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# Revisal Problems (Past 13 Years) JEE Main (Chemical Kinetics)

## 27 Questions MCQ Test Chemistry for JEE Advanced | Revisal Problems (Past 13 Years) JEE Main (Chemical Kinetics)

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This mock test of Revisal Problems (Past 13 Years) JEE Main (Chemical Kinetics) for JEE helps you for every JEE entrance exam. This contains 27 Multiple Choice Questions for JEE Revisal Problems (Past 13 Years) JEE Main (Chemical Kinetics) (mcq) to study with solutions a complete question bank. The solved questions answers in this Revisal Problems (Past 13 Years) JEE Main (Chemical Kinetics) quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Revisal Problems (Past 13 Years) JEE Main (Chemical Kinetics) exercise for a better result in the exam. You can find other Revisal Problems (Past 13 Years) JEE Main (Chemical Kinetics) extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

### Direction (Q. Nos. 1-24) This section contains 24 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct. At a given temperature k1 = k2 for the reaction, For which case, a steady state is obtained?

Solution:

A steady state is obtained when

QUESTION: 2

### In the reaction,  the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with the reaction time as shown in the figure. The overall order of the reaction is   [JEE Advanced 2013]

Solution:

Time taken for 75% reaction if P is twice the time taken for 50% reaction of P.
Thus, w.r.t. P, order = a = 1.
Concentration of Q decreases linearly with time. Thus, w.r.t. Q
rate = k[Q]° = k = constant
Thus, w.r.t. Q, order = zero
Total order = 1 + 0 = 1

QUESTION: 3

### The time for half-life period of a certain reaction, A → Products is 1 h. When the initial concentration of the reactant A is 2.0 mol L-1, how much time does it take for its concentration to come from 0.50 to 0.25 mol L-1, if it is a zero order reaction? [AIEEE 2010]

Solution:

For zero order reaction,

QUESTION: 4

A radioactive substance decays 20% in 10 min. If at the start, there are 5 x 1020 atoms present, after what time will the number of atoms be reduced to 5 x 1017 atoms?

Solution:

Radioactive substance decays 20% in 10 min.

We know for first order reaction, time of 99.9% reaction = 10 x t 50
t = 10 x 31 = 310 min

QUESTION: 5

Rate constant of a reaction with a virus is 3.3 x 10-4 s-1. Time required for a virus to become 75% inactivated is

Solution:

Virus inactivation follows first order kinetics based on unit of M (s-1).

QUESTION: 6

For the first order reaction,   total pressure is 350 mm after 30 min and 500 mm after complete reaction. Thus, rate constant is

Solution:

QUESTION: 7

t1/A can be taken as the time taken for the concentration of a reactant to drop to 3/4 of the initial value. If the rate constant for a first order reaction is k, then t 1/4 can be written as

Solution:

Initial concentration = a

QUESTION: 8

For the following first order gaseous phase reaction,

time taken for acetone vapour pressure to drop to 0.30 atm from 2 atm is

Solution:

QUESTION: 9

The rate constant for a first order reaction is 60 s-1. How much time will it take to reduce the initialconcentration of the reactant to its (1/16)th value?

Solution:

For first order kinetics,

QUESTION: 10

The half-life of a reaction was doubled when initial concentration was doubled. Thus, order of the reaction is

Solution:

For nth order reaction,

QUESTION: 11

Rate law of the reaction A → product is

rate = k[A]

Graphically, it is represented as

Hence, rate constant is

Solution:

Rate = k [A ]
Thus, it is first order reaction.
When [A] = 0.1 M from graph
Rate = 1x 10-3 = k(0.1)
∴  k = 1 x 10-2 s-1

When [A] = 0.3 M from graph
3 x 10-3 = k(0.3)
∴ k = 1 x 10-2 s-1

QUESTION: 12

For the non-stoichiometric reaction, the following kinetic data were obtained in three separate experiments, all at 298 K.

Q.

The rate law for the formation of C is

Solution:

QUESTION: 13

A hydrogenation reaction is carried out at 500 K.

Q.

If rate remains constant, then Ea is

Solution:

Rates are equal,hence

QUESTION: 14

Rate of a particular reaction increases by a factor 2 when the temperature is increased from 27°C to 37°C. Hence, activation energy of the reaction is

Solution:

QUESTION: 15

Interconversion of boat form to chair form of cyclohexane is first order (in both sides).

Equilibrium constant, Kc =104

Also, energy of activation of conversion of chair to boat form is 42 kJ mol-1 in the following equation

Q.

Observed reaction rate constant at 298 K is

Solution:

QUESTION: 16

In the reaction,

if we start with [A]0 = 10 M , then after one natural life time, concentration of A decrea sed to

Solution:

QUESTION: 17

Number of natural life times (Tav ) required for a first order reaction to achieve 99.9% level of completion is

Solution:

Time required for 99.9% reaction is 10 times of half-life period.

QUESTION: 18

For the reaction,

rate law is

Q.

At the start, pressure is 100 mm and after 10 min, pressure is 120 mm. Thus, rate constant (in min-1) is

Solution:

It is a gaseous phase reaction.

Thus, total pressure at timet = pi - x + x + x = (pi + x)

QUESTION: 19

Q.

HA and HB are two strong acids. Relative strength of strong acids is

Solution:

In both acids, first order kinetics is followed.
Rate constant = k1 with HA = 2 x 10-3 min-1
Rate constant = k2 with HS = 1 x 10-3min-1

QUESTION: 20

Initial concentration of A = 2M and after 10 min, reaction is 10% completed. Thus, half-life period is

Solution:

Thus, it follows first order kinetics.

QUESTION: 21

The decomposition of N2O to N2 and O2 follow s rate law in gaseous phase as

Q.

At 1000 K, half-life of the reaction is 3.58 x 103 min. Starting with initial pressure of N2O as 2.20 atm, total pressure after one half-life is

Solution:

QUESTION: 22

The rate of a first order reaction is 0.04 mol L-1 s-1 at 10 min and 0.03 mol L-1 s-1 at 20 min. Thus , half-life of the reaction is

Solution:

QUESTION: 23

For the non-equilibrium process,

the rate is first order w.r.t. A and second order w.r.t. B.
When [A]= 1.0 M, [B]= 1.0 M, rate = 1.0x 10-2 M s-1

When 50% of each of reactant has been converted into products, rate is

Solution:

QUESTION: 24

For a reaction of the type

[X]0 is the concentration of X at time t = 0 and [X] is the concentration of X at time t = t

Thus, correct rate expression is

Solution:

QUESTION: 25

Direction (Q. Nos. 25-27) This section is based on Statement I and Statement II. Select the correct answer from the codes given below.

Q.

Statement I : The rate of chemical reaction generally increases rapidly for a small temperature increase.

Statement II : Increase in temperature increases fraction of molecules with energies in excess of activation energy.

Solution:

Thus, u (rate) ∞ T
Hence, rate increases with increase in temperature.
Distribution of molecules (N) with velocity (u) at two temperatures is shown as

When temperature is increased, fraction of molecules with energy excess of Ea is increased.
Thus, both Statements I and II are correct and Statement II is the correct explanation of Statement I.

QUESTION: 26

Statement I : For a certain reaction, large fraction of molecules has energy more than the threshold energy,yet the rate of reaction is very slow.

Statement II : Collisions are only effective if they have proper orientation. In the absence of proper orientation rate becomes, slow.

Solution:

There are collisions among molecules which increase the rate of reaction. These collisions are only effective if they have proper orientations. In the absence of proper orientations rate becomes slow. Thus, Statements I and II both are correct and Statement II is the correct explanation of Statement I.

QUESTION: 27

Statement I : Certain reactions are thermodynamically feasible but their rates are very slow.

Statement II : Energy of activation may be high for such reactions.

Solution:

This reaction is thermodynamically feasible due to (ΔG < 0). But Ea is very high.
Thus, rate becomes slow. Thus, Statements I and II are correct and Statement II is the correct explanation for Statement I.