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Test: Chemical Kinetics(10 Nov) - JEE MCQ


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15 Questions MCQ Test - Test: Chemical Kinetics(10 Nov)

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Test: Chemical Kinetics(10 Nov) - Question 1

Rate of reaction can be expressed by Arrhenius equation as k = Ae–E/RT , In this equation, E represents

[AIEEE 2006]

Detailed Solution for Test: Chemical Kinetics(10 Nov) - Question 1

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

Test: Chemical Kinetics(10 Nov) - Question 2

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 Test: Chemical Kinetics(10 Nov) - Question 2

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.
 

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Test: Chemical Kinetics(10 Nov) - Question 3

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 Test: Chemical Kinetics(10 Nov) - Question 3

The correct answer is Option C

Test: Chemical Kinetics(10 Nov) - Question 4

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 Test: Chemical Kinetics(10 Nov) - Question 4

The correct answer is Option C.

Now 2.303log10 
2.303log102 = 

=53598.6 J 
=53.6 kJmol−1

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