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Page 1: Dr. Sangeeta Khanna Ph.D 1sangeetakhanna.com/wp-content/uploads/2016/07/2-Grand-test-3.pdfDr. Sangeeta Khanna Ph.D Dr. Sangeeta Khanna Ph.D 4 CHEMISTRY COACHING CIRCLE D:\Important

Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Physical\Test\Grand Test\Grand Test-3\+2 Grand test-3.doc

Page 2: Dr. Sangeeta Khanna Ph.D 1sangeetakhanna.com/wp-content/uploads/2016/07/2-Grand-test-3.pdfDr. Sangeeta Khanna Ph.D Dr. Sangeeta Khanna Ph.D 4 CHEMISTRY COACHING CIRCLE D:\Important

Dr. Sangeeta Khanna Ph.D

Dr. Sangeeta Khanna Ph.D 2 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Physical\Test\Grand Test\Grand Test-3\+2 Grand test-3.doc

Test Date – 2.7.2016 Topic: Redox Reaction, Electrochemistry, Chemical Kinetics & Nuclear Chemistry

READ INSTRUCTIONS CAREFULLY 1. The test is of 2 hour duration.

2. The maximum marks are 290.

3. This test consists of 69 questions.

4. Keep your mobiles switched off during Test in the Halls.

(Single Correct Choice Type) Negative Marking [-1]

This Section contains 45 multiple choice questions. Each question has four choices A), B), C) and D) out

of which ONLY ONE is correct. Marks: 45 × 4 = 180

1. Reaction 2A + B C occurs in a single step. Starting with 3 mol of A and 2 mol of B, when 1 mol of C has been formed, the ratio of present rate to the original rate will be:

a. 18

1 b.

9

1 c.

3

1 d.

6

1

A Sol. r = k[A]2[B], r1 = k[3]2[2] = 18k When 1 mol of C has been formed, 2 mol of A and 1 mol of B have been consumed. rf = k[1]2[1] = k

18

1

k18

k

r

r

i

f

2. For the reaction:

H2 + I2 2HI; dt

]HI[d

2

1 = (1.6 × 10-3) [H2][I2] – (2.1 × 10-2) [HI]2

What is the rate constant for the forward reaction?

a. 1.6 × 10-3 b. 2.1 × 10-2 c. 1.6 × 10-3 × 2.1 × 10-2 d. 2

3

101.2

106.1

A Sol. Compare with

dt

]HI[d

2

1 = k1[H2][I2] – k2[HI]2

k1 = 1.6 × 10-3 3. Which of the following graph represents the first order reaction? a. b.

Conc.

Slope = k

Rate

Con

c.

Time

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Dr. Sangeeta Khanna Ph.D 3 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Physical\Test\Grand Test\Grand Test-3\+2 Grand test-3.doc

c. d. All represent 1st order reaction

D 4. (i) P + Q A(fast)

(ii) A + R B (slow) (iii) B + Q S + T (fast) are the elementary steps of the reaction P + 2Q + R S + T The rate law of the reaction is:

a. r = kl[P] [Q] b. r = k [P]2 [Q] [R]3 c. r = k[P]1/2[Q] [R]1/3 d. r = k[P] [Q] [R] D Sol. From slow step (ii), A + R B r = k2[A][R] ...(i) From first step (i)

[Q] ]P[

]A[= k1 (Equilibrium constant)

[A] = k1[P] [Q] ...(ii) From equations (i) and (ii), r = k2k1[P][Q][R] i.e., r = k[P][Q][R] 5. For the reaction, A B + C + D, starting with initial pressure 400 atm. if the total pressure after 2

hours is 800 atm, the value of rate constant is: (consider the reaction to be of 1st order):

a. 0.643 hr –1 b. 0.463 hr –1 c. 4.63 hr –1 d. 0.347 hr –1 D

Sol. A B + C + D At t = 0, 400 atm After 2Hr, 400 – x x x x 800 = 400 – x + x + x + x x = 200

1-

A

A hr 347.0200

400log

2

303.2

P

Plog

2

303.2k

6. When KMnO4 acts as an oxidising agent and ultimately forms MnO 24

, MnO2, Mn2O3 and Mn2+ in

separate reactions then the number of electrons transferred in each case respectively is:

a. 4, 3, 1, 5 b. 1, 5, 3, 7 c. 1, 3, 4, 5 d. 3, 5, 7, 1 C

Sol. MN+7 of 4MnO changes to:

Mn+6 of MnO 24

by the gain of le–, Mn+4 of MnO2 by the gain of 3e–,

Mn+3 of Mn2O3 by the gain of 4e–, Mn+2 by the gain of 5e–

7. Which of the following reactions is possible at the anode?

a. 2Cr3+ + 7H2O 272OCr + 14H+ + 6e– b. F2 + 2e– 2F–

c. OHe2H2O2

122 d. Cu+2 + 2e– Cu

A

Sol. Cr3+ is oxidised to Cr6+ (in Cr2O2

7 ). Oxidation takes place at anode.

log

[A

]

Time

log[A]0

Intercept constant

303.2

kSlope

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Dr. Sangeeta Khanna Ph.D 4 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Physical\Test\Grand Test\Grand Test-3\+2 Grand test-3.doc

8. KCl, BaCl2 and AlCl3 are used as electrolytes in three different cells connected in series. If 3F charge is passed, the molar ratio of masses of K, Ba and Al deposited at cathode will be:

a. 1 : 1 : 1 b. 1 : 2 : 3 c. 3 : 2 : 1 d. 3 : 15 : 1 D

Sol. Valencies of K, Ba and Al are 1, 2 and 3 respectively. 3F charge will produce 3 mol K, 3/2 mol Ba and 1 mol Al.

Molar ratio of masses = 3 : 2

3 : 1 or 3 : 1.5 : 1

9. 965 coulomb of electricity is passed through a solution of AlCl3. The mass of Al deposited on cathode is (At. mass of Al = 27):

a. 27g b. 2.7 g c. 0.27 g d. 0.09 g D

Sol. 96500 charge = 1 Eq. of Al = g3

27

965C charge = g09.096500

965

3

27

10. 0.5 L solution of 1.0 M molarity of NaCl is electrolysed by passing a steady current of 5.0 A for 965 seconds. The pH of solution is: [Assume volume remain constant]

a. 4.7 b. 1 c. 10.7 d. 13 D

Sol. 96500 C 1 mol Na+ 1 mol OH–

965 × 5C = -2 OH mol 10596500

5965

in 0.5 lit ; 10-1 mol/lit

pOH = 1 pH 13 11. What are x and y in the following reaction?

xBr2 + 6OH– BrO 3 + yBr– + 3H2O

a. x = 3, y = 5 b. x = 5, y = 3 c. x = 1, y = 1 d. x = 2, y = 3 A

12. -12-1ClNH

mol cm 1304

-12-1NaOH mol cm 220

-12-1NaCl mol cm 110

If m of NH4OH at a given concentration is 12.0 -1 cm2 mol-1, what is its percentage dissociation? a. 1% b. 2% c. 3% d. 5% D

Sol. NaCl -NaOH ClNH OHNH mm4m4m

= 130 + 220 – 110 = 240S cm2 mol–1

%5100240

12100%

m

cm

13. A solution of 50 moles of NaOH in 1 m3 solution offered a resistance of 50 ohm using a cell of cell constant 100 m-1. Molar conductance of the solution is:

a. 1.0 × 10-2 S m2 mol–1 b. 2.0 × 10-2 S m2 mol–1 c. 4.0 × 10-2 S m2 mol–1 d. 5.0 ×10-2 S m2 mol–1 C

Sol. C1000

1

AR

1m

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1-22 mol m S 104

1000

501000

1

1

100

50

1

(using 1m3 = 1000 L or dm–3 C = 1000

50mol dm–3)

14. The conductivity of 0.25 M solution of a weak electrolyte XY is 0.0125 S cm – 1. m of the electrolyte

is 500 S cm2 mol–1. The Ostwald dilution constant of the electrolyte is:

a. 2.5 b. 2.5 × 10 – 1 c. 2.78 × 10 – 3 d. 2.78 × 10 – 4 C

Sol. 1.050025.0

0125.010001

C

1000

mm

m

32

1078.29.0

1.01.025.0

1

CK

15. In the arrangement shown in the figure, on pressing the key: a. Current will flow from Ni to Ag

b. current will flow from Ag to Ni

c. Ag+ ions will move towards Ni much faster

d. no current will flow D Sol. Ni metal is in direct contact with AgNO3 (aq), giving a direct

redox reaction. Only heat will be produced and not the current. 16. In the cell that utilizes the reaction,

Zn(s) + 2H+ (aq) Zn2+ (aq) + H2(g) (E = emf) Addition of H2SO4 to cathode department will: a. lower the E & shift the equilibrium to the left

b. increase the E & shift the equilibrium to the left c. increase the E & shift the equilibrium to the right

d. lower the E & shift the equilibrium to the right C

Sol. H+ ions are shown as cathode ions. Increase in concentration of ions increases EMF. Hence, EMF of cathode will increase by addition of H2SO4. This will increase the EMF of the cell and shift the reaction towards the product side, i.e., the right side.

17. For the given three cells, which of the following is correct?

(a) Zn | Zn2+ (1.0 M) || Cu2+ (1.0M)| Cu; E1 (b) Zn | Zn2+ (1.0M) || Cu2+ (10.0M) | Cu; E2 (c) Zn | Zn2+ (10.0M) || Cu-2+ (0.1M) | Cu; E3 a. E1 > E2 > E3 b. E3 > E2 > E1 c. E2 > E1 > E3 d. E1 = E2 = E3 C Sol. In the cell number (ii), the concentration of ions at cathode has increased which will increase the EMF

of the cell. In the cell number (iii), the concentration of ions at anode has increased which will decrease the EMF of the cell.

18. For the cell, Pt; H2 (1 atm) | HCl (0.1M)|| CH3COOH (0.1M)|H2 (1 atm); Pt, the emf will not be zero because

a. the temperature is constant b. acids used in two half cells are different in basicity c. emf depends upon molarities of acids used d. pH of 0.1 M HCl & 0.1 M CH3COOH is not the same D

e–

Key

AgNO3(aq)

Ag Ni

G

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Sol. CH3COOH (0.1 M) will not give 0.1 M H+ -ion concentration because of being weak acid while HCl (0.1 M) will give 0.1 M H+ -ion concentration.

19. In H2 – O2 fuel cell, which of the following is incorrect when acid is used as electrolyte?

a. anode reaction : H2(g) 2H+ + 2e–

b. Cathode reaction : 2H+ + 2e– + )g(O2

12 H2O()

c. Net reaction: H2(g) + 2

1O2(g) H2O ()

d. Hydrogen act as oxidising agent D 20. Which of the following is not correct w.r.t. rusting of iron?

a. Rust is FeOFe2O3xH2O b. Presence of less reactive metal as impurity increases the process of rusting of iron c. Presence of salts in water help rusting d. During rusting, iron forms anode. A

Sol. Chemically, rust is Fe2O3.xH2O. 21. In which of the following reactions H2O2 acts as a reducing agent?

a. 2FeCl2 + 2HCl + H2O2 2FeCl3 + 2H2O b. Cl2 + H2O2 2HCl + O2

c. 2HI + H2O2 2H2O + I2 d. H2SO3 + H2O2 H2SO4 + H2O B

Sol. The oxidation state of oxygen changes from – 1 to 0 22. A catalyst lowers the activation energy of a reaction from 20 kJ mole–1 to 10 kJ mole–1. The

temperature at which the uncatalysed reaction will have the same rate as that of the catalysed at 270C is:

a. -1230C b. 3270C c. 3720C d. +230C B

Sol. 21

a

T

Ea

T

E

2T

20

300

100 T2 = 600 K = 327°C

23. The reaction: 2NO + 2H2 N2+2H2O has been assigned to follow following mechanism: I. NO + NO N2O2(fast)

II. N2O2 + H2 N2O + H2O(slow)

III. N2O + H2 N2 + H2O (fast) The rate constant of step II is 1.2 x 10 -4 mole–1 L. min–1 while equilibrium constant of step I is 1.4 × 10–2

mol–1L min–1. What is the rate of reaction when concentration of NO and H2 each is 0.5 mole L– 1. a. 2.1 x 10–7 mole L–1 min –1 b. 3.2 x 10–6 mole L–1 min–1 c. 3.5 x 10–4 mole L–1 min –1 d. 1 × 10–6 mole L–1 min–1 A Sol. Kobs = KI . KII = 1.2 × 10-4 × 1.4 × 10-2 = 1.68 × 10-6 mol-1 L min-1 Rate = Kobs. [NO]2 [H2]

= 1.68 × 10-6 × [0.5]2 [0.5] = 2.1 × 10-7

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24. For reaction 3A product, it is found that the rate of reaction increases 4 fold when concentration of A is

increased 16 times keeping the temperature constant. The order of reaction is?

a. 2 b. 1 c. 3 d. 0.5

D

Sol. (D) Rate A

i.e. Rate [A]1/2

Thus order = 0.5

25. The kinetic data for the reaction :

2A + B2 2AB are as given below:

[A] (mol/L) [B2](mol/L) Rate(mol/L/min)

0.5 1.0 2.5 x 10-3

1.0 1.0 5.0 x 10-3

0.5 2.0 1 x 10-2

Hence the order of the reaction with respect to A and B2 are, respectively,

a. 1 and 2 b. 2 and 1 c. 1 and 1 d. 2 and 2

A

Sol. 2.5 × 10-3 = K[0.5] [1.0] …(i)

5 × 10-3 = K [1.0] [1.0] …(ii)

1 × 10-2 = K[0.5] [2.0] …(iii)

Dividing (i) by (ii) we get

2

1

2

1 = 1

Dividing (i) by (ii) we get

4

1 =

2

1 = 2

26. An artificially produced radioactive isotope decomposes according to the first order rate law with a half-

life period of 15 minutes. In what time will 80% of the sample be decomposed? (log 5 = 0.6989,

log 4 = 0.6021)

a. 48.43 min b. 40.43 min c. 30.48 min d. 34.84 min

D

Sol. t1/2 = K

693.0

K = 15

693.0 = 0.0462 min-1

K = xa

alog

t

303.2

t =

20

100log

0462.0

303.2

= 0462.0

303.2× log 5

= 0462.0

303.2× 0.6989

t = 34.84 min

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27. Rate constant of reaction (k) is 175 litre2 mol–2 sec–1. What is the order of reaction? a. first b. second c. third d. zero

C Sol. Given; Rate constant

K = 175 L2 mol–2 sec–1 Rate = K[A]3

S

L mol -1

= K (mol L–1)3

K = mol-2 L2 s–1

28. Consider this gas phase reaction.

Cl2(g) + CHCl3(g) HCl (g) + CCl4 (g) The reaction is found experimentally to follow this rate law. Rate = k [CHCl3] [Cl2]

1/2 Based on this information, what conclusions can be drawn about this proposed mechanism? Step 1. Cl2 (g) 2Cl (g)

Step 2. Cl (g) + CHCl3 (g) HCl (g)+CCl3 (g)

Step 3 . Cl (g) + CCl3 (g) CCl4 (g)

a. step 1 is the rate determining step b. step 2 is the rate determining step c. step 3 is the rate determining step d. the rate determining step cannot be identified. B 29. The rate of the gaseous reaction is given by the expression k[A] [B]. if the volume of the reaction

mixture is reduced to 4

1initial volume, the reaction rate relating to original rate will be

a. 10

1 b.

8

1 c. 8 d. 16

D Sol. On reducing the volume of gaseous reactant the conc. increases 4 times.

Given R = K [A] [ B] Let the initial conc. of reactant = a R = K [a] [a] R = K[a]2 ....(i)

When volume reduces 4

1 times the conc. increases 4 times

R = K [4a] [4a]

R = K 16[A]2 ….(ii) Or from (i) & (ii) we get

R = R × 16

30. A catalyst a. increase the free energy change in the reaction b. decreases the free energy change in the reaction c. does not increase or decrease the free energy change in reaction d. can either decrease or increase the free energy change depending on what catalyst we use. C 31. The rate constant of reaction increases with the increase of temperature because

a. the activation energy increases b. the population of activated molecules increases c. the activation energy decreases d. the population of activated molecules decreases. B

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32. 4MnO ion oxidise I- according to the following reaction

OH8I5Mn2H16I10MnO2 222

4

The rate of appearance of I2 is 1.14 × 10-2 M s-1. What is the rate of disappearance of 4MnO ?

a. 1.14 × 10-2 M s-1 b. 1.14 × 10-3 M s-1 c. 4.56 × 10-2 M s-1 d. 4.56 × 10-3 M s-1

D

Sol. Given that dt

]I[d 2 = 1.14 × 10-2 ms–1

Also ]I[dt

d

5

1]MnO[

dt

d

2

124

or ]I[dt

d

5

2]MnO[

dt

d24

Rate of disappearance of MnO4– =

5

2×1.14×10–2

= 0.456 × 10–2 ms–1 = 4.56 × 10-3 ms–1 33. The rates of many catalysed reactions follow the profile shown in

the graph. Why does the reaction rate level off? a. The reactant is used up b. The reverse reaction becomes dominant c. The catalyst decomposes as the reaction proceeds d. The active sites on the catalyst are occupied. D 34. The rate for a certain reaction is given below: Rate = [A]x [B]1/y [C]x/y. The order of reaction is:

a. y

xyx b.

y

x1x

c.

z

1

y

1x d. z(x+y)

B 35. The rate of a reaction can be increased in general by all the following factors except a. by increasing the temperature b. using a suitable catalyst c. by an increase in activation energy d. by increasing the concentration of the reactants. C 36. If reaction is 2nd order

)I(

utesmin4

)II(

t

Time taken for the (II ) change will be:

a. 4 minutes b. 8 minutes c. 16 minutes d. 3 minutes B 37. A piece of wood recovered in excavation has 25% as much C14 as ordinary wood has today. If t1/2 of

C14 is 5760 years, then piece of wood remained buried for a. 5760 years b. 11520 years c. 2880 years d. 1440 years B Sol. 25% C – 14 present means it remained buried for two half-lives.

Rea

ctio

n r

ate

[Reactant], M

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38. One micro-gram of radioactive sodium 2411

Na with half-life of 903 minutes was injected into a living

system for a bioassay. How long will it take for the radioactivity to fall by 90% of the initial value?

a. 22.5 hours b. 50 hours c. 30 hours d. 3.75 hours

B

Sol. 10logt

303.22log

t

303.2

9.02/1

t1/2 = 0.3010 × t0.9

hence t0.9 = 3010.0

903

3010.0

t5.0 = 3000 min = 50 hr

39. A radioactive isotope having a half-life of 3 days was received after 12 days. It was found that there

was 3 g of the isotope in the container. The initial weight of the isotope when placed was

a. 12 g b. 24 g c. 36 g d. 48 g

D

Sol. n = 3

12 = 4, Cn =

n

0

2

C or 3 g =

42

1[A0] or [A0] = 48 g

40. Half life period of a radioactive element A is 10 days. The amount of A left after 11th day starting with 1

mol A is :

a.

2

1 b.

11/12

2

1

c.

11/1

2

1

d.

10/11

2

1

D

Sol. Amt,. of A after 10 days = 0.5

mol 1 of

2

1

Amt. of A left after 11 day (1 day)

N = N0

y

2

1

Here y =

5.0t

time total

N0 is amt. on 10th day i.e., 2

1

N =

10/1110/1

2

1

2

1

2

1

41. The probability (P) of survival of a radioactive nucleus for one average life is

a. e

1 b.

e

11 c. n

e

2 d.

e

2 n1

A

Sol. P(survival) = t

0

t0

0

eN

eN

N

)t(N

For t = ,1

probability is e–1 or

e

1

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42. Which of the following is true for a zero order reaction when (a – x) is reactant concentration ?

a. b. c. d. A 43. An exothermic chemical reaction proceeds by two stages.

reactants 1 stage

intermediate 2 stage

products The activation energy of state 1 is 50 kJ mol–1. The overall enthalpy change of the reaction is

– 100 kJ mol–1. Which diagram could represent the energy level diagram for the reaction? a. b. c. d. C 44. A certain zero-order reaction has k = 0.025 M s–1 for the disappearance of A. What will be the

concentration of A after 15 seconds if the initial concentration is 0.50 M? a. 0.50 M b. 0.32 M c. 0.125 M d. 0.060 M C Sol. For zero order reaction Kt = [A]0 – [A]t

0.025 × 15 = [0.50] – [A] [A] = 0.50 – 0.375 = 0.125

45. The rate of the reaction A + 2B 3C gets increased by 72 times when the concentration of A is tripled and that of B is doubled. The order of the reaction with respect to A and B are ……. and …. Respectively:

a. 1, 2 b. 2, 3 c. 3, 2 d. 2, 2

B

x

t

log (a – x)

t

)xa(1

t

2)xa(

1

t

50

100

150

200

R

P

Progress of reaction

E

50

100

150

200

R

P

Progress of reaction

E

50

100

150

200

R

P

Progress of reaction

E

50

100

150

200

R

P

Progress of reaction

E

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SECTION – B (ASSERTION & REASON) This Section contains 5 multiple choice questions. Each question has four choices A), B), C) and D)

out of which ONLY ONE is correct. (5 × 4 = 20 Marks)

(A) If both A and R are true and R is the correct explanation of A

(B) If both A and R are true and R is not the correct explanation of A (C) If A is true but R is false (D) Both A & R are false

1. Assertion: Zn + 3NO

4NH + Zn+2 + 3H2O is not a Balanced redox reaction

Reason: Number of electrons lost is not equal to number of electrons gained.

a. (A) b. (B) c. (C) d. (D) A 2. Assertion: Oxidation number of nitrogen in NH4NO3 is – 3 & +5

Reason: OH2ONNONH 2234

is conproportionation

a. (A) b. (B) c. (C) d. (D)

B 3. Assertion:- At the end of electrolysis using platinum electrodes, an aqueous solution of copper

sulphate turns colourless. Reason: Copper in copper sulphate is converted to copper hydroxide during the electrolysis. a. (a) b. (b) c. (c) (d). d C 4. Assertion: In electrolysis, the quantity of electricity needed for depositing 1 mole of silver is different

from that required for 1 mole of copper. Reason: The molecular weights of silver and copper are different. a. (a) b. (b) c. (c) (d). d B 5. Assertion (A): The activity of a catalyst is enhanced in the presence of a promoter. Reason (R): The promoter make the surface of the catalyst more uneven and thereby increases the

number of active centres. a. (a) b. (b) c. (c) d. (d) A

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SECTION – C (Paragraph Type)

This Section contains 1 paragraph. Each of these questions has four choices A), B), C) and D) out of which ONLY ONE is correct. 4 × 4 = 16 Marks

Passage – 1 Radioactive decay follows first order kinetics.

N

Nlog

t

303.2 010

Here, is decay constant; N0 denotes initial number of nuclei and N is remaining number of nuclei after time ‘t’.

Radioactive decay is independent of all external conditions; -emission )He(42

decreases atomic

number by 2 and mass number by 4. However, -emission increases atomic number by one while mass number remains same. Emission of these particles changes the position of daughter nuclei in the periodic table.

Half-life corresponds to the time in which 50% of the radioactive element will decay. 1. Time for which of the following percentage decay corresponds to average life of the radioactive

element:

a. 100% b. 99.9% c. 63.2% d. 36.8% C

Sol. Average Life of nuclide is the time taken to reduce initial number of atoms to 37% 2. Half-life of the radioactive element is 5 days, what percentage of the element will decay after 10 days?

a. 100% b. 75% c. 50% d. 90% B

Sol. After two half life 4

1th of the initial Nucleid will remain

3. 16 g sample of a radioactive element was taken in a sealed tube on 1st of Feb 2008 and it was found that 0.25 g of element was remained on 1st July 2008. The half-life of the radioactive element will be:

a. 20 days b. 25 days c. 30 days d. 35 days B

4. The analysis of a rock shows that the relative number of 206Pb and 238U atoms is Pb/U = 0.25. If t1/2 for the reaction 238U 206Pb is 4 × 109 years, the age of rock in years is

a. 4

5log)104(

693.0

303.2 9 b. 4

1log)104(

693.0

303.2 9

c. 1

4log)104(

693.0

303.2 9 d. 5

4log)104(

693.0

303.2 9

A Sol. 238U 206Pb

1

25.0

U of Atoms

Pb of Atoms

Initially these atoms of Pb were present as atoms of uranium. a0 = Initial amount (no. of atoms) of U = 0.25 + 1 = 1.25 (a0 – x) = No. of atoms of U present now = 1

= xa

alog

t

303.2

0

0

and =

2/1t

693.0

00.1

25.1log

t

303.2

104

693.09

4

5log)104(

693.0

303.2t 9

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SECTION – D (More than One Answer Type) No Negative Marking

This Section contains 8 multiple choice questions. Each question has four choices A), B), C) and D) out of which One or More than one answer may be correct. 8 × 5 = 40

1. For the reaction: [Cu(NH3)4]

2+ + H2O [Cu(NH3)3H2O]2+ + NH3 The net rate of reaction at any time is given by:

net rate = 2.0 × 10–4 ]O[H ])NH(Cu[ 22

43 - 3.0 × 105 ]NH[ ]OH)NH(Cu[ 3

2233

Then correct statement is (are): a. Rate constant for forward reaction = 2 × 10–4 b. Rate constant for backward reaction = 3 × 105

c. Equilibrium constant for the reaction = 6.6 × 10–10 d. Forward & backward reaction, both are 1st order A,B,C

2. The rate constant is numerically the same for three reactions of first, second and third order respectively. Which of the following is correct: a. if [A] = 1 then r1 = r2 = r3 b. if [A] < 1 then r1 > r2 > r3 c. if [A] > 1 then r3 > r2 > r1 d. can’t predicted A,B,C

3. Listed in the table are forward and reverse rate constants for the reaction 2NO(g) N2(g) + O2(g)

Temperature (K) Kf(M–1S – 1) Kb (M

–1S–1 )

1400 0.29 1.1 x 10–6

1500 1.3 1.4 x 10–5

Select the incorrect statement:

a. Reaction is exothermic & value of equilibrium constant (Keq) at 1400 K is 3.79 x 10–6 b. Reaction is endothermic and value of Keq at 1400 K is 2.63 x 105 c. Reaction is exothermic and value of Keq at 1400 K is 2.63 x 105 d. Reaction is endothermic and value of Keq at 1500 K is 9.28 x 104

A, B, D

Sol. Keq at 1400= ;5

1063.26

101.1

29.0

Keq at

1500 = 9.28 x 104; value of Keq is reduces with rise in temperature so reaction is exothermic. 4. Which expression is correct for first order reaction?

a.

t

0

A

Alog

t

303.2 b.

t

0

A

Alog

303.2

t

c.

0

t

A

Alog

t

303.2 d. Rate = k[A]

A,C,D 5. Which of the following is/’are correct for the given graphs?

Rate

Conc.

(I)

Time

Conc.

(II)

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a. (I) represents zero order reaction b. (II) represents zero order reaction

c. (III) represents first order reaction d. (IV) represents zero order reaction A,B,C

Sol. (i) Rate of zero order reaction remains constant, it does not change with the concentration of the reactant.

(ii) Concentration of reactant decreases linearly with passage of time (for first order reaction)

6. For a strong electrolyte, increases slowly with dilution and can be represented by the

equation: 2/10 AC

Which electrolyte (s) have the same value of the constant A? a. NaCl b. CaCl2 c. ZnCl2 d. AlCl3 B,C 7. The lead storage cell:

a. Is a galvanic cell at the time of discharging and electrolytic cell at the time of charging b. It cannot be recharged c. concentration of H2SO4 decreases on discharge d. Needs no salt bridge A,C,D

Sol. As reagents are solids and concentrated solutes, the cell has relatively constant potential. And the solid nature of oxidizing and reducing agents prevents direct contact, no matter which way the reactions are run. Thus, all the reagents can be placed in the same vessel, no use of salt bridge.

8. Mark out the correct statements among the following:

a. Aqueous AgNO3 solution can be stored in a copper bowl b. Aq CuSO4 solution can be stored in a silver bowl c. Cu, Ag can release hydrogen gas from dil HCl d. H2 can reduce Cu2+ and Ag+ in the form of metals

B,D

SECTION – E (Matrix Type) No Negative Marking

This Section contains 2 questions. Each question has four choices (A, B, C and D) given in Column I and five statements (p, q, r, and s) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. 8 × 2 = 16 Marks

1. Match the following:

Column I (Half-life) Column II (Order of reaction) (A) t1/2 = constant (p) first order

(B) t1/2 a (q) Pseudo first order

(C) t1/2 a

1 (r) Second order

(D) t1/2 p

1 (s) Zero order

Sol. A p, q; B s; C r; D r

Conc.

(III)

Rate

Time

(IV)

Rate

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Half-life of a reaction depends on initial concentration or initial pressure of the reaction as,

1n2/11n2/1

p

1t or

a

1t

where n = order of reaction 2. Match Column – I with Column – II

Column – I Column – II

(A) Order of reaction = 1 (p) A B elementary single step process

(B) Order of reaction > 1 (q) A B a process which occurs only in presence sunlight

(C)

dt

)B(d

dt

)A(d

(r) A2 + 2B 2 AB having mechanism A2 A + A (fast)

2A + 2B 2A B (slow)

(D)

dt

)B(d

2

1

dt

)A(d 2 (s) A2 + 2B 2AB

having mechanism

A2 A + A (slow)

2A + 2B 2A B(fast)

(t) A2 + 2B 2 A B if mechanism of reaction is

B + A A + B A B + A B

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

(P) A B elemental single step reaction hence order = 1

dt

)B(d

dt

)A(d

(Q) A B occur in presence of sunlight means photochemical reaction hence zero order

dt

)B(d

dt

)A(d

(R) A2 + 2B 2 AB from mechanism rate law is r = K[A2]

1/2 [B] order = 1.5

dt

)B(d

2

1

dt

)A(d 2

(S) A2 + 2B 2 AB From mechanism Rate law is r = K[A2]

1 Order = 1

dt

)dB(

2

1

dt

)A(d 2

(T) A2 + 2B 2 AB From mechanism rate law is r = k[A2] [B]2 Order = 3

dt

)B(d

2

1

dt

)A(d 2

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SECTION – F (Integer Type) No Negative Marking

This Section contains 6 questions. The answer to each question is a single digit integer ranging from 0 to 9. 6 × 3 = 18

1. When a platinum rod is dipped in 11,200 ml of H2 gas at STP. Volume of H2 reduces to 5600 ml. The no. of mole of H2 adsorbed in pt is 25 × 10–x. Calculate x?

Sol. 2 gas absorbed = 5600 ml 22400 ml = 1 mole

5600 ml = 4

1

22400

5600

mole = 0.25 = 25 × 10–2

2. The ratio of molar conductivity and equivalent conductivity.

]Hg[K for 42eq

m I

is:

Sol. 2 Charge = +2 or – 2

3. Among the following the total number of compounds in which molar conductivity (m) in similar to

equivalent conductivity eq KCl, BaCl2, [Co(NH3)5Cl]Cl2, K2[NiCl4], KMnO4, [Co(NH3)4Cl2]Cl, AgNO3,

Sol. 4 Those compounds in which cationic charge = + 1 KCl, KMnO4, [Co(NH3)4Cl2]Cl, AgNO3,

4. atm 1

H Pt 2

NaOH M 10

H

x

Value of ‘x’ so that oxidation potential of this electrode becomes 0.4728 V at 25°. Sol. 6

For hydrogen electrode (oxidation) at 1 atmospheric;

-2 e HH

2

1

Eoxd = E° - )(H log1

059.0 E = E°

Eoxd = 0.059 (pH) E = E° - 0.059 log H+ E = 0.059 pH Eoxi = 0.059 (pH)

8059.0

4728.0pH

pH = 8, pOH = 6, [OH–] = 10–6

5. From the following reactions, the total number of reactions which follow first order kinetics.

(a) 0 1-

147

146

N C

(b) 4PH3(g) P4(g) + 6H2(g) having value of rate constant 2 × 10-3 sec–1 and 4 × 10–3 sec–1 at 300 K and 500 K respectively.

(c) 2AgBr h

2Ag + Br2

(d) 2H2O2(aq) 2H2O + O2(g)

(e) CH3COOC2H5 + H2O OH

CH3COO– + C2H5OH

(f) 2A + B C. If rate law of reaction is rate = K[B]

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(g) CH3COOC2H5 + H2O H

CH3COOH + C2H5OH

(h) Decomposition of NH3 at high pressure

(i) For the reaction A B, at 300 K the following graph was obtained

Sol. 5 (a) Radioactivity is of first order. (b) The unit of rate constant = sec–1 which is a first order unit

(c) Photochemical reactions are of zero order (d) Decomposition of H2O2 is a unimolecular reaction, follows first order kinetics. (e) Hydrolysis of ester is a 2nd order reaction r = k (ESTER) [OH–]

(f) r = k[B] first order (g) Hydrolysis of ester in acidic medium is pseudo-unimolecular reaction, follows first order kinetics (h) Decomposition of NH3 at high pressure

22cataly st Pt

K 11303 H3NNH2

r = k [NH]0 Zero order reaction (i) Zero order graph.

(For first order in place of [A] log [A]

6. The total number of ‘’ and ‘’ particles emitted in the nuclear reaction U23892

21482

Pb is:

Sol. 8

64

24

4

214238

= Z2 – (Z1 - 2) = 82 – (92 – 2 × 6) = 82 – (80)

= 2

+ = 6 + 2 = 8

[A]

t