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JEE • CHEMISTRY • CHEMICAL KINETICS

Chemical kinetics practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-60 of 103 questions

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31 Easy
31 of 103
Understanding +4 | -1
In the reaction A+B→productsA + B \to products, if B is taken in excess, then it is an example of
View answer and explanation
Correct option: C
Pseudo unimolecular reaction
The correct answer is C (Pseudo unimolecular reaction) because when one reactant is in excess, the reaction order appears to be determined only by the other reactant.
32 Easy
32 of 103
Understanding +4 | -1
Which of the following statement is incorrect about the molecularity of a reaction?
View answer and explanation
Correct option: D
There is no difference between order and molecularity of a reaction
Option D (There is no difference between order and molecularity of a reaction) is correct because order and molecularity are different concepts.
33 Easy
33 of 103
Remembering +4 | -1
The unit of rate constant for a zero order reaction is
View answer and explanation
Correct option: A
mol.L−1.S−1mol.{L^{ - 1}}.{S^{ - 1}}
Option A (mol.L-1.S-1) is correct because the rate constant for a zero-order reaction has units of concentration per time.
34 Easy
34 of 103
Application +4 | -1
For A+B→C+D,  ΔH=−20  kJ  mol−1A + B \to C + D,\;\Delta H = - 20\;\mathrm{kJ}\;\mathrm{mo}{\mathrm{l}^{ - 1}} the activation energy of the forward reaction is 85  kJ  mol−1.85\;\mathrm{kJ}\;\mathrm{mo}{\mathrm{l}^{ - 1}}. The activation energy for backward reaction is …kJ  mol−1\mathrm{kJ}\;\mathrm{mo}{\mathrm{l}^{ - 1}}.
View answer and explanation
Correct option: B
105
Option B (105) is correct because the activation energy for the backward reaction is the sum of the activation energy for the forward reaction and the enthalpy change.
35 Easy
35 of 103
Analyzing +4 | -1
The rate constant K1{K_1} of a reaction is found to be double that of rate constant K2{K_2} of another reaction. The relationship between corresponding activation energies of the two reactions at same temperature (E1  and  E2)({E_1}\;and\;{E_2}) can be represented as:
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because without knowing the pre-exponential factors (A1 and A2), we cannot definitively relate E1 and E2 even with the given rate constant relationship.
36 Easy
36 of 103
Application +4 | -1
Calculate the overall order of a reaction which has the rate expression

(a) Rate =k[ A]1/2[ B]3/2=k[\mathrm{~A}]^{1 / 2}[{~B}]^{3 / 2}

(b) Rate =k[ A]3/2[ B]−1=k[\mathrm{~A}]^{3 / 2}[{~B}]^{-1}
View answer and explanation
Correct option: A
2, 1/2
Option A (2, 1/2) is correct because the overall order of a reaction is the sum of the exponents in the rate expression.
37 Easy
37 of 103
Understanding +4 | -1
The given reaction,

2FeCl3+SnCl2→SnCl4+2FeCl22\mathrm{FeC}{\mathrm{l}_3} + \mathrm{SnC}{\mathrm{l}_2} \to \mathrm{SnC}{\mathrm{l}_4} + 2\mathrm{FeC}{\mathrm{l}_2} is an example of:
View answer and explanation
Correct option: B
Third order reaction
Option D (None of these) is correct because the order of a reaction cannot be determined solely from the stoichiometry of the balanced chemical equation; it must be determined experimentally.
38 Easy
38 of 103
Analyzing +4 | -1
During a particular reaction 20%20 \% of the reactant decomposes in 1 h, 40%1 \mathrm{~h},\,40 \% in 2 h,60%2 \mathrm{~h}, 60 \% in 3 h3 \mathrm{~h} and so on. The unit of rate constant is
View answer and explanation
Correct option: C
mol L−1 h−1\mathrm{mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1}
Option C (mol L−1 h−1\mathrm{mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1}) is correct because the amount of reactant decomposing is directly proportional to time, indicating a zero-order reaction.
39 Easy
39 of 103
Analyzing +4 | -1
Which of the following is an incorrect statement about first-order reaction?
View answer and explanation
Correct option: B
Half-life is the half of the time of complete reaction
Option B is correct because the half-life of a first-order reaction is independent of the initial concentration and is not half of the time for complete reaction (which is theoretically infinite).
40 Easy
40 of 103
Remembering +4 | -1
In Arrhenius equation K=A  e−Ea/RTK = A\;{e^{ - {E_a}/RT}}, the quantity −Ea/RT- {E_a}/RT is referred as:
View answer and explanation
Correct option: A
Boltzmann factor
Option A (Boltzmann factor) is correct because in the Arrhenius equation, the term −EaRT\frac{{ - {E_a}}}{{RT}} is known as the Boltzmann factor, representing the fraction of molecules with sufficient energy to react.
41 Easy
41 of 103
Remembering +4 | -1
In Arrhenius equation, the fraction of effective collisions is given by
View answer and explanation
Correct option: C
e−Ea/RT\mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}
Option C (e−Ea/RT\mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}) is correct because this term represents the fraction of molecules with sufficient energy to overcome the activation energy barrier.
42 Easy
42 of 103
Understanding +4 | -1
The half life of a reaction is—of temperature
View answer and explanation
Correct option: C
Decreased with increase
Option C (Chain isomers) is correct because the two compounds have the same molecular formula but differ in the arrangement of the carbon chain.
43 Easy
43 of 103
Application +4 | -1
Half-life of a reaction is found to be inversely proportional to the cube of initial concentration. The order of reaction is
View answer and explanation
Correct option: A
4
Option A (4) is correct because the half-life equation t1/2∝1an−1{t_{1/2}} \propto \frac{1}{{{a^{n - 1}}}} shows that when n=4, half-life is inversely proportional to the cube of initial concentration.
44 Easy
44 of 103
Remembering +4 | -1
The slope in Arrhenius plot, is equal to:
View answer and explanation
Correct option: A
−Ea2.303R-\frac{E_a}{2.303 R}
Option A (−Ea2.303R-\frac{E_a}{2.303 R}) is correct because the Arrhenius equation in logarithmic form shows that the slope of the plot of log k vs 1/T is equal to −Ea2.303R-\frac{E_a}{2.303 R}.
45 Easy
45 of 103
Understanding +4 | -1
For the reaction, H2+I2→2HI{H_2} + {I_2} \to 2HI, the differential rate law is
View answer and explanation
Correct option: A
−d[H2]dt=−d[I2]dt=12d[HI]dt- \frac{{d\left[ {{H_2}} \right]}}{{dt}} = - \frac{{d\left[ {{I_2}} \right]}}{{dt}} = \frac{1}{2}\frac{{d\left[ {HI} \right]}}{{dt}}
Option A (−d[H2]dt=−d[I2]dt=12d[HI]dt- \frac{{d[ {{H_2}} ]}}{{dt}} = - \frac{{d[ {{I_2}} ]}}{{dt}} = \frac{1}{2}\frac{{d[ {HI} ]}}{{dt}}) is correct because it represents the correct stoichiometric relationship between the reactants and products in the rate law.
46 Easy
46 of 103
Application +4 | -1
The rate of a reaction is doubled when temperature increases by 10∘C10^\circ C. If temperature is increased by 100∘C100^\circ C, then rate of reaction will become
View answer and explanation
Correct option: D
1024 times
Option D (1024 times) is correct because the rate increases by a factor of 2 for every 10°C increase, leading to 2102^{10} times increase for 100°C.
47 Easy
47 of 103
Understanding +4 | -1
The unit of rate constant of second order reaction is
View answer and explanation
Correct option: B
L / Mol s
Option B (L / Mol s) is correct because the unit of the rate constant for a second-order reaction is L/mol-s.
48 Easy
48 of 103
Understanding +4 | -1
Rate equation for a second order reaction is:
View answer and explanation
Correct option: C
K=1t⋅xa(a−x)K = \frac{1}{t} \cdot \frac{x}{{a\left( {a - x} \right)}}
The correct answer is C (K=1t⋅xa(a−x)K = \frac{1}{t} \cdot \frac{x}{{a(a - x)}}) because it represents the integrated rate law for a second-order reaction where rate = k[A]2.
49 Easy
49 of 103
Application +4 | -1
For a first order reaction, the half life period is 5 minutes. The time required for the concentration of the substance to fall from x M   to   x32  Mx\,M\;\,\mathrm{to}\,\,\,\frac{x}{{32}}\,\,M is
View answer and explanation
Correct option: A
25 minutes
Option A (25 minutes) is correct because it is calculated by determining the number of half-lives required for the concentration to fall to x/32.
50 Easy
50 of 103
Understanding +4 | -1
The rate of chemical reaction depends on the nature of chemical reactions, because:
View answer and explanation
Correct option: A
The threshold energy level differs from one reaction to another
The correct answer is A (The threshold energy level differs from one reaction to another) because the activation energy, which is related to the threshold energy, varies for different reactions, affecting their rates.
51 Easy
51 of 103
Understanding +4 | -1
A chemical reaction proceeds following formula

k=PZ  e−Ea/RTk = PZ\;{e^{ - {E_a}/RT}}

Which of the following process will increase the rate of reaction?
View answer and explanation
Correct option: A
Lowering of Ea{E_a}
Option A (Lowering of Ea{E_a}) is correct because decreasing the activation energy (EaE_a) increases the rate constant (kk) according to the Arrhenius equation.
52 Easy
52 of 103
Application +4 | -1
For a given reaction, pressure of catalyst reduces the energy of activation by 2  k  cal  at  27∘C.2\;\mathrm{k}\;\mathrm{cal }\;\mathrm{at}\;27^\circ C. The rate of reaction will be increased by:
View answer and explanation
Correct option: C
28 times
Option C (28 times) is correct because the rate constant increases exponentially with a decrease in activation energy, as described by the Arrhenius equation.
53 Easy
53 of 103
Application +4 | -1
Half-life of a reaction is found to be inversely proportional to the cube of initial concentration. The order of reaction is
View answer and explanation
Correct option: A
4
Option A (4) is correct because the half-life equation t1/2∝1an−1{t_{1/2}} \propto \frac{1}{{{a^{n - 1}}}} shows that when n=4, half-life is inversely proportional to the cube of initial concentration.
54 Easy
54 of 103
Understanding +4 | -1
For the elementary step,

(CH3)3⋅CBr(aq)→(CH3)3C+(aq)+Br−(aq){\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3} \cdot \mathrm{CBr}\left( {aq} \right) \to {\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}{\mathrm{C}^ + }\left( {aq} \right) + \mathrm{B}{\mathrm{r}^ - }\left( {aq} \right) the molecularity is:
View answer and explanation
Correct option: B
1
Option B (1) is correct because molecularity is the number of reactant molecules involved in an elementary step, which is one in this case.
55 Easy
55 of 103
Understanding +4 | -1
The decomposition of phosphine (PH3)\mathrm{(P}{\mathrm{H}_\mathrm{3}}\mathrm{)} on tungsten at low pressure is a first – order reaction. It is because the
View answer and explanation
Correct option: D
Rate is independent of the surface coverage
Option D (Rate is independent of the surface coverage) is correct because a first-order reaction's rate depends only on the concentration of the reactant, not the surface coverage in this specific scenario.
56 Easy
56 of 103
Understanding +4 | -1
A reaction involving two different reactants can never be
View answer and explanation
Correct option: D
Unimolecular reaction
Option D (Unimolecular reaction) is correct because a reaction involving two different reactants must have at least two molecules participating in the rate-determining step, making it impossible for it to be unimolecular.
57 Easy
57 of 103
Application +4 | -1
For A+B→C+D,  ΔH=−20  kJ  mol−1A + B \to C + D,\;\Delta H = - 20\;\mathrm{kJ}\;\mathrm{mo}{\mathrm{l}^{ - 1}} the activation energy of the forward reaction is 85  kJ  mol−1.85\;\mathrm{kJ}\;\mathrm{mo}{\mathrm{l}^{ - 1}}. The activation energy for backward reaction is …kJ  mol−1\mathrm{kJ}\;\mathrm{mo}{\mathrm{l}^{ - 1}}.
View answer and explanation
Correct option: B
105
Option B (105) is correct because the activation energy for the backward reaction is the sum of the activation energy for the forward reaction and the enthalpy change.
58 Easy
58 of 103
Analyzing +4 | -1
For the reaction, N2O5→2NO2+12O2{\mathrm{N}_2}{\mathrm{O}_5} \to 2\mathrm{N}{\mathrm{O}_2} + \frac{1}{2}{\mathrm{O}_2}

Given, −d[N2O5]dt=K1[N2O5]- \frac{{d\left[ {{\mathrm{N}_2}{\mathrm{O}_5}} \right]}}{{dt}} = {K_1}\left[ {{\mathrm{N}_2}{\mathrm{O}_5}} \right]

d[NO2]dt=K2[N2O5]\frac{{d\left[ {\mathrm{N}{\mathrm{O}_2}} \right]}}{{dt}} = {K_2}\left[ {{\mathrm{N}_2}{\mathrm{O}_5}} \right]

and    d[O2]dt=K3[N2O5]\mathrm{and}\;\;\frac{{d\left[ {{\mathrm{O}_2}} \right]}}{{dt}} = {K_3}\left[ {{\mathrm{N}_2}{\mathrm{O}_5}} \right]

The relation in between K1,K2  and  K3{K_1},{K_2}\;and\;{K_3} is:
View answer and explanation
Correct option: A
2K1=K2=4K32{K_1} = {K_2} = 4{K_3}
Option A (2K1=K2=4K32{K_1} = {K_2} = 4{K_3}) is correct because it correctly relates the rate constants based on the stoichiometry of the reaction.
59 Easy
59 of 103
Understanding +4 | -1
The decomposition of phosphine (PH3)\left( {P{H_3}} \right) on tungsten at low pressure is a first order reaction. If is because the
View answer and explanation
Correct option: A
Rate is proportional to the surface coverage
Option A (Rate is proportional to the surface coverage) is correct because for a first-order reaction on a surface, the rate is directly proportional to the surface coverage of the reactant.
60 Easy
60 of 103
Application +4 | -1
The rate law of the reaction,

A+2B→Product.A + 2B \to \mathrm{Product}\mathrm{.}

Product is given by d[P]dt=K[A]2⋅[B].\frac{{d\left[ P \right]}}{{dt}} = K{\left[ A \right]^2} \cdot \left[ B \right].. If AA is taken in large excess, the order of the reaction will be:
View answer and explanation
Correct option: B
1
Option B (1) is correct because when reactant A is in large excess, its concentration remains nearly constant, making the reaction's rate dependent only on the concentration of reactant B, thus making it a first-order reaction.
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