. This indicates that
more NO2 is formed. Thus, the reaction of formation
of NO2 is endothermic, for which ΔH will be
positive.
Soln. 25(d)
Kp is not affected by change in volume of the
container.
N2O4(g) ⇌ 2NO2 (g)
At t = 0
1
0 (α = degree of
At equilibrium (1 – α)
2α
dissociation)
If total pressure = P atm
Then, total moles at equilibrium = 1 – α + 2α = 1 + α
∵ PA = XA x Ptotal (XA = Mole fraction of A)
1− 𝛼
2𝛼
∴ PN2 O4 = 1+ 𝛼 x P and PNO2 = 1+ 𝛼 x P
(PNO )2
∴ Kp = (P
or, α2 =
2
N2 O4
)1
=
2𝛼 P
1+ 𝛼
1 –α
P
1+α
4𝛼 2 𝑃
=2(1− 𝛼 2 )
KP
4P +KP
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Solution of Chemical Equilibrium
∵α∝
Soln. 31(b)
KP
α=
4P + KP
QC =
1
𝑃2 (𝑔) 2
𝑃2 (𝑔)
1 2
2
= (3/2) = 3
P
∵ KP is constant, as the volume of container is
reduced to its half, the pressure will become double
and α will be decreased.
2
[A] = 3 ;
[𝐶]2
Soln. 26(b)
Given K1 =
Soln. 32(a)
Given KC = 49
[𝑁𝑂2 ]2
[𝑁𝑂]2 [𝑂2 ]1
Q = [𝐴]2 [𝐵] =
[𝑁 𝑂4 ]
and K2 = [𝑁𝑂2
and K3 = [𝑁𝑂]2 [𝑂
1
2]
Multiplying K2 by K1, we get
[𝑁 𝑂4 ]
2
2]
[𝑁𝑂 ]2
x [𝑁𝑂]2 2[𝑂
1
2]
[𝑁 𝑂 ]
= [𝑁𝑂]22 [𝑂4
1
2]
3
3
2
1
x 2
3
3
3
[C] = 3
27
3= 4
∵ Q < KC, thus the reaction must proceed in forward
direction.
2
2]
[𝑁2 𝑂4 ]
K2 x K1 = [𝑁𝑂2
1
[B] = 3 ;
= K3
Soln. 27(a)
[𝑁𝑂]2
Given K1 = [𝑁
Soln. 33(d)
Concentration of all gases increases and equilibrium
shifts toward less no. of moles but new equilibrium
concentration of every gas would be higher than
earlier.
2 ][𝑂2 ]
and
K2 =
⇒ 𝐾22 =
[𝑁2 ]1/2 [𝑂2 ]1/2
[𝑁𝑂]
[𝑁2 ][𝑂2 ]
1
=𝐾
[𝑁𝑂]2
1
1 2
or, K1 =
𝐾2
Soln. 28(a)
Kp is independent of pressure for reaction 2NH3(g) ⇌
N2(g) + 3H2(g); ∆ng> 0. Thus, degree of dissociation
of NH3 will decrease as pressure increases and thus
the concentration of NH3 will be changed.
Further, equilibrium concentration of N2 will be less
than the equilibrium concentration of H2.
Soln. 29(c)
The value of equilibrium constant K decreases with
increase in temperature. This indicates that the
forward reaction is exothermic. Hence, energy of
HI > total energy H2 + I2
∵
1
Stability ∝ Energy
∴
HI will be relatively less stable than that of H2
and I2
Soln. 30
Unit of Kc = [Unit of concentration]n
= [mol. dm – 3] (n =  1)
= dm3 mol. – 1
Soln. 34(d)
At constant volume, there is no effect of inert gas
addition.
Soln. 35(a)
Equilibrium will shift in backward direction when
volume in increased, thus number of moles of H2 will
increase.
Soln. 36(b)
Addition of inert gas will shift the equilibrium
towards higher volume direction.
PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)
Thus, addition on inert gas at constant pressure will
increase the dissociation of PCl5.
Soln. 37(a)
In the reaction C(s) + H2O(g) ⇌ CO(g) + H2(g)
ng> 0, hence the reaction will shift in forward
direction by increasing volume (or lowering the
pressure).
Soln. 38(a)
The reaction includes only solid species hence
Le Chatelier’s principle is not applicable to it.
Soln. 39(a)
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Solution of Chemical Equilibrium
Soln. 40(c)
Addition Cu(s) will not change the equilibrium
because:
Fe 3+
K=
2
Soln. 45(b)
Let initial moles of H2(g) is 1
Fe2O3(s) + 3H2(g) ⇌ 2Fe(s) + 3H2O(g)
At eqm
–
1 – 3x
–
3x
3𝑥 3
𝑉
1−3𝑥 3
𝑉
Cu 2+ Fe 3+ 2
Kc =
Soln. 41(a)
1
CO2 (g) ⇌ CO (g) + 2 O2 (g)
1 
teq.
𝛼 1/2
2
𝛼
K=
1−𝛼

=
𝛼
2
𝛼 3/2
1−3𝑥
 x = 0.22
% of H2 unreached
=
2
3
3𝑥
8=
1−3 ×0.22
1
× 100 = 34
(<< 1)
Soln. 46(b)
Soln. 42(c)
N2O4 (g) ⇌ 2NO2 (g)
t0
1
0
teq.
1 
2
Kp =
2𝛼 2
×
1−𝛼
𝑃
n = 1  + 2 = 1 + 
ng = 2 – 1 = 1
1
1+𝛼
2NOCl(g) ⇌ 2NO(g) + Cl2(g)
At eqm
P – 2x
2x
x
P + x =1; P – 2x = 0.64, x = 0.12
0.24 2 0.12
Kp =
0.64 2
= 168.875 × 10 – 3 atm.
4α2
Kp = 1−α2 P
Soln. 47(c)
1/2
𝐾𝑝
𝑃
𝐾𝑝
4+
𝑃
𝛼=
1 ×1
Qc = 1 ×1 = 1
Soln. 43(c)
The gaseous mixture contains 40% Cl2 and 40%
PCl3, since they are produced in 1:1 mole ratio. The
PCl5 % is 20.
For ideal gases mole % ≡ volume %
PCl 2 = PPCl 3
 2 × 0.40 = 0.80 atm
PPCl 5 = 2 × 0.2 = 0.40 atm
 Kp =
=
𝑃 𝑃𝐶𝑙 3 .𝑃 𝐶𝑙 2
0.80 ×0.80
Soln. 44(a)
For an ideal gas mole % ≡ volume %
2
=
0.55 2
0.45 2
1+𝑥
1+𝑥
1−𝑥
1−𝑥
10
10
10
10
 x = 0.333
1+𝑥
[A2(g)] = 10 =
1.333
10
= 0.133
Soln. 48(b)
2
𝑛 𝐴𝐵
2
2
𝐴 2 .𝑛 𝐵 2
= 5 × 32 ×
0.40
𝑃𝐻2
0.25 =
1−𝑥 2
10
1+𝑥 2
10
22
= 1.6 atm
Kp =
Conc. eqm
Kc = 𝑛
𝑃 𝑃𝐶𝑙 5
𝑃𝐻2 2 𝑂
∵ Qc> Kc so reaction will proceed in backward
direction
A2(g) + B2(g) ⇌ C2(g) + D2(g)
×𝑉 =𝑛
2
𝑛 𝐴𝐵
2
2
𝐴 2 .𝑛 𝐵 2
×
𝑛 𝑡𝑜𝑡𝑎𝑙 𝑅𝑇
𝑃
10 ×0.0821 ×300
8.21
Soln. 49(a)
Kc =
𝐻2 𝑆(𝑔)
𝐻2 (𝑔)
𝑥
= 1.49
 8 × 10 – 2 = 0.3−𝑥
 x = 0.022
[H2S(g)] =
0.022
2
= 0.011 M
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Solution of Chemical Equilibrium
Soln. 50(b)
Initial conc.
At eqm
5=
𝑥+1
4
CO + H2O ⇌ CO2 + H2
3
3
x
0
2
2
(x + 1) 1
= 20 = x + 1
 x = 19
Soln. 51(a)
A2(g) + B2(g) ⇌ 2AB(g)
Initial conc
10/2
15/2
5/2
At eqm.
5–x
7.5 – x 2.5 + 2x
∵ 2.5 + 2x = 7.5
 x = 2.5
7.5 2
Kc = 2.5 ×5 = 4.5
Soln. 52(d)
2NH3 (g) ⇌ N2 (g) + 3H2 (g)
t=0
2
0
0
teq.
1
1/2 3/2
Kc =
=
Soln. 56(d)
AB(g) ⇌ A(g) + B(g)
1
0
0
1
1
1
- 3
+ 3
+ 3
2
1
1
3
3
3
2
1
1 4
(Σn)eqlm = 3 + 3 + 3= 3
Kp =
1 3 3
×
2 2
12
= 1.685 = 1.7
Soln. 53(d)
Fe3+ + SCN –⇌ [FeSCN]2+
t0
3.1
3.2
0
teq.
0.1
0.2
3
3
K = 0.1×0.2 = 150
𝑝 AB
=
1
1
3
3
4 𝑝4 𝑝
3
3
2
3
4 𝑝
3
1
= 8 p.
∴ p = 8Kp.
Soln. 57(b)
NH4COONH2(s) ⇌ 2NH3(g) + CO2(g)
2
1
𝑝3 𝑝
3
2
Kp = 𝑝NH
𝑝
=
3 CO2
2
3
𝑝
2 1
3
4
𝑝 = 27 𝑝3
4
= 27 x (2)3 = 1.185
Soln. 58(a)
𝑁2 𝐻2 3
𝑁𝐻3 2
𝑝 A𝑝 B
𝑃𝑆4
Kp = 𝑃
2(𝑔)
=
𝑆 8(𝑔)
4 ×0.3 4
0.70
= 2.96
Soln. 59(c)
Kp = 𝑃 4
1
𝐻 2 𝑂(𝑔)
PH 2 O
g
=
1
1/4
Kp
– 12 1 / 4
 (10 ) = 10 – 3 atm
 Kp = 10 – 3 × 760 = 0.76 torr.
Soln. 54(a)
Soln. 55(d)
A(g) + B(g) ⇌ 2C(g)
𝒶
3
0
𝒶− 𝓍 3–𝓍
2𝓍
From the question, 2𝓍 = 3 ⇒ 𝓍 = 1.5
3=
Soln. 60(b)
XY ⇌ 2Y
t0
1
0
teq.
1  2
Kp 1 =
(2𝑥)2
𝑎−𝑥 (3−𝑥)
Substituting 𝓍 = 1.5, we get 𝒶 = 3.5
2
2α
P
1+α 1
1−α
P
1+α 1
4𝛼 2 𝑃1
=
1−𝛼 2
Kp1
Kp2
1
=
9
𝑃1
𝑃2
=
Kp 2 =
Z⇌ P+Q
1
0 0
1   
2
α
P
1+α 2
1−α
P
1+α 2
𝛼2𝑃
= 1−𝛼22
4𝛼 2 𝑃 1
1−𝛼 2
𝛼2𝑃 2
1−𝛼 2
4𝑃1
𝑃2
1
= 36
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Solution of Chemical Equilibrium
Soln. 61(d)
For equilibrium,
2NH3(g) ⇌ N2(g) + 3H2(g)
At t = 0
2
0
0
At equilibrium
1
½
3/2
∴ Kc =
1
𝑁2 [𝐻2 ]3
x
3
27
3 = 1.6875 ≈ 1.7
= 2(1)22 = 16
[𝑁𝐻3 ]2
Soln. 62(a)
For equilibrium I2⇌ 2I−
(at 1000 K)
At t = 0
1 0 (α = degree of dissociation)
At equilibrium (1 – α) 2α
[I2]eq =
1− 𝛼
1
; [I−]eq = 2α
∴ KC = 10-6 =
[2𝛼]2
KC =
[𝐶]2 [𝐷]
𝐴 [𝐵]2
0.2 2 (0.1)
1.0 (1.8)2
= 0.00123
Soln. 65(d)
2AB3(g) ⇌ A2(g) + 3B2(g)
At t = 0
8
0
0
At equilibrium (8 – 2x)
x
3x
Given [A2] = 2 (at equilibrium)
∴ Moles of AB3 at equilibrium = 8 – 2 x 2 = 4
Moles of B2 = 3 x 2 = 6
4
∴ [AB3]eq = 1
(Volume: 1 dm3 = 1 L)
[A2] = 2
[B2] = 6
[𝐴2 ][𝐵2 ]3
Kc =
1− 𝛼
=
[𝐴𝐵3 ]2
2 x(6)2
=
42
= 27
4𝛼 2
10-6 = (1− 𝛼)
Soln. 66(b)
4𝛼 2
1
or, 10 6 = (1− 𝛼)
Kp = (𝑃
or, 4 x 106α2 = 1 – α
taking 1 – α ≈ 1
1
Kp = (𝑃 ′
[𝑆𝑂3 ]2
[𝑆𝑂2 ]2 [𝑂2 ]
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠
𝑣
𝑥
10
= 0.2 x 4.5 = 0.4 atm
Soln. 67(d)
A(g) + B(g) ⇌ C(g) + D(g)
At=0
1
2
3
2
At equilibrium (1 – x)
(2 – x) (3 + x) (2 + x)
∴ KC =
1
x
10
or, x = 0.1
x=
Soln. 64(a)
A + 2B ⇌
At t = 0
1.1
2.2
At equilibrium (1.1 – x) (2.2 – 2x)
Given 2x = 0.2
or, x = 0.1
𝐶 [𝐷]
𝐴 [𝐵]
(3+x)(2 + x)
20(1 – x)(2 – x) = (3 + x)(2 + x)
20(2 – 2x – x + x2) = 6 +2x + 3x + x2
20(2 – 3x + x2) = x2 + 5x + 6
or,19x2 – 65x + 34 = 0
∵ [SO3] = [SO2]
∴ 100 =
0.6 x 0.6
20 = (1 – x)(2 – x)
[𝑆𝑂3 ]2
[𝑆𝑂2 ]2 x
0.2 x (𝑃𝐵′ )
2
(𝑃𝐵′ 2 )
Total volume = V litre.
Suppose number of moles of O2 = x, then
∴ 100 =
= 4.5
′
(𝑃𝐴𝐵
)2
or, 4.5 =
Soln. 63(b)
For reaction, 2SO2(g) + O2(g) ⇌ 2SO3(g)
𝑂2 = 10 𝐴𝑐𝑡𝑖𝑣𝑒 𝑚𝑎𝑠𝑠 =
(0.3)2
0.1 x 0.2
′
𝐴 2 ) x (𝑃𝐵 2 )
(0.6)2
⇒ α = 5 x 10-4
∴ [I2]eq = 1 – α ≈ 1
and indicates, [I2(g)]eq> [I−(g)]eq
𝑥
=
When the volume is made half of the initial volume,
(Kp remains unchanged)
or, α2 = 4 x 106
KC = 100 =
(𝑃 𝐴𝐵 )2
1
1
2
𝐴 ) x (𝑃 𝐵 2 )
2C + D
0
0
2x
x
65 ± (−65)2 − 4 x 19 x 34
2 x 19
=
65±40.5
38
x = 2.77 or, x = 0.64
2.77 is not possible because x > [A]initial
Therefore, x = 0.64
Thus, the equilibrium concentration of
B = 2 – x = 2 – 0.64 = 1.36
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Page 9
Solution of Chemical Equilibrium
Soln. 68(d)
Soln. 70(b)
2SO3(g) ⇌ 2SO2(g) + O2(g)
At=0
1
0
0
At equilibrium (1 – 2x) 2x = 0.6
x
2x = 0.6
∴ x = 0.3
[SO3]eq = 1 – 2x = 1 – 0.6 = 0.4
[SO2]eq = 0.6; [O2]eq = 0.3
∴K=
[𝑆𝑂2 ]2 [𝑂2 ]
[𝑆𝑂3 ]2
=
(0.6)2 x 0.3
(0.4)2
= 0.675
Soln. 69(c)
Suppose, the initial concentration of A is a mole/L.
Then [B]initial = 3a
A + 2B ⇌ 2C + D
At t = 0
a
3a
0
0
At equilibrium a – x
3a – 2x
2x
x
But, [A]equ = [C]equ
∴ a – x = 2x or, a = 3x
thus [A]equ = a – x = 3x – x = 2x
[B]equ = 3a – 2x = 3.3x – 2x = 7x
[C]equ = 2x
[D]equ = x
∴ KC =
[𝐶]2 [𝐷]
𝐴 [𝐵]2
(2𝑥)2 x 𝑥
= 2𝑥
x (7𝑥)2
4𝑥 3
= 98𝑥 3 = 0.04
Soln. 71(a)
P Trans −2−butene
P Cis −2−butene
0.290
Kp =
= 0.085 = 3.4
Soln. 72(a)
∆rG = ∆rH – T. ∆rS
= – 30 – 300 × 0.1 = 0
∆rG = – 2.303 RT log K
K=1
Soln. 73(d)
Ptotal = 0.2 bar
At equilibrium PCO 2 = PH 2 O = 0.1 bar
Kp = (0.1)2 = 0.01
∆rG = – RT ln Kp
= – 8.314 × 420 in (0.01)
= 16083.6 J / mol or 16.083 kJ / mol.
Soln. 74(a)
K = 𝐴𝑛𝑡𝑖𝑙𝑜𝑔
= 𝐴𝑛𝑡𝑖𝑙𝑜𝑔
−∆𝐺°
2.303 𝑅𝑇
−78×1000
2.303 ×8.314 ×1000
= 8.4  10 – 5
Cool MCQ Level 2
Soln. 1(b)
Concentration of [NO2] will decrease with increase in
concentration [N2O4].
Soln. 2(b)
∵ PV = nRT
or,
p
n
=
RT V
=
number of moles
volume in litre
= Active mass
Soln. 4(c)
Equilibrium constant KC =
when concentration of [Ca2+] is increased by 4 times.
To keep KC constant, the concentration of F− ions
must be decreased to half of its original
concentration.
4[Ca2+ ]
∴ Kc =
Soln. 3(a)
I: CaCO3(s) ⇌ CaO(s) + CO2(g) ; 𝐾𝑝′ = 𝑃CO2
II: N2(g) + O2(g) ⇌ 2NO(g) ; Kp =
1
2
III: H2(g) + O2(g) ⇌ H2O(l) ;
(P𝑁𝑂 )2
𝑃𝑁 2 . 𝑃𝑂 2
1
𝐾𝑝′ =
𝑃𝐻 2 .(𝑃𝑂 2 )1/2
[𝐶𝑎 2+ ][𝐹 − ]2
[𝐶𝑎𝐹2 ]
[F−]2
(2)2
[CaF2 ]
Soln. 5(d)
Using the equation
log
(𝐾𝑝 )40 °C
(𝐾𝑝 )25 °C
=
we get, log 4 =
∆𝐻
1
2.303𝑅 𝑇1
∆𝐻
−
1
𝑇2
,
1
2.303 x 8.314 273 + 25
−
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1
273 + 40
.
Page 10
Solution of Chemical Equilibrium
∴ ∆H = 71.67 kJ mol-1.
K3 =
Soln. 6(a)
Kp = PCO 2
log
𝐾2
=
𝐾1
0.34
Suppose K =
∆𝐻
1
2.303 𝑅 𝑇1
∆𝐻
−
log 0.14 = 2.303 ×8.314
∵
1
𝑇2
1
1
𝐾2𝐾33
𝐾1
∴K=
− 420
400
H = 64 kJ / mol.
Soln. 7(b)
Soln. 8(a)
According to van’t Hoff equation:
∆𝐻°
log K = −
2.303 𝑅
Y=
1
∆𝑆°
𝑇
𝑅
× +
[𝑁𝑂]2 [𝐻2 𝑂]3
[𝑁𝐻3 ]2 [𝑂2 ]5/2
[𝑁𝑂][𝐻2 𝑂]3
[𝑁𝐻3 ]2 [𝑂2 ]5/2
𝐾2 𝐾33
𝐾1
Soln. 14(a)
Lesser the value of equilibrium constant, less
favourable is the forward reaction and greater the
stability of reactant.
Soln. 15(a)
x +C
M
=
[𝐻2 𝑂]
[𝐻2 ][𝑂2 ]1/2
1
When log K is plotted against 𝑇 , we get straight line
∆𝐻°
having slope equal to − 2.303 𝑅 .
For exothermic reaction, slope of the line will be
positive.
QC =
6 2
2
2 4 3
2 2
=
9
8
Qc<Kc so reaction will proceed in forward
direction.
Soln. 16(d)
At point ‘A’,
Q = tan 60 = 1.732
 Q = K.
Soln. 9(a)
∆𝐻°
Slope = − 2.303 𝑅
∆𝐻°
tan 45° = − 2.303 ×8.314
H =  2.303  8.314 =  19.14 J mol. – 1
Soln. 10(a)
Hg(Ɩ) ⇌ Hg (g) ng = 1
Kp =
0.002
760
Soln. 17(b)
ng> 0, therefore by decreasing the pressure and by
introducing inert gas at constant pressure, the
equilibrium will shift in forward direction.
= 2.63 × 10−6 𝑎𝑡𝑚.
Kc = RTp = 0.0821 ×300 = 1.068 × 10−7 𝑀
Soln. 18(a)
Reaction is exothermic, therefore, its yield will
decrease with increase in temperature.
Soln. 11(d)
Soln. 19(b)
Kp = Kc (RT)n
K
2.63×10 −6
Soln. 20(c)
Soln. 12(c)
Stability 𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚
1
𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑓𝑜𝑟 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛
Soln. 13(d)
From given equations,
We have, K1 =
K2 =
[𝑁𝐻3 ]2
[𝑁2 ][𝐻2 ]3
[𝑁𝑂]2
[𝑁2 ][𝑂2 ]
Soln. 21(b)
For ideal gas mole % = volume %
N2O4(g) ⇌ 2NO2(g)
Initial moles
a
0
At eqm.
a(1 – a)
2a
As per given original volume
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Solution of Chemical Equilibrium
=
75
100
Soln. 24(b)
× Volume at eqm
1
𝑛
Kp =
A + B ⇌ C +
At t = 0
1
1
0
At equilibrium 1 – α
1–α
α
(α = degree of dissociation)
If total volume of the solution = V litres
and [C] = [D] =
or, K =
⇒ 𝐾
𝐶 [𝐷]
𝐴 [𝐵]
α α
x
𝑉 𝑉
1 –α
1–α
x
𝑉
𝑉
α2
∴
𝐾
1+ 𝐾
𝑃
𝛼
2
1+
𝛼 3/2 𝑝 1/2
1−𝛼 2+𝛼 1/2
∆H
∆S
=T=
30 x 10 3
60
= 500 K
20
⇒ 2.303 x 0.0831 x 300 = -logKC
N2O4(g) ⇌ 2NO2(g)
At t = 0
1
0
At equilibrium (1 – α)
2α
Total mole at equilibrium = 1 – α + 2α = 1 + α
PN2 O4 = XN2 O4 x P =
)2
(PN2 O4 )
=
1+α
2𝛼 P
1+2𝛼
1 –α
1 +α
xP
P
⇒ 100.35 =
4𝛼 2 P2
(1+ 𝛼)2
x
(1+ 𝛼 )
(1− 𝛼 )P
K𝑐
or, 1 x 0.35 = log
log100.35 = log
=
1
or, 0.35 = +log
1–α
xP
1+α
2𝛼
and PNO2 = XNO2 x P =
4𝛼 2 𝑃
(1− 𝛼 2 )
×
𝑛
1/2
Soln. 27(c)
∆G = -2.303RT logKC
Soln. 23(d)
(PNO2
1−𝛼
×
Soln. 26(b)
∵ ∆G = ∆H – T∆S
∆G = 0 at equilibrium
∆H = 30 kJ = 30 x 103 J
∆S = 60 JK-1
(1 – α)2
𝛼
=
(1− 𝛼)
∴ Kp =
𝑛𝐻 2𝑂
2
∆𝑛 𝑔
𝑃
Soln. 25(a)
G =  nFE
…..(i)
When E is negative, G will be positive.
G =  2.303 RT log10 K
……(ii)
When G is positive, then the value of K will be
less than 1.
⇒ 𝐾(1 − 𝛼) = α
⇒ 𝐾−α 𝐾 =α
⇒ 𝐾=α+α 𝐾
⇒α=
1
𝛼

1/2
𝐻 2 ×𝑛 𝑂
2
𝛼 1/2
𝛼×
2
=
𝑉
then, equilibrium constant K =
or, K =
=
D
0
α
1–α
α
𝑉
1 
teq.
Soln. 22(a)
then, [A] = [B] =
3
𝐻2 𝑂 𝑔 ⇌ 𝐻2 𝑔 + 2 𝑂2 𝑔 ∆𝑛𝑔 = 2 − 1 = 2
at constant T and P: V  n
 a = 0.75 × a (1 + a)
 = 0.33
=
1
K𝑐
(log10 = 1)
1
K𝑐
1
K𝑐
⇒ KC = 10-0.35 ≈0.46
Soln. 28(d)
∆G° = -2.303RT logKC
-3000 = -2.303 x 2 x 300 logKC
or,
3 x 10 3
2.303 x 2 x 300
= logKC
or, 2.171 = logKC
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Solution of Chemical Equilibrium
or, 2.171 x 1 = logKC
or, 2.171 log10 = logKC
or, log102.171 = logKC
or, KC = 102.171≃ 102.2
Soln. 36(c)
Soln. 29(a)
G =  2.303 RT log10 KP
=  2.303  8.314  298 log (6.1  10 – 5)
= + 24048 J / mol. = + 24 kJ / mol.
Soln. 37(a)
For association of molecules,
Soln. 30(a)
G =  2.303RT log10K
=  2.303  8.314  298 log (4.42  104)
=  26506 J mol. – 1
=  26.5 kJ / mol.
𝑑𝑅𝑇
= 57.47
𝑃
99−57.47
= 0.72
57.47
MO =
=
α=
𝐷𝑂 − 𝐷𝑇
𝐷𝑂 1− 1𝑛
where n = number of molecules which associate to
form a big (associated) molecule
for dimerisation n = 2
α=
𝐷𝑂 − 𝐷𝑇
𝐷𝑂 1− 12
=2
𝐷 𝑂 − 𝐷𝑇
𝐷𝑂
Soln. 38(a)
Lesser is the percentage dissociation; greater is the
vapour density.
Soln. 31(a)
Soln. 32(a)
Soln. 33(a)
K=
𝐵 𝐶
𝐴
=
0.15×0.05
1×10 −3
= 7.5
G =  2.303RT log10 K
=  2.303  8.134  298 log 7.5
=  4992 J mol. – 1
=  5 kJ mol. – 1
Soln. 39(d)
The reaction is:
N2O4⇌ 2NO2 (g)
20
n = 2 and  = 100 = 0.2
D = 4.6, initial vapour density
d = vapour density at equilibrium
=
𝐷−𝑑
𝑛−1 𝑑
46−𝑑
0.2 =
2−1 𝑑
Soln. 34(c)
Average kinetic energy depends only on temperature,
Since, both (liquid or vapour) phases are at same
temperature. Thus, average kinetic energy of
molecules in both the phases will equal.
d = 38.3
Molar mass equilibrium = 2  38.3 = 76.6
Pm = dRT Here, d = density of gas mixture
Soln. 35(a)
Soln. 40(a)
𝑀𝑇 −𝑀𝑂
𝑛−1 𝑀𝑂
92− 𝑀𝑂
0.2 =
𝑀𝑂
=

RT
1×76.6
= 0.0821 ×300 = 3.12 𝑔/𝐿
Soln. 42(b)
Kc =
dmixture=
=
Pm
Soln. 41(d)
 MO = 76.66
𝑃𝑀 𝑚𝑖𝑥𝑡𝑢𝑟𝑒
𝑅𝑇
1 ×76.66
= 3.11
0.821 ×300
d(mix) =
g / litre.
𝑛 𝐵 𝑛 𝐶3
𝑛 𝐴2
×
1
𝑉2
2 × 23
23 × 𝑉 2
1
2
 16 =
V=
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Solution of Chemical Equilibrium
Soln. 43(b)
H2(g) + CO2(g) ⇌ H2O(g) + CO(g)
Moles of eqn. 0.8 – x 0.8 – x
x
x
Conc. at eqm
0.8−𝑥 0.8−𝑥
5
5
𝑥
5
𝑥
5
∵ ∆ng = 0
2=
PNO 2 = 0.02
 x = 0.533
0.8−0.533
5
Soln. 48(c)
Soln. 44(a)
H2(g) + S(s) ⇌ H2S(g)
0.5 – x –
x
Conc. at eqn.
Kc =
𝐻2 𝑆
𝐻2
 7 × 10 – 2 =
n H 2S
V
0.0821 ×1000
1/4 2
Kc =
0.5
4
= 256
0.5 2
4
1
3
2
K1c
RT
Soln. 45(d)
Moles of CO2 present at equilibrium
=
Initial moles
At eqm
Where x = 0.5
NH3(g) ⇌ N2(g) + H2(g)
= 0.0327 × 0.0821 × 360 = 0.966
1.642 ×50
N2(g) + 3H2(g) ⇌ 2NH3(g)
1
2
0
1–x
2 – 3x
2x
equilibrium constant for the reaction
𝑥
0.5−𝑥
 x = 0.0327
PH 2 S =
0.5 2 0.25
2
𝑃𝑁𝑂
2
 156.25 =
𝑥
0.8−𝑥
[CO2(g)] =
2𝑥 2 𝑥
𝑃−2𝑥 2
Kp =
𝑥 2
5
0.8−𝑥 2
5
Kc =
Soln. 47(b)
Let P is initial pressure of NO2
2NO2(g) ⇌ 2NO(g) + O2(g)
At eqm.
P–x
2x
x
As per given x = 0.25
=1
1
=
𝐾𝑐
2
1
= 16
Soln. 49(b)
NH4COONH2(s) ⇌ 2NH3(g) + CO2(g)
2
3
1
𝑝3 𝑝
2
Kp = 𝑝NH
𝑝
=
3 CO2
2
3
𝑝2
1
3
𝑝 =
4 3
𝑝
27
4
Mole % of XCO3 decomposed
1
4
= × 100 = 25%
Hence, 75% remains undecomposed.
= 27 x (2)3 = 1.185
Soln. 50(a)
For the equilibrium,
1
2
H2O(g) ⇌ H2(g) + O2(g)
Soln. 46(c)
3
1
AB3(g) ⇌ AB2(g) + B2(g)
2
At eqn. 800 – x x
x/2
𝑥
800 – x + x + = 900  x = 200
2
% dissociated =
200
800
× 100 = 25%
Kp =
1
𝛼 2𝑝 2
2
p = 1 atm.
α = ( 2 · 𝐾𝑝 )
2
3
= 0.0205 ≈ 2%.
Soln. 51(b)
XY2(g) ⇌
At t = 0
600 mm
At equilibrium (600 – P) mm
According to question
600 – P + P + P = 800
600 + P = 800
XY(g) + Y(g)
0
0
P mm
P mm
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Solution of Chemical Equilibrium
∴
P = 200 mm
(Pxy2 )eq = 600 – 200 = 400 mm
Soln. 54(c)
(Pxy)eq = 200 mm
(Py)eq = 200 mm
Equilibrium constant = K =
Pxy x Py
Pxy2
200 x 200
=
400
= 100
Soln. 52(d)
Given PCH3 OH = 2 atm, PCO = 1 atm and PH2 = 0.1
atm
∴ Kp =
PCH3 OH
PCO x (PH )2
2
=
2
1 x (0.1)2
= 200
∴ Equilibrium constant for decomposition reaction
(reverse reaction)
CH3OH(g) ⇌ 2H2(g) + CO(g) is
𝐾𝑃′ =
1
𝐾𝑃
=
1
200
A2(g) + B2(g) ⇌ 2AB(g)
At t = 0
a
(100 – a)
0
At equilibrium (a – x) (100 – a – x)
2x
Given [A2]eq = [AB]eq
∴ a – x = 2x
or, a = 3x
Thus [A2]eq = a – x = 2x
[B2]eq = 100 – a – x = 100 – 4x
[AB]eq = 2x
[𝐴𝐵]2
KC =
[𝐴2 ][𝐵2 ]
∴ 100 =
4x2
(2x)(100 – 4x)
⇒ (100)·2x(100 – 4x) = 4x2
⇒ 20000x – 800x2 = 4x2
⇒ 20000x = 804x2
⇒
20000
804
Soln. 55(b)
PNH 3 = PH 2 S
2
 Kp = PNH 3
2
64 = PNH 3
PNH 3 = 8 atm.
Total pressure = PNH 3 + PH 2 S
= 8 + 8 = 16 atm.
= 5 x 10-3 atm2
Soln. 53(d)
∴
2NH3(g) ⇌ N2(g) + 3H2(g)
At t = 0
2
0
0
At equilibrium (2 – 2x)
x
3x
x = 27% = 0.27
∴ [NH3]eq = 2 – 2 x 0.27 = 1.46
[N2]eq = x = 0.27
[H2]eq = 3x = 3 x 0.27 = 0.81
= x2
⇒ x = 24.87
∴ [B2]eq = 100 – a – x = 100 – 4x
[B2]eq = 100 – 99.50 [a = 3x]
[B2]eq = 0.50
Soln. 56(c)
A2 (g) + B2 (g) ⇌ 2AB (g)
t0
2
4
0
2−𝑠 4−𝑠 2𝑠
teq.
4
4 4
AB 2
Kc =
4=
A2 B2
2x 2
4
2−𝑥 4−𝑥
4
4
4x 2
4 = x 2 −6x+8
x 2 − 6x + 8 = x 2
8
4
x=6=3
(AB) =
2𝑥
4
2
4
= 4 × 3 = 0.66
Soln. 57(c)
C (g) + D (g) ⇌ A (g) + B (g)
t0
1
1
1
1
teq.
1 
1 
1+ 1+
4=
2=
1+α 2
1−α 2
1+𝛼
1−𝛼
2  2 = 1 + 
1
 = 3 = 0.33
[A] =
1+𝛼
10
=
1+0.33
10
= 0.133 𝑀
Soln. 58(b)
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Solution of Chemical Equilibrium
Soln. 59(a)
N2 (g) + 3H2 (g) ⇌ 2NH3 (g)
t0
1
3
0
teq. 1 – 0.5 3 – 1.5
1 n = 0.5 + 1.5 + 1 = 3
PNH 3 = xNH 3 × P
1
P
= 3×P = 3
Soln. 66(c)
SO2(g) + NO2(g) ⇌ SO3(g) + NO(g)
Initial moles
2
2
2
2
At eqm
2–x
2–x
2+x
2+x
Total no. of moles of gases at equilibrium
= 8 + 2 = 10
𝑃 𝑆𝑂 3 .𝑃 𝑁𝑂
Soln. 60(b)
Kp = 𝑃
Soln. 61(b)
 25 =
5=
Soln. 62(a)
pNH3 = pH2S = 0.33 atm.
K p = pNH 3 × pH s S
𝑆𝑂 2 .𝑃 𝑁𝑂 2
2
2+𝑥
×𝑃
10
2
2−𝑥
×𝑃
10
2+𝑥
2−𝑥
 x = 1.33
= 0.33  0.33 = 0.109 atm.2
Partial pressure of NO2 =
=
0.666
10
2−𝑥
10
×2
Soln. 63(b)
= 0.133 atm
Soln. 64(a)
Soln. 67(a)
2AB(s) ⇌A2(g) + B2(g)
0.5 + x x
KP = PA 2 . PB 2
2NOBr(g) ⇌ 2NO(g) + Br2(g)
Initial pressure
P
0
0
At eqm
P – 2x
2x
x
where 2x = 0.40 P
 x = 0.20 P
 1.20 P = 0.30
 P = 0.25 atm
Kp =
=
 0.06 = (0.5 + x)x
= x2 + 0.5x – 0.06
 x = 0.1
Ptotal = PA 2 + PB 2 = 0.6 + 0.1 = 0.70 atm
2 .𝑃
𝑃𝑁𝑂
𝐵𝑟 2
2
𝑃𝑁𝑂𝐵𝑟
0.4𝑃 2 0.2𝑃
0.6𝑃 2
= 0.0222
Kp for the reaction
2NO(g) + Br2(g) ⇌ 2NOBr(g)
1
is0.0222 = 45
Soln. 68(a)
Sb2S3(s) + 3H2(g) ⇌ 2Sb(s) + 3H2S(g)
0.01 – x 0.01 – 3x 2 x
3x
where 3x = 0.005
H2S + Pb2+ → PbS + 2H+
no. of moles of PbS formed
=
Soln. 65(c)
∵ Initially only A is present so at eqm B and C
should be present in 2:1
2A(g) ⇌ 2B(g) + C(g)
At eqm
400 mL 200 mL 100 mL
For ideal gases volume % = mole %
Kp =
× Ptotal
2 100
200
×10
×10
700
700
2
400
×10
700
1.19
238
= 0.005 mole
At eqm [H2] –
Kc =
0.005 3
0.005
0.005
25
;
=1
10
= 28
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Solution of Chemical Equilibrium
At eqm 0.1 – x a – x
Soln. 69(b)
Kc =

𝑐𝑜𝑚𝑝𝑙𝑒𝑥
𝐻3 𝐵𝑂3 𝑔𝑙𝑦𝑐𝑒𝑟𝑖𝑛
𝑐𝑜𝑚𝑝𝑙𝑒𝑥
80
𝐻3 𝐵𝑂3
= 0.90
x = 0.1 × 100 = 0.08;
= 4.44 M
a – x = 4.44
 a = 4.44 + 0.08 = 4.52 M
or Initial moles = 4.52
= 20
 [glycerin] =
x
80
80
20 ×0.9
H3BO3 + glycerin ⇌ (H3BO3 – glycerin)
Cool MCQ With More Than One Correct
Option
Soln. 1(b, c, d)
Soln. 7(a, b, c, d)
Soln. 2(a, b, c, d)
Soln. 8(a)
Soln. 3(a, b, c, d)
Soln. 9(b, c, d)
Soln. 4(a, b)
Soln. 10(a, b)
Soln. 5(a, b, c)
Soln. 11(a, c)
Soln. 6(a, c, d)
Soln. 12(a, b, c)
Soln. 13(a, d)
MCQ Handle With Cool Brain
Soln. 1c)
Use van’t Hoff equation.
Soln. 3(a)
Soln. 2(a)
For the reaction, CaCO3 (s) ⇌ CaO(s) + CO2 (g); K
= PCO 2
∆𝐻°
1
We know log10 K = log10 A − 2.303 𝑅 × 𝑇
∆𝐻°
1
log10 PCO 2 =log10 A − 2.303 𝑅 × 𝑇
The reaction is endothermic, therefore, log10 PCO 2
will linearly decrease with increase in the value of
1
𝑇
.
Soln. 4(d)
Reaction: 2A + B ⇌ C + D
n ×n
n
B
A
∆n g
P
Kp = n C2 ×nD ×
ng = 2 – 3 =  1
n ×n
Kp = n C ×nD
A
B
n
P
…..(i)
PV = nRT
𝑉
𝑅𝑇
=
n
P
… . (ii)
From eqn. (i) and (ii)
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Solution of Chemical Equilibrium
n ×n
V
Soln. 10(c)
C2H5OH(l) + CH3COOH(l) ⇌
CH3COOC2H5(l) + H2O
At t=0
1
1
0
0
At equilibrium 1/3
1/3
2/3
2/3
Kp = n C ×nD × RT
A
B
Soln. 5(a)
pNH3 = pH2S =
20
= 10 𝑎𝑡𝑚.
2
Kp = pNH3 pH2S
= 100 atm.2
Kp = Kc (RT)n
100 = Kc (0.0821  400)2
Kc = 0.092 M2
Soln. 6(a)
C(s) + CO2(g) ⇌ 2CO(g);
….(1)
14
KP1 = 10 atm
2CO(g) + 2Cl2(g) ⇌ 2COCl2(g);
KP2 = (6 × 10 – 3)2 atm – 2
Add (1) and (2)
C(s) + CO2(s) + 2Cl2(g) ⇌ 2COC2(g);
KP = 1014 × 36× 10–6
= 36 × 108
For given reaction ∆ng = – 1
 Kc = Kp(RT)
= 36 × 108 × 0.0821 × 1120
Kc = 3.31 × 1011 M – 1
Kc =
[𝐶𝐻3 𝐶𝑂𝑂𝐶2 𝐻5 ][𝐻2 𝑂]
[𝐶2 𝐻5 𝑂𝐻][𝐶𝐻3 𝐶𝑂𝑂𝐻]
=
2/3 x 2/3
1/3 x 1/3
=4
∵ ∆G = -2.303RT logKC
∴ ∆G = -2.303 x 8.314 x 298 x log4 ≈ -3435 J
Soln. 11(a)
𝑑𝐺 𝑥
𝑑𝐺 𝐸𝑞
= ∆G (minimum)
𝑃,𝑇
….(2)
Soln. 12(b)
Soln. 13(d)
α=
𝑀TH − 𝑀obs
𝑀obs (𝑛−1)
Molar mass of N2O4 = 92 g mol-1.
Here, n = 2.
α=
92.00 − 77.70
77.70(2 − 1)
= 0.184 = 18.4%.
Soln. 14(c)
Soln. 7(c)
Kp, Kc are independent of pressure but Kx is related
to pressure.
𝐷𝑇 − 𝐷𝑂
(𝑛−1)𝐷 𝑂
α=
Soln. 15(d)
Molecular mass
2
Soln. 8(b)
N2(g) + 3H2(g) ⇌ 2NH3 + Q cal
Since, the preparation of ammonia is an exothermic
reaction. Thus, the % yield will be increased, as the
temperature decreases.
∴ (DT )NH3 =
Soln. 9(c)
For dissociation of molecule
(n = 2, because 1 molecule of NH3 gives 2 molecules
of N2 + H2)
𝛼=
=
𝑑−𝐷
1
𝑛
𝑑 1−
2 𝑑−𝐷
𝑑
=
We know that vapour density =
17
2
= 8.5
Vapour density of gaseous mixture (VD) = 4.5
∴α=
𝐷 𝑇 − 𝐷𝑂
𝐷 𝑂 (𝑛−1)
=
8.5−4.5
4.5 x (2 – 1)
=
4.0
4.5
= 89%
𝑑 −𝐷
1
2
𝑑 1−
Soln. 16(d)
NH2 COONH4 2NH3 (g) + CO2 (g)
D=
=
𝑀𝑤
=
2
𝐷−𝑑
𝑛−1 𝑑
79
2
=
= 39
39−13
3−1 13
=1
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Solution of Chemical Equilibrium
Soln. 17(d)
=
M 0 −M
Soln. 22(c)
92−77.7
𝑟 𝑚𝑖𝑥
= (2−1)×77.7 0.184
n−1 M
𝑟𝑂2
% dissociation = 18.4
32
=
𝑀 𝑚𝑖𝑥
 Mmix = 40 g / mol
O2(g) ⇌
Soln. 18(a)
=
𝐷−𝑑
Initial moles
𝑛−1 𝑑
40−𝑑
0.5 =
1.5−1 𝑑
32
40
𝑑
𝑛−1 𝑑
𝐷−𝑑
3

1
𝐷−𝑑
=
𝑛−1 𝑑
𝐷
𝐷−𝑑
2−1 𝑑
𝐷
Thus, (1 + ) will linearly increase with increase in 𝑑 .
3Se2(g) ⇌ Se6(g)
𝑎
a (1– ) 3
Moles at eqm
𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑛𝑜 .𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑎𝑡 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑛𝑜 .𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
𝑎𝛼
𝑎 1− 𝛼 +
3
𝑎
𝑀
= 𝑀𝑇
𝑂
𝑀𝑇
= 𝑀 =  = 0.315
(MO) =
𝑤𝑅𝑇
𝑃𝑉
Soln. 24(d)
For reaction,
2A+(aq) + B(s) ⇌ B2+(aq) + 2A(s)
At equilibrium
n2
n1
∴ Equilibrium constant
Kc =
Kc =
= 1 ×2.463 × 10 −3 = 200
[𝐴2+ ]22
⇒ Kc x
=
𝑛1
(𝑛 2 )2
([A+] = n2 [B+]1 = n1)
n2
4
=
[𝐵 2+ ]2
n2
2
2
= [B2+]2
𝐾𝑐 x n2 [𝐵 2+ ]1 n1
=
=
4
4
4
n
n
1
2+
[𝐵 ]2 = which is less than 1
4
3
[B2+]2 =
6HCHO ⇌ C6H12O6
𝐶
 (1 – a)
Conc. at eqm
𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑚𝑜𝑙𝑒𝑠 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛
𝑀
30
= 𝑀 𝑇 = 150
𝑂
or,
6
=
𝑀𝑇
𝑀𝑂
MT = Theoretical molar mass of HCHO
  = 0.96
[𝐵 2+ ]2
0.02 ×24.63
Soln. 21(d)
𝐶𝛼
+
6
[𝐵 2+ ]1
[𝐴 + ]21
or, n1 = Kc xn22 or, [B2+]1 = Kc x n22
when volume of the solution is made three times then
𝑂
where molar mass of Se2
(MT) = 79 × 2 = 158
and molar mass of mixture
𝐶
1–
(1 + ) = 𝑑
2−1 𝑑
Soln. 20(b)
𝐶 1− 
0
2
2
1−+ 
3
=
=
=1

O3(g)
Soln. 23(a)
PCl5 (g)⇌ PCl3 (g) + Cl2 (g)
𝐷−𝑑
At ‘A’,  = 0
𝐷
3
  = 0.6
 %  = 60
Soln. 19(c)
0=
2
1
At eqm
0.5  0.5d = 40 – d
1.25d = 40
d = 32
=
= 0.4 5
Soln. 25(b)
N2O4 (g) ⇌ 2NO2 (g)
At t = 0
1
0
At equilibrium (1 – 0.2) = 0.8 2 x 0.2 = 0.4
P1 = 1 atm ;
P2 = ?
n1 = 1 ;
n2 = 0.8 + 0.4 = 1.2
T1 = 300 K ;
T2 = 600 K
∵
PV = nRT
for constant volume
P1
P2
=
n1 T1
n2 T2
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Solution of Chemical Equilibrium
∴P =
1 x 300
1.2 x 600
or
P2 = 2.4 atm
1
2
=
 PV = nRT  3 × 5 = n × 0.0820 × 310
 n = 0.59 i.e. (a – 0.2) = 0.59
 initial moles of NH3 = 0.79
1
2.4
Soln. 26(a)
K=
Soln. 30(c)
NH2COONH4(s) ⇌ 2NH3(g) + CO2(g)
Let partial pressure at equilibrium of CO2 be P, then
PNH 3 = 2P and total pressure at equilibrium = 3P
Kp = (2P)2 × P = 4P3
….(1)
If NH3 is added and the pressure of NH3 after
addition at equilibrium is 3P
−
RNH +
3 𝑂𝐻
8  10
𝑅𝑁𝐻2
𝑥×𝑥
–6
=
0.5
x = 2  10 – 3
[OH – ] = 2  10 – 3 M
pOH =  log 2  10 – 3
= 2.7

PCO
=9 𝑃
2

𝑃𝐶𝑂
2
 ratio of 𝑃
Soln. 28(d)
Ptotal = PHNO 3 + PNO 2 + PH 2 𝑂 + PO 2
∵ PNO 2 = 4PO 2 and PH 2 O = 2PO 2
=
Kp =
=4
7
4
𝑃𝑁𝑂
𝑃
.𝑃
4. 𝐻 2𝑂 𝑂2
4
𝑃𝐻𝑁𝑂
3
4
4 ×4 × 2 ×4 2 ×4
24
Kp = Kc RT
 Kc =
220
32 3
∆n g
x=
= 220
𝑃𝐶 2 𝐻 6
⇒
𝑥2
1−𝑥
= 5 × 10 – 2
−0.05+
0.05 2 + 4 ×0.05
2
= 0.20 atm
0.8
 mole % of C2H6 = 1.2 × 100 = 66.66
1
𝐾𝑝 = 9 𝑎𝑡𝑚
−2
Let initial moles of NH3 is a for completion of
reaction.
1
𝑁𝐻 3
𝑃 𝐶 2 𝐻 4 .𝑃 𝐻 2
Partial pressure of C2H6 = mole fraction × total
pressure
 0.80 = mole fraction × 1.2
= Kc (0.08 × 400)3
= 32.
𝐾
C2H6(g) ⇌ C2H4(g) + H2(g)
1–x
x
x
x2 + 0.05x – 0.05 = 0
Soln. 29(d)
LiCl. 3NH3(s) ⇌ LiCl. NH3(s) + 2NH3(g)
[Kp = 9 atm2]
Therefore,
LiCl. NH3(s) + 2NH3(g) ⇌ LiCl.
3NH3(s);
Initial moles
0.1
a
0
Final moles at eqn. 0
(a – 0.2)
0.1
At eqn Kp =
4
=9
At eqm
 30 – 2 =PO 2 × 7
Kp =
𝐶𝑂 2
Soln. 31(c)
 Ptotal =PHNO 3 + 7PO 2
28

× 𝑃𝐶𝑂
2
4
Soln. 27(b)
PO 2 =
2

Kp = 4P3 = 𝑃𝑁𝐻
3
1
2
𝑜𝑟 9 =
Soln. 32(b)
2SO2(g) + O2(g) ⇌ 2SO3(g)
Kc =
0.12 2
0.12 2 ×5
= 0.2
Another vessel
Moles of eqm
2SO2(g) + O2(g) ⇌ 2SO3(g)
0.5 – 2x y – x
2x
20
As per given 2x = 100 × 0.5 = 0.1
Kc =
0.1 2
0.4 2 𝑦−0.05
= 0.20
y = 0.3625 mole
 mass of O2 added = 11.6 g
1
𝑃
2
𝑁𝐻 3

PNH
= 3 atm
3
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Solution of Chemical Equilibrium
Soln. 33(b)
Let initial mole of N2O4(g) is one
N2O4(g) ⇌ 2NO2(g)
Initial moles
1
0
At equilibrium 1 – 
2
Total no. of moles at equilibrium = 1 + 
PN 2 O 4 =
PNO 2 =
1− 
1+
2
×P
2
𝑃𝑁𝑂
2
Hence, Kp = 𝑃
 4.5 =
4 2
1− 2
𝑁2𝑂4
=
42
1− 2
×𝑃
×2
1−
XN 2 O 4 = 1+ = 0.25
XNO 2 = 0.75
Average molar mass of mixture
= 0.25 × 95 + 0.75 × 46 = 57.5
Soln. 34(a)
2A(g) + B(g) ⇌ C(g) + D(g)
Initial moles
1
1
7
3
At eqm
2 – 2x 1 – x
7+x 3+x
Due to very high value of Kc we can assume that
reactant almost converted into products so
1 – x = y;
2 – 2x = 2y  x ≃ 1
8 ×4
2𝑦 2 𝑦
 y3 = 8 × 10 – 12
 equilibrium concentration of
A = 2y = 4 × 10 – 4
Soln. 35(c)
2SO3(g) ⇌ 2SO2(g) + O2(g)
Moles at eqm
1 – 2x
2x
x
Only SO2 (O.No. = 4) will oxidized so equivalent of
SO2 = equivalent of KMnO4
2x × 2 = 0.2 × 5

2x = 0.5
Kc =
0.5 2 0.5
2
2
0.5 2
2
=
𝑟𝑠𝑜 2
𝑀 𝑆𝑂 2
𝑀 𝑚𝑖𝑥
64
 2.56 = 𝑀
𝑚𝑖𝑥
 Mmixture = 25
Let mole fraction of F2 is x
38 ×𝑥+ 1−𝑥 ×19
1
x = 0.315;
 = 0.6
Mole fraction of N2O4:
1012 =
𝑟 𝑚𝑖𝑥
25 =
×𝑃
1+
Soln. 36(d)
= 0.125
𝑃2
 Kp = 𝑃 𝐹 =
𝐹2
0.685𝑃 2
0.315 𝑃
≃ 1.49 atm
Soln. 37(b)
𝑥2
8 × 10 – 6 = 0.5
 x = 2 × 10 – 3
 pOH = 2.7
So, pH = 11.3
Soln. 38(d)
N2O5(g) ⇌ N2O3(g) + O2(g)
4–x
x–y x+y
N2O3⇌ N2O + O2
x–y
y
y+x
∵ [O2] = x + y = 2.5
and 2.5 =
𝑥+𝑦 𝑥−𝑦
4−𝑥
 x = 2.166
[N2O5] = 4 – x = 1.846
Soln. 39(b)
Let x is partial pressure of A and y is partial pressure
of C when both equilibrium simultaneously
established in a vessel
X(s) ⇌ A(g) + 2B(g)
X (2x + 2y);
2
K P 1 = PA . PB(total
)
Y(s) ⇌ C(g) + 2B(g)
Y (2y + 2x);
K p 2 = PC . PB2(total )
𝐾𝑝 1
𝐾𝑝 2
𝑥
= 𝑦  x = 2y
KP1 = x(2x + 2y)2
 x = 0.1 atm
 y = 0.05 atm
Total pressure of gases = PA + PB + PC
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Solution of Chemical Equilibrium
10 −6
= 3 (x + y)
= 0.45 atm
= 0.5 ×
Soln. 40(c)
∆rG = 0 – 77.1 × 2 = – 154.2 kJ / mol
∆rG = – 154.2 +
Q=𝑃
𝐻+
10 −10
= 2 × 104
∆G = ∆rG + RT ln Q
8.314 × 300 ln 2 × 10 4
1000
= – 129.5 kJ / mol
2
+ 2
𝐻 2 . 𝐴𝑔
Cool Assertion Reason
Soln. 1(b)
Catalyst alters the rate of reaction. In reversible
reaction, the rate of forward and backward reactions
is affected equally. Thus, catalyst does not alter the
position of equilibrium.
Soln. 2(b)
∆G = ∆G + RT loge Kc
At equilibrium, ∆G = 0
 ∆G = – RT loge Kc
Soln. 3(c)
N2(g) + 3H2(g) ⇌ 2NH3(g)
1
2
3
N2(g) + 2 H2(g) ⇌ 2NH3(g)
K1 =
K2 =
𝑁𝐻3 2
𝑁2 𝐻2 3
𝑁𝐻3
𝑁2 1/2 𝐻2 1/2
1/2
K1
K2 =
Thus, equilibrium constant changes
stoichiometric coefficient of the reaction.
with
the
Soln. 4(c)
When Qc < Kc, then the reaction will be fast in
forward reaction. Thus, statement – 2 is wrong.
Soln. 5(d)
For the reaction: PCl5(g) ⇌ PCl3(g) + Cl2(g)
the equilibrium will change by changing the volume
of vessel but the equilibrium constant Kc will be
unaffected.
Soln. 6(b)
PCl5(g) ⇌ PCl3 + Cl2(g)
∆ng > 0, volume4 of vessel is not constant.
Thus, addition of inert gas will shift the equilibrium
toward higher volume direction i.e., in forward
direction.
Soln. 7(d)
A+B⇌C+D
C+D⇌A+B
K1
K2
1
K2 = 𝐾 i.e. equilibrium constant will be reciprocated
1
when the reaction is reversed.
Cool Comprehension Based Problems
Paragraph – 1
Paragraph – 2
Soln. 1(b)
Soln. 1(c)
Soln. 2(a)
Soln. 2(a)
Soln. 3(c)
Soln. 3(b)
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Solution of Chemical Equilibrium
Paragraph – 3
Soln. 4(c)
Soln. 1(a)
Soln. 5(a)
Soln. 2(b)
Soln. 3(a)
Soln. 4(b)
Soln. 5(c)
Soln. 6(a)
Cool Matrix Match Type Problems
Soln. 1(a – q) (b – p, r) (c – r) (d – s)
Soln. 2(a – q, s) (b – p, r) (c – q, r) (d – p, s)
Soln. 3(a – p) (b – q) (c – r) (d – s)
Soln. 4(a – p, q, s) (b – q, r) (c – q) (d – p, s)
Soln. 5(a – s) (b – p) (c – q) (d – r)
Soln. 6(a – s) (b – p) (c – q) (d – r)
Soln. 7(a – s) (b – q, r) (c – q) (d – p)
Soln. 8(a – p, r , s), (b – q, r), (c – q, r), (d – q, r)
Soln. 9(a – s) (b – p) (c – q) (d – r)
Cool Single Digit Type Problems
Soln. 1
2HI(g) ⇌ H2(g) + I2(g)
Kc = 0.25
For: H2(g) + I2(g) ⇌ 2HI(g)
1
Kc = 0.25 = 4
Soln. 2
A(g) + B(g) ⇌ C(g) + D(g)
t0
4
4
0
0
teq 4 – 2 4 – 2
2
2
Kc =
𝐶 𝐷
𝐴 𝐵
2 ×2
= 2 ×2 = 1
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Solution of Chemical Equilibrium
x = 5 atm
Total pressure of equilibrium = 1 + 5 + 2 = 8 atm
Soln. 3
AB(g) ⇌ A(g) + B(g)
1
0
0
t0
1
1−3
teq
Kp =
=
𝑃
𝐾𝑝
𝑛𝐴 × 𝑛 𝐵
𝑛 𝐴𝐵
1 1
×
3 3
2
3
×
×
𝑃
𝑃
1
1
3
∆𝑛 𝑔
3
2
1
1
4
n = 3 + 3 + 3 = 3
𝑛
1
4/3
Soln. 7
PCl5(g) ⇌ PCl3(g) + Cl2(g)
Kp =
1
=6×
3𝑃
4
𝑃
=8
=
𝑛 𝑃𝐶𝑙 3 × 𝑛
2 ×2
2
𝐶𝑙 2
𝑛 𝑃𝐶𝑙 5
×
3 1
6
×
𝑃
∆𝑛 𝑔
𝑛
=1
=8
Soln. 4
Equilibrium constant does not change on changing
the concentration of reactants.
Soln. 5
A(g) + B(g) ⇌ C(g) + D(g)
t0
a
a
0
0
teq a – x a – x
x
x
x = 2(a – x)
3x = 2a
2
x=3a
2
1
[A] = [B] = 𝑎 − 3 𝑎 = 3 𝑎
2
[C] = [D] = 3 𝑎
Kc =
𝐶 𝐷
𝐴 𝐵
2
2
3
1
𝑎× 𝑎
3
3
= 31
𝑎× 𝑎
=4
Soln. 6
SO2Cl2(g) ⇌ SO2(g) + Cl2(g)
1 atm
x atm
2 atm
Kp =
𝑥 ×2
1
Soln. 8
NH4Cl(s) ⇌ NH3(g) + HCl(g)
Kp = PNH 3 × PHCl
81 = p2
p = 9 atm
Total pressure = pNH 3 × pHCl = 18 atm
Thus, total pressure is 2 times the pressure of NH3.
Soln. 9
PV = nRT
1 × 8.96 = n × 0.0821 × 473
n = 0.23
NH2COONH4(s) → 2NH3(g) + CO2(g)
nNH 2 COONH 4 =
0.23
3
Mass of NH2COONH4 = 0.0767 × 74
 6g
Soln. 10
∆G = 2.303 RT log10Kp
When Kp = 1, ∆G = 0
=10
Cool Previous Year Problems
IITJEE / IIT ADVANCE
Soln. 1(d)
MX
Let s be the solubility of MX
+
⊖
MX ⇌ M + 𝑋
S × s = Ksp of MX = 4 × 10 – 8
MX2
MX2 ⇌ M + 2 + 2 𝑋 ⊖
Ksp of MX2 = x × (2x)2 = 3.2 × 10 – 14
4x3 = 32.0 × 10 – 15
x3 = 8 × 10 – 5
x = 2.0 × 10 – 5 mol / litre
For
M3X ⇌ 3M + 1 + 𝑋 ⊖
3
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Solution of Chemical Equilibrium
Ksp of M3X = x × (3x)3 = 27x4 = 2.7 × 10 – 15
27x4 = 27 × 10 – 16
x4 = 10 – 16
x = 1 × 10 – 4 mol / litre
2.4 × 10 – 4 > 1.0 × 10 – 4 > 2.0 × 10 – 5
MX > M3X > MX2
Soln. 4(b)
N2 + 3H2 ⇌ 2NH3
At equilibrium, addition of a reactant or any catalyst
has no effect. However, the catalyst brings the
equilibrium faster by increasing both forward and
backward reactions.
Soln. 2(c)
During neutralization of mass acidic base (BOH)
with HCl,
BOH + HCl → BCl + H2O
At equilibrium point
Meq of BOH = Meq of BCl
M1 × V1 = M2 × V2
Soln. 5(b)
Ag + NH3 ⇌ [Ag(NH3)]
2𝑀
2.5 ×
5
= 𝑉2 ×
2𝑀
18
𝑜𝑟 𝑉2 = 7.5 mL
One mol of BCl is formed.
Total volume of solution after neutralization = 2.5 +
7.5 = 10 mL
1
6.8 × 10 – 3 = K1 =
1.6 × 10 – 3 = K2

K=
10 −14
[𝐵 +]
10 −12
𝑐𝑕 ×𝑐𝑕
𝑐𝑕 2
= 𝑐(1−𝑕 ) = (1−𝑕)
10 – 2 =
0.1 × 𝑕 2
(1−𝑕 )
1
1
1
or pH = 2 𝑝𝐾𝑤 − 2 𝑝𝐾𝑏 − 2 log 𝑐
1
1
1
= 2 × 14 − 2 × 12 − 2 log 10−1
pH = 7 – 6 + 0.5 = 1.5
[H+] = 3.2 × 10 – 2 M.
Soln. 3(a)
CO2 + H2O ⇌ H2CO3
Solution contains CO2, H2CO3
H2CO3 ⇌ 𝐻 ⊕ + 𝐻𝐶𝑂3⊝
𝐻𝐶𝑂3⊝
⇌ 𝐻 ⊕ + 𝐶𝑂3−2
Hence, the solution contains the species
CO2, H2CO3, 𝐻𝐶𝑂3⊝ , 𝐶𝑂3−2
…(2)
[𝐴𝑔 (𝑁𝐻3 )2 ]⊕
𝐴𝑔 + ×[𝑁𝐻3 ]2
K = K1 × K2 =
[𝐴𝑔 (𝑁𝐻3 )+ ]
𝐴𝑔 + ×[𝑁𝐻3 ]
[𝐴𝑔 (𝑁𝐻3 )2 ]1
𝐴𝑔 + × [𝑁𝐻3 ]2
Soln. 6(b)
𝐵𝑂𝐻 ×[𝐻 + ]
𝐴𝑔 𝑁𝐻3 ⊕×[𝑁𝐻 3 ]
Ag + 2NH3 ⇌[Ag(NH3)2]
BCl is a salt of weak base a strong acid
BCl + H2O  BOH + HCl
or 𝐵⨁ + 𝐻2 𝑂 ⇌ 𝐵𝑂𝐻 + 𝐻 ⊕
Initial conc. 0.1
(c = 0.1)
h = degree of hydrolysis 0.1 – 0.1 h
𝑏
[𝐴𝑔(𝑁𝐻3 )2 ]⊕
+
K=
𝐾
…(1)
𝐴𝑔 + [𝑁𝐻3 ]
[Ag(NH3)] + NH3 ⇌ [Ag(NH3)2]
conc. of BCl = [BCl] = 10 = 0.1 M
KH = 𝐾𝑤 =
[𝐴𝑔 (𝑁𝐻1+)]
×
[𝐴𝑔(𝑁𝐻3 )2 ]⊕
𝐴𝑔 (𝑁𝐻3 )⊕ ×[𝑁𝐻3 ]
= 6.8 × 10−3 × 1.6 × 10−3
= 1.08 × 10−5

CH3NH2 + HCl ⇌ CH3N H3𝐶𝑙 ⊖
Initial conc.
1.0
0.08
0.0
At neutralization 0.02
0.00
0.08
point conc. left
CH3NH2 is a base and forms salt with HCl. HCl is the
limiting reagent. So a buffer is formed with base
conc.
= 0.02 mol
Salt conc = 0.08 mol
pOH = 3.30 + 0.6021 = 3.9021
 pH = 14 – pOH = 14 – 3.9021 = 10.098
10.098 = – log[H+]
[H+] = 8.0 × 10 – 11 M
Soln. 7(c)
Na2SO4 ⇌ 2Na + SO−2
4
0.004
Let x be the degree of dissociation
x
2x
x
conc left 0.004 – x
2x
x
No. of species left in soln. = 0.004 – x + x + 2x
= 0.004 + 2x
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Solution of Chemical Equilibrium
The conc. of isotonic
c1 = c2
0.004 + 2x = 0.01
2x = 0.006
X = 0.003 = 3 × 10 – 3 M
Soln. 11(a)
ApBq ⇌ pAq+ + qB p−.
ps
qs
Solubility product Ls = (pS)p × (qS)q
= S p + q pp qq
3 × 10 −3
0.003
% of dissociation = 4 × 10 −3 × 100 = 0.004 × 100
= 75%
Soln. 8(b)
Since HX is a weak acid and NaX is its salt with a
strong base, degree of hydrolysis of salt of weak acid
and strong base
h=
𝐾𝑤
𝐾𝑎 ×𝑐
=
1 × 10 −14
1 × 10 −5 ×0.1
= 1 × 10−8 = 10 – 4
% of hydrolysis = 100 × 10 – 4 = 10 – 2 % = 0.01%
Soln. 9(d)
N2O4(g) ⇌ 2NO2 (g) (∆n = 1)
Kp > Kc
Since the temperature is kept constant hence both Kp
and Kc remain unchanged. However, when volume is
halved, pressure doubles up (Boyle’s law). So in
order to keep Kp unchanged  must decrease.
N2O4 (g) ⇌ 2NO2
Initial conc.
a
0
 = Degree of

2
dissociation
a–
2
conc. left
Total no. of moles at equilibrium = a –  + 2
= (a + )
Let the pressure at equilibrium = P by (a + ) mole
2
Partial pressure of NO2 = PNO 2 = a+  P
a− 
Partial pressure of PN 2 O 4 = a+  P
Kp =
(𝑃 𝑁𝑂 2 )2
𝑃𝑁 2 𝑂 4
𝐾
1.44 × 10 −5
Kc = (𝑅𝑇)𝑝−2 = (0.082 ×773)−2
Soln. 13(a)
Equilibrium is affected by temperature and pressure
due to change in heat as well as change in volume of
substance.
Soln. 14(b)
Solution of NaCl will be neutral pH = 1. But HCl ⇌
H + Cl⊖ is highly acidic and pH = 1 for 0.1 M sol.
NH4Cl + H2O ⇌ NH4OH + HCl
Strong base Weak acid
NaCN being the salt of a strong base and a weak acid
NaCN is alkaline. Hence, increasing order of pH is
HCl < NH4Cl < NaCl < NaCN
Soln. 15(d)
A2X3 ⇌ 2A + 3 + 3 X – 2
2y
3y
let y be the solubility
Ksp = (2y)2 × (3y)3 = 108 y5
44 𝑃
= 𝑎 2 −2
Since Kp is unchanged

Soln. 12(b)
N2(g) + 3H2(g) ⇌ 2NH3 (g)
∆n = 2 – (1 + 3) = – 2
R = 0.082
T = 500 + 273 K
= 773 K
Kp = Kc × (RT) – 2
1
𝑃
As P increases  decreases or changes.
Soln. 10(d)
Kp is not based upon pressure and concentration.
Soln. 16(c)
pKb = 10.83
pKw = pKa + pKb
pKa = 14 – 10.83
pKa = 3.17
Ka = 6.75 × 10 – 4
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Solution of Chemical Equilibrium
Soln. 17(a)
Acid strength of oxy acids of halogens are related to
(i) electronegativity of halogen (ii) oxidation state
negativity of halogens is Cl > Br > I. Hence the acid
strength order is HClO > HBrO > HOl.
Volume = 100 mL
8
(OH –) = 100 = 8 × 10−2
pH = 11.1
Volume of
Soln. 18(c)
AlCl3 is hydrolysed by water as
AlCl3 + 3H2O ⇌ Al(OH)3 + 3HCl
and is the most acidic of all.
𝑀
5
5
HCl ≡ 25 𝑚𝐿 𝑜𝑓
𝑀
5
NaOH
HCl left = 50 mL
𝑀
= 100 mL of 10 = 10 millimol
Volume = 75 + 25 = 100
10
[H+] = 100 = 10 – 1 pH = 1.0
Soln. 19(a)
Soln. 22(d)
H2O ⇌ H  + OH ⊖
Initial conc.
c
0
0
 = Degree of dissociation c
c
c
c(1 – ) c
c
Density of H2O = 1.0 g / c c
Weight of one litre of H2O = 1000 g
Now, c = no of moles / litre =
[H+] = c = 55.56 ×
1.9 × 10 −7
100
1000
18
= 55.56 mol / litre
= 1.05 × 10−7
= 1.055 × 10 – 7 mol / litre
Kw = Ionic product of water = [H+] × [OH –]
= [1.055 × 10 – 7]2
= 1 × 10 – 14
Soln. 20(b)
CN ⊝ B.order = 3 C ≡ N ⊖
N2 B. order = 3(N ≡ N)
But due to absence of bond polarity it is inert.
Soln. 21(d)
𝑀
They neutralize each other and pH = 7.0
b → After neutralization
𝑀
10
= 45 𝑚𝐿 of
𝑀
10
HCl + CH3COOH
Acid1
Base2
Cl + CH3COOH2
Base1
Acid2
According to Lowry – Bronsted theory acid donates
the proton and base accepts the proton.
H
HCl
Acid1
Cl
+H
Conjugate Base1
Soln. 23(a)
a → Fe2O3 + 6HCl → 2FeCl3 + 3H2O
FeCl3 is not hydrolysed.
b → NH3 + H2O +NaCl → NH4Cl + NaOH
NaOH reacts with NH4Cl to give back the reactants.
c → SnCl4 + HgCl2 → 2HgCl2 + SnCl2
The reverse of it is a redox reaction.
d → 2CuI + I2 + 4H → 2Cu + 2 + 4HI
Cu+ is not easily oxidized to Cu + 2 in acidic
conditions using I2.
𝑀
a → N1V1 = N2V2, 100 × 10 = 100 × 10
45 mL of
𝑀
d → 25 mL of
NaOH
Soln. 24(d)
Solubility of hydroxides of II group increases down
the group. So, Be(OH)2 has least solubility and
lowest value of Ksp.
Amount of HCl left
𝑀
1
= 10 mL of 10 = 10 × 10 = 1 m mol
Soln. 25(a)
Total volume = 55 +45 = 100
+
1
(H ) = 100 = 10
–2
H
ClO4
HClO4
M
Strong Acid
+H
Weak Conjugate Base
pH = 2.0
𝑀
𝑀
c → 10 mL of 10 ≡ 10 mL of 10 NaOH
80 mL of
𝑀
10
NaOH is left = 8 millimol
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Solution of Chemical Equilibrium
Soln. 26(d)
OCOCH3
OCOCH3
COOH + HCl
COOH
Unionised
Aspirin
OCOCH3
Soln. 29(c)
(a) CaO + 2CH3COOH → (CH3COO)2 Ca + H2O
Cal. acetate
(b) CaCO3 + 2CH3COOH → (CH3COO)2 Ca + CO2
+ H2O
(c) CaC2O4 + CH3COOH → do not dissolve as no
reaction takes place.
(d) Ca(OH)2 + 2CH3COOH → (CH3COO)2 Ca +
2H2O.
OH
COO +
COOH + NaOH
In alkaline
medium
Ionised
Aspirin remains unionized in acidic medium and
ionized in the alkaline medium.
Soln. 27(a)
Precipitation takes place only when the value o ionic
product > Ksp.
𝑉
Conc. of [Ag+] on mixing = 2𝑉 × 10−4 = 10 – 5 × 5
Both have equal volumes (V each)
Soln. 30(d)
CH3COOHBF3 with 6 electrons in outermost shell  Lewis acid
AlCl3 with 6 electrons in outermost shell  Lewis
acid
BeCl2 with 4 electrons in outermost shell  Lewis
acid
SnCl4 is not a Lewis acid as its octet is complete.
Soln. 31(a)
+H
NH2
NH3
H
Base
Conjugated acid
A Bronsted base can accept a proton to give a
conjugate acid.
𝑉
Ksp = 1.8 × 10 – 10 (Cl –) = 2𝑉 × 10−4 = 5 × 10 – 5
I.P. = 5 × 10 – 5 × 5 × 10 – 5 = 2.5 × 10 – 9
I.P. > Ksp. Hence, ppt. ion take place.
Soln. 24(d)
(a) CH3COONH4 + H2O ⇌ CH3COOH + NH4OH
Weak acid Weak
base
It is neutral. pH = 7
(b) NH4Cl + H2O ⇌ NH4OH + HCl
Weak base Strong acid
Solution is acidic in nature
(pH < 7.0)
(c) (NH4)2 SO4 + 2H2O ⇌ 2NH4OH + H2SO4
Weak base Strong acid
Solution is acidic (pH < 7)
(d) CH3COONa + H2O ⇌ NaOH + CH3COOH
Strong base Weak acid
Solution is alkaline or basic (pH > 7)
Soln. 32(d)
Since the acid is weak and base is strong the
equivalence point will be basic (as salt formed will be
basic). That indicator whose pH range lies in basic
medium (8 to
9.6) will
be preferred
(Phenolphthalein).
Soln. 33(d)
Those reactions in which insoluble precipitates are
formed or the gaseous products which are insoluble
in water are formed are irreversible in nature. Only
(d) is reversible.
Soln. 34(c)
For reaction between weak acid and a strong base
BOH
Strong base
1
𝐾𝐻
=
+ HA
BA + H2O
Weak acid
𝐾𝑤
𝐾𝑎
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Solution of Chemical Equilibrium
 KH = 1010
Hence, HClO is the weakest acid and its conjugate
base is strongest.
Soln. 35(a)
Apply Henderson’s equation [X –] = [HX]
pOH = pKb + log
(𝑆𝑎𝑙𝑡 )
(𝐴𝑐𝑖𝑑 )
(Kb = 10 – 10)
H
ClO
HClO
Acid
+H
Conjugate Base
pKb = – log Kb
[𝑋 −]
pOH = 10 + log [1+𝑋]
[𝑋 − ]
[1+𝑋]
= 1, log 1 = 0
pH = 14 – pOH = 14 – 10 = 4
Soln. 36(a)
As temperature is constant, Kp does not change or
remains unchanged.
Soln. 37(c)
Both liquid and vapour are in equilibrium and
temperature is remaining constant. Thus the total
energy remains the same for two phases.
Soln. 38(b)
Let V be the volume of Ca + 2 and F ⊖ ions.
The precipitation takes place only when I.P > Ksp
Ksp = 1.7 × 10 – 10
Type MX2,
Ksp = x × (2x)2 = 4x3
𝑉
[Ca +2] of mixing = 2𝑉 × 10−2 = 5 × 10 – 3 M
𝑉
[F ⊖ ] of mixing = 2𝑉 × 10−3 = 5 × 10 – 4 M
I.P. = 5 × 10 – 3 × (5 × 10 – 4)2 = 125 × 10 – 11
I.P = 1.25 × 10 – 9
I.P > Ksp. Hence, precipitation takes place.
Soln. 39(b)
Kw = [H+] × [OH –] at 90C[H3+O] = 10 – 6 mol / litre
= 10 – 6 × 10 – 6 = 10 – 12
Soln. 40(a)
If acid is strong, the conjugate base must be weak.
Acid strength decreases in the order:
HClO4 > HClO3 > HClO2 > HClO
Soln. 41(d)
H2(g) + I2(g) ⇌ 2HI(g)
Since, the no. of gaseous products are equal to no. of
gaseous reactants Kp = Kc
Kp depends only on temperature
Soln. 42(b)
2SO2(g) + O2 (g) ⇌ 2SO3 + heat
Using Le – Chatelier’s principle.
Since, the reaction is exothermic; hence it is favoured
only by low temperature.
2SO2(g) + O2(g) ⇌ 2SO3(g)
∆n = No. of moles of gaseous products – No. of
moles of gaseous reactants
=2–3=–1
Soln. 43(d)
[H+]Total = [H+]Acid + [H+]Water
H2O ⇌ H+ + OH –
10 – 8 + x x
(10 – 8 + x)x = 10 – 14
Solve for x, we get x = 9.5 × 10 – 8
H+ = 10 – 8 + 9.5 × 10 – 8 = 10.5 × 10 – 8
pH = – log (10.5 × 10 – 8)
 pH lies between 6 and 7.
Soln. 44(a)
An acid buffer is an equimolar mixture of weak acid
and its salt with strong base or conjugate base of acid.
CH3COOH is a weak acid and CH3 COO⊖ given by
ammonium acetate is its conjugate base, constitute
the buffer.
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Solution of Chemical Equilibrium
AIEEE / IIT MAINS
Soln. 1(a)
Kp = Kc(RT)∆ng
∆ng = 1 – 1.5 = – 0.5
Kc =
[𝑁2 ]1/2 [𝑂2 ]1/2
[𝑁𝑂]
= (1/𝐾𝑐 ) =
1/4 × 10−4 = 50
𝐾
Kp = Kc(RT) – ½ = (𝑅𝑇)𝑐1/2
Soln. 8(a)
For endothermic reaction, ∆H = +ve
∆H = Ef – Eb , it means Eb < Ef
𝐾
So, 𝐾𝑐 = (RT)1/2..s
𝑝
Soln. 2(c)
Soln. 9(a)
Kp = Kc(RT)∆n
∆n = 3 – 2 = 1
Kp = Kc (0.0831 × 457)1.
1.8 × 10 −3
𝐾
(Kc) = (𝑅𝑇)𝑝∆𝑛 = (8.134 ×700)1
= 3.1 × 10 – 7
Soln. 3(c)
∆ng = – ve → takes place with decrease in number of
moles or pressure, so increase in pressure shifts
equilibrium in forward side.
∆H = – ve → takes place with evolution of heat or
increase in temperature, so decrease in temperature
shifts this equilibrium in forward side.
Soln. 4(c)
[𝑁𝑂2 ]2
Kc = [𝑁
2 𝑂4
= 3 × 10
=
]
−3
(1.2 × 10 −2 )2
4.8 × 10 −2
mol L – 1
Soln. 5(b)
(Kp) = Kc (RT)∆n ……….(i)
(where ∆n = 1 – 2 = – 1)
Substituting this value of ∆n in equation (i),
Kp = KC (RT) – 1
𝐾𝑝
𝑐
Soln. 6(b)
In the expression for equilibrium constant species in
solid state are not written (that is their mar
concentration are taken as 1)
P4(s) + 5O2(g) ⇌ P4O10(s)
So, Kc = 1/[O2]5
Kc =
[𝑁𝑂]2
𝑁2 [𝑂2
= 4 × 10 – 4
]
NO ⇌ 1 2 N2(g) + 1 2 O2(g)
NH4HS(s) ⇌ NH3(g) + H2S(g)
Initial
0
0.5
0
pressure
At equilibrium 0
0.5 + x
x
Total pressure = 0.5 + 2x = 0.84
So, x = 0.17 atm
Kp = PNH 3 × PH 2 S = 0.11 atm2
Soln. 11(d)
Cl2(g) + 3F2(g) ⇌ 2 ClF3(g); ∆H = – 329 kJ.
Favourable conditions
1. Decrease in temperature
2. Addition of reactants
3. Increase in pressure, that is, decrease in volume
1
or 𝐾 = 𝑅𝑇
Soln. 7(d)
N2(g) + O2(g) ⇌ 2NO(g)
Soln. 10(d)
Soln. 12(a)
PCl5 ⇌ PCl3 + Cl2
1
0
0
1–x
x
x
PPCl 5 = Total pressure × mole fraction of PCl3
=P
𝑥
1+𝑥
Soln. 13(a)
K c 1 = 4.9 × 10 – 2 =
Kc 2 =
𝑆𝑂2 𝑂2 1/2
[𝑆𝑂3 ]
[SO 3 ]2
[SO 2 ]2 [O 2 ]
(1/Kc 1 )2
Kc 2 =
= 416.5
= (1 / 4.9 × 10 – 2)2
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Solution of Chemical Equilibrium
Soln. 14(a)
X ⇌ 2Y
1
0
(1 – x) 2x
KP1 =
(2𝑥)2
(1− 𝑥)
Soln. 16(d)
The volume occupied by water molecules in vapour
phase is (1 – 10 – 4)dm3, that is approximately 1 dm3.
Pvap V = nH 2 O RT
3170 (Pa) × 1 × 10 – 3 (m3) =
nH 2 O (mol) × 8.314 (J K – 1 mol – 1) × 300 (K)
(P1 / 1 + x)1
Z⇌P+Q
1
0 0
(1 – x) x x
nH 2 O =
(𝑥)2
𝑃2
8.314 ×300
= 1.27 × 10 – 3
Soln. 17(d)
CO2(g) + C(s) ⇌ 2CO(g)
0.5 – x
2x
0.5 – x + 2x = 0.8
x = 0.3
KP2 = (1−𝑥) (P2 / 1 + x)1
4 × 𝑃1
3170 × 10 −3
1
=9
On solving, we get
P1: P2 = 1: 36
K=
Soln. 15(c)
As equation (c) = equation (a) + equation (b)
So K3 = K1K2
0.6 ×0.6
0.2
= 1.8 atm
Soln. 18(d)
As 2nd eq is reverse and half of 1st one so
K c =
1
Kc
=
1
4 × 10 −4
1
= 2 × 10 −2 = 50
AIPMT/ CBSE – PMT/NEET
Soln. 1(c)
NaCl is a salt of strong acid and strong base hence its
aqueous solution will be neutral i.e. pH = 7. NaHCO3
is an acidic salt hence pH < 7. Na2CO3 is a salt of
weak acid and strong base. Hence its aqueous
solution will be strongly basic ie. pH > 7.
NH4Cl is salt of weak base and strong acid, hence its
aqueous solution will be strongly acidic i.e. pH > 7.
Soln. 2(a)
For reaction (1)
K1 =
𝑁𝑂 2
𝑁2 𝑂2
and for reaction (2)
K2 =
𝑁2
1
2 𝑂2
𝑁𝑂
1
2
1
therefore K1 = 𝐾 2
2
Soln. 4(a)
Sodium borate is a salt of strong base (NaOH) and
weak acid (H3BO3). Hence its aqueous solution will
be basic.
Soln. 5(d)
According to Le – Chatelier’s principle” whenever a
constraint is applied to a system in equilibrium, the
system tends to readjust so as to nullify the effect of
the constraint.
Soln. 6(b)
AgBr has the highest solubility in 10 – 3 M NH4OH.
All other solvents will dissolve AgBr poorly.
Moreover bromides of Ag+, Hg22+ and
Cu22+ are water insoluble.
Soln. 3(b)
An aqueous solution of acetic acid dissociates as
CH3COOH + H2O ⇌ CH3COO – + H3O+
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Solution of Chemical Equilibrium
Soln. 7(d)
Solid ⇌ Liquid
It is an endothermic process. So when temperature is
raised, more liquid is formed. Hence adding heat will
shift the equilibrium in the forward direction.
Soln. 8(c)
In presence of the given salts, the solubility of AgCl
decreases due to common ion effect.
Soln. 9(b)
For Bi2S3. Ksp – = (2s)2. (3s)3
= 4s2.27s3 = 108s5
or s = 5
𝐾𝑠𝑝
108
=
5 1 × 10 −17
108
Soln. 13(b)
According to equation
2HI
⇌
H2 + I2
At t = 0 (2 moles)
0
0
At equilibrium
 moles  moles
(2 – 2) moles
Total moles at equilibrium = 2 – 2 +  +  = 2
moles
Soln. 14(b)
A2 ⇌ 2A Equilibrium constant is given by
Kc =
𝐴2
𝐴2
𝑐𝑜𝑛𝑐 .𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
= 𝑐𝑜𝑛𝑐 .𝑡𝑒𝑟𝑚𝑠
𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
Since the value given is very small, hence conc. of
products is less. It means the reaction is slow.
For MnS.Ksp = s2
or s = 𝐾𝑠𝑝 = 7 × 10−16
for CuS s = 𝐾𝑠𝑝 = 8 × 10−37
For Ag2S Ksp = 2s2 .s = 4s3
or s =
3 𝐾𝑠𝑝
4
thus MnS has maximum solubility.
Soln. 10(a)
Molarity (M) = 10M. HCl is strong acid and it is
completely dissociated in aqueous solutions as:
HCl(10) ⇌ H+(10) + Cl –.
Therefore [H+] = [HCl] or [H+] = 10.
pH = – log [H+] = – log [10] = – 1.
Soln. 15(d)
CuS ⇌ Cu2+ + S2
s
s
2
Ksp = S or S = 𝐾𝑠𝑝
For Binary salts like CuS & HgS, solubility,
S = 𝐾𝑠𝑝
 SCuS = 10−31 , SHgS = 10−54
For Ag2S→ 2 Ag+ + S2
2s
s
Ksp = 4s3 or SAg 2 S =
K sp
4
=
3 10 −44
4
 The order is CuS > Ag2S > HgS
Soln. 16(d)
Soln. 11(d)
Rate constant of forward reaction (Kf) = 1.1 × 10 – 2
and rate constant of backward reaction (Kb) = 1.5 ×
10 – 3 per minutes. Equilibrium constant (Kc) =
𝐾𝑓
𝐾𝑏
1.1 × 10 −2
= 1.5 × 10 −3 = 7.33
Soln. 12(d)
The buffer system present in serum is H2CO3 +
NaHCO3 and as we known that a buffer solution
resist the change in pH therefore pH value of blood
does not change by a small addition of an acid or a
base.
[𝑆𝑎𝑙𝑡 ]
pH = pKa + log10[𝐴𝑐𝑖𝑑 ]
For small concentration of buffering agent and
maximum buffer capacity
𝑆𝑎𝑙𝑡
𝑎𝑐𝑖𝑑
≅1
 pH = pKa
Soln. 17(d)
A2(g) + B2(g) ⇌ 3C(g) ⇌ D(g)
step – 1 step – 2
since the step 1 and 2 are exothermic hence low
temperature will favour both the reactions. In step – 1
moles are increasing hence low pressure will favour
it. In step – 2 moles are decreasing, hence high
pressure will favour it.
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Solution of Chemical Equilibrium
Soln. 18(d)
For the reaction
XeF6(g) + H2O(g) ⇌ XeOF4(g) + 2HF(g)
K1 =
𝑋𝑒𝑂𝐹4 𝐻𝐹 2
…(1)
𝑋𝑒𝐹6 𝐻2 𝑂
and for the reaction
XeO4(g) + XeF6(g) + XeO3F2(g)
K2 =
𝑋𝑒𝑂𝐹4 𝑋𝑒𝑂 3 𝐹2
…(2)
𝑋𝑒𝑂 4 𝑋𝑒𝐹6
Since the value of pH > 7, therefore the solution is
basic.
Soln. 23(a)
MgCO3(s) → MgO(s) + CO2(g)
MgO & MgCO3 are solid and they do not exert any
pressure and hence only pressure exerted is by CO2.
Therefore KP = PCO 2
For reaction:
K=
Soln. 24(b)
Because NH3 after losing a proton (H+) gives NH2−
NH3 + H2O ⇌ NH2− + H3O+
(Conjugate acid – base pair differ only by a proton)
𝑋𝑒𝑂 3 𝐹2 𝐻2 𝑂
𝑋𝑒𝑂 4 𝐻𝐹 2
 From eq. no. (1) and (2)
K = K2 / K1
Soln. 19(c)
Lewis acid is that compound which have electron
deficiency. Eg. BF3, SnCl2.
Soln. 20(d)
4
No. of moles of NaOH = 40 = 0.1
[Molecular weight of NaOH = 40]
No. of moles of OH – = 0.1
–
0.1
Soln. 25(b)
In polyprotic acids the loss of second proton occurs
much less readily than the first. Usually the Ka values
for successive loss of proton from these acids differ
by at least a factor of
10 – 3 i.e. K a 1 > Ka 2
H2X ⇌ H+ + HX – K a 1
HX – ⇌ H+ + X 2 – K a 2
Concentration of OH = 1 𝑙𝑖𝑡𝑟𝑒 = 0.1 Moles / L
As we know that, [H+] [OH –] = 10 – 14
 [H+] = 10 – 13 ∵ 𝑂𝐻 − = 10−1
Soln. 21(d)
BA2 → B + 2A
x 2x
Solubility product = [x] [2x]2 = 4x3
4 × 10 – 12 = 4x3 or x =
 x = 10
3 4 × 10 −12
4
–4
Soln. 22(a)
Given: Hydroxyl ion concentration
[OH –] = 0.05 mol L – 1. We know that the [H+][OH –]
= 1 × 10 – 14
Or [H+] =
1 × 10 −14
0.05
Soln. 26(b)
CH3COOH is weak acid while NaOH is strong base,
so one equivalent of NaOH can not be neutralized
with one equivalent CH3COOH. Hence the solution
of one equivalent of each does not have pH value as
7. Its pH will be towards basic side as NaOH is a
strong base hence conc. of OH – will be more than the
conc. of H+.
= 2 × 10 – 13 mol L – 1
We also know that
pH = – log [H+] = – log[2 × 10 – 13] = – log 2 – log 10
– 13
= – log 2 – (– 13) log 10
= – 0.3010 + 13.0000 = 12.6990.
Soln. 27(a)
M2S ⇌ 2 M+ + S – –
2x
x
Solubility product = (2x)2 (x) = 4x3
= 4 (3.5 × 10 – 6)3 = 1.7 × 10 – 16
Soln. 28(c)
CH3COOH ⇌ CH3COO – + H+
Ka =
𝐶𝐻3 𝐶𝑂𝑂 − 𝐻 +
𝐶𝐻3 𝐶𝑂𝑂𝐻
Given that,
[CH3COO –] = [H+] = 3.4 × 10 – 4 M
Ka for CH3COOH = 1.7 × 10 – 5
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Solution of Chemical Equilibrium
CH3COOH is weak acid, so in it [CH3COOH] is
equal to initial concentration. Hence
1.7 × 10 – 5 =
3.4 × 10 −4 3.4 × 10 −4
[CH3COOH] =
[CH3COOH] =
𝐶𝐻3 𝐶𝑂𝑂𝐻
3.4 × 10 −4 ×3.4 × 10 −4
𝐶𝐻3 𝐶𝑂𝑂𝐻
3.4 × 10 −4 ×3.4 × 10 −4
1.7 × 10 −5
= 6.8 × 10 – 3 M
Soln. 29(c)
Strong bases higher tendency to accept the proton.
Increasing order of base and hence the order of
accepting tendency of proton is
I – < HS – < NH3 < RNH2
Soln. 30(c)
For the reaction
BaO2(s) ⇌ BaO(s) + O2(g); ∆H = + ve.
At equilibrium kp = PO 2
[For solid and liquid concentration term is taken as
unity]
Hence, the value of equilibrium constant depends
only partial pressure of O2. Further on increasing
temperature formation of O2 increasing as this is an
endothermic reaction.
Soln. 31(d)
Given s = 0.5 × 10 – 4 moles / lit
[MX2 ⇌ M2+ + 2X –]
∵ For MX2, Ksp = s × (2s)2 = 4s3
Ksp = 4 × (0.5 × 10 – 4)3 = 4 × 0.125 × 10 – 12
= 0.5 × 10 – 12 = 5 × 10 – 13
𝑆𝑎𝑙𝑡
 14 – pH – log 𝐵𝑎𝑠𝑒 = pKb
0.1
14 – 9.25 – log0.1 = pKb
14 – 9.25 – 0 = pKb
pKb = 4.75
Soln. 34(d)
For reaction to proceed from right to left
Q
>
Kc
(backward rate)
(forward rate) i.e. the
reation will be fast in backwared direction i.e r b > rf.
Soln. 35(c)
The higher is the tendency to donate proton, stronger
is the acid. Thus the correct order is R – COOH >
HOH > R – OH > CH ≡ CH depending upon the rate
of donation of proton.
Soln. 36(b)
B(OH) does not provide H+ ions in water instead it
accepts OH – ion and hence it is Lewis acid.
B(OH)3 + H2O ⇌ [B(OH)4] – + H+
Soln. 37(c)
Ksp for AgI = I × 10 – 16
In solution of KI, I – would be due to the both AgI and
KI, 10 – 4 solution KI would provide = 10 – 4 I –
AgI would provide, say = x I – (x is solubility of AgI)
Total I – = ( 10 – 4 + x), Ksp of AgI = (10 – 4 + x)x
 Ksp = 10 – 4 x + x2
as x is very small
 x2 can be ignored
 10 – 4 x = 10 – 16
or x =
10 −16
10 −4
= 10 – 12 (mol – 1)
Soln. 32(b)
Na2CO3 is a salt of weak acid H2CO3 and strong base
NaOH, therefore, its aqueous solution will be basic
hence has pH more than 7.
Na2CO3 + 2H2O → 2NaOH + 2H2CO3
Strong base weak base
(solubility)
Soln. 33(b)
Soln. 39(d)
For an acid – base indicator
Hln ⇌ H+ + ln –
𝑆𝑎𝑙𝑡
pOH = pKb + log 𝐵𝑎𝑠𝑒
𝑠𝑎𝑙𝑡
or pKb = pOH – log 𝐵𝑎𝑠𝑒
but pOH + pH = 14
Soln. 38(c)
For AX2; Ksp = 4s3  3.2 × 10 – 11 = 4 s3
or s = 3
 Kln =
or pOH = 14 – pH
3.2 × 10 −11
4
𝐻 + 𝑙𝑛 −
[𝐻𝑙𝑛 ]
= 2 × 10−4
𝐻𝑙𝑛
or [H+] = Kln × [𝑙𝑛 − ]
𝐻𝑙𝑛
or log H+ = log Kln + log 𝑙𝑛 −
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Solution of Chemical Equilibrium
Total [H+] = 10 – 7 + 10 – 8 = 10 × 10 – 8 + 10 – 8
10 – 8 (10 + 1) = 11 × 10 – 8
Taking negative on both sides
𝐻𝑙𝑛
– log [H+] = – log Kln – log 𝑙𝑛 −
ln⁡−
or we can write pH = pKln + log [𝐻𝑙𝑛 ]
or log
𝑙𝑛 −
𝐻𝑙𝑛
= pH – pKln
Soln. 40(a)
IVth group needs higher S 2 – ion concentration. In
presence of HCl, the dissociation of H2S decreases
hence produces less amount of sulphide ions due to
common ion effect, thus HCl decrease the solubility
of H2S which is sufficient to precipitate II nd group
radicals.
Soln. 41(a)
The solution formed from isomolar solutions of
sodium oxide, sodium sulphide, sodium selenide
H2O, H2Se & H2Te respectively. As the acid strength
increases from H2O to H2Te thus pH decreases and
hence the correct of pH is
pH1 > pH2 > pH3 > pH4.
Soln. 42(d)
Given Kb = 1.0 × 10 – 12
[BPH] = 0.01 M
[OH] = ?
t=0
BOH ⇌ B+ + OH – 1
t
c
0
0
eq
c(1 – x) cx
cx
Kb = 𝑐
𝑐 2𝑥 2
1−𝑥
𝑐𝑥 2
= (1−𝑥)
0.01𝑥 2
 1.0 × 10 – 12 = 0.01(1−𝑥)
On calculation, we get, x = 1.0 × 10 – 5
Now [OH –] = cx
= 0.01 × 10 – 5
= 1 × 10 – 7 mol L – 1
Soln. 45(d)
HNO2 is a weak acid and strong base (NaOH).
Soln. 46(a)
Given Ka = 1.00 × 10 – 5, C = 0.100 mol for a weak
electrolyte, degree of dissociation
() =
𝐾𝑎
𝐶
=
1 × 10 −5
0.100
= 10 – 2 = 1%
Soln. 47(d)
Given,
N2 + 3H2 ⇌ 2NH3; K1
N2 + O2 ⇌ 2NO; K2
…(i)
…(ii)
1
…(iii)
H2 + 2 O2 ⇌ H2O; K3
We have to calculate
4NH3 + 5O2 → 4NH + 6H2O; K = ?
or 2NH3 + 5O2 → 2 NO + 3H2O
For this equation, K =
but K1 =
& K3 =
𝑁𝐻3 2
𝑁2 𝐻2
=
=
𝑁2 𝑂2
𝑁𝐻3 2 𝑂5 5/2
𝐾 =
𝐻2 𝑂 3
𝐻2 3 𝑂2 3/2
𝐾2 .𝐾33
𝐾1
𝐻2 𝑂 3
.
𝐻2 3 𝑂2 3/2
𝑁𝑂 2 𝐻2 𝑂 3
𝑁2 𝑂2
𝑜𝑟 𝐾3 =
𝐻2 𝑂2 1 2
×
𝑁𝑂 2
, 𝐾2 =
𝐻2 𝑂
Now operate,
𝑁𝑂 2
3
𝑁𝑂 2 𝐻2 𝑂 3
𝑁𝐻3 2 𝑂2 5/2
𝑁2 𝐻2 3
𝑁𝐻3 2
=K
𝐾2 .𝐾33
𝐾1
Soln. 48(a)
Given [H2O+] = 1 × 10 – 10 M
at 25 [H3O+] [OH –] = 10 – 14
10 −14
Soln. 43(a)
First option is incorrect as the value of KP given is
wrong. It should have been
KP =
𝑃 𝐶𝑂 2
𝑃 𝐶𝐻 4 × 𝑃 𝑂 2
 [OH –] = 10 −10 = 10−4
Now, [OH –] = 10−p
 pOH = 4
OH
= 10 – 4 = 10−p
CH
2
Soln. 44(a)
For a solution of 10 – 8 M HCl [H+] = 10 – 8 [H+] of
water = 10 – 7
Soln. 49(b)
[H3O]+ for a solution having pH = 3 is given by
[H3O]+ = 1 × 10 – 3 moles / litre
[ [H2O]+ = 10 – pH]
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Solution of Chemical Equilibrium
Similarly for solution having pH = 4,
[H3O]+ = 1 × 10 – 4 moles / litre and for pH = 5
[H3O+] = 1 × 10 – 5 moles / litre
Let the volume of each solution in mixture be IL,
then total volume of mixture solution L = (1 + 1 + 1)
L=3L
Total [H3O]+ ion presnt in mixture solution = (10 – 3 +
10 – 4 + 10 – 5) moles
Then [H3O]+ ion concentration of mixture solution
=
10 −3 + 10 −4 + 10 −5
3
𝑀=
0.00111
3
M
= 0.00037 M = 3.7 × 10 – 4 M.
Since x is very small so can be neglected in
denominator
Thus, we get
PAB 2 = (1 – x) × P
PAB = x × P
𝑥
PB 2 = × 𝑃
2
=
K=
𝐹𝑒 3+ 𝑂𝐻 − 3
𝐹𝑒 𝑂𝐻 3
= (Fe3+) (OH –)3
[ [solid] = 1]
If (OH –) is decreased by
1
4
times then for reaction
equilibrium constant to remain constant, we have to
increase the concentration of [Fe3+] by a factor of 43
i.e 4 × 4 x= 64. Thus option (c) is correct answer.
Soln. 51(b)
The highest pH will br recorded by the most basic
solution. The basic nature of hydroxides of alkaline
earth metals increase as we move from Mg to Ba and
thus the solution of BaCl2 in water will be most basic
and so it will have highest pH.
𝑥
𝑥 2 × 𝑃 2 .𝑃×
2
1−𝑥 2 ×𝑃 2
𝑥 3 .𝑃 3
[ 1 – x ≃1]
= 2 ×1 × 𝑃 2
=
Soln. 50(c)
For this reaction Keq is given by
𝑃 𝐴𝐵 2 𝑃 𝐵 2
2
𝑃 𝐴𝐵 2
Now, KP =
𝑥 3 .𝑃
2
1
Kc =
𝐴𝐵 2 𝐵2
𝐴𝐵2 2
or Kc =
2𝑥 2 ×𝑥
2 1−𝑥 2
3
= x [(1 – x) can be neglected in denominator (1 – x)
≃ 1]
The partial pressures at equilibrium are calculated on
the basis of total number of moles at equilibrium.
Total number of moles
= 2 (1 – x) + 2x + x = (2 + x)
 PAB 2 =
2𝑥
2(1−x)
(2+x)
× P where P is the total pressure.
𝑥
PAB = (2+𝑥) × 𝑃, 𝑃𝐵2 = (2+𝑥) × 𝑃
𝑃
𝑜𝑟 𝑥 =
2𝐾𝑃 3
𝑃
Soln. 53(c)
Given reaction are
X⇌Y+Z
…(i)
and A ⇌ 2 B
….(ii)
Let the total pressure for reaction (i) and (ii) be P 1
and P2 respectively, then
𝐾𝑃 1
𝐾𝑃 2
9
=1
(given)
After dissociation,
X⇌Y+Z
At equilibrium (1 – )  
[Let 1 mole of X dissociate with  as degree of
dissociation]
Total number of moles = 1 –  +  + 
= (1 + )
1− 
Thus PX =
Soln. 52(b)
For the reaction
2AB2(g) ⇌ 2AB(g) + B2(g)
At equi. 2(1 – x)
2x
x
2.𝐾𝑃
𝑜𝑟 𝑥 3 =
PZ =
. P1; PY =
1+

P1 ;
.P1
1+ 

 K P1 =
. P1 ×
1+ 
….(i)
Similarly for
At equilibrium
We have,
2𝑃2 2
KP2 =

1+ 
1+ 

1+ 
. 𝑃1 /
1− 
1+ 
. 𝑃1
A ⇌ 2B
(1 – )
2
1− 
/
1+ 
𝑃2
….(ii)
Dividing (i) by (ii), we get
𝐾𝑃 1
𝐾𝑃 2
2 .𝑃
𝐾
1 𝑃
= 4 2 .𝑃1 𝑜𝑟 𝐾𝑃 1 = 4 . 𝑃1
2
1 𝑃
or 9 = 4 . 𝑃1
2
𝑃1
or 𝑃 =
2
36
1
𝑃2

𝐾𝑃 1
𝐾𝑃 2
2
9
=1
or P1: P2 = 36:1.
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Solution of Chemical Equilibrium
i.e. Option (c) is correct answer.
Soln. 54(c)
Given, CH3COOH ⇌ CH3COO – + H+; K a 1 = 1.5 × 10
–5
……(i)
HCN ⇌ H+ CN – ⇌ HCN;
K a 2
1
=𝐾
𝑎2
1
….(ii)
= 4.5 × 10 −10
 From (i) and (ii), we find that the equilibrium
constant (Ka) for the reaction,
CN – + CH3COOH ⇌ CH3COO – + HCN, is
Ka = K a 1 × K a 2
=
1.5 × 10 −5
1
4.5 × 10 −10
= × 105 = 3.33 × 104.
3
Soln. 55(d)
(CH3)3 B – is an electron deficient, thus behave as a
lewis acid.
Soln. 56(d)
Ammonium chloride is a salt of weak base and strong
acid. In this case hydrolysis constant Kh can be
calculated as
Kh =
𝐾𝑤
𝐾𝑏
1 × 10 −14
= 1.77 × 10 −5 = 5.65 × 1010
Soln. 57(d)
−
Ba(OH)2(s) → Ba2+
aq + 2OH aq
pH = 12 or pOH = 2
[OH –] =10 – 2 M
Ba(OH)2 → Ba2+
+ 2OH –
0.5 × 10 – 2
10 – 2
2+
[ Concentration of Ba is half of OH –]
Ksp = [Ba2+][OH –]2
= [0.5 × 10 – 2][1 × 10 – 2]2
= 0.5 × 10 – 6 = 5 × 10 – 7 M3
Soln. 59(d)
2C(s) + O2(g) ⇌ 2 CO2(g)
∆n = 2 – 1 = + 1
 Kc and Kp are not equal.
Soln. 60(d)
Kb = 10 – 10 ; Ka = 10 – 4 or pKa = 4
For the buffer solution containing equal
concentration of B – and HB
pH = pKa + log 1
pH = pKa = 4
The octahedral complex ion [COCl2(NH3)4]+ i.e. tetra
amminedichloro cobalt (III) ion exists as cis and trans
isomers.
Soln. 61(c)
Boron in B2H6 is electron deficient.
Soln. 62(c)
2A(g) + B(g) ⇌ 3C(g) + D(g)
2
1
3
1
1
1
0
0
Mole ratio
Molar conc. at
t=0
Equilibrium
0.50
Molar concentration
Kc =
𝐶 3 𝐷
𝐴2 𝐵
=
0.75
0.75 0.25
0.75 3 0.25
0.50 2 0.75
Soln. 63(b)
Given [NH3] = 0.3 M, 𝑁𝐻4+ = 0.2 M, Kb = 1.8 × 10
–5
.
pOH = pKb + log
𝑠𝑎𝑙𝑡
𝑏𝑎𝑠𝑒
[pKb = – log Kb ; pKb = – log
–5
1.8 × 10 ]
 pKb = 4.74
0.2
= 4.74 + log 0.3 = 4.74 + 0.3010 – 0.4771 = 4.56
PH = 14 – 4.56 = 9.436
Soln. 58(d)
pH = p Ka + log
𝑆𝑎𝑙𝑡
𝐴𝑐𝑖𝑑
Log[H+] = log Ka – log
+
Lod[H ] = log Ka + log
+
[H ] = Ka
𝑆𝑎𝑙𝑡
Soln. 64(c)
BF3 behaves as lewis acid.
𝐴𝑐𝑖𝑑
𝐴𝑐𝑖𝑑
𝑆𝑎𝑙𝑡
𝐴𝑐𝑖𝑑
𝑆𝑎𝑙𝑡
0.1
= 1.8 × 10 – 5 × 0.2 = 9 × 10 – 6 M
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Solution of Chemical Equilibrium
Ksp = [Pb+2][Cl –]2
1.7 × 10 – 5 = [Pb+2][0.1]2
[Pb+2] = 1.7 × 10 – 3 M.
Soln. 65(c)
Ksp = [Ag+][Cl –]
1.8 × 10 – 10 = [Ag+][0.1]
[Ag+] = 1.8 × 10 – 9 M
AIIMS
Soln. 1(c)
K does not depend upon the initial concentration of
the reactants.
Soln. 2(b)
Apply Le – Chatelier’s principle.
Soln. 3(d)
K for the reaction is given by K = [CO2]
Because concentration of solids taken to be unity.
Since [CaCO3] and [CaO] do not figure in the
expression. There will be no effect of addition of
CaCO3.
Soln. 4(c)
The given reaction, in the forward direction, is
endothermic and proceeds with an increase in number
of moles gaseous constituents. As such high
temperature and low pressure will favour the forward
process. An increase in concentration of I or decrease
in concentration of I2 will shift the equilibrium in the
backward direction.
Subjective Questions From IITJEE
Soln. 1
Conc. of NaOH = 0.1 M
Conc. of Ha = 0.1 M
NaOH + HA → NaA + H2O
Since, at the end point all the acid and base neutralize
each other and form the salt.
Conc. of NaA =
0.1
2
= 0.05 M
⊖
(As volume of NaA doubled)
A⊝ + H2O ⇌ HA + OH ⊖
HA ⇌ H + A⊖
Ka =
𝐻 + × [𝐴]⊖
[𝐻𝐴]
= 5.6 × 10 – 5
– log [H+] = log
= – log
𝐾𝑤 × 𝐾𝑎
𝑐
= – log
𝐾𝑤 × 𝐾𝑎
– log [H+] = pH = 9
Soln. 2
Molal elevation in boiling point constant is the
measure of strength of intermolecular forces. If Kb is
high, the bond is polar and stronger dipole – dipole
interaction leads to higher boiling point. Thus higher
the value of Kb more is the boiling point.
X
Y
Z
Boiling Point
100C
27C
253C
Kb
0.68
0.53
0.98
𝑐
1 × 10 −14 ×5.6 × 10 −5
0.05
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Solution of Chemical Equilibrium
10 – 5 = Conc. =
Soln. 3
No. of moles of acetic acid =
500 × 0.2
1000
500 × 0.2
No. of moles of HCl =
1000
= 0.1 M
= 0.1 M
Since, HCl is a strong acid and due to common ion
effect, the dissociation of CH3COOH will be
CH3COOH ⇌ H (aq) + Cl⊖
Initial conc.
c
0
0
 = degree of
c
c
c
ionization
at equilibrium c – c c + 0.1 M
c
( = Degree of dissociation and is small)
Ka =
Ka =
𝐻 + ×[𝐶𝐻3 𝐶𝑂𝑂 − ]
[𝐶𝐻3 𝐶𝑂𝑂𝐻]
=
𝑐  +0.1 ×(𝑐 )
𝑐 (1− )
𝑐 2 + 0.1
(1− )
As  is small, 1 –   1 neglecting c2
Ka = 0.1
𝐾
 = 0.1𝑎 =
1.75 × 10 −5
0.1
= 1.75 × 10 – 4
[H+] = c + 0.1
Since c is very small.
[H+] = 0.1 M or pH = 1
(ii) Amount of NaOH added = 6g
𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑆𝑂2 𝑖𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1.3653
Moles of SO2 in solution = 1.365 × 10 – 5
1
pH = 2 (pKa – log c)
1
pH = 2 [1.92 – (log 1.365 × 10 – 5)]
1
= 2 (1.92 + 5 – log 1.365)
1
= 2 (1.92 + 5 – 0.135)
1
1
= 2 (6.92 – 0.135) = 2 (6.7849)
pH = 3.3925
Soln. 5
Pb(OH)2 ⇌ Pb +2 + 2OH ⊖
S
S
2S
Let S be the solubility
Ksp = Pb(OH)2 = 4S3
= 4 × (6.7 × 10 – 6)3 = 1.203 × 10 – 15
pH = 8
pOH = 14 – 8 = 6
[OH –] = 10 – 6
Pb +2 × [OH –]2 = Ksp
[Pb +2] =
1.203 × 10 −15
(10 −6 )2
= 1.203 × 10 – 3 mol / litre
6
No. of moles of NaOH = 40 = 0.15 M
NaOH is a strong base and HCl is a strong acid. They
neutralize each other
0.1 M NaOH = 0.1 M HCl
Conc. of NaOH left = 0.15 – 0.1 = 0.05 M
NaOH + CH3COOH ⇌ CH3COONa +
H2O
0.05 0.1 M
0.05 M
0.05
Conc. left 0
0.05 M
0.05 M
0.05
They form a buffe of sod. acetate + CH3COOH
Apply Henderson’s equation
[𝑆𝑎𝑙𝑡 ]
pH = pKa + log [𝐴𝑐𝑖𝑑 ]
= − log 1.75 × 10
−5
(log 1 = 0)
0.05
+ log 0.05
pH = 5 – log 1.75 = 4.75
Soln. 4
SO2 + H2O → H2SO3
Solubility of SO2 in water at 298 K = 1.3653 mol /
litre
Average conc. of SO2 in atmosphere = 10 ppm
= 10 × 10 – 6 = 10 – 5 mol / litre.
Soln. 6
Amount of NH4HS = 3.06 g
Mol. wt of NH4HS = 14 + 4 + 32 +1 = 51
No. of moles of NH4HS =
3.06
51
= 0.06 mol
Conc. of NH4HS in solution =
0.06
2
= 0.03 mol / litre
whose volume is 2 litre.
NH4HS is 30% dissociated.
Conc. of dissociated NH4HS
30
= 100 × 0.03 = 9 × 10 – 3 = 0.009 mol / litre
NH4HS(s) ⇌ NH3 (g) + H2S(g)
Kc =
𝑁𝐻3 ×[𝐻2 𝑆]
[𝑁𝐻4 𝐻𝑆 ]
= [NH3] × [H2S]
–5
Kc = 8.1 × 10 = 0.009 × 0.009
∆n = No. of moles of gaseous products – No. of
moles of gaseous reactants
=2–0=2
T = 27C = 300K
Kp = Kc × (RT)∆n = Kc × (RT)2
Kp = 8.1 × 10 – 5 × (0.082 × 300)2 = 4.9 × 10 – 2 atm
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Solution of Chemical Equilibrium
Since, NH4HS is solid , thus addition of solid NH4HS
is involved neither in Kc nor in Kp. Hence, it has no
effect.
Soln. 7
Volume of HCl of pH (2) = 200 mL
Volume of NaOH = pH (12) = 300 mL
Since, pH = 2, [H+] = 0.01 M
pH = 12, [OH] = 10 – 2 M = 0.01 M
So 200 mL of HCl of 0.01 M will be neutralized by
200 mL of NaOH of same molarity (0.01 M)
Volume of NaOH left Molarity = 300 – 200 = 100
No. of millimoles of NaOH = 100 × 10 – 2 = 1.0
Total volume = 500 mL
1
–
[OH ] = 500 = 2 × 10
–3
M
pOH = 3 – log 2 = 3 – 0.03010 = 2.7
pH = 14 – 2.7 = 11.3010
𝐴𝑔 + × [𝑁𝐻3 ]2
[𝐴𝑔 (𝑁𝐻3 )2 ]+
= 6.2 × 10 – 8
Ag + 2NH3 ⇌ [Ag(NH3)2]
1
1
K = 𝐾 = 6.2 × 108 =
+
[Ag ] =
10 8
6.2
=
[𝐴𝑔 (𝑁𝐻3 )2 ]⊕
[𝐴𝑔 ]+ [𝑁𝐻3 ]2
[𝐴𝑔 (𝑁𝐻3 )2 ]⊕
10 8
×[𝑁𝐻3 ]2
6.2
Since, K is very high, whatever Ag + is formed from
dissociation of AgCl changes to [Ag(NH3)2].
Hence, [Cl –] = [Ag(NH3)2]+ formed = c moles each
Let the conc. of 𝐶𝑙 ⊖ formed = c mole / litre
Ksp of AgCl = [Ag +] × [Cl –]
Ksp of AgCl =
[𝐴𝑔 (𝑁𝐻3 )2 ]+
10 8
× [𝑁𝐻3 ]2
6.2
= 1.8 × 10 – 10 10 8
6.2
1.8 × 10
– 10
×
10 8
6.2
× 𝑐 𝑜𝑓 𝐶𝑙 ⊖
6.2
Log [H+] = log Ki + log
=
1.8 × 10 −2
6.2
= 0.29 × 10 – 2
c = (0.29 × 10−2 ) = 0.0538 M
Soln. 9
HIn ⇌ 𝐻 ⊕ + 𝐼𝑛⊝
Km = [H+] × [In –] / [HIn]
𝐻𝐼𝑛
𝐼𝑛 −
[𝐼𝑛 −]
– log [H+] = – log Ki + log
pH = Ki + log
[𝐻𝐼𝑛 ]
𝐼𝑛 −
𝐻𝐼𝑛
To get the sharp colour change that must be visible to
naked eye.
[𝐼𝑛 −]
= ±10
[𝐻𝐼𝑛 ]
[𝐼𝑛 −]
If [𝐻𝐼𝑛 ] = + 10
pH1 = pKi + log 10 = 5 + 1 = 6
[𝐼𝑛 −]
when [𝐻𝐼𝑛 ] = −10
pH2 = pKi – log 10 = 5 – 1 = 4
pH1 = 6, pH2 = 4.
8.2 × 10 −12
[Ag+]2 =
1.5
[Ag+] = 2.338 × 10 – 6 = 2.34 × 10 – 6 mol / litre
𝐶𝑙 ⊖ = 0.0026 g
Atomic wt. of Cl = Mol. wt. of Cl = 35.5
[Cl –] =
0.0026
35.5
= 7.32 × 10 – 5 M
Ksp of AgCl = [Ag+] × [Cl –]
= 2.34 × 10 – 6 × 7.32 × 10 – 5
= 1.71 × 10 – 1 0
Soln. 11
N2(g) + O2(g) ⇌ 2NO(g)
Kc =
× [𝑁𝐻3 ]2 = c2
1.8 × 10 −10 × 10 8
[𝐼𝑛 ]
Taking logarithm of eqn. (2)
𝑐 ×𝑐
×[𝑁𝐻3 ]2
[NH3] = 1 M
c2 =
[𝐻𝐼𝑛 ]
Soln. 10
Ag2CO3 ⇌ 2Ag (aq) + CO−2
3 (aq)
Ksp = [Ag+]2 × [CO−2
]
=
8.2
× 10 – 12
3
But [CO−2
3 ] = 1.5 given
Soln. 8
[Ag(NH3)2]+ → Ag+ + 2NH3
K=
H = Ki ×
[𝑁𝑂]2
𝑁2 ×[𝑂2 ]
N2(g) + O2(g) ⇌ 2NO(g)
Let the total no. of moles of O2 + N2 in air = 100
Let the no. of moles of N2 = a
N2(g) + O2(g) ⇌ 2NO(g)
Initial
a
100 – a
0
Let x of
x
x
2x
N2 combines
Conc. left
a – x 100 – a – x 2x
At equilibrium mole % of NO is 1.8
2𝑥
100
1.8
= 100 or x = 0.9
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Solution of Chemical Equilibrium
Kc =
(2𝑥)2
𝑎−𝑥 (100 –𝑎−𝑥)
1.8 ×1.8
𝑎−𝑥 (100−𝑎−𝑥)
=
= 2.1 × 10 – 3
1.8 ×1.8
𝑎−0.9 (100 – 𝑎−0.9)
= 2.1 × 10 – 3
 a ≃ 7.9
Thus % of N2 in air = 79%
O2 = 100 – 79 = 21%
Soln. 12
Conc. of HCO⊖
3 = 183 ppm = 183 g / mL
= 183 mg / litre
Amount of HCO⊖
3 present in 1000 kg of water (=
1000 L)
=
183 ×1000
1000
= 183 g / 1000 kg of water
Mol. wt. of HCO⊖
3 = 61
No. of moles of HCO⊖
3 =
183
61
= 3 moles of HCO⊖
3
CaO + H2O → Ca(OH)2
Ca(OH)2 + Ca(HCO3)2 → 2CaCO3 + 2H2O
From equation thus one mole of CaO requires 2
moles of HCO⊖
3 ion.
No. of moles of CaO = 1.5 mol
−2
After removal of HCO⊖
3 , the solution contains SO3
2H2O + CaSO4 → Ca(OH)2 + H2SO4
96
= No. of moles Ca +2 = 40 = 2.4 moles
+2
Conc. of Ca in solution = 96 ppm
CaSO4 + 2H2O ⇌ Ca(OH)2 + H2SO4
Conc. of Ca +2 ≡ Conc. of (H +) = 2.4 mol / 1000 L
2.4
= 1000 = 2.4 × 10 – 3 mol / litre
pH = – log (H +) = – log (2.4 × 10 – 3)
= 3 – log 2.4 = 3 – 0.38 = 2.62
Soln. 13
NaCN is a salt of a week acid (HCN) and a strong
base (NaOH)
NaCN + H2O ⇌ NaOH + HCN
Or
CN ⊖ + H2O ⇌ 𝑂𝐻⊖ + 𝐻𝐶𝑁
Initial conc.
c
0
0
0
h = Degree of
ch
ch
ch
hydrolysis
Conc. left
c – ch
ch
ch
Kb =
Kb =
𝑂𝐻 − × [𝐻𝐶𝑁 ]
[𝐶𝑁 − ]
𝑂𝐻 − × [𝑂𝐻 − ]
[𝐶𝑁 − ]
𝑐𝑕 ×𝑐𝑕
= 𝑐 (1−𝑕)
=
[𝑂𝐻 −]2
….(i)
[𝐶𝑁 − ]
Taking logarithm of eqn. (1)
log Kb = log [OH –]2 – log [CN –]
– log Kb = – log [OH]2 + log [CN –]
pKb = – 2 log (OH –) + log (0.5)
4.70 = – 2 log (OH –) + log (0.5) = 2 pOH + log 0.5
2 pOH = 4.70 log 0.5 = 4.70 + 0.30
pOH = 2.5
[OH –] = 10 – 2.5
𝐾
[H +] = [𝑂𝐻𝑤− ] =
1 × 10 −14
10 −2.5
= 10 – 11.5
pH = 11.5
Soln. 14
Ammonium acetate is a salt of a weak acid and a
weak base
1
pH = (pKw + pKa – pKb) [pKa = 3.8, pKb = 4.8]
2
1
= 2 [14 + 3.8 – 4.8] = 6.5
Soln. 15
[Ag(CN)2]⊖ ⇌ 𝐴𝑔+ + 2𝐶𝑁 ⊖
Conc. of [Ag +] = Conc. of AgNO3 = 0.03 M
1 mole of Ag + requires two moles of CN ⊖
Then, CN ⊖ needed = 2 × 0.03 = 0.06 M
Actual conc. of CN ⊖ = 0.1 M
Conc. of CN ⊖ left = 0.1 – 0.06 = 0.04
[Ag(CN)2]⊖ ⇌ 𝐴𝑔+ + 2𝐶𝑁 ⊖
4.0 × 10 – 19 = K =
𝐴𝑔 + × [𝐶𝑁 − ]2
[𝐴𝑔 (𝐶𝑁2 )⊖
[Ag +] = 4.0 × 10 – 19 ×
=
[𝐴𝑔 (𝐶𝑁2 )]⊖
[𝐶𝑁 − ]2
4.0 × 10 −19 × 0.03
(0.04)2
[Ag ] = 7.5 × 10 – 19 M
+
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Page 41