JEE  >  Chemistry for JEE Advanced  >  Chemical Kinetics MCQ - 1 (Advanced) Download as PDF

Chemical Kinetics MCQ - 1 (Advanced)


Test Description

25 Questions MCQ Test Chemistry for JEE Advanced | Chemical Kinetics MCQ - 1 (Advanced)

Chemical Kinetics MCQ - 1 (Advanced) for JEE 2022 is part of Chemistry for JEE Advanced preparation. The Chemical Kinetics MCQ - 1 (Advanced) questions and answers have been prepared according to the JEE exam syllabus.The Chemical Kinetics MCQ - 1 (Advanced) MCQs are made for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Chemical Kinetics MCQ - 1 (Advanced) below.
Solutions of Chemical Kinetics MCQ - 1 (Advanced) questions in English are available as part of our Chemistry for JEE Advanced for JEE & Chemical Kinetics MCQ - 1 (Advanced) solutions in Hindi for Chemistry for JEE Advanced course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Chemical Kinetics MCQ - 1 (Advanced) | 25 questions in 60 minutes | Mock test for JEE preparation | Free important questions MCQ to study Chemistry for JEE Advanced for JEE Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you?
*Multiple options can be correct
Chemical Kinetics MCQ - 1 (Advanced) - Question 1

Ozone decomposes according to the equation

2O3(g) → 3O2(g) Mechanism of the reaction is

Step I : O3(g)  O2(g) + O(g) (fast)

Step II : O3(g) + O(g) → 2O2(g) (slow)

Which of the following is correct ?

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


Slowest step is rate determing step
rate  = K[O3][O]  .........(1)
Since, [O] is not in the original reaction hence it has to be eliminated.
From first step

Putting the value of [O] in the eq. (1)

*Multiple options can be correct
Chemical Kinetics MCQ - 1 (Advanced) - Question 2

 Consider the following case of competing 1st order reactions After the start of the reaction at t = 0, with only P, concentration of Q is equal to R at all times. The time in which all the three 

concentration will be equal is given by

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


Since [Q] = [R] all the times,


Taking ln of both side
–2K1t = – ln3

As K1 = K2

*Multiple options can be correct
Chemical Kinetics MCQ - 1 (Advanced) - Question 3

A substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as

The % distribution of B and C is ................ and .................. respectively.

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 3


K1 = 1.26 × 10–4 sec–1
K2 = 3.8 × 10–5 sec–1

*Multiple options can be correct
Chemical Kinetics MCQ - 1 (Advanced) - Question 4

Which of the following concepts are correct

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 4

Energy of activation can never be negative Option (D) is incorrect.

*Multiple options can be correct
Chemical Kinetics MCQ - 1 (Advanced) - Question 5

The basic theory behind Arrhenious equation is that

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 5

Only ans will be given

*Multiple options can be correct
Chemical Kinetics MCQ - 1 (Advanced) - Question 6

Which of the following statements are correct about half life period ?

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 6

(A) For zero order reaction

⇒ t1/2 α a0 (initial concentration)
(B) For first order reaction
Average life = 1.44 × half life
(C) For IInd order reaction



t99.9  = 100 min

*Multiple options can be correct
Chemical Kinetics MCQ - 1 (Advanced) - Question 7

For a reaction : 2A + 2B → products, the rate law expression is r = k[A]2 [B]. Which of the following is/are correct?

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 7

Rate = K[A]2[B]
⇒  rate w.r.t B = 1
Overall order of the reaction = 2 + 1 = 3
for a complex reaction the rate determing step is determined experimentally.

*Multiple options can be correct
Chemical Kinetics MCQ - 1 (Advanced) - Question 8

 

Which of the following statements are correct ?

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 8

In SN1 reaction, formation of carbocation takes place which is planar.
⇒ equimolar mixture of (I) and (II) will be formed in case of SN1.
In SN2 reaction Nucleophile will attack from the opposite to leaving group
⇒ Product (II) will be formed in case of SN2 reaction.

*Multiple options can be correct
Chemical Kinetics MCQ - 1 (Advanced) - Question 9

Rate constant k varies with temperature by equation log10k(min-1) = 5 – 2000/T. We can conclude

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 9



log A = 5 ⇒ A = 105

Ea = 2000 × 2.303 × 8.314 cal
Ea = 9.212 Kcal

*Multiple options can be correct
Chemical Kinetics MCQ - 1 (Advanced) - Question 10

A reaction is catalysed by H+ ion. In presence of HA, rate constant is 2 × 10-3 min-1and in presence of HB rate constant is 1 × 10-3 min-1. HA and HB being strong acids, we may conclude that

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 10


Rate = K[H+][A]
Rate = K’[A]
If K’ has greater value for HA
⇒ [H+] has to be large
⇒ for that acid has be stronger
⇒ [HA] is strongers that HB
Also Relative strength = 2

Chemical Kinetics MCQ - 1 (Advanced) - Question 11

The elements which are good catalysts and have the ability to change their oxidation number are

*Multiple options can be correct
Chemical Kinetics MCQ - 1 (Advanced) - Question 12

Statement-1 : The time of completion of reactions of type A → product (order <1) may be determined.

Statement-2 : Reactions with order ³ 1 are either too slow or too fast and hence the time of completion can not be determined.

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 12

eg. Zero order reaction completion time can be determind for order > 1 time can be determined

*Multiple options can be correct
Chemical Kinetics MCQ - 1 (Advanced) - Question 13

Statement-1 : Temperature coefficient of an one step reaction may be negative.

Statement-2 : The rate of reaction having negative order with respect to a reactant decreases with the increase in concentration of the reactant.

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 13

Ea > 0
If order

*Multiple options can be correct
Chemical Kinetics MCQ - 1 (Advanced) - Question 14

Statement-1 : The overall rate of a reversible reaction may decrease with the increase in temperature.

Statement-2 : When the activation energy of forward reaction is less than that of backward reaction, then the increase in the rate of backward reaction is more than that of forward reaction on increasing the temperature.

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 14

*Multiple options can be correct
Chemical Kinetics MCQ - 1 (Advanced) - Question 15

Statement-1 : In a reversible endothermic reaction, Eact of forward reaction is higher than that of backward reaction

Statement-2 : The threshold energy of forward reaction is more than that of backward reaction

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 15


∴ ΔH > 0

*Multiple options can be correct
Chemical Kinetics MCQ - 1 (Advanced) - Question 16

A hypothetical elementary reaction  where  Initially only 2 moles of A are present.

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

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 16


2 – x – y = 1
⇒ x + y = 1  ............(1)


⇒ [B]  = [C]
⇒ 2x = y  .............(2)
Solving (1) and (2)

Total moles after 50 % of the reaction
= 2 – x – y + 2x + y
= 2 + x = 7/3

*Multiple options can be correct
Chemical Kinetics MCQ - 1 (Advanced) - Question 17

A hypothetical elementary reaction  where ​ Initially only 2 moles of A are present.

Number of moles of B are

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 17

No. of Moles of B = 2x
= 2/3
= 0.666

Chemical Kinetics MCQ - 1 (Advanced) - Question 18

The rate of a reaction increases significantly with increase in temperature. Generally, rates of reaction are doubled for every 10º rise in temperature. Temperature coefficient gives us an idea about the change in rate of a reaction for every 10º change in temperature.

Temperature coefficient 

Arrhenious gave an equation which describes rate constant k as a function of temperature is k = A –Ea/RT

where k is a rate constant

A is frequency factor or pre exponential factor

Ea is activation energy

T is temperature in kelvin and

R is universal gas constant

Equation when expressed in logarithmic form becomes

Q.

For a reaction Ea = 0 and k = 3.2 × 105 s-1 at 325 K. The value of k at 335 K would be

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 18


When Ea = 0, k = 3.2 ×105 & it does not depend on T

Chemical Kinetics MCQ - 1 (Advanced) - Question 19

The rate of a reaction increases significantly with increase in temperature. Generally, rates of reaction are doubled for every 10º rise in temperature. Temperature coefficient gives us an idea about the change in rate of a reaction for every 10º change in temperature.

Temperature coefficient 

Arrhenious gave an equation which describes rate constant k as a function of temperature is k = A –Ea/RT

where k is a rate constant

A is frequency factor or pre exponential factor

Ea is activation energy

T is temperature in kelvin and

R is universal gas constant

Equation when expressed in logarithmic form becomes

Q.

 For which of the following reactions k310/k300 would be maximum ?

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 19

When Ea = is max,

Chemical Kinetics MCQ - 1 (Advanced) - Question 20

The rate of a reaction increases significantly with increase in temperature. Generally, rates of reaction are doubled for every 10º rise in temperature. Temperature coefficient gives us an idea about the change in rate of a reaction for every 10º change in temperature.

Temperature coefficient 

Arrhenious gave an equation which describes rate constant k as a function of temperature is k = A –Ea/RT

where k is a rate constant

A is frequency factor or pre exponential factor

Ea is activation energy

T is temperature in kelvin and

R is universal gas constant

Equation when expressed in logarithmic form becomes

Q.

Activation energies of two reaction are Ea and Ea with Ea > Ea. If the temperature of the reacting systems is increased from T1 to T2 (k` are rate constants at higher temperature).

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 20




Chemical Kinetics MCQ - 1 (Advanced) - Question 21

The rate of a reaction increases significantly with increase in temperature. Generally, rates of reaction are doubled for every 10º rise in temperature. Temperature coefficient gives us an idea about the change in rate of a reaction for every 10º change in temperature.

Temperature coefficient 

Arrhenious gave an equation which describes rate constant k as a function of temperature is k = A –Ea/RT

where k is a rate constant

A is frequency factor or pre exponential factor

Ea is activation energy

T is temperature in kelvin and

R is universal gas constant

Equation when expressed in logarithmic form becomes

Q.

For the reactions, following data is given

Temperature at which k1 = k2 is

Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 21



Chemical Kinetics MCQ - 1 (Advanced) - Question 22

Match the column


*Answer can only contain numeric values
Chemical Kinetics MCQ - 1 (Advanced) - Question 23

The activation energy of a reaction at a given temperature is found to be 2.303 RT J mol–1. The ratio of rate constant to the Arrhenius factor is


Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 23 K=Ae^(-Ea/RT), K/A=10^(-Ea/2.303RT), now put the value of activation energy, k/A=10^-1 , so, k/A=0.1
*Answer can only contain numeric values
Chemical Kinetics MCQ - 1 (Advanced) - Question 24

An optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C by pseudo first order kinetics. The observed rotation of the mixture after 20 min was 5° while after completion of the reaction it was –20°. If optical rotation per mole of A, B & C are 60°, 40° & –80°. Calculate half life of the reaction.


Detailed Solution for Chemical Kinetics MCQ - 1 (Advanced) - Question 24


Optical rotation reaction no. of moles

At, t = 20 min
60 (a – x) + 40 x – 80 x = 5
60a – 100 x = 5
30 – 100 x = 5

No. of moles of A is becoming half after 20 min
∴ t1/2 = 20 min

*Answer can only contain numeric values
Chemical Kinetics MCQ - 1 (Advanced) - Question 25

 Consider the following first order decomposition process:

Here, "t" corresponds to the time at which (1/6)th of reactant is decomposed. The value of "n" is


12 videos|53 docs|118 tests
Use Code STAYHOME200 and get INR 200 additional OFF
Use Coupon Code
Information about Chemical Kinetics MCQ - 1 (Advanced) Page
In this test you can find the Exam questions for Chemical Kinetics MCQ - 1 (Advanced) solved & explained in the simplest way possible. Besides giving Questions and answers for Chemical Kinetics MCQ - 1 (Advanced), EduRev gives you an ample number of Online tests for practice