Chemistry Exam  >  Chemistry Tests  >  Physical Chemistry  >  Test: Chemical Kinetics - 3 - Chemistry MCQ

Test: Chemical Kinetics - 3 - Chemistry MCQ


Test Description

30 Questions MCQ Test Physical Chemistry - Test: Chemical Kinetics - 3

Test: Chemical Kinetics - 3 for Chemistry 2024 is part of Physical Chemistry preparation. The Test: Chemical Kinetics - 3 questions and answers have been prepared according to the Chemistry exam syllabus.The Test: Chemical Kinetics - 3 MCQs are made for Chemistry 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Chemical Kinetics - 3 below.
Solutions of Test: Chemical Kinetics - 3 questions in English are available as part of our Physical Chemistry for Chemistry & Test: Chemical Kinetics - 3 solutions in Hindi for Physical Chemistry course. Download more important topics, notes, lectures and mock test series for Chemistry Exam by signing up for free. Attempt Test: Chemical Kinetics - 3 | 30 questions in 90 minutes | Mock test for Chemistry preparation | Free important questions MCQ to study Physical Chemistry for Chemistry Exam | Download free PDF with solutions
Test: Chemical Kinetics - 3 - Question 1

The energy of activation of a forward reaction is 50Kcal. The energy of activation of a backward reaction is:

Test: Chemical Kinetics - 3 - Question 2

The rate constant is numerically the same for three reactions of first, second and third orders respectively. When concentration is less than unity, the rate of reactions are, in order:

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Chemical Kinetics - 3 - Question 3

The number of molecules of the reactants involved in a single stage of the reaction indicates

Detailed Solution for Test: Chemical Kinetics - 3 - Question 3

the molecularity of reaction

Test: Chemical Kinetics - 3 - Question 4

Rate of a reaction. A + B → Product is given below as a function of different initial concentrations of A and B.

The rate of the reaction is:

Test: Chemical Kinetics - 3 - Question 5

In a reaction: 2N2O5 → 4NO2 + O2 , the rate is expressed as

The relation among K1, K2 and K3 is:

Detailed Solution for Test: Chemical Kinetics - 3 - Question 5

Correct Answer :- c

Explanation : As we know, for a reaction: 2N2O5 → 4NO2 + O2

=> -1/2 d[N2O5]/dt = 1/4 d[NO2]/dt = d[O2]/dt

So, 2k1 = k2 = 4k3

Test: Chemical Kinetics - 3 - Question 6

Calculate order of reaction A → product , from the following data:

Detailed Solution for Test: Chemical Kinetics - 3 - Question 6

By doubling the concentration the rate becomes half from (i) to (ii) also from (ii) to (iii).

Hence, B is correct.

Test: Chemical Kinetics - 3 - Question 7

A first order reaction is 50% complete in 30mins at 270C and in 10mins at 470C. At 270C, the activation energy of the reaction in KJ/mole is:

Detailed Solution for Test: Chemical Kinetics - 3 - Question 7

Test: Chemical Kinetics - 3 - Question 8

What is the rate of simple reaction 2NO + O2 → 2 NO2 ; when the volume of the reaction vessel is doubled?

Test: Chemical Kinetics - 3 - Question 9

Which among the following procedures will lead to a change in rate constant ‘k’ of a reaction?

(I) A change in pressure
(II) Change in temperature
(III) Change in volume of the reaction vessel
(IV) An introduction of a catalyst

Test: Chemical Kinetics - 3 - Question 10

The concept of t1/2 is useful for the reaction of:

Test: Chemical Kinetics - 3 - Question 11

For a given set of reactions:

The rate constant for the decay of N2O5 depends on:

Test: Chemical Kinetics - 3 - Question 12

The given reaction (pre-equilibrium mechanism)

Test: Chemical Kinetics - 3 - Question 13

The gas phase decomposition of NOBr is second order in [NOBr], with k = 0.810 M–1 s–1 at 100C. We start with in a flask at 100C. How many seconds does it take to use up

2NOBr(g) → 2NO(g) + Br2(g)

Rate = k[NOBr]2

Test: Chemical Kinetics - 3 - Question 14

The concentration of a reactant undergoing decomposition was 0.1, 0.08 and 0.067 mol L–1 after 1.0, 2.0 and 3.0 hr respectively. The order of the reaction is:

Detailed Solution for Test: Chemical Kinetics - 3 - Question 14

Correct Answer :- c

Explanation : A ----->k P

-d[A]/dt = k[A](0 to n)

-{[A]0 - [A]}/(t2 - t1) = k[A](0 to n)

{[A] - [A]0}/(t2 - t1) = k[A](0 to n)

[A]0 = concentration at t1

and [A] = concentration at t2

{[A] - [A]0}/(t2 - t1) = k[A](0 to n)

(0.1-0.08)/(2-1) = k[0.1]n

k[0.1]n = 0.02......(1)

(0.08-0.067)/(3-2) = k[0.08]n

k[0.08]n = 0.013.........(2)

Dividing (1) by (2)

k[0.1]n/k[0.08]n = 0.02/0.013

{[0.1]/[0.08]}n = 1.5385

[1.25]n = 1.5385

[1.25]n = (1.25)2

n = 2

Test: Chemical Kinetics - 3 - Question 15

Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows Arrhenius equation is:

Test: Chemical Kinetics - 3 - Question 16

A reaction of first-order completed 90% in 90 minutes, hence, it is completed 50% in approximately.

Test: Chemical Kinetics - 3 - Question 17

The unit of catalytic constant is equal to:

Test: Chemical Kinetics - 3 - Question 18

At 350K, the effective rate constant for a gaseous reaction A → P , which has a Lindeman-Hinshelwood mechanism are and  respect ively. The rate constant for activation step in the mechanism is:

Test: Chemical Kinetics - 3 - Question 19

In the temperature range of 250 K to 450 K, the pre-exponent ial factor, A, for the reaction. Cl(g) + H2(g) → HCL(g) + H(g) is found to be equal to 1.20 x 1010 dm3 mol-1 s-1 . If M (Cl) = 35.453 g mol-1, M[H2] = 2.016 g mol-1, d(Cl) = 200 pm and d[H2] = 150 pm, the value of the Steric factor, P is:

Test: Chemical Kinetics - 3 - Question 20

The t1/2 of a reaction is doubled as the initial concentration of a reactant is increased 4 times. The order of the reaction is:

Detailed Solution for Test: Chemical Kinetics - 3 - Question 20

Correct Answer :- c

Explanation : A → P . For zero order reaction, rate= k. So, unit of k is

moles/sec.

- d[A]/dt = k

Upon integration between limits, [ A ] − [ A0 ] = − kt 

 At t1/2, [A]=[A0]/2.

Hence, t1/2 = [A0]/2k

Increased by 4 times, so it will become double

 

Test: Chemical Kinetics - 3 - Question 21

The relaxation time for the fast reaction:

and equilibrium constant is 1.0 x 10–3. The rate constant of the backward reaction is:

Detailed Solution for Test: Chemical Kinetics - 3 - Question 21

Correct Answer :- d

Explanation : Kb = (10 * 10-6) * 10-3

= 10-8 s-1

Test: Chemical Kinetics - 3 - Question 22

For which of the following reactions, the rate constant will be independent of ionic strength.

(I) H+ + Br + H2O2
(II) C2H5COOCH3 + OH
(III) S2O32– + I

Test: Chemical Kinetics - 3 - Question 23

Which of the following plots is correct for an ionic reaction:

Test: Chemical Kinetics - 3 - Question 24

For the first order isomerization of an organic compound at 1300C, the activation energy is 108.4 KJ mol–1 and the rate constant is 9.12 x 10–4 s–1. The standard entropy of activation for this reaction is:

Detailed Solution for Test: Chemical Kinetics - 3 - Question 24

For the first order isomerization of an organic compoundat 130°c,the activation energy is 108.4KJ mol-1 and the rate constant is 9.12×10^-4s-1. calculate the standard entropy of activation for this reaction. I know the answer is
- 45.2 JK–1 mol–1 .

Test: Chemical Kinetics - 3 - Question 25

For reaction system,

2NO(g) + O2 (g) → 2 NO2 (g)

Volume is suddenly reduced to half of its value by increasing the pressure on it. If the reaction is of first order with respect to O2 and second order with respect to NO, the rate of reaction will.

Detailed Solution for Test: Chemical Kinetics - 3 - Question 25

Test: Chemical Kinetics - 3 - Question 26

Rate = K[A][B]; If the volume of reaction vessel is suddenly reduced 1/4th of initial value. The new rate will be effected by:

Test: Chemical Kinetics - 3 - Question 27

Under the same reaction conditions, initial concentration of 1.386 mol dm–3 of a substance becomes half in 40 s and 20 s through first order and zero order kinetics, respectively., Ratio  of the rate constants for the first order k1 and zero order (k0) of the reaction is 

Test: Chemical Kinetics - 3 - Question 28

T50 (Half-life period) of first-order reaction is 10 minute. Starting with 10 mol L–1. Rate after 20 minute is:

Test: Chemical Kinetics - 3 - Question 29

If the fermentation of sugar in an enzymatic solution that in 0.12 M. the concentration of the sugar is reduced to 0.06 M in 10 h and to 0.03 M in 20 h. What is the order of the reaction:

Test: Chemical Kinetics - 3 - Question 30

In a zero-order reaction for every 10° rise of temperature, the rate is doubled. If the temperature is increased from 10°C to 100°C, the rate of the reaction will become 

83 videos|142 docs|67 tests
Information about Test: Chemical Kinetics - 3 Page
In this test you can find the Exam questions for Test: Chemical Kinetics - 3 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Chemical Kinetics - 3, EduRev gives you an ample number of Online tests for practice
83 videos|142 docs|67 tests
Download as PDF