For the reaction $2 \mathrm{N}_2 \mathrm{O}_5(\mathrm{g}) \rightarrow 4 \mathrm{NO}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}),$ the concentration of $\mathrm{NO}_2$ increases by $2.4 \times 10^{-2} \mathrm{M}$ in $6 \mathrm{~s}.$ What will be the average rate of appearance of $\mathrm{NO}_2$ and the average rate of disappearance of $\mathrm{N}_2 \mathrm{O}_5?$

$2 \times 10^{-3} \mathrm{Ms}^{-1}, 4 \times 10^{-3} \mathrm{Ms}^{-1}$

$2 \times 10^{-3} \mathrm{Ms}^{-1}, 1 \times 10^{-3} \mathrm{Ms}^{-1}$

$2 \times 10 \mathrm{Ms}^{-1}, 2 \times 10^{-3} \mathrm{Ms}^{-1}$

The activation energies of two reactions are ${E_a}$ and ${E'_a}$ with ${E_a} > {E'_a}.$ If the temperature of the reacting systems is increased from ${T_1}$ to ${T_2}$, predict which alternative is correct. $K'$ are rate constants at higher temperature. Assume $A$ being same for both the reactions:

$\frac{{{{K'}_1}}}{{{K_1}}} = \frac{{{{K'}_2}}}{{{K_2}}}$

${K_1} > {K_2}$ and ${K'_1} > {K'_2}$

$\frac{{{{K'}_1}}}{{{K_1}}} < \frac{{2{{K'}_2}}}{{{K_2}}}$

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JEE • CHEMISTRY • CHEMICAL KINETICS

Chemical kinetics practice · Medium

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Showing 1-30 of 144 questions

1 Medium

1 of 144

Understanding +4 | -1

Which of the following is a zero order reaction?

View answer and explanation

Correct option: A

2{\text{N}}{{\text{H}}_3}\left( {\text{g}} \right)\xrightarrow[{{\text{Pt catalyst}}}]{{1130\,{\text{K}}}}{{\text{N}}_{\text{2}}}\left( {\text{g}} \right) + 3{{\text{H}}_{\text{2}}}\left( {\text{g}} \right)

(Reactant

\mathrm{NH}_3

is strongly adsorbed

Option A (

2{\text{N}}{{\text{H}}_3}\left( {\text{g}} \right)\xrightarrow[{{\text{Pt catalyst}}}]{{1130{\,\text{K}}}}{{\text{N}}_{ ext{2}}}\left( {\text{g}} \right) + 3{{\text{H}}_{ ext{2}}}\left( {\text{g}} \right)

(Reactant

\mathrm{NH}_3

is strongly adsorbed) is correct because the decomposition of ammonia on a platinum surface at high pressure is a zero-order reaction due to the surface being saturated with ammonia.

2 Medium

2 of 144

Application +4 | -1

The energy of activation for an uncatalysed reaction is

100 \mathrm{~kJ/mol}.

Presence of a catalyst lowers the energy of activation by

75 \%

. The

\log _{10} \frac{K_2}{K_1}

of the ratio of rate constant of catalysed and uncatalysed reactions at

27^{\circ} \mathrm{C}

is?

Assume the frequency factor is same for both reactions. (Given

2.303 \times 8.314=19.147

)

View answer and explanation

Correct option: A

13.05

Option A (13.05) is correct because it is the result of calculating the ratio of rate constants using the Arrhenius equation and the given activation energies.

3 Medium

3 of 144

Application +4 | -1

A first order reaction has a rate constant

1.15 \times {10^{ - 3}}{S^{ - 1}}

. How long will 5g of this reactant take to reduce to 3g (given log 1.667 = 0.2219)

View answer and explanation

Correct option: B

444 S

Option B (444 S) is correct because the time is calculated using the first-order kinetics equation with the given rate constant and initial/final amounts.

4 Medium

4 of 144

Application +4 | -1

For the reaction,

\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})

. The initial pressure was

\mathrm{P}_0

while pressure after time '

\mathrm{t}

' was

\mathrm{P}_{\mathrm{t}}

. The rate constant

\mathrm{k}

will be given by:

View answer and explanation

Correct option: C

\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{P}_0}{2 \mathrm{P}_0-\mathrm{P}_{\mathrm{t}}}

Option C (

\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{P}_0}{2 \mathrm{P}_0-\mathrm{P}_{\mathrm{t}}}

) is correct because it correctly applies the integrated rate law for a first-order gas-phase reaction.

5 Medium

5 of 144

Application +4 | -1

For the reaction

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

rate & rate constant are

1.02 \times 10^{-4}

and

3.4 \times 10^{-5} \mathrm{sec}^{-1}

respectively then concentration of

\mathrm{N}_2 \mathrm{O}_5

at that time will be:

View answer and explanation

Correct option: B

Option B (3) is correct because the concentration of N2O5 is calculated directly from the rate law, given the rate and rate constant.

6 Medium

6 of 144

Application +4 | -1

The value of rate constant for the above reaction is:

View answer and explanation

Correct option: D

None of these

The correct answer is None of these because the rate constant cannot be determined without more information about the rate of reaction.

7 Medium

7 of 144

Analyzing +4 | -1

STATEMENT-1: Active complex is an intermediate product.

STATEMENT-2 : Active complex is unstable with high vibrational energy.

View answer and explanation

Correct option: D

If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE

Option D is correct because Statement-1 is false, as the active complex is an intermediate product, and Statement-2 is true, as the active complex is unstable with high vibrational energy.

8 Medium

8 of 144

Analyzing +4 | -1

Statement I For the reaction

\begin{aligned} & 2 \mathrm{KClO}_3 \longrightarrow 2 \mathrm{KCl}+3 \mathrm{O}_2 \\ & \frac{-d\left[\mathrm{KClO}_3\right]}{d t}=\frac{2}{3} \frac{d\left[\mathrm{O}_2\right]}{d t} \end{aligned}

Statement II If average rate of formation of

\mathrm{O}_2

0.01 \mathrm{~mol}

\mathrm{min}^{-1}

, rate of disappearance of

\mathrm{KClO}_3

0.03 \mathrm{~mol} \mathrm{~min}^{-1}

View answer and explanation

Correct option: C

Statement I is correct but Statement II is incorrect

The correct answer is C because Statement I is correct based on the stoichiometry of the reaction, but Statement II is incorrect as the rate of disappearance of KClO₃ should be 0.02 mol min⁻¹ when the rate of formation of O₂ is 0.03 mol min⁻¹.

9 Medium

9 of 144

Analyzing +4 | -1

For the reaction

A+B \rightarrow 2 C+D,

which of the following statement are incorrect?

View answer and explanation

Correct option: A

\text{Rate of disappearance of }B =2 \times \text{rate of appearance of }C

The correct answer is A (Rate of disappearance of B = 2 x rate of appearance of C) because, according to the stoichiometry of the reaction, the rate of disappearance of B should be half the rate of appearance of C.

10 Medium

10 of 144

Analyzing +4 | -1

Assertion: Acid catalyzed of ethyl acetate is a first order reaction.

Reason: Water does not take part in the reaction.

View answer and explanation

Correct option: C

If Assertion is True but the Reason is False

Option C (If Assertion is True but the Reason is False) is correct because the acid-catalyzed hydrolysis of ethyl acetate is pseudo-first order, but water does participate in the reaction.

11 Medium

11 of 144

Understanding +4 | -1

The order of a reaction with rate equal to

k\;C_A^{3/2}\;C_B^{ - 1/2}

View answer and explanation

Correct option: A

Option A (1) is correct because the order of the reaction is the sum of the exponents of the concentration terms in the rate law, which in this case is 3/2 + (-1/2) = 1.

12 Medium

12 of 144

Understanding +4 | -1

STATEMENT 1 : Product can form only when the required orientation and energy conditions are met
STATEMENT-2 : All collisions between reactants yield the desired product.

View answer and explanation

Correct option: C

If STATEMENT-1 is TRUE and STATEMENT-2 is FALSE

Option C (If STATEMENT-1 is TRUE and STATEMENT-2 is FALSE) is correct because product formation requires specific orientation and energy, but not all collisions lead to product formation.

13 Medium

13 of 144

Application +4 | -1

For a first order reaction rate constant is given as,

\log \,\,K = 14 - \frac{{1.2 \times {{10}^4}}}{T}

, then what will be value of

'T'

if its half-life is

6.93 \times {10^{ - 3}}\,\,\min

View answer and explanation

Correct option: B

1000 K

Option B (1000 K) is correct because the temperature is calculated using the Arrhenius equation and the given half-life.

14 Medium

14 of 144

Application +4 | -1

If rate constant of a reaction is

2 \times 10^{-4} \,{~s}^{-1}

then the time required for the completion of

90 \%

of the reaction is

View answer and explanation

Correct option: A

1.15 \times 10^4\, {~s}

Option A (

1.15 \times 10^4\, \mathrm{~s}

) is correct because it is the time calculated using the first-order kinetics equation for 90% completion of the reaction.

15 Medium

15 of 144

Application +4 | -1

The first order rate constant for the decomposition of ethyl iodide by the reaction

{C_2}{H_5}I\left( g \right) \to {C_2}{H_4}\left( g \right) + HI\left( g \right)

at 600 K is

1.60 \times {10^{ - 5}}\,\,{s^{ - 1}}.

Its energy of activation is 209 kJ/mol, T the rate constant of the reaction at 700 K is.

View answer and explanation

Correct option: B

6.36 \times {10^{ - 3}}\,\,{s^{ - 1}}

Option B (

6.36 \times {10^{ - 3}}\,{s^{ - 1}}

) is correct because it is calculated using the Arrhenius equation relating rate constants at different temperatures.

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16 Medium

16 of 144

Application +4 | -1

Half life of a first order reaction:

A(

aq.

) \rightarrow 2 B(a q)

is 40 min at

27^{\circ} \mathrm{C}

and 20 min at

37^{\circ} \mathrm{C}

. The activation energy of reaction is

\left(\ln 2=0.7, R=2.0 \mathrm{~cal} \mathrm{~mol}^{-1} K^{-1}\right)

View answer and explanation

Correct option: B

13020 cal/mol

Option B (13020 cal/mol) is correct because it is the calculated activation energy using the Arrhenius equation and the given half-lives at different temperatures.

17 Medium

17 of 144

Application +4 | -1

For the reaction

2 \mathrm{N}_2 \mathrm{O}_5(\mathrm{g}) \rightarrow 4 \mathrm{NO}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}),

the concentration of

\mathrm{NO}_2

increases by

2.4 \times 10^{-2} \mathrm{M}

6 \mathrm{~s}.

What will be the average rate of appearance of

\mathrm{NO}_2

and the average rate of disappearance of

\mathrm{N}_2 \mathrm{O}_5?

View answer and explanation

Correct option: D

4 \times 10^{-3} \mathrm{Ms}^{-1}, 2 \times 10^{-3} \mathrm{Ms}^{-1}

Option D (

4 \times 10^{-3} \mathrm{Ms}^{-1}, 2 \times 10^{-3} \mathrm{Ms}^{-1}

) is correct because the average rate of appearance of NO2 and disappearance of N2O5 are calculated based on the stoichiometry of the reaction.

18 Medium

18 of 144

Application +4 | -1

The half-life of first order decomposition of

\mathrm{NH}_4 \mathrm{NO}_3

2.10 \mathrm{hr}

288 \mathrm{~K}

temperature

\mathrm{NH}_4 \mathrm{NO}_3(a q) \longrightarrow \mathrm{N}_2 \mathrm{O}(g)+2 \mathrm{H}_2 \mathrm{O}(l)

, if

6.2 \mathrm{~g}

\mathrm{NH}_4 \mathrm{NO}_3

is allowed to decompose, The time required for

\mathrm{NH}_4 \mathrm{NO}_3

to decompose

90 \%

is :

View answer and explanation

Correct option: A

6.978 \mathrm{~hr}

Option A (

6.978 \mathrm{~hr}

) is correct because it accurately calculates the time required for 90% decomposition of

\mathrm{NH}_4 \mathrm{NO}_3

using first-order kinetics.

19 Medium

19 of 144

Understanding +4 | -1

Statement I: The overall order of the reaction is the sum of the exponents of all the reactants in the rate expression

Statement II: There are many higher-order reactions

View answer and explanation

Correct option: B

If statement I and statement II are correct but statement II is not the correct explanation of statement I

Option B (If statement I and statement II are correct but statement II is not the correct explanation of statement I) is correct because both statements are true, but the existence of higher-order reactions does not explain the definition of overall order.

20 Medium

20 of 144

Understanding +4 | -1

Nuclear reaction accompanied with emission of neutron(s) is:

View answer and explanation

Correct option: A

{ }_{13}^{27} {Al}+{ }_2^4 {He} \longrightarrow{ }_{15}^{30} {P}+{ }_0^1 {n}

Option A (

{ }_{13}^{27} {Al}+{ }_2^4 {He} \longrightarrow{ }_{15}^{30} {P}+{ }_0^1 {n}

) is correct because it shows the emission of a neutron (

{ }_0^1 {n}

) as a product of the nuclear reaction.

21 Medium

21 of 144

Application +4 | -1

{ }_{89} A c^{231}

after emission of some

\alpha

and

\beta

-particles gives

{ }_{82} P b^{207}

. The number of such

\alpha

and

\beta-

particles are respectively

View answer and explanation

Correct option: B

6, 5

Option B (6, 5) is correct because the number of alpha particles is calculated from the mass number difference, and beta particles from the atomic number difference.

22 Medium

22 of 144

Understanding +4 | -1

When a graph is plot for

\log [\mathrm{A}]

against time (t), then the slope and intercept of the line for first order reaction is equal to

View answer and explanation

Correct option: C

\frac{-\mathrm{k}}{2.303}, \log [\mathrm{A}]_0

Option C (

\frac{{-\mathrm{k}}}{{2.303}}, \log [\mathrm{A}]_0

) is correct because it matches the slope and intercept derived from the integrated rate law for a first-order reaction when plotting

\log[A]

vs. time.

23 Medium

23 of 144

Application +4 | -1

For a first order reaction,

A \longrightarrow

products, the initial concentration of

\mathrm{A}

decrease from

0.1 \mathrm{~M}

0.05 \mathrm{~M}

in 1 minutes. The initial rate of the reaction (in

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

) is

View answer and explanation

Correct option: C

1.15 \times 10^{-3}

Option C (

1.15 \times 10^{-3}

) is correct because the rate of a first-order reaction is directly proportional to the concentration of the reactant, and the rate constant can be calculated from the half-life.

24 Medium

24 of 144

Analyzing +4 | -1

Assertion : The kinetics of the reaction

m A+n B+p C \longrightarrow m^{\prime} X+n^{\prime} Y+p^{\prime} Z

obey the rate expression as

\frac{d X}{d t}=k[A]^m[B]^n

.

Reason : The rate of the reaction does not depend upon the concentration of

C

View answer and explanation

Correct option: A

If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.

Option A is correct because the rate expression given is consistent with the rate depending only on the concentrations of A and B.

25 Medium

25 of 144

Application +4 | -1

If the initial concentration of reactants are doubled half life becomes half. Then order of the reaction will be

View answer and explanation

Correct option: B

Option B (2) is correct because the half-life is inversely proportional to the initial concentration raised to the power of (n-1), where n is the order of the reaction.

26 Medium

26 of 144

Application +4 | -1

On doubling the concentration of reactant, if half-life also gets doubled then the order of reaction will be

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because for a zero-order reaction, the half-life is directly proportional to the initial concentration of the reactant.

27 Medium

27 of 144

Analyzing +4 | -1

Two reactions

A \to products

and

B \to products

have rate constants

{K_A}

and

{K_B}

at temperature

T

and activation energies

{E_A}

and

{E_B}

respectively. If

{K_A} > {K_B}

  and

{E_A} < {E_B}

and assuming that

A

for both the reactions is same, then:

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because the given conditions satisfy the Arrhenius equation, and the relative magnitudes of rate constants change with temperature as described.

28 Medium

28 of 144

Understanding +4 | -1

Assertion (A): Molecularity of a reaction cannot be more than three.
Reason (R): Probability of simultaneous collision between more than three particles is very less.

View answer and explanation

Correct option: A

Both (A) and (R) are true (R) is the correct explanation of (A)

Option A is correct because both the assertion and the reason are true, and the reason correctly explains why molecularity cannot exceed three.

29 Medium

29 of 144

Understanding +4 | -1

The slope of the line obtained on plotting

\log _{10} \mathrm{k}

against

\frac{1}{\mathrm{T}}

is equal to

View answer and explanation

Correct option: D

\frac{-\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}

Option D is correct because, according to the Arrhenius equation, the slope of the plot of log10 k against 1/T is equal to -Ea/(2.303R).

30 Medium

30 of 144

Analyzing +4 | -1

The activation energies of two reactions are

{E_a}

and

{E'_a}

with

{E_a} > {E'_a}.

If the temperature of the reacting systems is increased from

{T_1}

{T_2}

, predict which alternative is correct.

K'

are rate constants at higher temperature. Assume

A

being same for both the reactions:

View answer and explanation

Correct option: B

{K_1} < {K_2}

and

{K'_2} < {K'_1}

The correct answer is B (

{K_1} < {K_2}

and

{K'_2} < {K'_1}

) because a higher activation energy implies a smaller rate constant, and the effect of temperature is more pronounced for reactions with higher activation energies.

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