CHEMICAL EQUILIBRIUM(E).doc

CHEMISTRY CHAPTER WISE_QUESTIONS

CHEMICAL EQUILIBRIUM (E)

29. PCl5_{ }g ‡ ˆˆˆ ˆ† PCl3_{ }g Cl2_{ }g ;COCl2_{ }g ˆ ˆ†‡ ˆˆ CO_{ }g Cl2_{ }g Simultaneous equilibrium is

established for above reactions in a vessel. Removing some amount of PCl3 from

reaction mixture, at constant volume results in (after reestablishment) 1) Decrease in concentration of PCl5 than initial equilibrium

2) Decrease in concentration of CO than initial equilibrium 3) Increase in concentration of COCl2 than initial equilibrium

4) All of these ANS:4

SOL:

42. If 20% of N O2 4molecules have dissociated in a sample of gas at 270C and 760 mm Hg,

the density of equilibrium mixture is

1) 1.48 g/L 2) 1.84 g/L 3) 2.21 g/L 4) 3.11 g/L ANS:4

SOL: NaCN HCl  NaCl HCN

8.5 H P  Ka 9.39 P  log H Ka NaCN P P HCN     _{ }  









8.5 9.39 log NaCN HCN  









0.9 log HCN NaCN  8 0.1 x x   0.8 8 x  x 0.8 9 0.8 0.089 9 x  x  moles 2 8.8 10  

22. Ammonia gas at a pressure of 20 atm and 0

27 C is heated in a constant volume in a

container to a temperature of 0

following equilibrium 2NH3( )g ˆ ˆ†‡ ˆˆ N2( )g 3H2( )g is established. Percentage dissociation

of ammonia at this temperature is _X2_{. The value of X is}

ANS:5 SOL: 3( ) 2( ) 2( ) 2NH _{g ‡ ˆˆ}ˆ ˆ† N _g 3H _g 3 1 2 2     1 Totalmoles  PV nRT 20 1 300 50 600 V R V  n R 1.25 n 0.25  % 25 . . i e x5

50. The equilibrium constant K for the reaction p 2SO2 O2‡ ˆˆˆ ˆ† 2SO3 is 900 atm1 at 800 K . A mixture containing SO and 3 O having initial pressure of 1 atm and 2 atm 2 respectively is heated at constant volume to equilibriate. The partial pressure (atm) of

2 O gas at 800 K is 1) 0.9764 2) 2.01 3) 0.0236 4) 0.0118 ANS:2 SOL: 2SO3‡ ˆ ˆˆ ˆ† 2SO2 O2 Int.Pressure 1 0 2 Pressure at eq.



1 x



x 2 2 x 





2 2 3 2 2 2 2 2 2 1 O SO SO p x x p p k p x  _     _{ }    1 small 1 900 p k  x





2 2 1 2 900 ₁ p x k x    1 2 , 0.0236 30 1 x x x    3 '_SO 1 1 0.0236 0.9764 atm p    x  2 '_SO 0.0236 atm p  x 2 2 ₂ 2.0118 atm O x p   

2. 2 moles of N O2 4 is heated to form NO and O2.As soon as NO and O2 are formed, they

react to form N O2 5. Two equilibria are simultaneously established

2 4 2 2 2 5

3

2 2

2

N O ƒ NO O NO O ƒ N O

at equilibrium degree of dissociation of N O2 4 was found to be 50% which of the

following is correct at equilibrium

A) [NO] [ ] [+O2 = N O2 4] [+ N O2 5] B) [ ]2 [ 2 5] [ ] 1 1 O N O NO 2 2 + = C) [ ] [ ]2 1 3 NO O 2 =2 D) [ 2 4] [ ] [ ] [2 2 5] 3 2 N O NO O N O 2 = = + ANS:B SOL:

66. At temperature T, a compound AB2(g) dissociates according to the reaction

 

 

 

2 2

2AB g ƒ 2AB g B g With a degree of dissociation, x, which is small compared with unity. Deduce the expression for x in terms of the equilibrium constant, Kp and total pressure P.

1) 4 p K x P  2) 1 3 2K_p x P    _ _ 3) 1 3 4 p K x P     _{ } 4) 1 2 8 p K x P     _{ } ANS:2 SOL: + -ˆ -ˆ† ‡ ˆˆ 2 2 2AB 2AB B x 1 x x ₂

1. The value of kp for the reaction CO2 g Cgra ‡ ˆˆˆ ˆ† 2CO g is 3 bar at 1000 k. If initially

2

CO

P _{is 0.48 bar ,}P_{CO is zero and pure graphite is present. The partial pressure of} CO_{2 at}

equilibrium is

A) 0.33 bar B) 0.15 bar C) 0.30 bar D) 0.2 bar

ANS:B

SOL: NCERT Page No. 197

      2 0.48 2 2 g s g x x CO C CO   _{‡ ˆˆ}ˆ ˆ† 2 2 4 3 4 1.44 3 0.48 x x x x     2 4x 3x 1.44 0     3 9 23 3 32 0.33 8 8 x          2 0.48 0.33 0.15 CO P    bar 23. In the equilibrium _{   }2 _{ } 2s aq 2 aq

CaF _{‡ ˆˆ}ˆ ˆ† Ca   F _{, if the concentration of F} _{at new}

2

Ca  _{at the new equilibrium will be how many times less than the concentration of Ca}2

at the old equilibrium. ANS:4

SOL:

10. H (g) I (g)2  2 ƒ 2HI(g). When 46.0 g of I2 and 1.00g of H2 are heated to equilibrium

at _{470 C}o _{, the equilibrium mixture contains} 2

1.9g I _{. Which of the following options are}

correct regarding moles of a species at equilibrium and the value of equilibrium constant.(I=127 g/mole)

a) moles of I2 0.0075 b) moles of H2 0.326 c) moles of HI 0.348 d) Keq = 500 ANS:ABC

SOL: Since the same number of moles of gas appears on both sides of the equation, the equilibrium constant

expression can be stated as a ratio of moles instead of concentrations. 2 2 2 1mol I (46.0gI ) 0.181 254g I   _  

  mol I initially present 2 2 2 2 2 1mol H (1.00g H ) 0.500 mol H 2.00g H   _     2 2 2 2 1mol I (1.9g I ) 0.0075mol I 254g I   _     at equilibrium

Initial Produced Used up Equilibrium

2 [I ] 0.181 0.174 0.0075 2 [H ] 0.500 0.174 0.326 [HI] 0 2(0.174) 0.348 2 2 2 2 [HI] (0.348) K 50 [I ][H ] (0.326)(0.0075)   

1. At constant temperature, the equilibrium constant (K_p) _{for the decomposition}

reaction. N O2 4 ƒ 2NO2 is expressed by Kp 4x P2 / (1x2) where P is pressure, x is

extent of decomposition. Which of the following statement is true? 1) Kp increases with increase of P 2) Kp increases with increase of x

3) Kp increases with decrease of x 4) Kp remains constant with change in P or x

ANS:4 SOL:     2 4g 2 2g N O ƒ NO t=0 1 0 tequi 1-x 2x Total moles = 1 x 2x 1 x 2 4 2 1 2 1 1 N O NO x x P P P P x x      _{ } _{ }       2 4 N O Q is 25% dissociated

 x0.25and P 1atm 2 4 1 0.25 1 0.6 1 0.25 N O P _  _   atm    2 2 0.25 1 0.4 1 0.25 NO P _  _   atm    2 2 4 1 2 0 0.4 0.4 0.267 0.6 NO p N P K atm P    

Suppose the degree of dissociation of N O at 0.1 atm is 2 4 

2 4 2 1 2 0.1 0.1 1 1 N O NO P  P       _{ }        2 2 4 0.1 0.267 1 p K       63.2%   

2. At a certain temperature equilibrium constant

 

Kc is 16 for the reaction.

       

2g 2g 3g g

SO NO ƒ SO NO _{. If we take one mole of each of all the four gases in one}

litre container. The % of moles of NO2 at equilibrium is?

1) 30 2) 20 3) 10 4) 40 ANS:3 SOL:

 

3

22

ggg

SO



N

O



SO

N

O

t=0 tequi 1-x 1-x 1+x 1+x 1 1 1 1

  















2 3 2 2 2 1 16 1 c SO NO x K SO NO _x      3 0.6 5 x moles mole    At equi nSO2 0.4nNO2 0.4 nSO3 1.6 nNO 1.6 2 0.4 % 100 10% 4 NO of n   

6. The preparation of SO g3

 

by reaction 2_{ } 2_{ } 3_{ }

1 2

g g g

SO  O ƒ SO is an exothermic reaction. If the preparation follows the following temperature - pressure relationship for its %yield, then the temperature T T and T1, 2 3. The correct option is

1) T3T2 T1 2) T1T2 T3 3) T1T2 T3

4) nothing could be predicated about temperature through given information

2

T

1

T

T

3

1 2 ₃ 4 10 20 30 40 50 % yiel d press (atm) ANS:2 SOL:

7. Pure ammonia is placed in a vessel at a temperature where its dissociation constant

 

 is appreciable. At equilibriumN2 3H2 ƒ NH3

1) Kp does not change significantly with pressure

2)  does not change with pressure

3) concentration of NH3does not change with pressure

4) concentration of hydrogen is less than that nitrogen ANS:1 SOL: _{N O2 5 ƒ N O2 3O2} 4 x x y x y  2 3 2 2 N O ƒ N O O x y y y x

 

O2   x y 2.5 Q



 



4 c x y x y K x     2.166 x



N O2 5



  4 x 1.846

8. At a certain temp, Kp for dissociation of solid CaCO3 is 4 10 2atm and for the reaction

 s 2( )g 2 ( )g

C CO ƒ CO _{is 2.0 atm respectively. The partial pressure of CO at this}

temperature when solid C CaO CaCO, , 3 are mixed and allowed to attain equilibrium is?

ANS:4

SOL: For dissociation of CaCO3

      3s s 2g CaCO ƒ CaO CO Given that 2 2 4 10 p CO K    P

For the formation of CO,

 s 2 s 2 2 g C CO ƒ CO

 

2 2 2.0 CO CO P Kp P  

Now both are at equi and thus P_CO2  4 102 which should remains constant

 

2 2 2.0 0.28 4 15 CO CO P P    

9. The equilibrium constants for the reactions are: m 1 3 4 2 4; K H PO _{‡ ˆ ˆˆ}ˆ ˆ ˆ† HH PO K₁ 2 2 2 4 4 K H PO_{ˆ ˆ ˆ†} H_HPO  ‡ ˆ ˆˆ ; K2 3 2 3 4 4 K HPO ˆ ˆ ˆ†_{‡ ˆ ˆˆ} HPO _; K₃

The equilibrium constant for

3 3 4 3 4 H PO _ƒ H_PO  _{will be:} 1) K K K1/ 2 3 2) K1K2K3 3) K2/K K1 3 4) K1K2K3 ANS:2 SOL: HI disscociates as

2H

IH



22

I

at t = 0 1 0 0 at equi

1







/2



/2





2 2 4 4 1 c K     

Now for the association equilibrium

2 2 2 H I ƒ HI Initial 2x 2x 2x





2 1 2 1 4 1 2 4 5 2 c c x K x K x       Thus I left = 2 2 8 8 2 2 5 5mole 2     equivalent mcq of Na S O2 2 3 mcq of I2left 8 2 2 1000 5 v     V=1.6 lt

10. The equilibrium constants K and Kp1 P2 for the reactions xƒ 2y and zƒ P Q

respectively are in the ratio of ratio of 1:9. If the degree of dissociation of x and z be equal then the total pressure at these equilibrium is_____

1) 1:36 2) 1:1 3) 1:3 4) 1:9 ANS:1 SOL: _{xƒ 2y} 1 0 1-x 2x

 

1 2 1 2 . 1 1 ' p x P K x x    _{ }     zƒ P Q 1 0 0 1-x x x 2 2 2 . 1 1 ' p x P K x x    _{ }     1 1 2 2 4 1 1 9 36 P P P P    

11. The equilibrium constant

 

Kc for the reaction N2 g O2 g 2NO g at temp T is 4 10 4.

The value of Kc for the reaction _{ } 2_{ } 2_{ }

1 1 2 2

g g g

NO  N  O at the same temp is 1) _{2.5 10} 2 _{2) 4 10} 4 _{3) 50 4) 0.02} ANS:3 SOL: 4 2 2 2 ; c 4 10 N O  NO K    1 2 2 2 1 1 2 10 2N 2O NO Kc Kc       '' 2 2 ' 1 1 1 50.0 2 2 c _c NO N O K K    

12. The thermal dissociation of CaCO3 g is studied under different conditions

     

3s s 2g

CaCO ƒ CaO CO _{. For this equilibrium the incorrect statements is}

1) H is dependent on T

2) Equilibrium constant (K) is independent of the initial amount of CaCO3

3) Equilibrium constant (K) is independent on the pressure of CO2 at a given T

4) H is independent of the catalyst, if any ANS:3

SOL: k depends upon the temperature. But at constant temperature it depends upon products and reactants concentration and pressures

13. For the chemical equilibrium CaCO3 s ƒ CaO s CO2 g Ho_r can be determined from

which one of the following plots

1) 2)

2

log

e

PC

O

T 3)

2

log

e

PC

O

log T ₄₎

2

PC

O

1/T ANS:1 SOL:   3_g 2 CaCO ƒ CaO CO 2 CO K P 0_/ r H RT K _ Ae 0 2.303 r P H Log K LogA RT    2 0 2.303 r CO H Log P LogA RT    18. G0_for 2 1 N2(g) + 2 3

H2(g) NH3(g) is – 16.5 kJ/mol. Then the G0 for N2 + 3H2 2NH3 at 25°C. is (nearly)

1) -33 kJ/mole 2) -53 kJ/mole 3) -73 kJ/mole 4)-93 kJmole ANS:1 SOL: 0 p 2.303 RT logK G  



3



p 1.6 10 2.303 8.314 298log K      1 779.4atm P K    Also K for N'p 23H2 ƒ 2NH3

 

2





2 1 _779.4 p p k  k  5 2 6.06 10 atm   Also G0  2.303RTLogK1p



5



2.303 8.314 298log 6.07 10 32.989KJ mole/

     

19. For the reaction : N2( )g 3H2( )g ‡ ˆˆˆ ˆ† 2NH3( )g ;

H ve

   , the correct statement is:

1) Addition of catalyst does not change Kp but changes H

2) At equilibrium, 2GNH3 GN2 3GH2(G is Gibbs energy)

3) At higher temperature, the rate of forward and backward reaction increases by a factor 2

4) At 400K, addition of catalyst increases rate of forward reaction by 2 times and backward by 1.7 times. ANS:2 SOL: _{2 4    } 3 2 2 _{g g} NH COONH  NH CO



1



D d n d    

X  no. of moles of product formed by dissociation of 1 mole of rectant Initial vapour density . 78 39

2 2 M Wt D  



3 1 1339 13



1      

21. An acid-type indicator, HIn, differs in colour from its conjugate base . The human eye is sensitive to colour differences only when the ratio _ In_ / HIn





  is greater than 10 or

smaller than 0.1. What should be the minimum change in the pH of the solution to observe a complete colour change ?



5



1.0 10 a K    1)4 2)3 3)2 4)1 ANS:3 SOL: _{HInƒ H In} 





a H In k HIn            When





10; In HIn     _{   H}_{  1 10 /10 10}5 _ 6   pH = 6 When





101 ; In HIn     _{   H}_{  1 10}5_{ 10 10}4   pH = 4  minimum change in pH = 6 – 4 =2

30. An equilibrium mixture at 300K contains N O2 4 and NO2at 0.28 and 1.1 atmosphere

respectively. If the volume of container is doubled. Calculate the new equilibrium pressure of two gases (in atm)

N O2 4 NO2 (a) 0.095 0.64 (b) 0.64 0.095 (c) 0.17 0.32 (d) 0.32 0.17 ANS:A SOL:

26. The equilibrium, 2CO + O2 ƒ 2CO2 heat, is subjected to the following combination

of stresses. In which one of the following cases, it is not possible to predict the effect of stress on the direction of reaction?

(a) Addition of CO and removal of CO2

(b) Increase in temperature and decrease in volume (c) Addition of O2 and decrease in volume

(d) Addition of a catalyst and decrease in temperature. ANS:B

SOL: As both the stresses drive the reaction in opposite directions in (B), it is not possible to determine its effect on the direction of the reaction

38. 0.2 mol of each A2(g) and B2(g) are introduced in a sealed flask and heated to 2000 K where following reaction is established:

A2(g) + B2(g) ƒ 2 AB(g)

At equilibrium, moles of AB is 0.3. At this stage, 0.1 mol of C2(g) is added and the following equilibrium is also established:

A2(g) + C2(g) ƒ 2 AC(g)

At the new equilibrium, the moles of AB become 0.24. If the equilibrium constant for the second reaction is 3x, the value of x is

ANS:6 SOL:

24. Observe the following reaction.

A(g) + 2B(g) ƒ _{AB2(g); ∆H = -230 kJ mol}-1

Incorrect statement(s) regarding this reaction is/are

(A) As temperature increases the rate of forward reaction decreases while rate of backward reaction increases.

(B) Energy of activation for backward reaction is more than energy of activation of forward reaction.

(C) As temperature increases that rate of forward reaction increases while the rate of backward reaction decreases.

(D) Addition of an appropriate catalyst at the same temperature speeds up the reaction by making the reaction more exothermic.

ANS:ACD SOL:

25. Ethyl acetate is synthesized from ethanol and acetic acid in a nonreacting solvent by the following reaction:

3 2 5 3 2 5 2

CH COOH( )+C H OH( )l l ƒ CH COOC H ( ) H O( );K=2.2l  l

If the reaction is set up taking [CH3CO2C2H5] = 0.22 M, [H2O] = 0.10 M,

[CH3CO2H] = 0.010 M, [C2H5OH] = 0.010 M, the changes which take place as the system attains equilibrium is/are

(A) Concentrations of both ethyl acetate and water decrease.

(B) Concentration of ethyl acetate decreases while that of water remains constant.

(C) Concentrations of both ethanol and acetic acid increase.

(D) No change in concentrations of reactants as well as product as the given concentrations correspond to their equilibrium values.

ANS:AC SOL:

39. Match the reactions in List I with appropriate effects in List II.

List I List II

(P) _{Fe ( ) SCN ( )}3 _{aq }  _{aq ƒ [Fe(SCN)] ( )}2 _{aq (1) High pressures favour}

forward reaction.

(Q) H2 (g) + I2(g) ƒ 2HI (g) (2) Change in pressure has no effect on the equilibrium (R) 2NaNO3(s) ƒ 2NaNO2(s) + O2(g) (3) Lower pressures favour

forward reaction

(S) H2O (s) ƒ H2O (l) (4) Addition of oxalate ion drives the reaction in backward direction P Q R S (A) 4 2 3 1 (B) 1 2 4 3 (C) 3 2 1 4 (D) 4 3 2 1 ANS:A SOL:

11. Consider the following equilibria:

(1) N2 (g) + 3H2 (g)  2NH3 (g) (2) N2(g) + O2(g)  2NO (g) (3) PCl5(g)  PCl3 (g) + Cl2 (g)

A) Addition of an inert gas at constant volume has no effect on all the three equilibria. B) Addition of an inert gas at constant pressure favours the forward reaction in (3), backward reaction in (1) and has no effect on (2).

C) Addition of an inert gas at constant pressure has no effect on equilibrium (2), but favours the forward reaction in (1) and backward reaction in (3).

D) Addition of inert gas has no effect on all the three equilibria at constant temperature and also at constant pressure.

ANS:AB SOL:

82. For the reaction A g

 

ƒ B g

 

C g

 

1) 3 p K  p 2) 2



₁



p P K  K  p 3) 2





p P K  K  p 4) 2 P p K p K p      _{ }   ANS:3 SOL:

_

_{ }







2 2 2 2 2 ; 1 ; ; 1 p p p p p K  K   p K  K p       

55. For the reaction 2NO g₂

( )

1O g₂

( )

N O g_{2 5}

( )

2

+ _{‡ ˆ ˆˆ}ˆ ˆ ˆ† _{, if the equilibrium constant is} K_p , then the equilibrium constant for the reaction

( )

( )

( )

2 5 2 2 2N O g _{‡ ˆ ˆˆ}ˆ ˆ ˆ† 4NO g +O g _{would be} 1) K2p 2) p 2 K 3) 2 p 1 K 4) p 1 K ANS:3 SOL:

64. For a reversible reaction PCl

_{ }

5

_{   }

PCl3 Cl2

g g g

 ƒ

. The position of equilibrium at constant temperature can be shifted towards right side by

1) Addition of inert gas at constant volume 2)Addition of catalyst

3) Addition of inert gas at constant pressure 4) increasing pressure

ANS:3

SOL: On addition of inert gas at constant pressure, volume of vessel increases. So position of equillibrium shifted towards more number of moles

19. A vessel of 250 litre was filled with 0.01 mole of Sb S2 3and 0.01 mole of H2to attain the

equilibrium at 4400_{C as}

 

 

 

 

2 3 3 2 2 3 2

Sb S s  H g ƒ Sb s  H S g _{after the}

with excess of _Pb2_{to give 1.195g of PbS(m.wt=239) precipitate. The value of K}

c of the reaction at 4400_{C is______}

ANS:1 SOL:

1. The equation for the reaction in the figure below is:

 

 

 

2 2 2 H g I g Heatƒ HI g 0.8 0.6 0.4 0.2 0 Time (min) 1 2 3 IH₂2 HI conc .( mm ol/L )

At the instant 3 min , what change was imposed into the equilibrium ? A) pressure was increased

B) Temperature was increased C) Iodine was added to the system D) Hydrogen was added to the system ANS:B

SOL: Given equilibrium reaction is endothermic

10. The equilibrium constants K and K1 2 for the reactions X‡ ˆˆˆ ˆ† 2Y and Z‡ ˆˆˆ ˆ† P Q ,

respectively are in the ratio of 1:9. If the degree of dissociation of X and Z be equal then the ratio of total pressure at these equilibria is

1) 1 : 3 2) 1 : 9 3) 1 : 36 4) 1 : 1

ANS:3

SOL: The initial moles of X and Z taken are ‘a’ and ‘b’ respectively

_



_{ }



1

_

1 2 2 1 1 T p a P K a a       Moles at equilibrium b(1Z)ƒ bP Q_b

 



 

2



2 2 1 1 T p b P K b b       1 1 1 2 2 2 4 _{1 1} ; 9 36 P T T P T T K P P K  P  P 

33. If 6.25 gram of a sample of NH C4 l (solid) is placed in an evacuated four litre container

at 0

27 C and after equilibrium the inside pressure becomes 0.821 atmosphere then the amount of solid NH C4 l left in the vessel is -  NH C4 l s ƒ NH3 g HCl vap

A)2.69gram B)3.125gram C) 1.765gram D) 2.856gram

ANS: A SOL: 0.821 4 0.133 0.0821 300 gas mix n    mole  3 0.133 0.0666 2 HCl nNH n   moles



4



0.0666 n NH C decomposedl 





6.25 0.1168 53.5

n initial moles of NHCl   moles



 

0.1168 0.0666



0.05 n NHC leftl   



4



0.05 53.5 2.65 W NH C leftl    grm 34. 500mL of 0.2M of HCl and 500mL of 0.2M of CH COOH3





5 2 10 Ka   are mixed together and to it six gram of NaOH is added then the pH of resulting solution

(approximately) is____ A) 5.3 B) 7 C) 6.7 D) 4.7 ANS: D SOL:





3 3 2 150 100 100 0 0 50 50

NaOH CH COOH CH COONa H O

in mole     ƒ

 

 

log WA PH pKa salt  



5 log 2



log 50 4.7 50

 

   _{  } 

1. Which of the following statement is/are correct

A) If K1be the equilibrium constant for Aƒ B and K2 be the equilibrium constant for

Cƒ D _{then equilibrium constant for the reaction}

A D+ ƒ B C+ is 1 2

K K

B) If K1be the equilibrium constant for Pƒ Q and K2 be the equilibrium constant for

Rƒ S then equilibrium constant for the reaction

P R ƒ Q S _is 1

2

K K

C) If K1be the equilibrium constant for A B ƒ C D the equilibrium constant for

nA nB ƒ nC nD is

 

1 1/ , n

K n can be fraction also

D) The equilibrium constant for the gaseous reaction C H2 4H2 ƒ C H2 6 is mol dm2 3

ANS: A

SOL: i.e. A D ƒ B C _is 1 2

K K

2. Calculate the percentage dissociation of H S2  g If 0.1 mole of H S2 is kept in 0.5 L vessel

at 1000K. The value of Kc for the reaction       2 2 2 2H S_g ƒ 2H _g S _{g is} 7 1.0 10  A) 0.1 B) 0.01 C) 1 D) 10 ANS: C SOL: 2H S2 ƒ 2H2 g S2 g

Let x be the degree of dissociation

2 7 2 0.1 0.1 2 10 0.1 0.1 c x x v v K x v                       3 6 10 2 x v   x = 0.01 Degree of dissociation

 

 0.1 0.1 x x   1%  dissociation of H S2

3. For the reaction CO2 g H2 g ƒ CO g H O2  g , Keq is 0.63 at 727 C0 and 1.26 at 927 C0 .

Then calculate the value of Keq at 1227 C0 ( use log 2 0.310  )

A) 8.32 B) 2.52 C) 7.78 D)5.04

ANS: B

SOL: Using the vant’ Hoff equation

8.32 /

H kcal mol  

Now, K2 be the equilibrium const at T2 1500K 1 1000 T  K and then K10.63

 

3 2 10 8.32 10 1500 1000 log 0.63 2.303 2 1500 1000 K     _       _{  }   2 10 10 log 0.6 log 4 0.63 K   2 0.63 4 2.52 K    

4. For the equilibrium

 

3 2 2 3 g

CH CH CH CH ƒ H₃C C

CH₃ CH₃(g)

H equilibrium constant is found to be 1.732 at 298 K. Now if in a vessel at 298 K, a mixture of these two gases be taken as represented by the point P in the fig. predict what will happen

P 750 conc. of n-butane co nc . o f is o-bu ta ne

A) Immediately, above equilibrium will be setup

B) Above reaction will go in the forward direction till it attains equilibrium C) Above reaction will go in the backward direction till it attains equilibrium D) Nothing can be said

ANS: C

SOL: From given information

0 . tan tan 75 3 . tan C conc of isobu e Q K conc of n bu e      So backward reaction.

5. For the chemical equilibrium , CaCO s3

 

ƒ CaO s

 

CO2 g_{ }

0

Hf

 can be determined from which one of the following plots?

A) 1/T log_eP_CO2 B) 1/T logePCO2 C) log T log_eP_CO2 D) log T log₂P_CO2 ANS: A SOL: Kp PCO2 0 log log 2.303 p HF K A RT    0 2 1 log log 2.0303 Hf PCO A R T    …. (i)

So, graph (a) respresents (i) and its slope will be used to determine the heat of the reaction 11. N2O2 ƒ 2NO K; 1 2 2 2 1 1 ; 2N 2O ƒ NO K 2 2 3 2NOƒ N O K; 2 2 4 1 1 2 2 NOƒ N  O K 1 3 , 1 4 , 3 2 K K x K K Y K K Z

What will be the value of XYZ ANS: 1

SOL:

2. If 20% of N O2 4molecules have dissociated in a sample of gas at 270C and 760 mm Hg,

the density of equilibrium mixture is

A ) 1.48 g/L B) 1.84 g/L C) 2.21 g/L D) 3.11 g/L ANS:D

SOL: The reaction is N O g2 4 ˆ ˆ†‡ ˆˆ 2NO g2  20

2 0.2

100

n and  

D =46, initial vapoue density, d = Vapoue density at equilibrium



1



0.2



462 1



38.3 D d d d n d d          

M P dRT M p d RT  1 76.6 ( ) 3.12 / 0.0821 300 M P d mix g l RT      Paragraph-II:(31-32) Complex-ion equilibria  aq 3 aq



3



_{ aq} 1 Ag NH ƒ _ Ag NH _  Kf



3



3 aq



3



2 _{ aq} 2 Ag NH NH  Ag NH   Kf     ƒ   1. 2 Kf Kf Kf

Overall formation of the complex _ Ag NH



_{3 2}



_   aq 2 3

 



_{3 2}



_{ aq} Ag _ NH aq _ Ag NH _    ƒ





2 ₇ 3 2 / 3 1.7 10 Kf _ Ag NH _  _ Ag _{ } NH_  





2 3 _{3 2} 1 / Kd Ag NH Ag NH Kf          _{     } 31. What is the concentration of Ag aq

 

ion in 0.01M AgNO3 that is also 1.00M NH3?

A) _{1.7 10} 7 _{B) 1.6 10} 10 _{C) 6.1 10} 5 _{D) 6.1 10} 10

ANS:D

SOL: Kf of _ Ag NO



_{3 2}



_  is very high we assume Ag_{ }aq 

reacts completely to form _ Ag NH



_{3 2}



_ ;

0.01M

 Ag aq 

1 lt. reacts completely giving 0.01M _ Ag NH



_{3 2}



_ leaving



1.00 2 0.01 



mol. 3 NH



3



2 _{ aq}  aq 2 3 aq Ag NH  Ag NH      ƒ 0.01 0 0.98 (0.01-x) x 0.98+2x









2 7 0.98 2 1 1 0.01 1.7 10 x x Kd Kf x       





2 8 . 0.98 5.9 10 0.01 x _   _{ x 6.1 10} 10

32. The Molar solubility of AgCl in 1 M NH3 at 250C is





10 1.8 10 sp K of AgCl   A) 0.01 B) 0.05 C) 0.08 D) 0.001 ANS:B SOL:







3 2



2 3 . C f sp Ag NH K K K Ag Cl Ag NH         _{   }   _{  } _    







3 2



7 10 3 2 3 1.7 10 1.8 10 3.1 10 C f. sp Ag NH Cl K K K NH  _               

 



3



_{ }   3 2 2 _{aq aq aq} AgCl NH ƒ _ Ag NH _ Cl 1.0 0 0 (1.0-2x) x x





2 3 2 3.1 10 1.0 2 x x      x 0.05

32) For the reaction, SnO s2

 

2H g2

 

‡ ˆˆˆ ˆ† 2H O g2

 

Sn l

 

at 900 K, the equilibrium steam –

hydrogen mixture was found to be 40% H2 by volume. The Kp is

1)1.15 2)2.25 3)7.75 4)10 ANS:2 SOL: 2 / 2 60 / 40 3 / 2 1.5 H O H P P   

 

2 2 2 2 2 2 2 2 2 2 1.5 2.25 H O H O P H H P P K P P    _ _    

1. The degree of dissociation ‘ ’ of the reaction : N O g2 4

 

ƒ 2NO g2

 

can be related to p K _as: (A) 4 p p K P K P    (B) 4 p p K K    (C) 1/2 / 4 / p p K P K P  _{ } _      (D) 1/2 4 p p K K  _{ } _      ANS:C SOL:

 

 

2 4 2 2 N O g ƒ NO g 0 t 1 0



n  1  2  1  eq t _{1 2}    n_g 2 1 1

 





2 1 2 1 1 p P K      _{ }     2 2 4 1 p K  P    1/2 / 4 / p p K P K P    _      2.

When A2 and B2 are allowed to react, the equilibrium constant of the reaction at 27 C is

found



Kc 4



.

 

 

 

2 2 2

A g B g ƒ AB g

What will be the equilibrium concentration of AB?

ANS:C SOL:

 

 

 

2 2 2 A g B g ƒ AB g 0 t 2 4 0 eq t 2 4 x  4 4 x  2 4 x

 

  

2 2 2 c AB K A B 





2 2 / 4 4 2 4 4 4 x x x             2 2 4 4 6 8 x x x    2 _{6 8} 2 x  x x 8 4 6 3 x 

 

2 2 4 0.66 4 4 3 x AB    

3. Consider the following reactions in which all the reactants and products are in gaseous state. 5 2 2 1 2PQƒ P Q K ; 2.5 10 3 2 2 1 ; 5 10 2 PQ R ƒ PQR K   

The value of K3 for the equilibrium : 2 2 2

1 1 1 2P 2Q 2R ƒ PQR is : (A) _{2.5 10} 3 _{(B) 2.5 10} 3 _{(C) 1.0 10} 5 _{(D) 5 10} 3 ANS:C SOL: 2 2 2 1 1 1 2P 2Q 2R ƒ PQR K ? 2 2 1 1 2P 2Q ƒ PQ 5 1 1 1 ' 2.5 10 K K    2 1 2 PQ R ƒ PQR 3 2 5 10 K    3 2 ₅ 1 1 1 5 10 2.5 10 K K K        5 1 10  

7. Decomposition of N O g2 4

 

takes place as follows : N O g2 4

 

ƒ 2NO g2

 

. D = Initial

vapour density, d = Vapour density of mixture at any instant. We get following graph when the degree of dissociation ‘’ is plotted against D

d

     

What is the value of D d at A? (A) 0 (B) 0.5 (C) 1 (D) 1.5 ANS:C SOL:  

_

_nD d_₁

_

_d At ‘A’,  0





0 2 1 D d d     1 D d 

8. Degree of dissociation for a reversible reaction at equilibrium is calculated as



1



D d

n d

   

D = Initial vapour density, d = Vapour density at equilibrium. The above relation is correctly matched for which of the following reactions :

(A)

 

5

 

 

6 6 n n X g ƒ Y g  Z g (B)

 

 

2

 

3 3 n n A g ƒ B g  C g (C)

 

4

 

 

5 5 n n X g ƒ Y g  Z g _(D)

 

 

 

2 2 n n X g ƒ Y g  Y g ANS:ABCD

SOL: In the reaction :  

_

_nD d_₁

_

_d

n Number of gaseous moles of products formed by 1 mol of reactant

In all the given reactions, n moles of products are formed by 1 mol of reactant.

9. Which is/are correct? (A) 2.303logK H S RT R        (B)    G 2.303RTlogK (C) 2.303log 2 H S K RT R         (D) 2.303log K 1



H S



RT       ANS:AB

SOL: We know     G G 2.303RTlog10Q At equilibrium  G 0 and Q K

10 0 G 2.303RTlog K     

 

10 2.303 log 1 G RT K      10 2.303 log H T S RT K       

 

10 2.303 log 2 H S RT K RT R         Passage-2: (15 to 16)

Mass action ration or reaction quotient Q for a reaction can be calculated using the law of mass action,

 

 

 

 

A g B g ƒ C g D g

  

  

C D Q A B 

The value Q decides whether the reaction is at equilibrium or not. At equilibrium Q K ; For non-equilibrium process, Q K

When Q K , reaction will favour backward direction and when Q K , it will favour forward direction.

Thus, the relative value of Q with respect to equilibrium constant decided the direction to which the reversible reaction will proceed.

15. In a reaction mixture containing H N2, 2 and NH3 at partial pressure of 2 atm, 1 atm and

3 atm respectively, the value of Kp at 725 K is 4.28 10 atm 5 2. In which direction the net

reaction will go?

 

 

 

2 3 2 2 3

N g  H g ƒ NH g

(A) Forward (B) Backward

(C) No net reaction (D) Direction cannot be predicted ANS:B

SOL: The reaction is :

 

 

 

2 3 2 2 3 N g  H g ƒ NH g 5 4.28 10 p K    2 2 3 3 3 2 3 2 1 p H p NH Q PN p     = 4.5 ; p p

Q K hence this reaction will proceed in backward direction to attain equilibrium i.e., reaction will be fast in backward direction

16. Consider the following equilibrium in a closed container

 

 

2 4 2 2

N O g ƒ NO g

At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements hold true regarding the equilibrium constant

 

Kp

and degree of dissociation

 

 ? A) Neither K norp  changes

B) Both Kp and  change

C) Kp changes but  does not change

D) Kp does not change but  changes

ANS:D

17. For the equilibrium AB g

 

ƒ A g

 

B g

 

, _{at a given temperature} 1

3rd of AB is

dissociated, then

p

P

K will be numerically equal to _______.

ANS:8 SOL: AB g

 

ƒ A g

 

B g

 

0 t 1 0 0 . eq t 1 1 3  1 3 1 3 2 1 1 4 3 3 3 3 n   



18. SO Cl2 2 and Cl2 are introduced into a 3L vessel. Partial pressure of SO Cl2 2 and Cl2 at

equilibrium are 1atm _and 2 atm _{respectively. The value of} Kp for the following reaction

 

 

 

2 2 2 2

SO Cl g ƒ SO g Cl g _{is 10. The total pressure in atm at equilibrium would be}

___. ANS:8

SOL: SO Cl g2 2

 

ƒ SO g2

 

Cl g2

 

1 atm x atm 2 atm

2 10 1 p x K    5 x atm

19. Given the hypothetical reaction :

 

 

 

2A s nB g ƒ 3C g _0.0105 x _0.45( 1₎x

p c

K _ atm and K _ mol L

at 250 C . What is the value of coefficient ‘n’? (log2.33=0.3674,log4.289=0.6322) ANS:4 SOL:

 

ng p c K K RT 





0.0105 0.45 0.082 523  ng





0.0233 0.082 523 ng





0.0233 42.886 ng log 0.0233 nglog 42.886 1.632 n_g 1.6323     1 g n   





3   2 n 1 2 n 4 2 n

21. For the reaction CaCO s3

 

‡ ˆˆˆ ˆ† CaO s

 

CO g2

 

Kp = 1.16 atm at 1073K. If 15.0 gm of 3

CaCO _{was put into 10lit. flask heated to 1073K what % age of} CaCO3would remain

unreacted at equilibrium?

A) 88% B) 12% C) 10% D) 90%

SOL: KP P CO₂

 

g

22. H g2

  

0.3mol



was injected into one litre vessel having the following composition at

equilibrium

  2   2  2 

0.1 0.1 0.40 0.1

g g g g

CO H O CO H

mol mol mol mol

 _{‡ ˆˆ}ˆ ˆ† 

Calculate the new concentration of CO2 when the equilibrium establishes again

A) 0.2 mol/L B) 3.33 mol/L C) 0.33 mol/L D) 0.022 mol/L ANS:C

SOL:

23. 0.96 gm of HI was heated till the equilibrium 2HI g

 

‡ ˆˆˆ ˆ† H g2

 

I g2

 

is reached, The

reaction mixture was suddenly cooled and the amount of iodine produced required

15.7 10

N

ml _{hypo solution. The Kp of the reaction is}

A) _{7.85 10} 4 _{B) 7.5 10} 3 _{C) 5.93 10} 3 _{D) 17.52 10} 3

ANS:D SOL:

24. For the reversible reaction N g₂

 

3H g₂

 

ƒ 2NH g₃

 

at _{500 C}0 _{, the value of} 5

p

K is 1.44 10_ 

when partial pressure is measured in atmosphere. The corresponding value of Kc with concentration in mol/L is

A)

_

_

5 2 1.44 10 0.082 500     B)





5 2 1.44 10 8.314 773     C)

_

_

   5 2 1.44 10 0.082 773 D)





5 2 1.44 10 0.082 773     ANS:D SOL: Passage-1: (29to30)

Consider the decomposition of a pare solid of CaCO3in a closed vessel. Let  be the

extent of the reaction

 

 

3 2

CaCO _{‡ ˆˆ}ˆ ˆ† CaO s CO g

3 2

d dnCaCO dnCaO dnCO

    



2 3



, CaO CO CaCO T P G dG    d G        _{ }      29. At equilibrium A)

 

G T P, 0 B) , 0 T P G     _  _    C) , 0 T P G     _  _    D) , 0 T P G     _  _    ANS:B SOL: At equilibrium , 0 T P G H        _{ } _

30. Pure ice can be made to melt slightly below 0 C

B) By increasing the pressure C) By adding more ice

D) By increasing volume of the vessel ANS:B

SOL:

35. KNO3(s) dissociates on heating as: KNO3(s) ƒ KNO2 (s) + ½ O2 (g). At equilibrium in a closed vessel:

a) Addition of KNO3(s) favours forward reaction. b) Addition of KNO2 (s) favours reverse reaction c) Increasing temperature favours forward reaction d) Decreasing pressure favours forward reaction ANS:CD

SOL:

39. If logKc log 1 0

Kp RT  then above is true for the following equilibrium reaction

A) 3

 

2

 

2

 

1 3 2 2 NH g _{‡ ˆˆ}ˆ ˆ† N g  H g _B)

 

   

3 2 CaCO Cao CO s s g  ˆ ˆ† ‡ ˆˆ C) 2NO g2

 

‡ ˆˆˆ ˆ† N O g2 4

 

D) H g2

 

I g2

 

ˆ ˆ†‡ ˆˆ 2HI g

 

ANS:AB SOL:

40. If two gases AB2 and B C2 are mixed the following equilibrium are readily established.

 

 

 

 

2 2 3

AB g B C g ƒ AB g BC g

 

2

 

3 2

 

BC g B C g ƒ B C g

If the reaction is started only with equal moles of AB2and B C2 , then which of the

following is necessarily true at equilibrium

A)



AB3 eq

  

 BC eq B)



AB3

 

eq  B C2



eq

C)



AB3

 

_eq  B C3 2



_eq D)



AB_{3 eq}

  

 BC _eq

ANS:CD SOL:

34. The rate of formation of 6 6 3 2 6 12

f

b K K

C H  H _{‡ ˆ ˆˆ}ˆ ˆ ˆ† C H _{for the forward reaction is first order}

with respect to C H6 6 and third order with respect toH2. Which one of the following

is/are correct a) eq f b K K K  b)







6

 

12 3 6 6 2 eq C H K C H H  c) rf =K C Hf

[

6 8

] [ ]

H2 3 d)



 

1 6 12 2 b b r K C H H  ANS:ABC SOL: rf K C Hf



6 6

 

H2 b b r K unknown At equilibrium rf rb

39. A chemical reaction occurs in three paths having rate constants k k1, &2 k3 respectively. If 1, 2

Ea Ea _and Ea3are 4,5 & 8kJ respectively and overall rate constant

1 3 2 . k k k k  _Assuming 1 3 2 av A A A A

 _{, the overall energy of activation in kJ is___.}

ANS:7

SOL: . E E E1 3 2/RT AV AV

k  A e   E_av    



4 8 5



7kJ

26. A vessel of volume of V lit contains an equilibrium gaseous mixture that consist of 2 mole each of PCl5, PCl3 and Cl2 . The equilibrium pressure is 3 atm and the temperature is T k . A

certain amount of chlorine is now introduced keeping the pressure and temperature constant , until the equilibrium volume is 2 V lit. Calculate the number of moles of Cl2 added . The given

equilibrium is PCl5 ƒ PCl3Cl2

a) 3.33 b) 5.00 c)6.67 d) 2.30

ANS:C SOL:

23. For the preparation of ammonia , N2 and H2are taken initially in 1:3 mole ratio. At

equilibrium 25% of each reactant reacted and showing total equilibrium pressure as 28 atm. Then what is the equilibrium pressure of NH3 at the same temperature ?

ANS:4 SOL:

24. For the reaction : N O2 4  2NO (g)2

This equilibrium is studied by measuring the vapour density of the equilibrium mixture. Which of the following statement is correct at moderately high pressure?

A) 1 KP 2 P    B) 1 2 P K 1 . 2 P     _ _ C)   1₄ K_PP D) 1 2 P K 1 . 4 P     _ _

ANS. . B

p p K K 4p   

At moderately high pressure, KP << 4P And so,





12 P 1 K / P 2  

39. When NaNO3 is heated in a closed vessel, oxygen is liberated and NaNO2 is left behind. At equilibrium

A) addition of NaNO2 favours reverse reaction B) addition of NaNO3 favours forward reaction C) increasing temperature favours forward reaction D) increasing pressure favours reverse reaction.

ANS. C, D

22. The adjoining Fig. contains two beakers (1 and 2) which are placed in a sealed container

12

2

100m

lof0.01M

aq.C

aC

lsoln.

100m

_aq.U

_reasoln.Fig

lof0.02M

The volume of the solution in 1 and 2 beakers left respectively at equilibrium are: A) 100ml, 100ml B) 120ml, 80m C) 80ml, 120ml D) 60ml, 140ml ANS C

SOL. 2 3

100- x =100+x

22. The equilibrium, 2CO + O2 ƒ 2CO2 heat, is subjected to the following combination

of stresses. In which one of the following cases, it is not possible to predict the effect of stress on the direction of reaction?

A) Addition of CO and removal of CO2

B) Increase in temperature and decrease in volume C) Addition of O2 and decrease in volume

D) Addition of a catalyst and decrease in temperature. ANS B

Conceptual

36. Which of the following represents “hydrolysis” process?

(a) NH4++H O2 ƒ NH3+H O3 + (b) NH4++2H O2 ƒ NH OH4 +H O3 + (c) HCO3- +H O2 ƒ H CO2 3+OH- (d) HCO3- +H O2 ƒ CO32- +H O3 + ANS. BC

37. The dibasic acid(s) is/are ____

a) chromic acid b) permanganic acid

c) phosphorous acid d) Peroxy di sulphuric acid ANS ACD

1. When NaNO3is heated in a closed vessel. Oxygen is liberated and NaNO2 is left behind.

At equilibrium

A) Addition of NaNO2favour reverse reaction

B) Addition of NaNO3 favour forward reaction

C) Increasing temperature favour forward reaction D) Increasing pressure favour forward reaction ANS C 3 2 2 1 ( ) ( ) ( ) 2 NaNo s ˆ ˆ†_{‡ ˆˆ} NaNo s  O g 3

NaNo _and NaNo2 are in solid state, changing their amount has no effect on equilibrium.

Increases temperature will favour forward reaction due to endothermic nature of reaction. Also, increase pressure will favour backward reaction in which some O g2( )

will combine with.

2( )

NaNo s _forming NaNo s₃( )_.

2. At 1100K, water vapour decomposes into H2 and O2 only to the extent of 5

10 % when pressure is maintained at 1 atm. The value of Kp for the decomposition of one mole of

2 H O _is A) _{5 10} 22 _{B) 5 10} 11 _{C) 5 10} 15 _{D) 3.3 10} 11 ANS. B 2 2 2 7 7 7 At equilibrium 1 ( ) ( ) ( ) 2 10 1 10 10 2 : H O s H g O g      ˆ ˆ† ‡ ˆˆ

Total moles = 1, Total pressure = 1 atm 2 2 2 Kp P p P H O H O  7 7 10 10 2 Kp 1   _   107 5 10 4  5 10 11

7. 0.5 mole of H g2

 

and 1.0 mole of HI g

 

( but noI2 ) are added to a 1.0 litre vessel

and allowed to reach equilibrium according to the following reaction

 

 

 

2 2 2

H g I g ƒ HI g _{If x is the equilibrium concentration of} I g₂

 

_{. Then correct}

expression for equilibrium constant is A)

_



_{  }



2 1 2 0.5 x x x   B)







  

2 1 2 0.5 x x x   C)







1 20.5



x x x   D)









2 0.5 1 2 x x x   ANS. A H g2

 

I g2

 

ƒ 2HI g

 









Initial con 0.5 mol/lit 0 mol/lit 1.0mol/lit

Equilibrium conc 0.5 x mol/lit x mol/lit 1 2x mol/lit       2 1 2 0.5 c x K x x   

8 For the reaction Ag CN

 

2 Ag 2CN ,

 _{ }

  

  ˆ ˆ†‡ ˆˆ the equilibrium constant at 25 C0 is 19

4 10 ._  _{If a solution is 0.1M in KCN and 0.03M in}

3

AgNO _{originally, at equilibrium,}

the conc. of Ag _is A) _{7.5 10 M} 16 _{B) 7.5 10 M} 18 _{C) 1.25 10 M} 19 _{D) 1.25 10 M} 17 ANS. B

 

19 2 1 2 10 4 Ag CN_{‡ ˆˆ}ˆ ˆ† Ag CN k   I conc 0.3 0.1 ---F conc X 0.04 0.03     19 2 1 0.03 10 4  0.04 X 18 7.5 10 X    M

9. Ammonia under a pressure of 20 atm. at _{127 C}0 _{is heated to 327 C}0 _{in a closed vessel.}

Under these conditions, ammonia is partially decomposed to N2 and H2according to the

reaction 2NH3ˆ ˆ†‡ ˆˆ N23H2

After decomposition at constant volume in the vessel, the pressure increases to 50 atm. which of the following statements are correct?

A) The degree of dissociation of NH3 in 2/3

B) The Kp of the reaction at _{327 C}0 _is 3 2 2.7 10 atm .

C) The pressure of N2 and NH3gas at equilibrium is 10 atm. each

D) The pressure of N2 at equilibrium is half that of NH3.

ANS. A,B,C 3 2 2 2 3 At 400 20atm 0 0 At 600 30 atm 0 0 NH N H k k  ˆ ˆ† ‡ ˆˆ

At constant value let 2x atm. NH3 disappear 30 2 x x 3x 30 2 50 2 20 10atm P Total  x x x 2 20 2 30 30 3 x     3 4 3 2 2 10 30 27 10 2.7 10 10 10 Kp       

10. Which of the following statements is/are wrong

A) At equilibrium, concentration of reactants and products becomes constant, because the reaction comes to an halt

B) Addition of inert gas at equilibrium for  n 0 at constant volume will favour forward reaction.

C) Addition of catalyst speeds up the forward reaction more than the backward reaction D) Equilibrium constant of an exothermic reaction increases with increase in temperature.

ANS. A,B,C,D

12. 2CaSO s4

 

ƒ 2CaO s

 

2SO g2

 

O g2

 

, H 0

Above equilibrium is established by taking sufficient amount of CaSO s4

 

in a closed

container at 1600K. Then which of the following may be correct option ( Assume that solid CaSO4 is present in the container in each case)

A) Mole of CaO s

 

_{will increase with the increase in temperature}

B) If the volume of the container is doubled at equilibrium then partial pressure of

 

2

SO g _{will change at new equilibrium}

C) If the volume of the container is halved partial pressure of O g2

 

at new equilibrium

will remain same

D) If two moles of the He gas is added at constant pressure then the moles of CaO s

 

will increase ANS. A,C,D

22. For the reaction, A g

 

B g

 

ƒ C g

 

D g

 

_{the initial concentrations of A and B are}

equal. The equilibrium concentration of C is two times the equilibrium concentration of A. The value of the equilibrium constant is _____.

ANS 4



AgBgC

_ax

_aa00

_



_ax

_



gD

_xx

g

0

t

_eq

t





2 x a x 3x2a 2 3 x a

   

A  B  a 2₃a1₃a

   

C  D  2₃a

  

  

2 2 3 3 ₄ 1 1 3 3 c a a C D K A B _{a a}      .

23. Equilibria, X ƒ 2Y and Z ƒ P Q _{are established with 1 mole of X and Z taken}

initially in separate vessels with equilibrium pressures of 1 atm and 36 atm respectively. If the degree of dissociations are same. Then KP2 nKP1. What is the value of ‘n’?

ANS 9

1) initial concentration of the reactants 2) pressure

3) temperature 4) all of these

ANS:3 SOL:

20. For the dissociation reaction N O g2 4

( )

ˆ ˆ†‡ ˆˆ 2NO g2

( )

, the degree of dissociation (a) in

terms of Kp and total equilibrium pressure P is :

1) 4 p p P K K a= + ₂₎ 4 p p K P K a = + 3) ₄ p K P a = 4) 4 p P K a = ANS:2 SOL: N O2 4‡ ˆˆˆ ˆ† 2NO2 1- 2aa 2 2 4 1 p K a p a = -2 2 1 4 p K p a a =

-21. Consider the partial decomposition of A as 2A g

( )

ˆ ˆ†‡ ˆˆ 2B g

( )

+C g

( )

At equilibrium 700mL gaseous mixture contains 100mL of gas C at 10 atm and 300K. What is the value of Kp for the reaction ?

1) 40 7 2) 1 28 3) 10 28 4) 28 10 ANS:3 SOL: 2Aˆ ˆ†‡ ˆˆ 2B C+ 400 200 100 2 2 2 .110 4 7 p K = 1. 2 ( ) 4

-CoCl aq _{is blue in colour while ( 2 )} 2 _{( )}

6 Co H O + aq

é ù

ê ú

ë û is pink. The colour of reaction

mixture ( ) _{( ) ( )} ( ) ( ) 2 2 2 6 4 4 6 2 + - -é _{ù + +} ê ú

ëCo H O û_aq Cl_aq ƒ CoCl aq H Oi is blue at room temperature while it is

pink when cooled hence

A) Reaction is exothermic in forward direction B) Reaction is endothermic in forward direction

C) Equilibrium will shift in forward direction on adding water to reaction mixture D) Equilibrium will not shift on adding water to equilibrium mixture

ANS:B