Solution Chemical Equilibrium Ionic Equilibrium Chemical Kinetics Online

ANDHERI / BORIVALI / DADAR / THANE / POWAI / CHEMBUR / NERUL / KHARGHAR IIT – JEE: 2017 CRASH COURSE (C - 2) DATE:18/11/16 TIME: 2HRS MAIN MARKS: 120 TOPIC: CHEMICAL EQUILIBRIUM, IONIC EQUILIBRIUM, CHEMICAL KINETICS

ANSWER KEY 1.

(D)

2.

(A)

3.

(B)

4.

(D)

5.

(B)

6.

(B)

7.

(C)

8.

(B)

9.

(B)

10.

(C)

11.

(C)

12.

(C)

13.

(D)

14.

(C)

15.

(C)

16.

(D)

17.

(C)

18.

(C)

19.

(D)

20.

(D)

21.

(A)

22.

(D)

23.

(C)

24.

(B)

25.

(B)

26.

(D)

27.

(D)

28.

(C)

29.

(C)

30.

(C)

ANDHERI / BORIVALI / DADAR / THANE / POWAI / CHEMBUR / NERUL / KHARGHAR IIT – JEE: 2017 CRASH COURSE (C - 2) DATE:18/11/16 TIME: 2HRS MAIN MARKS: 120 TOPIC: CHEMICAL EQUILIBRIUM, IONIC EQUILIBRIUM, CHEMICAL KINETICS

SOLUTION 1.

(D) Since given K1  K 2 1011 e3000/T  1010 e2000/T  10  e1000/T 1000 n10  T 1000 T K 2.303

2.

(A)

3.

(B)

 salt   pH  14  pOH  14  pKb  log   base   80  pK b  6.0 So, 7.40  14  log pKb  log 20  K b  1.0 106

KOH is a strong electrolyte, so it ionizes completely. Hence OH   0.01M , which remains constant, even at 500 C (assuming the volume of solution to be constant.) Hence pOH   log 0.01  2 , remains same. At 500 C , pKw  14, Since dissociation of water is endothermic So, pH at 500 C   pK w  2   12 4.

(D)

A t 0 2a t  eq 2a  x a  3x Given a-3x

=

3B a 2x 2x,

Hence % of B reacted 

5.

(B)

xa 5

3x 100  60% a

2C  0

D 0 x

 In   pH  pK In  log  pK In  1 (given)  HIn 

 In    log  1  HIn  % ionized 

6.

 In     0.1  HIn 

0.1100 10   9.1% 0.1  1 1.1

(B) NH2COONH4 (S)

2NH3  g   CO2  g  2P

P

2 K P  PNH  PCO2   2P   P  3.2 105 atm3 3 2

 P  2 102 atm  PNH3  2P  2  2 102  4 102 atm 7.

(C) For equal Ea, forward and Ea backward, the reactant and product molecules have equal energies. Hence, H  0; G  H  TS Or G  TS For spontaneity G must be negative and hence S  0

8.

(B)

H RT Since H is –ve for exothermic reaction, hence the plot (straight line) not passing through the origin  H  and will have a positive slope  i.e. . R   (B) For a 10K rise in temperature, collision frequency increases merely by 1 to 2 % but the no. of effective collisions increases by 100% to 200%.

nKp = Constant t 

9.

10.

(C)

2.0 103  4.0 K1 at 27 C  5 104 8 102  20 K 2 at 1270 C  4.0 103 K H T2  T1  log 2  K1 2.303R T1T2 0

T1  300K, H 

11.

(C) Ice

T2  400K

2.303  8.314  300  400  20.0   log  J 100  4.0 

water, P = 1 atm,

T = 273.15 K

For P  1 and T  273.15K, the equilibrium shifts completely in forward direction leading to reduction in volume of the system, i.e. ice melts completely. 12.

(C)  H  1 nK  constant     R  T Since Ea (forward) > Ea (backward), the reaction is endothermic, and so H  0 H  Slope   R

13.

(D) AgCl  s 

Ag   aq  Cl  aq 

H2O

S

S

Ksp  Ag   Cl 

(A) In pure water solubility  K sp (B) In 0.1 M NaCl (strong electrolyte), Cl is the common ion, and Sc due to common ion effect ionization of AgCl will be suppressed, so solubility is decreased. Ksp  Ag 1  Cl   S1 S1  0.1 where S1  0.1M  Ksp  S1  0.1

Ksp 0.1 (C) Here Cl   S2  0.2M  0.2M

 S1 

 Ksp  S2  0.2

Ksp M 0.2 (D) Ag  ion forms complex ion with NH3 as:

S2 



Ag   aq   2 NH3  aq 

Ag  NH3 2  Here solubility is increased (more than pure H 2 O , ) due to complehexation 14.

(C)

t  eq

PCl5  g 

PCl3  g  

Cl2  g 

1 





1  1  P  total pressure,

Mole reaction

KP 

12 P1  22 P2  1  12 1   22

Since 1  0.5,

 P2  0.75atm . 15.

(C)

  1  1    degree of dissociation

2  0.8,

P1  4atm

Data shows that t1/2 is inversely proportional to the initial pressure of the reactant. Hence, second order reaction. 16.

(D)

PC

PCl3  Cl2

5

0 0 t 0 1   T = eq. 1  Total moles at equilibrium  1        1   1 Molecular wt  moles Molecular weight initially n   Molecular weight at equilibrium 1

208.32  1    1.68 124   0.68



17.

(C) (I)  HCl in the mixture 

50  0.2  0.1M 100

 H   from HA (weak acid) is negligible due to common ion effect

 H    H 

HCl

 0.1M

 pH  1

(II) NaA  HCl  NaCl  HA 50  0.2  0.1M  HA  100 Here no HCl is left So  H   K a C  1.0 105  0.1  103  pH  3 18.

(C) The reaction is exothermic and takes place with a decrease in number of molecules of gaseous species. Obviously high P and low T are the favorable conditions for of shifting the equilibrium to product side. However, in ordinary flow process optimum elevated temperature is required to have more NH3 due to high activation energy of the reaction. ( N 2 is quite stable at low T)

19.

(D) NH4OH  aq   H  aq 

NH4  aq   H2O 

 NH 4  K1   1.8 109     NH 4OH H 

NH4  OH  aq 

NH4  aq   OH  aq 

 NH 4  OH   K2   NH 4OH  K2   H   OH    K w  1.0 1014 K1



K2 1.0 1014  1.8 109   1.0 1014  K1

 1.8 105 20.

(D) x y Rate  k  A  B 

 Rate    0.024   0.070  3 x y  Rate   0.024   0.035 2

x

y



8 103 8 1103

Or 2 y  8  y  3

 Rate 2 Also  Rate 1  2x  1 Or x  0  Rate 

21.

(A) t 0

8 103  0.024   0.070     2x 8 103  0.012  x  0.070  y

d  A dt

x

y

 K  B   A 3

0

NaCN  aq  

HCl  aq   HCN  aq   NaCl  aq 

50ml  0.1M  5mmol

50ml  0.2M  10m mol

HCl excess  10  5  5 m mol 5mmol  H     0.05M 100 ml

 pH   log 0.05   log  5 102 

 2  log5  2  0.6990  1.3 HCN practically remains unionized due to common ion effect of remaining HC 22.

(D) Each molecule of AB gets activated after absorption of one quantum of light. Hence, rate of formation of AB* is proportional to the rate of quantum absorbed, i.e., the intensity of light I.

23.

(C)

0.1 pH   log 0.05  1.301.  0.05M ; 2 (b) Complete neutralization of strong acid and strong base, pH  7

 a  H  

(c) Hydrolysis of the salt CH3COOK,

pH  7

(salt of weak acid and strong base) (d) Hydrolysis of the salt NH4 NO3 , pH  7 (salt of strong acid and weak base)

24.

(B) Ea , uncata

T1



Ea , uncat T2



T2 Ea , cat 75    0.75 T1 Ea ,uncat 100

300 300  0.75  T1   400 K  127o C T1 0.75

25.

(B)

Ksp Ca  OH 2   4.0 106  Ca 2  OH

2

 0.01 OH  

2

 OH   2 102 M

 pOH   log  2 102   2  log 2

 pH  14   2  log 2  12  log 2 26.

(D) Rate of SN2  5.0 105  OH    R  X 

% of SN 2   

27.

Rate of SN 2 100 Total rate  SN1  SN 2  5 105  OH    R  X 

5 105  OH    R  X   0.2 105  R  X 

5 10

5 10

5

5

102 

102    0.2 105 

100  20%

(D) K  Ae Ea / RT , K  2.3 103 min 1 e Ea / RT  fraction of molecules crossing over the energy barrier 0.001   105 100 Hence 2.3 103 min 1  A  1.0 105   A  230 min 1 Kmax  A when T  

Hence Kmax  230 min 1 28.

100

(C) CO2  H2O

HCO3  H

 HCO3   H   K1   4.5 107 CO2   HCO3  4.5 107 4.5 107 Or    11.25 4 108  H   CO2 

29.

(C)

Ag  NH3   2  The formation constant of the complex 108 This value is so high that we can practically assume that most of the Ag  present in the reaction will undergo complex formation Ag   2 NH3



i.e.  Ag  NH3 2   Ag   added  0.02 M

 NH3 free  0.14   2  0.02   0.1M  Ag    NH3   Ag     0.10  K d  10      0.02   Ag  NH3 2  2

8

 Ag    2 108 M

30.

(C)

4.606 103   K2 Ea log   2 K1 2.303RT 2.303  2  500 K  2  102  100 K1

2