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Rate Law, Molecularity & Order of a Chemical Reaction - NEET MCQ


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20 Questions MCQ Test - Rate Law, Molecularity & Order of a Chemical Reaction

Rate Law, Molecularity & Order of a Chemical Reaction for NEET 2024 is part of NEET preparation. The Rate Law, Molecularity & Order of a Chemical Reaction questions and answers have been prepared according to the NEET exam syllabus.The Rate Law, Molecularity & Order of a Chemical Reaction MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Rate Law, Molecularity & Order of a Chemical Reaction below.
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Rate Law, Molecularity & Order of a Chemical Reaction - Question 1

Direction (Q. Nos. 1-13) This section contains multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct

Q. In the following reaction, which has maximum rate w.r.t. rate of disappearance of NH3?
4NH3 + 50→  4NO + 6H2O

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 1

By stoichiometry of the reaction

Rate Law, Molecularity & Order of a Chemical Reaction - Question 2

The rate of formation of NO(g) in the reaction, 2NOBr(g) → 2NO(g) + Br2(g) was reported as 1.6 x 10-4 Ms-1. Thus, rate of the reaction is 

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 2

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Rate Law, Molecularity & Order of a Chemical Reaction - Question 3

Rate of formation of SO3 in the reaction below is 100 kg min-1.
2SO2 + O2 → 2SO3
Hence, rate of disappearance of SO2 will be:

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 3

Rate of disappearance of SO2 = Rate of formation of SO3 But quantity is to be expressed in terms of moles.
Rate of formation of SO3 = 100 kg SO3 min-1 

Rate Law, Molecularity & Order of a Chemical Reaction - Question 4

The following reaction can take place in both directions:

For the forward reaction, the rate varies with the concentration of A as and for the backward reaction,


Hence, net reaction rate is:

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 4

Rate Law, Molecularity & Order of a Chemical Reaction - Question 5

For a reaction, xA → yB, rate of disappearance of ‘A’ is related to the rate of appearance of 'B' by the equation 

Thus, x and y respectively are

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 5



Rate Law, Molecularity & Order of a Chemical Reaction - Question 6

For the reaction,

H2(g) + A2(g)  2H(g) + A2(g)

kf = 2.2 x 104 L mol-1s-1 and Kc = 1.00 x 10-4 at 3000 K. Thus, kb is 

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 6

Rate Law, Molecularity & Order of a Chemical Reaction - Question 7

For a reaction,

 

hen x, y and z are 

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 7

For the reaction: xA + yB → zC

Given that,

Multiplying equation 2 by 1/3, we get

On comparing equation 1 and 3,
x = 3, y = 3, z = 2.

Rate Law, Molecularity & Order of a Chemical Reaction - Question 8

For the reaction,

2N2O5(g) → 4NO2 (g) + O2(G)

Thus,

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 8

Rate Law, Molecularity & Order of a Chemical Reaction - Question 9

The decomposition of acetaldehyde is given by the following reaction:

CH3CHO(g) → CH4(g) + CO(g)

Rate of the reaction with respect to the pressure of the reactant is 

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 9

Rate Law, Molecularity & Order of a Chemical Reaction - Question 10

For the reaction, 

 then rate law is

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 10

Rate Law, Molecularity & Order of a Chemical Reaction - Question 11

For gaseous reaction, the rate can be expressed as

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 11





Rate Law, Molecularity & Order of a Chemical Reaction - Question 12

The initial rates of reaction for the equation, 2A + B → Products.

Products were determined under various initial concentrations of reactants.

Thus, rate law is equal to

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 12

Let order w.r.t. A = a, order w.r.t. B = b


Rate Law, Molecularity & Order of a Chemical Reaction - Question 13

For the reaction, A + 2B → Product, the reaction rate was halved on doubling the concentration of A. Thus, order w.r.t. A is

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 13

If concentration of the nth order reaction is made m time, rate becomes mn times.


Rate Law, Molecularity & Order of a Chemical Reaction - Question 14

Direction (Q. Nos. 14 and 15) This section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)

Passage 

The decomposition of NO2 at 400 K proceeds at a of rate of 5.4 x 10 -5 mol L-1 s-1 when [NO2] = 0.01 mol-1

2 NO2(g) → 2NO(g ) + O2(g).

Q. What is the rate law when observed rate is 1.35 x 10-5 mol L-1 s-1 at [NO2] = 0.005 mol L-1?

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 14

Correct answer is option (D).
The given information can be summarized as follows:

Decomposition of NO2:
- Reaction: 2 NO2(g) → 2NO(g) + O2(g)
- At 400 K: Rate = 5.4 x 10^(-5) mol L^(-1) s^(-1), [NO2] = 0.01 mol L^(-1)
- Observed rate: 1.35 x 10^(-5) mol L^(-1) s^(-1), [NO2] = 0.005 mol L^(-1)

Let's compare the rates and concentrations given in the problem:

Rate1 = 5.4 x 10^(-5) mol L^(-1) s^(-1), [NO2]1 = 0.01 mol L^(-1)
Rate2 = 1.35 x 10^(-5) mol L^(-1) s^(-1), [NO2]2 = 0.005 mol L^(-1)

Take the ratio of the two rates:

Rate2/Rate1 = (1.35 x 10^(-5))/(5.4 x 10^(-5)) = 1/4

Now, take the ratio of the two concentrations:

[NO2]2/[NO2]1 = (0.005)/(0.01) = 1/2

We can now evaluate each of the rate law options:

a. k[NO2]: (1/2) ≠ 1/4
b. k[NO2]^0: 1 ≠ 1/4
c. k[NO2]^3: (1/2)^3 ≠ 1/4
d. k[NO2]^2: (1/2)^2 = 1/4

Only option (d) satisfies the relationship between the rates and concentrations:
Rate Law:
- k[NO2]^2

Rate Law, Molecularity & Order of a Chemical Reaction - Question 15

The decomposition of NO2 at 400 K proceeds at a of rate of 5.4 x 10 -5 mol L-1 s-1 when [NO2] = 0.01 mol-1

2 NO2(g) → 2NO(g ) + O2(g).

Q. Rate constant of the reaction will be 

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 15


Rate Law, Molecularity & Order of a Chemical Reaction - Question 16

Direction (Q. Nos. 16-18) This section contains 3 questions. When worked out will result in one integer from 0 to 9 (both inclusive)

In the following reaction

Q. Where negative sign indicates rate of disappearance of the reactant. What is the value of x/y ?

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 16




Rate Law, Molecularity & Order of a Chemical Reaction - Question 17

The reaction rate is defined as the rate at which the concentration of the reactants __________ with time or the concentration of products ___________ with time.

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 17

The reaction rate is defined as the rate at which the concentration of the reactants decreases with time or the concentration of products increases with time.

Rate Law, Molecularity & Order of a Chemical Reaction - Question 18

In a certain polluted atmosphere containing O3 at a steady concentration of 2.0 x 10-6 M, the hourly production of O3 by all sources was estimated as 7.2 x 10-15 M. If the only mechanism for destruction of O3 in the second-order reaction is

2O3 → 3O2

then rate constant for destruction reaction, defined by the rate law for - Δ[O3]/Δf is x * 10-7 M-1 s-1 . What is the value of x?

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 18



Rate Law, Molecularity & Order of a Chemical Reaction - Question 19

Only One Option Correct Type

This section contains 2 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct

For a reversible reaction, net rate is 

hence given reaction is 

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 19

2A - B →(reversible) C

For a forward reaction, (dx/dt)f = k1[A]2 [B]-1
For a backward reaction, (dx/dt)b = k2[C]

Net rate = (dx/dt)f - (dx/dt)= k1[A]2 [B]-1 - k2[C]

Rate Law, Molecularity & Order of a Chemical Reaction - Question 20

Reaction kinetics deals with the study of

Detailed Solution for Rate Law, Molecularity & Order of a Chemical Reaction - Question 20

Reaction kinetics deals with the study of rate of reaction, their mechanism and the factors which affects the rate of reaction. It specifies all the general characteristics of a chemical reaction.

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