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Chemical Kinetics - 1 - JEE MCQ


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30 Questions MCQ Test Chemistry for JEE Main & Advanced - Chemical Kinetics - 1

Chemical Kinetics - 1 for JEE 2024 is part of Chemistry for JEE Main & Advanced preparation. The Chemical Kinetics - 1 questions and answers have been prepared according to the JEE exam syllabus.The Chemical Kinetics - 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Chemical Kinetics - 1 below.
Solutions of Chemical Kinetics - 1 questions in English are available as part of our Chemistry for JEE Main & Advanced for JEE & Chemical Kinetics - 1 solutions in Hindi for Chemistry for JEE Main & Advanced course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Chemical Kinetics - 1 | 30 questions in 60 minutes | Mock test for JEE preparation | Free important questions MCQ to study Chemistry for JEE Main & Advanced for JEE Exam | Download free PDF with solutions
Chemical Kinetics - 1 - Question 1
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A chemical reaction [2A] + [2B] + [C] → product follows the rate equation : then order of reaction is -            [AIEEE-2002]

Detailed Solution for Chemical Kinetics - 1 - Question 1

The correct answer is option C
 3A ➡ 2B

Thus, the order of reaction with respect to A = 1
The order of reaction with respect to B =1
Total order of reaction = 1+1 = 2

Chemical Kinetics - 1 - Question 2
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For the reaction system: 2NO(g) + O2(g) → 2NO2(g) volume is suddenly reduced to half its value by increasing the pressure on it. If the reaction is of first order with respect to O2 and second order with repect to NO, the rate of reaction will – 

  [AIEEE-2003]

Detailed Solution for Chemical Kinetics - 1 - Question 2

The correct answer is option A
rate=k[NO]2[O2]
As, C = n/V
On making the volume half, concentration doubles and rate of reaction becomes eight times.
 

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Chemical Kinetics - 1 - Question 3
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Consider an endothermic reaction X → Y with the activation energies Eb and Ef for the backward and forward reactions, respectively. In general   

[AIEEE-2005]

Detailed Solution for Chemical Kinetics - 1 - Question 3

The correct answer is option B
Here, Ef ​> Eb​

 

Chemical Kinetics - 1 - Question 4
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Rate of reaction can be expressed by Arrhenius equation as k = Ae–E/RT , In this equation, E represents

[AIEEE 2006]
Detailed Solution for Chemical Kinetics - 1 - Question 4

The correct answer is Option A.

E represents the energy of activation which implies it is the energy below which colliding molecules will not react Arrhenius equation gives the dependence of the rate constant k of a chemical reaction on the absolute temperature T (in Kelvin), where A is the pre-exponential factor (or simply the prefactor), Ea is the activation energy, and R is the Universal gas constant:
By Arrhenius equation, k=Ae-Ea/RT

Chemical Kinetics - 1 - Question 5
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The following mechanism has been proposed for the reaction of NO with Br2 to form NOBr :
NO(g) + Br2 (g)   NOBr2 (g)
NOBr2 (g) + NO (g) → 2 NOBr (g)
If the second step is the rate determining step, the order of the reaction with respect to NO (g) is -

[AIEEE 2006]
Detailed Solution for Chemical Kinetics - 1 - Question 5

The correct answer is option c
NO(g)+Br2(g)     ⇌NOBr2(g)
NOBr2(g)+NO(g) 2NOBr(g)[rate determining step]
Rate of the reaction (r)= K[NOBr2][NO]
Where [NOBR2]=Kc[NO][Br2]
r=K.Kc.[NO][Br2][NO]
r=K’[NO]2[Br2].
The order of the reaction with respect to
NO(g)=2.
 

Chemical Kinetics - 1 - Question 6
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The time for half life period of a certain reaction A → products is 1 hour. 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]
Detailed Solution for Chemical Kinetics - 1 - Question 6

The correct answer is Option C

Chemical Kinetics - 1 - Question 7
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The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be: (R = 8.314 JK–1 mol–1 and log 2 = 0.301)          

[IIT Mains 2013]
Detailed Solution for Chemical Kinetics - 1 - Question 7

The correct answer is Option C.

Now 2.303log10 
2.303log102 = 
∴
=53598.6 J 
=53.6 kJmol−1

Chemical Kinetics - 1 - Question 8
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Chemical Kinetics - 1 - Question 23
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For the reaction 2A + B → C, the values of initial rate at different reactant concentrations are given in the table below. The rate law for the reaction is :
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Chemical Kinetics - 1 - Question 24
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