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QUESTION: 1

Direction (Q. Nos. 1-21) This section contains 21 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.

**Q. For zeroth order reaction,**

**A → B**

**[A] _{0} = 0.01 M, [A]_{t} = 0.008Matter 10 min.**

**Thus, half-life is**

Solution:

For zeroth order reaction

When reaction is 50% completed.

QUESTION: 2

A Certain Zeroth Order reaction has k = 0.025 Ms^{-1 }for the disappearance of A. What will be the concentration of A after 15s, if the intial concentration is 0.50 M?

Solution:

x(product) formed after 15 s

= 0.025 Ms^{-1} x 15s

= 0.375 M

Then, reactant left

= 0.500 - 0.375

= 0.125 M

QUESTION: 3

For Zeroth Order Reaction, variation of x with time is shown as

**Q. At initial concenration of the reactant as 17.32 min dm ^{-1 }, half - life period is **

Solution:

QUESTION: 4

For the first order reaction, rate constant k is given by

kt = logC_{0 }- log C_{t}_{ }

where, C_{0} is the intial concenration and C_{t }is the concentration at time t. Graph between log C_{t }and time t is

Solution:

This represents a straight line

QUESTION: 5

In a first order reaction, the concentration of the reactant decreases from 0.8 M to 0.4 M is 15 min. The time taken for the concentration to change from 0.1 M to 0.025 M is

Solution:

For first order reaction,

time = 15 min to change its concentration from 0.8 m to 0.4 m Thus, 50% reaction.

Thus, t_{50} = 15 min

QUESTION: 6

For the First order reaction, concentration of the reactant after two averagelives is reduced to

Solution:

For first order reaction,

QUESTION: 7

The half-life period of a first order chemical reaction is 6.93 min. The time required for the completion of 99% of the chemical reaction will be (log2 = 0.3010)

**[AIEEE 2009]**

Solution:

∴

when, a = 100,

(a-x) = 100-99 = 1, t = t_{99}

QUESTION: 8

For the first order reaction,

A → Product

**Q. The concentration of A changes from 0.1 M to 0.025 M in 40 min. The rate of the reaction when the concentration of A is 0.01 M is**

**[AIEEE 2012]**

Solution:

For the first order kinetic,

QUESTION: 9

For the reaction,

A → Products,

Concentration of A at different time intervals are given :

**At 20 min, rate will be**

Solution:

QUESTION: 10

For the first order reaction with, C_{0} as the initial concentration and C at time t, (C_{0} - C) =

Solution:

QUESTION: 11

The radioisotope N-13 which has a half-life of 10 min is used to image organs in the body. If the injected sample has an activity of 40 μCi (microcurie), what is the activity after 30 min?

Solution:

Disintegration follows first-order kinetics

t = total time = 30 min

t _{50} = half-life = 10 min

QUESTION: 12

The following data were obtained during the first order decomposition of

2A(g) → B(g) + C(s)

at constant volume and at a particular temperature :

**Thus, rate constant is**

Solution:

QUESTION: 13

A G .M. Counter is used to study the process of first order of radioactive disintegration. In the absence of radioactive substance A, it counts 3 dps (disintegration per second). When A is placed in G.M Counter it records 23 dps at the start and 13 dps after 10 min. It records x dps after next 10 min. Half-life period of A is y min. Thus, x and y are

Solution:

When there is no radioactive substance in G.M. Counter, it records = 3 dps. Thus, it is the zero-error in the counter and in each it is added. Thus, initial counts = 23 - 3 = 20 dps.

Attimef (after 10min) = 13- 3 = 10dps

Thus, half-life is 10 min = y

Hence, after next 10 min, dps = 5

Hence, reading in G.M. Counter = 5 + 3 = 8 dps = x

QUESTION: 14

Inversion of sucrose (C_{12}H_{22}O_{11}) is a first order reaction and is studied by measuring angle of rotation at different interval of time

r_{0} = angle of rotation at the start, r_{t} = angle of rotation at time t

r_{∞} = angle of rotation at the complete reaction

There is 50% inversion, when

Solution:

r_{0} = rotation at the start, r_{t} = rotation at time t

r_{∞} = rotation at the end then (r_{∞} - r_{0}) = a and (r_{∞} - r_{t}) = (a - x)

When 50% inversion has taken place

∴ 2 (a - x) = a

2(r_{∞} - r_{t}) = r_{∞} - r_{0}

2r_{∞} - 2r_{t} = r_{∞} - r_{0}

∴ r_{0} = (2r_{t} - r_{∞})

QUESTION: 15

Following radioactive disintegration follows first order kinetics :

Starting with 1 mol of A in a 1L flask at 300 K, pressure set up after two half-lifes is approximately

Solution:

When B is formed from A, two a-particies (He) are formed.

A taken initially = 1 mol

This helium remains in closed vessel.

Thus, pressure set up is due to helium.

QUESTION: 16

For the first order reaction T_{av} (average life), T_{50} (half-life) and T_{75} (time 75% reaction) are in the ratio of

Solution:

QUESTION: 17

For the first order reaction,

Following equations are given :

**Select the correct equations.**

Solution:

For the first order reaction,

∴ (a-x) = a (1/2)^{y }(Equation II)

QUESTION: 18

Rate constant of a first order reaction (A → B) is 0.0693 min^{-1}. If initial concentration is 1.0 M, rate after 30 min is

Solution:

C_{0} = 1 M

∴

QUESTION: 19

Following reaction is pseudo-unimolecular w.r.t. C_{6}H_{5}N_{2}CI (A)

50 mL of 1M benzene diazonium chloride (A) is taken. After 1 h, 1.226 L of N_{2} gas at 1 atm and 300 K is obtained. Thus, half-life of the reaction is (log 250 = 2.40)

Solution:

Half-life(t_{50})=

QUESTION: 20

Minimum half-life of an isotope needed so that not more than 0.1 % of the nuclei undergo decay during a 3.0 h laboratory experiment is

Solution:

QUESTION: 21

Naturally occurring potassium consists of 0.01% at 40 K which has a half-life of 1.28 x 10^{9} yr. Activity of 1.00 g sample of KCl is

Solution:

Number of 40 K atoms in 1g KCI

QUESTION: 22

Direction (Q. Nos. 22-27) This section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d).

Passage l

Isotope of oxygen with mass number less than 16 undergoes β^{+} -emission. Assume an equimolar mixture of ^{14}O and ^{15}O, Also t_{50} (^{14}O) = 71s, t_{50} (^{15}O) = 124s.

**Q. Ratio of rate constants of ^{14}O to ^{15}O is**

Solution:

QUESTION: 23

**Passage l**

Isotope of oxygen with mass number less than 16 undergoes β^{+} -emission. Assume an equimolar mixture of ^{14}O and ^{15}O, Also t_{50} (^{14}O) = 71s, t_{50} (^{15}O) = 124s.

**Q. Ratio of ^{14}O to ^{15}O left after 1.00 h is**

Solution:

QUESTION: 24

**Passage II**

Consider acid-hydrolysis of ester which is first order reaction.

CH_{3}COOC_{2}H_{5} + H_{2}O CH_{3}COOH + C_{2}H_{5}OH

Kinetics is studied by titration of the acid with NaOH at a given time.

V_{0} = Volume of NaOH at start, V_{t} = Volume of NaOH at time t, V_{∞} = Volume of NaOH at the end of the complete hydrolysis.

**Q.
Rate constant k is given by**

Solution:

At time t = 0, volume of NaOH is due to H^{+ }(catalyst) = V_{0} and remains constant

at time t, volume of NaOH is due to H^{+} and CH_{3}COOH = V_{t}

At infinite time (complete reaction), volume of NaOH is due to total CH_{3}COOH (or ester) and H^{+} = a + V_{0}

If ester is 50% hydrolysed,

QUESTION: 25

**Passage II**

Consider acid-hydrolysis of ester which is first order reaction.

CH_{3}COOC_{2}H_{5} + H_{2}O CH_{3}COOH + C_{2}H_{5}OH

Kinetics is studied by titration of the acid with NaOH at a given time.

V_{0} = Volume of NaOH at start, V_{t} = Volume of NaOH at time t, V_{∞} = Volume of NaOH at the end of the complete hydrolysis.

**Q.**

**If ester is 50% hydrolysed, then **

Solution:

At time t = 0, volume of NaOH is due to H^{+ }(catalyst) = V_{0} and remains constant

at time t, volume of NaOH is due to H^{+} and CH_{3}COOH = V_{t}

At infinite time (complete reaction), volume of NaOH is due to total CH_{3}COOH (or ester) and H^{+} = a + V_{0}

If ester is 50% hydrolysed,

QUESTION: 26

**Passage III**

I, II and III are three vessels of equal volumes containing molecules of A as shown and in each A decomposes into products.

**Q.
Initial rate of decomposition of A will be in order in these three vessels as**

Solution:

Thus, ratio fo rates = 6 : 1 2 : 9 = 1 : 2 : 1 5

For first order reaction T_{50} is independent of initial concentration.

Thus, T_{50} will be equal to 1 :1 :1.

QUESTION: 27

**Passage III**

I, II and III are three vessels of equal volumes containing molecules of A as shown and in each A decomposes into products.

**Q.
Time of 50% decomposition is in order**

Solution:

Thus, ratio fo rates = 6 : 1 2 : 9 = 1 : 2 : 1 5

For first order reaction T_{50} is independent of initial concentration.

Thus, T_{50} will be equal to 1 :1 :1.

*Answer can only contain numeric values

QUESTION: 28

Direction (Q. Nos. 28-30) This section contains 3 questions. When worked out will result in one integer from 0 to 9 (both inclusive).This section contains 3 questions, when worked out will result in an integer value from 0 to 9 (both inclusive)

**Q. A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial activity is ten times the permissible value, it will be safer to enter the room after 10 ^{n} days. What is the value of n?**

Solution:

Permissible element = 10 times the initial value

*Answer can only contain numeric values

QUESTION: 29

For the first order reaction, variation of log_{10}(a - x)vs time (t) is shown below.

**Q.
Its half-life period is logx. What is the value of x?**

Solution:

*Answer can only contain numeric values

QUESTION: 30

For the first order reaction,

**Q.
If OA = 1.3010 then reactant left after 0.6021 min i s .......unit.**

Solution:

At time t = 0, x = 0

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