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JEE • CHEMISTRY • CHEMICAL KINETICS

Chemical kinetics practice · Hard

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Showing 1-30 of 54 questions

1 Hard

1 of 54

Analyzing +4 | -1

The reaction

\mathrm{A}_2+\mathrm{B}_2 \rightarrow 2 \mathrm{AB}

follows the mechanism

\begin{aligned} & {{\text{A}}_2}\mathop \rightleftharpoons \limits_{{{\text{k}}_{ - 1}}}^{{{\text{k}}_1}} {\text{A}} + {\text{A (fast) }} \\ & \mathrm{A}+\mathrm{B}_2 \xrightarrow{\mathrm{k}_2} \mathrm{AB}+\mathrm{B} \text { (slow) } \\ & \mathrm{A}+\mathrm{B} \rightarrow \mathrm{AB} \text { (fast) } \end{aligned}

The overall order of the reaction is:

View answer and explanation

Correct option: A

1.5

Option A (1.5) is correct because the rate law is determined by the slow step, and the concentration of the intermediate A is found using the equilibrium of the fast step, resulting in an overall order of 1.5.

2 Hard

2 of 54

Application +4 | -1

For a reaction of order

n,

the integrated form of the rate equation is

(n-1) \cdot K \cdot t=\left(C_0\right)^{1-n}-(C)^{1-n},

where

C_0

and

C

are the values of the reactant concentration at the start and after time '

t

'. What is the relationship between

t_{3 / 4}

and

t_{1 / 2},

where

t_{3 / 4}

is the time required for

C

to become

C_0 / 4.

View answer and explanation

Correct option: A

t_{3 / 4}=t_{1 / 2} \cdot\left[2^{n-1}+1\right]

The correct answer is

t_{3 / 4}=t_{1 / 2} \cdot\left[2^{n-1}+1\right]

because it is derived from the integrated rate equation for a reaction of order n.

3 Hard

3 of 54

Analyzing +4 | -1

The temperature coefficients of two reactions are 2 and 3 respectively. Which would be correct for these reactions?

View answer and explanation

Correct option: A

E_{a_1} < E_{a_2}

Option A (

E_{a_1} < E_{a_2}

) is correct because a higher temperature coefficient indicates a greater sensitivity to temperature changes, implying a higher activation energy.

4 Hard

4 of 54

Application +4 | -1

Half-life of two samples is 0.1 and 0.8 s. Their respective concentration is 400 and 50 respectively.

The order of reaction is

View answer and explanation

Correct option: B

Option B (2) is correct because the order of the reaction is calculated to be 2 using the half-life equation and the given concentrations.

5 Hard

5 of 54

Analyzing +4 | -1

In the following reaction;

xA \to yB

{\log _{10}}\left[ { - \frac{{d\left[ A \right]}}{{dt}}} \right] = {\log _{10}}\left[ {\frac{{d\left[ B \right]}}{{dt}}} \right] + 0.3010

A and B respectively can be

View answer and explanation

Correct option: C

{C_2}{H_4}

and

{C_4}{H_8}

Option C is correct because the stoichiometry of the reaction is 2A -> B, which corresponds to 2C2H4 -> C4H8.

6 Hard

6 of 54

Application +4 | -1

The thermal decomposition of a compound is of first order. If a sample of the compound decompose 50% in 120 min. What time will it take to undergo 90% decomposition?

View answer and explanation

Correct option: A

Nearly 400 min

The correct answer is Nearly 400 min because the time for 90% decomposition is calculated using the first-order kinetics equation.

7 Hard

7 of 54

Analyzing +4 | -1

If, at

\mathrm{t}=30 \mathrm{~sec},[\mathrm{A}]=100 \mathrm{~moles} / \mathrm{L}

,

at

\mathrm{t}=40 \mathrm{~sec},[\mathrm{A}]=80 \mathrm{~moles} / \mathrm{L}

,

at

\mathrm{t}=50 \mathrm{~sec},[\mathrm{A}]=64 \mathrm{~moles} / \mathrm{L}

For a reaction,

A \rightarrow B

, the order will be

View answer and explanation

Correct option: A

Option A (1) is correct because the rate constant is consistent throughout the given time intervals, indicating a first-order reaction.

8 Hard

8 of 54

Analyzing +4 | -1

The following reactions are occurring simultaneously in a vessel such that the temperature of the system is not changing.

\begin{aligned} & {A}+3 {B} \rightarrow {P}: \Delta {H}=-2 x \mathrm{~kJ} \\ & {M} \rightarrow 2 {Q}+{R}: \Delta {H}=+x \mathrm{~kJ} \end{aligned}

If the rate of disappearance of '

{B}

' is

y \mathrm{~M} \mathrm{s}^{-1},

the rate of appearance of '

{Q}

' (in

\mathrm{M} \mathrm{s}^{-1}

) is

View answer and explanation

Correct option: B

1.33 y \mathrm{~M} \mathrm{s}^{-1}

Option B (1.33 y M s⁻¹) is correct because the rate of appearance of Q is calculated based on the stoichiometry of the reactions and the given rate of disappearance of B.

9 Hard

9 of 54

Application +4 | -1

For a first order reaction

t_{3 / 4}

1200 \mathrm{~s}

. The specific rate constant is (in

\mathrm{s}^{-1}

)

View answer and explanation

Correct option: A

10^{-3}

Option A (

10^{-3}

) is correct because it is calculated using the formula for the half-life of a first-order reaction.

10 Hard

10 of 54

Understanding +4 | -1

Identify correct statements.

As the temperature increases.

A) Rate of reaction increases

B) Number of activated molecules increases

C) Collision frequency increases

View answer and explanation

Correct option: C

A, B, and C

Option C is correct because as temperature increases, the rate of reaction, the number of activated molecules, and the collision frequency all increase.

11 Hard

11 of 54

Application +4 | -1

The reaction,

{X}+2 {Y}+{Z} \rightarrow {N}

occurs by the following mechanism :-

(i)

{X}+{Y} \rightleftharpoons {M}

very rapid equilibrium

(ii)

{M}+{Z} \rightarrow {P}

slow

(iii)

{P}+{Y} \rightarrow {N}

very fast

What is the rate law for this reaction :-

View answer and explanation

Correct option: D

Rate

={k}[{X}][{Y}][{Z}]

The correct answer is Rate = k[X][Y][Z] because the rate law is determined by the slow step in the mechanism, and the concentrations of reactants in the slow step are multiplied by the equilibrium constant of the rapid equilibrium.

12 Hard

12 of 54

Application +4 | -1

The half-life periods of two first-order reactions are in the ratio

3: 2.

t_1

is the time required for

25 \%

completion of the first reaction and

t_2

is the time required for

75 \%

completion of the second reaction, then the ratio,

t_1: t_2,

(\log 3=0.48, \log 2=0.3)

View answer and explanation

Correct option: A

3: 10

Option A (

3: 10

) is correct because it represents the ratio of the time required for 25% completion of the first reaction to the time required for 75% completion of the second reaction, calculated using the half-life relationship for first-order reactions.

13 Hard

13 of 54

Application +4 | -1

In the following reaction,

A(g) \stackrel{k_1}{\underset{k_2}{\rightleftharpoons}} B(g)

\frac{-d[A]}{d t}=k_1[A]

for the forward reaction

\frac{-d[B]}{d t}=k_2[B]

for the backward reaction
Half-life period of the forward reaction is 10 min and that of backward reaction is 1 min. Thus, equilibrium constant is

View answer and explanation

Correct option: A

0.1

The correct answer is A (0.1). The equilibrium constant is the ratio of the forward rate constant to the backward rate constant, which can be calculated from the half-lives.

14 Hard

14 of 54

Analyzing +4 | -1

What will happen to the rate of nitration of benzene if large amount of

\mathrm{KHSO}_4

is added to the mixture responsible for nitration?

View answer and explanation

Correct option: B

Rate will decrease

Option B (Rate will decrease) is correct because adding

\mathrm{KHSO}_4

causes a common ion effect, reducing the dissociation of

\mathrm{H}_2\mathrm{SO}_4

and thus the formation of

\mathrm{NO}_2^+

15 Hard

15 of 54

Analyzing +4 | -1

In the reaction

a \mathrm{~A}+b \mathrm{~B}+c \mathrm{C} \rightarrow

Products,

(I) if concentration of

\mathrm{A}

is doubled, keeping concentration of

\mathrm{B}

and

\mathrm{C}

constant, the rate of reaction becomes doubled.
(II) if concentration of

B

is halved, concentration of

A

and

C

constant, the rate of reaction remains unaffected.

(III) if concentration of

\mathrm{C}

is made 1.5 times, the rate of reaction becomes 2.25 times.

The order of reaction is

View answer and explanation

Correct option: C

Option C (3) is correct. The rate law is determined to be Rate = k[A][C]², leading to an overall order of 3.

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16 Hard

16 of 54

Application +4 | -1

A flask contains a mixture of compounds

A

and

B.

Both compounds decompose by first-order kinetics. The half-life for

A

and

B

are 300 s and 180 s, respectively. If the concentrations of

A

and

B

are equal initially, the time required for the concentration of

A

to be four times that of

B

(in s) is (Use ln

2 = 0.693)

View answer and explanation

Correct option: D

900

Option D (900) is correct because it is the time calculated using first-order kinetics equations where the concentration of A is four times that of B.

17 Hard

17 of 54

Analyzing +4 | -1

The rate of the reaction

2 \mathrm{~N}_2 \mathrm{O}_5 \rightarrow 4 \mathrm{NO}_2+\mathrm{O}_2

can be written in three ways:

\begin{aligned} & \frac{-\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}={k}\left[\mathrm{~N}_2 \mathrm{O}_5\right] ; \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}={k}^{\prime}\left[\mathrm{N}_2 \mathrm{O}_5\right] \\ & \frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}={k}^{\prime \prime}\left[\mathrm{N}_2 \mathrm{O}_5\right] \end{aligned}

The relationship between

{k}

and

{k}^{\prime}

and between

{k}

and

{k}^{\prime \prime}

are:

View answer and explanation

Correct option: B

k^{\prime}=2 k ; k^{\prime \prime}=k / 2

Option B (

k^{\prime}=2 k ; k^{\prime \prime}=k / 2

) is correct because the rate constants are related by the stoichiometry of the reaction.

18 Hard

18 of 54

Analyzing +4 | -1

The rate equation for an autocatalytic reaction

\mathrm{A}+\mathrm{R} \xrightarrow{k} \mathrm{R}+\mathrm{R} \text { is } r_{\mathrm{A}}=-\frac{\mathrm{d} C_{\mathrm{A}}}{\mathrm{dt}}=k C_{\mathrm{A}} C_{\mathrm{R}}

The rate of disappearance of reactant

\text{A}

is maximum when

View answer and explanation

Correct option: B

C_{\mathrm{A}}=C_{\mathrm{R}}

Option B (

C_{\mathrm{A}}=C_{\mathrm{R}}

) is correct because the rate of an autocatalytic reaction is maximum when the concentrations of the reactant and product are equal.

19 Hard

19 of 54

Application +4 | -1

A flask contains a mixture of compounds

A

and

B

initially

\begin{aligned} & {[A]_0=[B]_0} \\ & A \xrightarrow{t_{50}=54 \mathrm{~min}} \text { product, }-\frac{d[A]}{d t}=k[A] \\ & B \xrightarrow{t_{50}=18 \mathrm{~min}} \text { product, }-\frac{d[B]}{d t}=k[B] \end{aligned}

After 54 minutes,

[A]=x[B]

what is the value of

x

View answer and explanation

Correct option: A

Option A (4) is correct because after 54 minutes, the concentration ratio [A]/[B] is calculated using the half-life information for first-order kinetics.

20 Hard

20 of 54

Application +4 | -1

Half-life of two samples is 0.1 and 0.8 s. Their respective concentration is 400 and 50 respectively.

The order of reaction is

View answer and explanation

Correct option: B

Option B (2) is correct because the order of the reaction is calculated to be 2 using the half-life equation and the given concentrations.

21 Hard

21 of 54

Analyzing +4 | -1

The reaction of

(+)

2-iodobutane and

\mathrm{NaI}^*

(

\mathrm{I}^*

is radioactive isotope of iodine) in acetone was studied by measuring the rate of racemization

\left(\mathrm{k}_{\mathrm{r}}\right)

and the rate of inversion

\left(\mathrm{k}_{\mathrm{i}}\right)

(+) \mathrm{CH}_3 \mathrm{CH}(\mathrm{I}) \mathrm{CH}_2 \mathrm{CH}_3+\mathrm{NaI}^* \rightarrow \mathrm{CH}_3 \mathrm{CH}\left(\mathrm{I}^*\right) \mathrm{CH}_2 \mathrm{CH}_3+\mathrm{NaI}

For this reaction, the relationship between

\mathrm{k}_{\mathrm{r}}

and

\mathrm{k}_{\mathrm{i}}

View answer and explanation

Correct option: B

\mathrm{k}_{\mathrm{i}}=0.5 \mathrm{k}_{\mathrm{r}}

Option B (

\mathrm{k}_{\mathrm{i}}=0.5 \mathrm{k}_{\mathrm{r}}

) is correct because the rate of racemization is twice the rate of inversion in this SN1 reaction.

22 Hard

22 of 54

Analyzing +4 | -1

For the reaction

2 \mathrm{NH}_3 \rightarrow \mathrm{N}_2+3 \mathrm{H}_2

-\frac{\mathrm{d}\left[\mathrm{NH}_3\right]}{\mathrm{d} t}=k_1\left[\mathrm{NH}_3\right],-\frac{\mathrm{d}\left[\mathrm{N}_2\right]}{\mathrm{d} t}=k_2\left[\mathrm{NH}_3\right], \frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{d} t}=k_3\left[\mathrm{NH}_3\right]

Then relation between

k_1, k_2

and

k_3

View answer and explanation

Correct option: A

1.5 k_1=3 k_2=k_3

Option A (

1.5 k_1=3 k_2=k_3

) is correct because it correctly relates the rate constants based on the stoichiometry of the reaction.

23 Hard

23 of 54

Application +4 | -1

After how many second will the concentration of the reactant in a first order reaction be halved if the rate constant is

1.155 \times {10^{ - 3}}\;{s^{ - 1}}?

View answer and explanation

Correct option: A

600

Option A (600) is correct because the half-life of a first-order reaction is calculated using the formula

t_{1/2} = \frac{{0.693}}{k}

, and substituting the given rate constant yields approximately 600 seconds.

24 Hard

24 of 54

Analyzing +4 | -1

Consider the two hypothetical reactions given below:

I

a A \rightarrow

Products,

k=x \mathrm{~mol}^{-1} \mathrm{~L} \mathrm{~min}^{-1}

b B \rightarrow

Products,

k=y \mathrm{~min}^{-1}

The half-lives of both the reactions are the same, equal to

1 \mathrm{hr}

when molar concentration of the reactant is

1.0 \mathrm{M}

in each case. If these reactions are started at the same time taking

1 \mathrm{M}

of the reactant in each case, the ratio

[A]/[B]

after

3 \mathrm{hr}

will be:

View answer and explanation

Correct option: D

Option D is correct. The ratio [A]/[B] after 3 hours is 2, calculated by determining the order of each reaction and then finding the concentration of A and B after 3 hours.

25 Hard

25 of 54

Application +4 | -1

The rate constant for a first-order reaction is

2.2 \times10^{-5} \mathrm{~s}^{-1}

320^{\circ} \mathrm{C}

. Calculate the time in which the reaction is

15 \%

complete.

View answer and explanation

Correct option: A

7391 \mathrm{~sec}

Option A (7391 sec) is correct because it is the time calculated using the first-order kinetics equation for 15% completion.

26 Hard

26 of 54

Application +4 | -1

Two substances A and B are present such that

\left[ A \right] = 4\left[ B \right]

and half-life of

A

is 5 minute and of

B

is 15 minute. If they start decaying at the same time following first order, how much time later will the concentration of both of them would be same?

View answer and explanation

Correct option: A

15 minute

Option A (15 minute) is correct because it calculates the time when the concentrations of A and B are equal using first-order kinetics.

27 Hard

27 of 54

Application +4 | -1

The rate law for a reaction between the substances

A

and

B

is given by

rate = k{\left[ A \right]^n}{\left[ B \right]^m}

. On doubling the concentration of

A

and halving the concentration of

B

, the ratio of the new rate to the earlier rate of the reaction will be as

View answer and explanation

Correct option: D

{2^{\left( {n - m} \right)}}

Option D (

{2^{\left( {n - m} \right)}}

) is correct because the new rate is the original rate multiplied by

2^n

(due to A) and

(1/2)^m

(due to B), resulting in a factor of

2^{n-m}

28 Hard

28 of 54

Application +4 | -1

t_{\mathrm{1/4} }

can be taken as the time taken for the concentration of a reactant to drop to

\frac{3}{4}

of the initial value. If the rate constant for a first order reaction is

k

, then

t_{1/4}

can be written as

View answer and explanation

Correct option: C

0.29 / k

Option C (

0.29 / k

) is correct because it accurately calculates the time for the reactant concentration to drop to 3/4 of its initial value in a first-order reaction.

29 Hard

29 of 54

Application +4 | -1

Consider the following reaction in aqueous solution

5 \mathrm{Br}^{-}(a q)+\mathrm{BrO}_3^{-}(a q)+6 \mathrm{H}^{+}(a q) \rightarrow 3 \mathrm{Br}_2(a q)+3 \mathrm{H}_2 \mathrm{O}(l)

If the rate of appearance of

\mathrm{Br}_2

at a particular time during the reaction is

0.025 \mathrm{~M} \mathrm{~sec}^{-1}

, what is the rate of disappearance (in

M \mathrm{sec}^{-1}

) of

\mathrm{Br}^{-}

at that time?

View answer and explanation

Correct option: B

0.042 \mathrm{~M} \mathrm{~sec}^{-1}

Option B (0.042 M sec⁻¹) is incorrect. Option D (0.125 M sec⁻¹) is correct because the rate of disappearance of Br⁻ is calculated based on the stoichiometry of the reaction and the rate of appearance of Br₂.

30 Hard

30 of 54

Application +4 | -1

An organic compound A decomposes the following two parallel first-order reactions

\mathrm{A} \xrightarrow{K_1} \mathrm{B}

and

\mathrm{A} \xrightarrow{K_2} \text{C}.

K_1

1.25 \times 10^{-5} \mathrm{~s}^{-1}

and

\frac{K_1}{K_2}=\frac{1}{9},

then the value of

\frac{[\mathrm{C}]}{[\mathrm{A}]}

after one hour of start of reaction by taking only

\mathrm{A},

(\ln 1.568=0.45)

View answer and explanation

Correct option: B

0.5112

Option B (0.5112) is correct because it is calculated using the integrated rate laws for parallel first-order reactions, considering the ratio of rate constants and the reaction time.

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