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Test: Chemical kinetics & The d & f block elements (December 15) - NEET MCQ


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15 Questions MCQ Test - Test: Chemical kinetics & The d & f block elements (December 15)

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Test: Chemical kinetics & The d & f block elements (December 15) - Question 1

A chemical reaction is catalyzed by a catalyst X.Hence X [1995]

Detailed Solution for Test: Chemical kinetics & The d & f block elements (December 15) - Question 1

A catalyst affects equally both forward and backward reactions, therefore it does not affect equilibrium constant of reaction.

Test: Chemical kinetics & The d & f block elements (December 15) - Question 2

The composition of ‘golden spangles’ is [1990]

Detailed Solution for Test: Chemical kinetics & The d & f block elements (December 15) - Question 2

PbI2 is yellow and is called golden spangles.

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Test: Chemical kinetics & The d & f block elements (December 15) - Question 3

A substance 'A' decomposes by a first or der reaction starting initially with [A] = 2.00 m and after 200 min, [A] becomes 0.15 m. For this reaction t1/2 is [1995]

Detailed Solution for Test: Chemical kinetics & The d & f block elements (December 15) - Question 3

Given initial concentration (a) = 2.00 m; Time taken (t) = 200 min and final concentration (a – x)    = 0.15 m. For a first order reation rate constant,

Further

Test: Chemical kinetics & The d & f block elements (December 15) - Question 4

Cinnabar is an ore of [1991]

Detailed Solution for Test: Chemical kinetics & The d & f block elements (December 15) - Question 4

Cinnabar (HgS) is an ore of Hg.

Test: Chemical kinetics & The d & f block elements (December 15) - Question 5

The rate of reaction depends upon the [1995]

Detailed Solution for Test: Chemical kinetics & The d & f block elements (December 15) - Question 5

The rate of a reaction is the speed at which the reactants are converted into products.
It depends upon the concentration of reactants. e.g for the reaction

Test: Chemical kinetics & The d & f block elements (December 15) - Question 6

If the rate of the reaction is equal to the rate constant, the order of the reaction is [2003]

Detailed Solution for Test: Chemical kinetics & The d & f block elements (December 15) - Question 6

∵ r = k[A]n
if n = 0
r = k[A]0 or  r = k thus for zero order reactions rate is equal to the rate constant.

Test: Chemical kinetics & The d & f block elements (December 15) - Question 7

Among the lanthanides the one obtained by synthetic method is [1994]

Detailed Solution for Test: Chemical kinetics & The d & f block elements (December 15) - Question 7

Pm is obtained by synthetic method.

Test: Chemical kinetics & The d & f block elements (December 15) - Question 8

K2Cr2O7 on heating with aqueous NaOH gives [1997]

Detailed Solution for Test: Chemical kinetics & The d & f block elements (December 15) - Question 8

Hence ion is obtained.

Test: Chemical kinetics & The d & f block elements (December 15) - Question 9

The lanthanide contraction is responsible for the fact that [1997]           (Atomic numbers : Zr = 40, Y = 39, Nb = 41, Hf = 72, Zn = 30)

Detailed Solution for Test: Chemical kinetics & The d & f block elements (December 15) - Question 9

We know that regular decrease in the size of the atoms and ions is called lanthanide contraction. In vertical column of transition elements there is a very small change in size and some times size is found same from second member to third member.The similarity in size of the atoms of Zr and Hf is evident due to the object of lanthanide contraction. Therefore Zr and Hf both have same radius 160 pm.

Test: Chemical kinetics & The d & f block elements (December 15) - Question 10

For the reaction rate and rate constant are 1.02 × 10–4 mol lit–1 sec–1 and 3.4 × 10–5 sec–1 respectively then concentration of N2O5 at that time will be [2001]

Detailed Solution for Test: Chemical kinetics & The d & f block elements (December 15) - Question 10

from the unit of rate constant it is clear that the reaction follow first order kinetics. Hence by rate law equation,    
 r = k [N2O5] where r = 1.02 × 10–4,
k = 3.4 × 10–5 1.02 × 10–4 = 3.4 × 10–5 [N2O5]
[N2O5] = 3M

Test: Chemical kinetics & The d & f block elements (December 15) - Question 11

Select the rate law that corresponds to data shown for the following reaction [1994]

Detailed Solution for Test: Chemical kinetics & The d & f block elements (December 15) - Question 11

From data 1 and 3, it is clear that keeping (B) const, When [A] is doubled, rate remains unaffected. Hence rate is independent of [A]. from 1 and 4, keeping [A] constant, when [B] is doubled, rate become 8 times. Hence [rate ∝ [ B]3].

Test: Chemical kinetics & The d & f block elements (December 15) - Question 12

In a reversible reaction the energy of activation of the forward reaction is 50 kcal. The energy of activation for the reverse reaction will be [1996]

Detailed Solution for Test: Chemical kinetics & The d & f block elements (December 15) - Question 12

Thus energy of activation for reverse reaction depend upon whether reaction is exothermic or endothermic If reaction is exothermic,  ΔH = +ve Ea(b) > Ea(f) If reaction is endothermic  ΔH =+ve Ea(b) < Ea(f )

Test: Chemical kinetics & The d & f block elements (December 15) - Question 13

Nitriding is the process of surface hardening of steel by treating it in an atmosphere of [1989]

Detailed Solution for Test: Chemical kinetics & The d & f block elements (December 15) - Question 13

When steel is heated in presence of NH3 , iron nitride on the surface of steel is formed which imparts a hard coating. This process is called nitriding.

Test: Chemical kinetics & The d & f block elements (December 15) - Question 14

The rate constants k1 and k2 for two different reactions are 1016 . e–2000/T and 1015 . e–1000/T, respectively. The temperature at which k1 = k2 is :

Detailed Solution for Test: Chemical kinetics & The d & f block elements (December 15) - Question 14

Given, 

when k1 and k2 are equal at any temperature T, we have

or 2.303 ×1×T=1000           [∴ log 10= 1]

Test: Chemical kinetics & The d & f block elements (December 15) - Question 15

Which of the following combines with Fe (II) ions to form a brown complex? [2000]

Detailed Solution for Test: Chemical kinetics & The d & f block elements (December 15) - Question 15

We know that when nitrogen oxide (NO) combines with Fe (II) ions, a brown complex is formed. This reaction is called brown ring test

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