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Chemical Kinetics MCQ - 2 (Advanced) - JEE MCQ


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30 Questions MCQ Test - Chemical Kinetics MCQ - 2 (Advanced)

Chemical Kinetics MCQ - 2 (Advanced) for JEE 2024 is part of JEE preparation. The Chemical Kinetics MCQ - 2 (Advanced) questions and answers have been prepared according to the JEE exam syllabus.The Chemical Kinetics MCQ - 2 (Advanced) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Chemical Kinetics MCQ - 2 (Advanced) below.
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Chemical Kinetics MCQ - 2 (Advanced) - Question 1

For a certain reaction A  →products, the t1/2 as a function of [A]0 is given as below : Which of the following is true :           

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 1

 

Chemical Kinetics MCQ - 2 (Advanced) - Question 2

By what factor does the rate of reaction of diacetone alcohol in water solution change if p0H is increased by 2 units other things remaining same ?

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 2

If [OH-] is decreased by 100 times pOH increases by 2 units. Hence rate decreases by 100 times

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Chemical Kinetics MCQ - 2 (Advanced) - Question 3

If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately

Chemical Kinetics MCQ - 2 (Advanced) - Question 4

In a reaction, → Product, rate is doubled when the concentration of is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (Aand B) are doubled, rate law for the reaction can be written as 

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 4

 

 

*Multiple options can be correct
Chemical Kinetics MCQ - 2 (Advanced) - Question 5

In the following gaseous phase first order reaction A(g) →2B(g) + C(g) initial pressure was found to be 400 mm of Hg and it changed to 1000 mm of Hg after 20 min. Then :

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 5

 

 

*Multiple options can be correct
Chemical Kinetics MCQ - 2 (Advanced) - Question 6

The polarimeter readings in an experiment to measure the rate of inversion of cane suger (1st order reaction) were as follows time (min) : 0 30 coangle (degree) : 30 20 -15 Identify the true statement (s) log 2 = 0.3, log 3 = 0.48, log 7 = 0.84 , loge 10 = 2.3

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 6

*Multiple options can be correct
Chemical Kinetics MCQ - 2 (Advanced) - Question 7

Identify the true statement(s)

*Multiple options can be correct
Chemical Kinetics MCQ - 2 (Advanced) - Question 8

The half – period T for the decomposition of ammonia on tungsten wire, was measured for different initial pressure P of ammonia at 25°C. Then P (mm Hg) 11 21 48 73 120 T (sec) 48 92 210 320 525

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 8

Chemical Kinetics MCQ - 2 (Advanced) - Question 9

The substance undergoes first order decomposition. The decomposition follows two parallel first orde reactions as : If the corresponding activation energies of parallel reaction are 100 and 120 kJ mol-1 then the net activation energy of A is / are :

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 9

*Multiple options can be correct
Chemical Kinetics MCQ - 2 (Advanced) - Question 10

A certain reaction obeys the rate equation (in the integrated form) [Co -") - Co' - = (n -1) kt where C, is the initial concentration and C is the concentration after time, t.

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 10

*Multiple options can be correct
Chemical Kinetics MCQ - 2 (Advanced) - Question 11

The ozone layer in the earth's upper atmosphere is important in shielding the earth from very harmful ultraviolet radiation. The ozone, 03, decomposes according to the equation 203(g) -› 302(g). The mechanism of the reaction is thought to proceed through an initial fast, reversible step and then a slow second step.

Step 1: Fast, reversible 03(g)⇔ 02(g) +0(g)

Step 2 : Slow 03(g) + 0(g) → 202(g)

The which of the following is correct ?

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 11

Slow step is rate determining step, from slow step –

Molecularity for step 1 is 1 and for step 2 is 2 as shown in above given reaction.

*Multiple options can be correct
Chemical Kinetics MCQ - 2 (Advanced) - Question 12

Consider the following case of COMPETING 1ST ORDER REACTIONS

After the start of the reaction at t = 0 with only A, the [C] is equal to the [D] at all times. The time in which all three concentrations will be equal is given by

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 12

K1=K2

*Multiple options can be correct
Chemical Kinetics MCQ - 2 (Advanced) - Question 13

Rate of a chemical reaction 2A (g) →B (g) is defined as 

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 13

Chemical Kinetics MCQ - 2 (Advanced) - Question 14

Consider the inter conversion of nitrosotnacetoamine into nitrogen phorone and water.

The reaction is 1st order in each direction, with an equilibrium constant of 104, the activation energy for the forward reaction is 11usactivation energy for the forward reaction is 57.45 kJ / mol. Assuming arrehemus preexponetial factor of 1012 S–-1

What is the expected forward constant at 300K , it we initiate this reaction starting with only reactant

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 14

Chemical Kinetics MCQ - 2 (Advanced) - Question 15

Consider the inter conversion of nitrosotnacetoamine into nitrogen phorone and water.

The reaction is 1st order in each direction, with an equilibrium constant of 104, the activation energy for the forward reaction is 11usactivation energy for the forward reaction is 57.45 kJ / mol. Assuming arrehemus preexponetial factor of 1012 S–-1

If the change in entropy of the reaction is 0.07 KJ. K-1 mol-1 at 1 atm pressure. Calculate up to which temperature the reaction would not be spontaneous. (For forward reaction)

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 15

Chemical Kinetics MCQ - 2 (Advanced) - Question 16

Consider the inter conversion of nitrosotnacetoamine into nitrogen phorone and water.

The reaction is 1st order in each direction, with an equilibrium constant of 104, the activation energy for the forward reaction is 11usactivation energy for the forward reaction is 57.45 kJ / mol. Assuming arrehemus preexponetial factor of 1012 S–-1

Calculate Kp of the reaction at 300 K

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 16

Chemical Kinetics MCQ - 2 (Advanced) - Question 17

Not all chemical reactions proceed to a stage at which the concentrations of the reactants become vanishingly small. Here we consider the kinetics of such reactions.Let a reaction be represented in general terms by the schemewhere k., and k_., represent the rate constants for the forwards and reverse reactions, respectively. The equilibrium constant for this reaction may be written asK = [B]∝ /[A]∝ = K1/K–1                   --(1)where the subscript 00 refers to a time t, sufficiently long to establish equilibrium at the given temperature. The initial concentration of species A is [A]O, and that of B is [B]O . After a time t, let the concentration of species A be [A/t and that of B be [Bat. The total rate of change of [At is given byd[A]t/dt= - K1[A]t + k-1[B]tIf, as is usual, [B]O is initially zero, it follows from a mass balance that at any time t,[Bat = [A]0 - [A]t, whenced[A]t/dt = –K1[A]t + k-1([A]O - [A]t)ord[A]t/dt = – (K1 + k-1) For the reaction  (having both 1ts order reactions), the concentration as a function of time are given for a certain experimental run along with a tangent to the graph at the origin. The ratio of the magnitude of the slopes of the graph of [A/ and [C/ at the origin would be :

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 17

And slope between conc and time is known as rate and they always equal at any time. 

Chemical Kinetics MCQ - 2 (Advanced) - Question 18

Not all chemical reactions proceed to a stage at which the concentrations of the reactants become vanishingly small. Here we consider the kinetics of such reactions.
Let a reaction be represented in general terms by the scheme

where k., and k_., represent the rate constants for the forwards and reverse reactions, respectively. The equilibrium constant for this reaction may be written as

K = [B] /[A] = K1/K–1                   --(1)
where the subscript 00 refers to a time t, sufficiently long to establish equilibrium at the given temperature. The initial concentration of species A is [A]O, and that of B is [B]O . After a time t, let the concentration of species A be [A/t and that of B be [Bat. The total rate of change of [At is given by

d[A]t/dt= - K1[A]t + k-1[B]t
If, as is usual, [B]O is initially zero, it follows from a mass balance that at any time t,

[Bat = [A]0 - [A]t, whence

d[A]t/dt = –K1[A]t + k-1([A]O - [A]t)

or

d[A]t/dt = – (K1 + k-1

At what time is the rate of change of concentration of A equal to rate of change of concentration of C in magnitude ?

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 18

Comparing the given equation with equation (i),

Chemical Kinetics MCQ - 2 (Advanced) - Question 19

There are many reactions which obey a first order rate equation although it reality they are bi- or ter-molecular. As an example of these may be taken the decomposition of Carbonyl sulfide in water, namely, COS + H20 → 002 + H2S
According to the law of mass action this reaction should be second order with the rate dependent on the concentration of both the carbonyl sulfide and the water. Actually however, the rate is found to be first order with respect to the carbonyl sulfide and independent of the water Reactions exhibiting such behaviour are said to be pseudo-molecular.
The pseudo-unimoecuar nature of this reaction is explainable by the fact that water is present in such excess that its concentration remains practically constant during the course of the reaction. Under these condition b x = b, and the rate equation becomes

On integration this leads to

which is the equation for a first order reaction. It is evident, however, that the now constant k is not independent of the concentration, as is the case with true first order constants, but may vary with b if the latter is changed appreciably, When such is the case, the true constant k2 can be obtained from k by dividing the latter by b. pseudo-molecular reactions are encountered whenever one or more of the reactants remain constants during the course of an experiment. This is the case with reactions conducted in solvents which are themselves one of the reactants, as in the decomposition of carbonyl sulfide in water, or in the esterification of acetic anhydride in alcohol

(CH3C0)20 + 2C2H5OH → 2CH3C00C2H5 + H20

Again, this is also true of reactions subject to catalysis, in which case the concentration of the catalyst does not change. The decomposition of diacetone alcohol to acetone in aqueous solution is catalysed by  hydroxyl ions, with the rate proportional to the concentration of the alcohol and that of the base. Since the concentration of the base does not change within any one experiment, however, the rate equation reduces to one of first order with respect to the alcohol. But the rate constant k obtained for various concentrations of base are not identical, as may be seen from table. To obtain from these the true second order velocity constant, the k's must be divided by the hydroxyl ion concentration. When this is done excellent k2 values result, as column 3 indicates

Table : Decomposition of diacetone alcohol in water at 25°C (Catalyst : NaOH)

Q.

Which of k or k2 is dependent on concentration ?

Chemical Kinetics MCQ - 2 (Advanced) - Question 20

For the reaction A + B → C ; r = k [A] [B] with [A]0 = 0.001 M and [B]0 = 10 M, the time in which the conc. of A becomes 0.005 M will be nearly

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 20

Chemical Kinetics MCQ - 2 (Advanced) - Question 21

Not all chemical reactions proceed to a stage at which the concentrations of the reactants become vanishingly small. Here we consider the kinetics of such reactions.
Let a reaction be represented in general terms by the scheme

where k., and k_., represent the rate constants for the forwards and reverse reactions, respectively. The equilibrium constant for this reaction may be written as

K = [B] /[A] = K1/K–1                   --(1)
where the subscript 00 refers to a time t, sufficiently long to establish equilibrium at the given temperature. The initial concentration of species A is [A]O, and that of B is [B]O . After a time t, let the concentration of species A be [A/t and that of B be [Bat. The total rate of change of [At is given by

d[A]t/dt= - K1[A]t + k-1[B]t
If, as is usual, [B]O is initially zero, it follows from a mass balance that at any time t,

[Bat = [A]0 - [A]t, whence

d[A]t/dt = –K1[A]t + k-1([A]O - [A]t)

or

d[A]t/dt = – (K1 + k-1

What is the Cc of the reaction ​?

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 21

Chemical Kinetics MCQ - 2 (Advanced) - Question 22

There are many reactions which obey a first order rate equation although it reality they are bi- or ter-molecular. As an example of these may be taken the decomposition of Carbonyl sulfide in water, namely, COS + H20 → 002 + H2S
According to the law of mass action this reaction should be second order with the rate dependent on the concentration of both the carbonyl sulfide and the water. Actually however, the rate is found to be first order with respect to the carbonyl sulfide and independent of the water Reactions exhibiting such behaviour are said to be pseudo-molecular.
The pseudo-unimoecuar nature of this reaction is explainable by the fact that water is present in such excess that its concentration remains practically constant during the course of the reaction. Under these condition b x = b, and the rate equation becomes

On integration this leads to

which is the equation for a first order reaction. It is evident, however, that the now constant k is not independent of the concentration, as is the case with true first order constants, but may vary with b if the latter is changed appreciably, When such is the case, the true constant k2 can be obtained from k by dividing the latter by b. pseudo-molecular reactions are encountered whenever one or more of the reactants remain constants during the course of an experiment. This is the case with reactions conducted in solvents which are themselves one of the reactants, as in the decomposition of carbonyl sulfide in water, or in the esterification of acetic anhydride in alcohol

(CH3C0)20 + 2C2H5OH → 2CH3C00C2H5 + H20

Again, this is also true of reactions subject to catalysis, in which case the concentration of the catalyst does not change. The decomposition of diacetone alcohol to acetone in aqueous solution is catalysed by  hydroxyl ions, with the rate proportional to the concentration of the alcohol and that of the base. Since the concentration of the base does not change within any one experiment, however, the rate equation reduces to one of first order with respect to the alcohol. But the rate constant k obtained for various concentrations of base are not identical, as may be seen from table. To obtain from these the true second order velocity constant, the k's must be divided by the hydroxyl ion concentration. When this is done excellent k2 values result, as column 3 indicates

Table : Decomposition of diacetone alcohol in water at 25°C (Catalyst : NaOH)

Q.

By what factor does the rate of reaction of diacetone alcohol in water solution change if p0H is increased by 2 units other things remaining same ?

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 22

If [OH-] is decreased by 100 times pOH increases by 2 units. Hence rate decreases by 100 times.

Chemical Kinetics MCQ - 2 (Advanced) - Question 23

From the following data answer the questions : Reaction :  A + B → P

The order w. r.t A is

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 23

Chemical Kinetics MCQ - 2 (Advanced) - Question 24

From the following data answer the questions : Reaction :  A + B→ P

The value of rate constant at 300 K is (M-2 sec-1)

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 24

Chemical Kinetics MCQ - 2 (Advanced) - Question 25

From the following data answer the questions : Reaction :  A + B → P

The energy of activation for reaction (KJ/mol) is

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 25

Chemical Kinetics MCQ - 2 (Advanced) - Question 26

 

The rate law expressed the relationship of the rate of a reaction to the rate constant and the concentration of the reactants raised to some powers for the general reaction aA + bB → cC + dD

Rate law takes the form r = k [A]X [B]y

where x and y are number that must be determined experimentally k is the rate constant and [A] and [B] are concentration of A & B respectively.

Q.

Gaseous reaction A → B + C follows first order kinetics concentration of A changes from 1M to 0.25M in 138.6 min. Find the rate of reaction when conc. of A is 0.1 M.

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 26

Chemical Kinetics MCQ - 2 (Advanced) - Question 27

The rate law expressed the relationship of the rate of a reaction to the rate constant and the concentration of the reactants raised to some powers for the general reaction aA + bB → cC + dD

Rate law takes the form r = k [A]X [B]y

where x and y are number that must be determined experimentally k is the rate constant and [A] and [B] are concentration of A & B respectively.

Q.

The initial rate of zero order reaction of the gaseous equation A (g) → 2B (g) is 10-2 M min1 if the initial conc. of A is 0.1 M what would be conc. of B after 60 sec.

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 27

Chemical Kinetics MCQ - 2 (Advanced) - Question 28

The rate law expressed the relationship of the rate of a reaction to the rate constant and the concentration of the reactants raised to some powers for the general reaction aA + bB → cC + dD

Rate law takes the form r = k [A]X [B]y

where x and y are number that must be determined experimentally k is the rate constant and [A] and [B] are concentration of A & B respectively.

Q.

The variation of concentration of 'A' with time in two experiments starting with two different initial concentration of 'A' is given by the following graph.The reaction is represented by A (aq) → B(aq) what is rate of reaction 1.5M (M/min) when conc. of A in aqueous solution was 1.8 M.

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 28

 

Chemical Kinetics MCQ - 2 (Advanced) - Question 29

For a hypotherical elementary reaction 

.Initially only 2 moles of A are present.

The total number of moles of A, B and C at the end of 50% reaction are:

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 29

When 1 mole of A is present then 2 mole of B and C will forms after completion of the reaction but when 50% is complete then the total mole = 1 + 2 * 1 = 3.

Chemical Kinetics MCQ - 2 (Advanced) - Question 30

For a hypotherical elementary reaction 

.Initially only 2 moles of A are present.

Number of moles of B are :

Detailed Solution for Chemical Kinetics MCQ - 2 (Advanced) - Question 30

Mole of B = 0.666.

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