<H1>Additional Objective Questions

<H2>Single Correct Choice Type

1.(C) Physical equilibrium do not include any chemical change.

2.(D) The equilibrium constant K for the reaction

N2 + 3H2 2NH3 is 1.64 10−4 atm at 400C.

3.(D) Since, the product in reaction given in option (D) is not much stable, therefore, the reaction is

reversible.

4.(A) For reaction2 22NO (g) 2NO(g) O (g)+ ; ng > 0

So, if volume increases, pressure decreases and n decreases, therefore, the reaction will move in the

direction where n increases, that is in the forward direction which in turn will favor the formation

of products.

5.(B) H2(g) + I2(g) 2HI(g)

Applying law of mass action, 2

c

2 2

[HI]=

[H ][I ]K

Given that [H2] = 8.0 mol L−1, [I2] = 3.0 mol L−1 [HI]=28.0 mol L−1

2

c

(28.0)32.66

(8.0) (3.0)K = =

6.(C) According to Le Chatelier’s principle, for an exothermic reaction, on lowering the temperature will

shift the reaction in forward direction. Also, as the number of gaseous moles is decreasing, so on

increasing the pressure, the reaction will shift in forward direction.

7.(A) For first reaction,

N2 + O2 2NO; K1

On reversing reaction, 2 2

1

12NO N O ;

K+

For second reaction 1/2

2 2 2

1

1 1NO N O ;

2K

K

+ =

Therefore,

1/2

2

1

1K

K

=

2

2

1

1K

K

=

Or 1 2

2

1K

K

=

8.(D) We know

2

o o o

NO NGG G G = − ( ) 1 152 87 kJmol 35kJmol− −= − = −

Since,o 2.303 log = −G RT K

1

1

35kJ mollog

8.301kJ mol 298K

−

−

−=−

K

On solving, we get

K = 1.365 106

9.(A) For + =2 2  3 C C

1SO O SO             (1/ );

2K K

For + = 2

2 2 3 C C2SO O 2SO           (1/ )K K

Therefore, 2

C 2

1

4.9 10K

−

=

410 10000     

4.9 4.9 24.01= =

 4.1649 100   416.49= =

10.(D) KC is a characteristic constant for the given reaction.

11.(C) This is because according to Le Chatelier’s principle, increase in H+ ions will shift the equilibrium

in the forward direction.

12.(B) We have

2 2CO Cl COCl

Initialconc. 2mol 3mol

Final conc. 1mol (3 1) 2mol

+

− =

Equilibrium constant can be calculated as 2C

2

[COCl ]

[CO][Cl ]K =

Therefore, ( )

( ) ( )C

1mol / 5L2.5

1mol / 5L 2mol / 5LK = =

13.(B) From the given reaction, we have

2

4 12

p H O 6.89 10K p −= = therefore 2

3

H O 1.62 10 atmp −=

14.(B) 2 2 32SO (g) O (g) 2SO (g)+

3

2 2

2 2SO

p 2 2

SO O

[ ] (0.331)= = =3.47

[ ] [ ] (0.559) (0.101)

pK

p p

SO2 and SO3 have same number of moles; their partial pressure will be equal

3 2 2SO SO SO

1, therefore 0.288atm

3.47p p p= =

15.(A) The answer should be

2A (g) 2A(g)

Initial moles 1 0

Moles at equilibrium 1 0.2 0.4−

So, 2A(g) A (g)

0.40.33 and 1 0.33 0.67

1.2x x= = = − =

Now,

2 2

2 2 2

A(g) A(g)

total

A (g) A (g)

( ) ( ) (1)(as 1)

( ) (1)p

p xK p

p x

= = =

Thus, we get Kp = 1/6

Now, 1ln 8.314 320 ln(6) 4766 J molpG RT K − = =− =−

16.(D) Since, equilibrium constant depends on temperature only, therefore, K remains unchanged but

degree of dissociation changes.

17.(D) For the given reaction

CH3COOH + C2H5OH CH3COOC2H5 + H2O

1 1 0 0

1 − 0.666 1 − 0.666 0.666 0.666

The equilibrium constant is

3 2 5 2

3 2 5

[CH COOC H ][H O] [0.666][0.666]

[CH COOH][C H OH] [0.333][0.333]CK = =

Let x mol of ester is formed from 1 mol of acid and 0.5 mol of alcohol, then 2

4 0.43(1 )(0.5 )

C

xK x

x x= = =

− −

18.(C) The reaction involved is

2 2 2

eq

2N (g) O (g) 2N O(g)

0 0.482 mol 0.933 mol 0

Charge 2 +2

(0.0482 2 ) (0.0933 ) 2

t

x x x

t t x x x

+

=

− −

= − −

The equilibrium constant is

( )

( ) ( )

2

37

C 2

22 10

0.0482 2 0.0933

xK

x x

−= =− −

As KC has small value, x should be low

0.0482 2 0.0482− x

0.0933 0.0933− − =x x

237 4

2 100.0482 0.0933

− =

x

On solving, we get

213.3 10x −=

As, 21

2N O 2 6.6 10−= = x

Therefore, 2 2N 0.0482M; O 0.0993M=

19.(B) When the denominator is increased by 2, 2

HCl 2=p

Therefore HCl 2 1.414= =p

20.(C) By writing the equation for the reverse reaction 1

2 5 2 22N O (g) 2NO (g) O (g)→ +

p

1as

Kand relating it to the desired reaction, the result can be obtained.

21.(A) Since, equilibrium is affected by temperature and pressure, therefore, at equilibrium, the amount of

X3Y is also affected by temperature and pressure.

22.(D) Concept based.

23.(A) (i) ( ) n

p CK K RT =

(ii) Therefore, 1n = and

p

C

1671.90

0.0821 1073

KK

RT= = =

Since 1n = , therefore

2p 4

C

1.8 104.4 10

0.082 500

KK

RT

−−

= = =

24.(C) The reaction is

8 2 S (g) 4S (g)

At 0 1 0

At equilibrium 1 0.29 0.71atm 4 0.29=1.16atm

t =

− =

The equilibrium constant is

2

8

4 4S 3

p

S

[ ] (1.16)2.55atm

[ ] (0.71)= = =

pK

p

25.(A)

( ) ( ) ( ) ( )

( )

2FeO s CO g Fe s CO g

0 1.4 atm 0.8 atm

At equilibrium 1.4 0.8

1

t

t t x x

x x

+ +

=

= + −

+ −

The reaction quotient is

0.8 40.55

1.4 7Q = = =

Since, ,Q K system is not at equilibrium, and to attain equilibrium, it will shift in backward direction

P

0.80.265

1.4

−= =

+

xK

x

On solving, we get x = 0.339

Therefore CO 1.739 atm=p and2CO 0.461 atm=p

26.(C) As KC is neither too high nor very low, reactants and products will be present in comparable amounts.

27.(C) Moles of I2 taken 46

0.181254

= =

Moles of H2 taken 1

0.52

= =

Moles of I2 remaining 1.9

0.0075254

= =

Moles of I2 used = 0.181 − 0.0075 = 0.1735

Moles of H2 used = 0.1735

Moles of H2 remaining = 0.5 − 0.1735 = 0.3265

Moles of HI formed = 0.1735 2 = 0.347

28.(B) The equilibrium constant for the given reaction is,

3

2

CH OH 1

2 2

CO H

2.0200 atm

1.0 (0.1)p

pK

p p

−= = =

29.(A) Given that Kp = 1.64 10−4atm, Kp2 =?; T1 = 400 + 273 = 673 K, T2 = 500 + 273 = 773 K; H =

−105185.8 J, R = 8.314 J K−1 mol−1

Applying the equation

2 1

2 1

1 2

log log2.303

p p

T THK K

R TT

−− =

2

4 105185.8 773 673log log1.64 10

2.303 8.314 773 673pK − − = −

or Kp2 = 0.144 10−4 atm.

30.(D) Increasing the amount of CO(g) in reactant part, the amount of CO2(g) in product can be increases.

31.(B) The reaction quotient for the given reaction is

3

2 2

2 2NH -1 -2

p 3 3

N H

( ) (3) 9atm 1.125 atm

( ) (1)(2) 8

pQ

p p= = = =

32.(A) For the given reaction 2CO CO ,p p hence

6

CO

1 12500

400 10pK

p= = =

oo 20700 11.97

ln ; lnp p

G TG RT K K

RT RT

+ = − = =

The equation when solved for T using R = 8.314 K−1 mol−1 gives T = 399 K.

33.(C) From the given equation, 3

p

6.36 10log 30.0 5.7 log

= − −K T

T

We know

o

p pln 2.303 log = − = −G RT K RT K

Therefore,

36.36 102.303 1.987 600 30.1 5.7log600

600G

= − − −

= −2.303 1.987 600[30.1 − 10.6 − 5.7 2.778]

= −[2.303 1.987 600 3.6654] cal mol−1

= −10063.9 cal mol−1 = −10.064 kcal mol−1

34.(B) We know

Vapor density 5Molecular weight of PCl=

2

208.5104.25

2= =

Applying the formula, 104.25

0.8(2 1)

d

d

−=

−

0.8 104.25d d= −

1.8 104.25d =

104.25

57.91.8

d = =

35.(B) According to Le Chatelier's principle, if CO2 escapes from the system, then carbonic acid will not

be form and as a result the hydrogen ion concentration will diminish.

36.(A) The reaction is

4 3 2

eq

NH HS(s) NH (g) H S(g)

0 0 0

atm atm

t

t t x x

+

=

=

Now, 2 22 1.12 0.56; 0.3136atmT pp x x K x= = = = =

37.(D) N2O4(g) 2NO2(g), molecular mass of N2O4 = (28 + 64) = 92

Vapor density, 92

466

= =D

Let the degree of dissociation be x. Applying the relationship,

(46 30) 160.533

30 30

D dx

d

− −= = = =

Degree of dissociation = 53.3%.

38.(A) For the given equilibrium, the equilibrium concentration is

2AB2(g) 2AB(g) + B2(g)

Equilibrium concentration c(1 − ), c, 2

c

Therefore, 2

2

2 32B AB T T

p 2 22AB

( )( ) ( / 2) ( )

( ) [ (1 )] [ (1 / 2)]2(1 ) 1

2

p p p pc cK

p c c

= = =

− + − +

As a is small compared with unity, so1 1and1 12

− + therefore

3

Tp

2

pK

=

39.(A) Melting of ice is the endothermic process, and on melting, volume decreases. Hence, high

temperature and high pressure is favorable.

40.(B) Kp = KC (RT)Δn where Kp = 1.44 10−5, n = 2 − 4 = −2. Hence,

5

2

1.44 10

(8.314 773)CK

−

−

=

41.(D) For the given reaction,

2HI(g) H2(g) + I2(g)

12 2

−

The equilibrium constant is 2

T

2 2

T

22or 2 or

(1 ) 1 1 2

p

p p

p

pK

K Kp K

= = =− − +

42.(A) The equilibrium is represented as:

A2(g) + B2(g) 2AB(g)

Initial concentration 1 2 0

Moles at equilibrium 1 − x 2 − x 2x

Total volume = 1 + 2 = 3 L

2 2

1 2 2[A ] ,[B ] , [AB]

3 3 3

− −= = =

x x x

2

2 2

[AB] 2 / 350

[A ][B ] (1 ) / 3(2 ) / 3

xK

x x= = =

− −

On solving, we get 23x2 − 75x + 50 = 0 x = 2.31 or 0.943.

As x cannot be more than 1, so x = 0.943

Number of moles of AB formed = 2 0.943 = 1.886 mol.

43.(A) The degree of dissociation cannot be calculated from the vapor density data because here, the number

of moles remains unchanged before and after reaching equilibrium.

44.(D) For the given reaction,

1 2 3

A(s) 2B(g) 3C(g)−

+x x x

The equilibrium constant is 2 3 5

C 4 27 108K x x x= =

2 6

A(s) 2B(g) 3C(g)

+x c x

5 2 2 3108 4 216= x x c x 2108 216 4= c

2 108

4 216=

c

1

2 2=c

45.(B)The ratio of number of moles will be the same as the ratio of volume.

According to Dalton's law, the partial pressure of a gas in a mixture is given by the product of its volume

fraction and the total pressure.

Therefore, the equilibrium pressure of each gas is,

3NH

9.650atm 4.8atm

100p = =

2N

22.650atm 11.3atm

100= =p

2H = 67.6 /100×50atm =33.9atmp

Totalpressure 50atm=

3

2 2

2

NHp

3

N H

[ ]

[ ][ ]

pK

p p=

Substituting the values of partial pressures,

25 2

3

(4.80atm)5.23 10 atm

(11.3atm)(33.9atm)pK − −= =

46.(C) The reaction is

2 2 3SO (g) NO (g) SO (g) NO(g)

Initialconcentration 1 1 1 1

At equilibrium 1 1 1 1x x x x

+ +

− − + +

The equilibrium constant is

( )( )

( )( )3

2 2

SO NO 1 1

SO NO 1 1

x xK

x x

+ += =

− −

( )

( )

( )

( )

21 1 x

16 4 or 0.61 1

+ += = =

− −

xx

x x

1

2[NO ] 1 1 0.6 0.4molL−= − = − =x

47.(D) Since, equilibrium constant (K) depends on temperature only, therefore, it remains constant with

change in p and x.

48.(B) The given reaction is,

22NO(g) + Br (g) 2NHOBr(g)

Initial 98.4 41.3 0

At equilibrium 98.4 − x 41.32

−x

x

Total pressure at equilibrium = 110.5 torr.

(98.4 − x) 41.32

+ − +

xx = 110.5

Solving, we get x = 58.4 torr (760 torr = 1 atm)

pNOBr = 58.4 torr = 7.68 10−2 atm.

pNO = 98.4 − 58.4 = 40 torr = 5.26 10−2 atm.

2

2

NOBrp 2

NO Br

=

pK

p p

2 2

2 2 2

(7.68 10 )134

(5.26 10 ) (1.59 10 )

−

− −

= =

49.(C) At 40% SO2 and 40% CO2 by volume2 2CO SO

40 40 and

100 100p p p p

= =

So, 2 2SO Cl

20

100p p

=

and given that p = 10 atm. Therefore,

Therefore, 2 2

2 2

CO SO

SO Cl

( )( ) (40 /100) (40 /100) 16 108 atm

( ) (20 /100) 2 10p

p p p pK

p p

= = = =

50.(B) For first equation

2 2N (g) O (g) 2NO(g)+

gp 0

C

( ) ( ) 1nK

RT RTK

= = =

For second equation

2 4 2N O (g) 2NO (g)

1p 31

C

( ) 24.6 dm atm mo l2K

RTK

−= =

For third equation

2 2 3N (g) 3H (g) 2NH (g)+

2

6 2 2p 2 3

C

1( ) 1.6 dm at5 10

( )m mol

KRT

K RT

− − − −= = =

<H2> Multiple Correct Choice Type

1.(B, C, D) The reaction is endothermic as H has a positive sign. Therefore, on increasing temperature,

more amount of heat is supplied to the system. So, by Le Chatelier’s principle, the system will try to release

heat and thus reaction will try to move in the reverse direction.

The no. of moles of gaseous substances reduces as we go from reactant to product side, so on increasing

the pressure, the system will try to reduce the pressure by moving in the forward direction. Addition of

catalyst will help in adsorption of the gaseous reactants on its surface and help lower the Eactivation for

conversion of reactants to products. On removing C, again by Le Chatelier’s principle, the system will try

to produce more and more C, by moving in the forward direction.

2.(A, C)

3.(A, B, C) The relation is expressed as

o

2 2 1

1 1 2

log2.303

K T TH

K R TT

−=

If T2 >T1 and H = 0, we have

log(K2/K1) = 0

K2/K1 = 1

K2 = K1

If T2 >T1 and H > 0, we have

log(K2/K1) > 0

K2/K1 > 1

K2 > K1

If T2 >T1 and H < 0, we have

log(K2/K1) < 0

K2/K1 < 1

K2 < K1

4.(A, B, C) Catalyst only speeds up the reaction.

5.(A, B, C) As the concentration of reaction (gaseous) is increased at equilibrium, reaction will go in the

forward direction. Also, the above reaction is endothermic. Therefore, an increase in temperature will favor

it.

6.(B, C) As Q varies as reaction proceeds, therefore, Q may be less, more or equal to KC.

7.(B, C, D) In accordance with Le Chatelier’s principle.

8.(A, C, D)

(A) As the reaction is endothermic, therefore it will go in the forward direction. Hence, moles of CaO will

increase.

(C) With the increase or decrease of volume, partial pressure of the gases will remain same.

(D) Due to the addition of inert gas at constant pressure, reaction will proceed in the direction in which

more number of gaseous moles are formed.

9.(B, C, D) The reaction is

N2O4 2NO2

t = 0 a 0

a(1 − ) 2a

Vapor density46

30.671

= =−

= 0.5 = 50%

Total pressure =1.5 1.5 0.082 300

6.67 atm8.2

= =

So,

2

2

49atm

1pK p

= =

−

and density of mixture 1 1138g L 16.83 g L

8.2

− −= =

10.(B, C) Since, in reactions given in option (B) and (C), the number of moles of products are more than

reactants, therefore, the application of high pressure does not increase the yield of product.

11.(A, C) With the increase in pressure, forward reaction is favored, because for forward reaction, lesser

number of molecules are involved. This is in accordance with Le Chatelier’s principle.

12.(B, C) For the given reaction

N2O4 2NO2

2 − x 2x

The equilibrium constant is

244

2 10 0.012

p

xK x−= = =

[NO] = 2x = 0.02 M, [N2O4] = 2/1 = 2 M

13.(A, B, D) On adding Cl− ion, the equilibrium will shift in the backward direction. So, [CoCl4]2− will

increase on dilution and hence the volume will increase.

Now, 2+ -

2 6

2

4

[CO(H O) ] Cl

c 4

CoCl

. 1= .

−

n nK

n V

So, Q will get decreased and equilibrium will shift in the forward direction. Although the amount of

[Co(H2O)]2+ will get increased, but its concentration cannot get doubled as volume has increased. As the

reaction is exothermic in nature, so increasing temperature will shift the equilibrium in backward direction.

14.(B, C) The pressure of NH3 will decrease due to addition of CO2, (backward, shifting Le Chatelier’s

principle). The pressure of CO2 will be more than 0.1 atm.

15.(A, B) At constant volume, the addition of inert gas does not affect equilibria. Since, the number of

moles of reactant is more in (I) and that of product is more in (III), while they are same in (II), therefore,

addition of an inert gas at constant pressure favors the forward reaction in (III), backward reaction in (I),

and has no effect on (II).

16.(A, B, C, D) At constant volume, the addition of inert gas does not affect equilibria.

<H2> Assertion–Reasoning Type

1.(A)

2.(C) The concentration of pure liquids and solids remain constant, and these terms are amalgamated (or

merged) in the equilibrium constant. That is why pure liquids and solids are ignored while writing the

equilibrium constant expressions.

In the first step of Ostwald’s process for synthesis of nitric acid, ammonium is oxidized to NO by the

reaction

4NH3(g) + 5O2(g) 4NO(g) + 6H = O(g); H = −905.6 kJ

3.(D) Reaction is exothermic, and low temperature favors forward reaction. High pressure favors forward

reaction; it is accompanied by a decrease in the number of moles.

4.(B) Using the equation,

2 2 110

1 1 2

Δlog =

2.303

K T TH

K R TT

−

5.(C) 3 2

5

1

PCl Cl

PCl T

×= ×p

n n pK

n n

with addition of more and more inert gas nT will increase that leads to decrease

in the value of p/nT, and hence to keep Kp constant, moles of the product would increase, that is, increases.

6.(B) Q < KC forward reaction; Q = Kc equilibrium; Q > KC backward direction.

7.(A) Since, the given reaction is a heterogeneous reaction, therefore, it will not affected by addition of

solid or liquid.

8.(C) In the exothermic reaction, equilibrium constant increases by decreasing the temperature.

<H2>Comprehension Type

1.(B) As n = 0, so KC = Kp = Kx

2

2

4 4H 4 5

4 4

H O

(2 / 2)18 1.05 10

(1 /18)x

xK

x= = = =

2.(C) For the given reaction, the equilibrium constant is

3 2

2 2 3

NH CO

1/3

p

( ) (2 ) 4

0.0667 atm4

pK p p p p p

Kp

= = =

= =

3.(C)

1/2

1 12 22 2 30

1N (g) O (g) NO(g)

2.4 10CK

+ =

( ) ( ) ( )12 22

NO g Br g NOBr g 1.4K+ =

( ) ( ) ( ) ( ) 161 1 12 2 2 c 22 2 2

N g O g Br g NOBr g 9.03 10−+ + = = K K K

4.(A) For the given reaction, the equilibrium constant is

3

2 2

2

SO

2

SO O

=.

p

pK

p p

Therefore 2O

1 1atm 0.29 atm

3.5p

pK

= = =

5.(B) More is the degree of dissociation, less is the vapor density of PCl5

6.(B) Clearly, concentration of Y is not changing with the time. Hence, it will be pure solid or liquid.

Concentration of X is decreasing. Hence, it will be in reactant and Z will be in product.

7.(C) The equilibrium constant for the reaction is

eq

eq

[Z(g)] 4 2

[X(g)] 6 3CK = = =

8.(D) From the graph, we can see that 4 mol L−1 of Z is formed. Therefore, 4 2 mol of solid will be

required.

9.(C) For reversible process, the Gibbs free energy and equilibrium relation is,

Go = −2.303RT log K

10.(D) We knowoΔ 2.303 logG RT K= −

log 1.665K = K = Antilog of 1.665 0.4625=

A + B C + D

Initial concentration 2 2 0 0

At equilibrium 2 2x x x x

− −

( )

2

e 20.4625

2

xK

x= =

−

C0.68

A 2= =

−

x

x

11.(D) We know oΔ 2.303 log pG RT K= − (1)

Substituting the values in Eq. (1), we get

oΔ 2.303×1.98 763 log45.9G = − oΔ 5.79 kcalG = −

<H2> Integer Answer Type

1.(1) Density of ammonium carbamate before dissociation

Molecular weight 7839

2 2D = = =

Density after dissociation (D) = 13.0 (given). Hence, 39 13 26

1( 1) (3 1) 13 26

D d

n d

− −= = = =

− −

So, ammonium carbamate will be 100% dissociated.

2.(4) 2 4

o 1

2 4 2 N ON O 2NO ; = 100kJ molG −

2

o 1

NOConcentration at = 0 5 5 = 50 kJ molt G −

Go for reaction = 2 2 4

o o

NO N O2 × = 2 50 100 = 0G G − −

G = Go + 2.303RT logQ

Now, 5

3 50 2.303 8.314 10 298log 3.99 kJ 4 kJ

5G − = + = +

3.(1) 3 2 2 24NH (g) 7O (g) 4NO (g) 6H O(g)+ +

Water and oxygen will not contribute to partial pressure. 4

2

4

3

( NO )1

( NH )

pKp

p= =

Answer should be -1.

4.(2) For 1 equilibrium 2 2 42NO N O

2 4

2

N O

2

NO

= 6.8( )

p

pK

p

= (1)

2 4N O =1.7 atm p therefore by Eq. (1) 2NO = 0.5 atmp .

The equilibria are maintained using NO and NO2 in the ratio of 1:2

For 2 equilibria

2 2 3NO + NO N O

Initial pressures p 2p

Pressures at equilibria (p − x) (2p − x − 3.4) x

3.4 atm of NO2 are used for 1 equilibrium to have 2 4N O =1.7 atmp

At equilibrium (p − x) 0.5 x

(Since2NO

p is same for both the equilibria, as both reactions are at equilibrium at a time)

Total pressure at equilibrium 2 2 3 2 4NO NO N O N O

= + + +p p p p 0.5 1.7= − + + +p x x

5.05 2.20= +p 2.85 atm=p

2 3.4 0.5− − =p x 2 2.85 3.4 0.5 − − =x 1.80 atm 2 atmx =

5.(8) Given that 1 700 273 973 K= + =T , 2 1000 273 1273 K= + =T , 1 20.63, 1.66= =K K

Using the van't Hoff equation o

2 2 1

1 1 2

Δlog

2.303

−=

K T TH

K R TT

oo 31.66 Δ 1273 973

log Δ 8.0 10 cal 8.0 kcal0.63 2.303 1.99 1273 973

HH

− = = =

Note: The units of R and H must be same.

6.(8) For the reaction

2

3 initial

2

2 2

[SO ]

[SO ] [O ]CK = (1)

When concentrations of SO2 and O2 are quadrupled, the expression becomes

2

3 final

2

2 2

[SO ]

[4SO ] [4O ]CK = (2)

Dividing Eq. (2) by Eq. (1), we get

2 2 2 2

3 final 3 initial 3 final 2 2

2 2 2 2

2 2 2 2 3 initial 2 2

[SO ] [SO ] [SO ] [4SO ] [4O ]16 4 64

[4SO ] [4O ] [SO ] [O ] [SO ] [SO ] [O ]= = = =

Hence, 3 final

3 initial

[SO ]8

[SO ]=

7.(4) The equilibrium constant is 3

2

3

3 2

[AB ]64

[A ][B ]CK = =

For the reaction

1

3 2 23A (g) B (g) AB (g)+

The new equilibrium is

1/32

1/3

3 2

[AB ]

[A ] [B ]C CK K = =

Therefore, KC = (64)1/3 = 4

<H2>Matrix–Match Type

1.(A)→ (r); (B) → (p); (C) → (q); (D) → (s, t)

The expression for relation between dissociation constant and equilibrium pressure is as follows:

(A)

2 2 3N 3H 2NH

Initial 3 0

Equilibrium 1 3(1 ) 2

+

− −

Total number of moles at equilibrium = 1 3 3 2 4 2 2(2 ) − + − + = − = −

Partial pressures of N2, H2 and NH3 are as follows:

2N T

1

2(2 )p p

−=

−

2H T

3(1 )

2(2 )p p

−=

−

3NH T

2

2(2 )p p

=

−

( )

( )3

2 2

2

2T 2 2

NH

3 3 4 2

TN H

T T

2

2(2 ) 16 (2 )

27(1 ) ( )1 3(1 )

2(2 ) 2(2 )

p

pp

Kpp p

p p

− − = = =

− − −

− −

Now, 1 – α 1, 2 – α 1so

2

T p T2

T

64 33

27( ) 8pK p K p

p

= =

(B)

2 2 32SO O 2SO

Initial 2 0

Equilibrium 2(1 ) (1 ) 2

+

− −

Total number of moles at equilibrium = 2 2 1 2 3 − + − + = −

Partial pressures of SO2, O2 and SO3 are as follows:

2SO T

2(1 )

3p p

−=

−

2O T

(1 )

(3 )p p

−=

−

3SO T

2

(3 )p p

=

−

( )

( )3

2 2

2

2T 2

SO

2 2 3

TO SO

T T

2

(3 ) (3 ) 1

(1 )1 2(1 )

(3 ) (3 )

p

pp

Kpp p

p p

− − = = =

− − −

− −

Now, 1 – α 1, 2 – α 1, 3 – α 1as so

2T p

T

T

3

3p

p KK p

p

= =

(C)

5 3 2PCl PCl Cl

Initial 0 0

Equilibrium (1 )

+

−

Total number of moles at equilibrium = 1 1 − + + = +

Partial pressures of PCl5, Cl2 and PCl3 are as follows:

5PCl T

(1 )

(1 )p p

−=

+

2Cl T(1 )

p p

=

+

3PCl T(1 )

p p

=

+

2 3

5

T T 2Cl PCl T

2

PCl

T

(1 ) (1 )

(1 )1

(1 )

p

p pp p p

Kp

p

+ + = = =

− −

+

Now, 1 – α 1, so p2

T

T T

1p

KK p

p p = =

(D)

2 2N O 2NO

Initial 0

Equilibrium (1 ) (1 ) 2

+

− −

Total number of moles at equilibrium = 1 1 2 2 − + − + =

Partial pressures of N2, O2 and NO are as follows:

2N T

(1 )

2p p

−=

2O T

(1 )

2p p

−= NO T

2

2p p

=

( )

2 2

2

2 T 2NO

2

N OT T

2

42

1 1 (1 )

2 2

p

pp

Kp p

p p

= = =

− − −

which is equal to KC

Now, 1 – α 1, so ( )02

T p T

14

2pK p K p = =

2.(A) → (r); (B) → (p, q); (C) → (p); (D) → (p, q)

(A) Backward reaction is favored.

(B) is affected by both temperature and pressure.

(C) KC is affected by temperature only.

(D) On increasing temperature and pressure, ice melts.

3.(A) → (s, t); (B) → (q, r); (C) → (q); (D) → (p)

4.(A) → (s); (B) → (r); (C) → (p); (D) → (q)

Using the relation Kp = KC(RT)Δn

5.(A) → (r),(s); (B) →(r); (C) → (q); (D) → (p),(s)

(A) As n > 0 therefore if p , reaction will go in the backward direction.

(B) As n < 0 therefore if V , p reaction will go in the direction in which more number of gaseous moles

are formed that is backward direction.

(C) As n = 0, hence no effect.

(D) If concentration of the product is decreased, reaction will go in the forward direction.