Skip to main content
Back to CHEMICAL KINETICS

Your Progress

0 / 301 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • CHEMICAL KINETICS

Chemical kinetics practice · Mixed difficulty

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

Practice format and chapter quiz

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.

Take a chapter quiz →
Showing 271-300 of 301 questions
271 Medium
271 of 301
Application +4 | -1
The rate of the reaction 3A+2B→Products3 \mathrm{A}+2 \mathrm{B} \rightarrow \text{Products} is given by the rate expression rate =K[A][B]2.=K[\mathrm{A}][\mathrm{B}]^2. If A\mathrm{A} is taken in large excess, then the reaction start obeying
View answer and explanation
Correct option: B
Second-order kinetics
Option B (Second-order kinetics) is correct because when A is in large excess, its concentration becomes constant, and the rate law simplifies to depend only on the concentration of B squared, making it second order.
272 Medium
272 of 301
Application +4 | -1
In a reaction for every 10∘C10^{\circ} \mathrm{C} rise of temperature, the rate is tripled. If the temperature is increased from 10∘C10^{\circ} \mathrm{C} to 50∘C50^{\circ} \mathrm{C}, the rate of the reaction will become
View answer and explanation
Correct option: A
81 times
Option A (81 times) is correct because the rate increases by a factor of 3 for every 10°C rise, and the temperature increases by 40°C.
273 Hard
273 of 301
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
2
The correct answer is 2 because the half-life is inversely proportional to the concentration for a second-order reaction.
274 Medium
274 of 301
Analyzing +4 | -1
Assertion (A):A plot of log⁡dXdt vs. log⁡(a−x)\log \frac{d X}{d t} \text{ vs. }\log (a-x) leads to value of KK as antilog (intercept) and slope equal to order of reaction.

Reason (R):dXdt=K(a−x)n\frac{d X}{d t}=K(a-x)^n
View answer and explanation
Correct option: C
Both A and R are correct and R is correct explanation of A
Option D (Both A and R are correct but R is not correct explanation of A) is incorrect. Option C (Both A and R are correct and R is correct explanation of A) is correct because the integrated rate law can be expressed in logarithmic form, where the intercept corresponds to the rate constant and the slope corresponds to the order of the reaction.
275 Medium
275 of 301
Analyzing +4 | -1
Statement I Kinetics of study of decomposition of N2O5(g)\mathrm{N}_2 \mathrm{O}_5(g) to NO2(g)\mathrm{NO}_2(g) and O2(g)\mathrm{O}_2(g) can be studied by measuring pressure of O2\mathrm{O}_2 gas in CCl4\mathrm{CCl}_4 solvent.

2 N2O5(g)⟶4NO2(g)+O2(g)2 \mathrm{~N}_2 \mathrm{O}_5(g) \longrightarrow 4 \mathrm{NO}_2(g)+\mathrm{O}_2(g)

Statement II At the start, volume of O2(g)\mathrm{O}_2(g) is zero.
View answer and explanation
Correct option: B
Both Statement I and Statement II are correct but
Statement II is not the correct explanation of Statement I
Option B (Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I) is correct because the kinetics can be studied by measuring the pressure of O2{}_2 in CCl4{}_4, and the initial volume of O2{}_2 is zero, but the second statement is not the reason for the first.
276 Hard
276 of 301
Application +4 | -1
For a reaction scheme, A→k1B→k2C,A\xrightarrow{{{k_1}}}B\xrightarrow{{{k_2}}}C, if the rate of formation of BB is set to be zero then the concentration of B is given by
View answer and explanation
Correct option: B
(k1k2)[A]\left( {\frac{{{k_1}}}{{{k_2}}}} \right)[A]
Option B is correct because at steady state, the rate of formation of B equals its rate of consumption, leading to [B] = (k1/k2)[A].
277 Easy
277 of 301
Application +4 | -1
The rate constant of a first order reaction is 3×10−63 \times {10^{ - 6}} per second and initial concentration is 0.10  M.0.10\;\mathrm{ }M. Then the initial rate of reaction is
View answer and explanation
Correct option: C
3×10−7  ms−13 \times {10^{ - 7}}\;m{s^{ - 1}}
The correct answer is C (3×10−7  ms−13 \times {10^{ - 7}}\;m{s^{ - 1}}) because it correctly applies the rate law for a first-order reaction.
278 Hard
278 of 301
Application +4 | -1
Two radio isotopes P{P} and Q{Q} of atomic mass 10 and 20 respectively are mixed in equal amount by mass. After 20 days, their weight ratio WPWQ\frac{{W}_{{P}}}{{W}_{{Q}}} is found to be 1:41: 4. Isotope Q{Q} has a half-life of 10 days. The half-life of isotope P{P} is:
View answer and explanation
Correct option: B
5 days
Option B (5 days) is correct because the half-life of isotope P is calculated using the given weight ratio and the half-life of isotope Q.
279 Easy
279 of 301
Analyzing +4 | -1
The rate equation for the reaction 2A+B→C2A + B \to C is found to be
Rate=k[A][B]Rate = k\left[ A \right]\left[ B \right]
The correct statement in relation to this reaction that the
View answer and explanation
Correct option: C
Rate of formation of CC is twice the rate of disappearance of AA
The correct answer is D because the rate of formation of C is half the rate of disappearance of A, as indicated by the stoichiometry of the reaction.
280 Hard
280 of 301
Analyzing +4 | -1
For the first-order parallel reactions A→K1B\mathrm{A} \xrightarrow{K_1} \mathrm{B} and A→K2C,K1=8 min−1\mathrm{A} \xrightarrow{K_2} \mathrm{C}, K_1=8 \mathrm{~min}^{-1} and K2=2 min−1K_2=2 \mathrm{~min}^{-1} at 300 K.300 \mathrm{~K}. If the activation energies for the formation of B\text{B} and C\text{C} are 2020 and 28.314 kJ/mol,28.314 \mathrm{~kJ} / \mathrm{mol}, respectively, then find the temperature at which B\text{B} and C\text{C} will be obtained in 2:12: 1 mole ratio. (ln⁡2=0.7)(\ln 2=0.7)
View answer and explanation
Correct option: C
379.75 K379.75 \mathrm{~K}
Option C is correct. The temperature at which B and C are obtained in a 2:1 mole ratio is calculated using the Arrhenius equation and the given activation energies and rate constants.
281 Medium
281 of 301
Application +4 | -1
The decomposition of N2O5{N}_2 {O}_5 is a first order reacton represented by N2O5⟶N2O4+12O2{N}_2 {O}_5 \longrightarrow {N}_2 {O}_4+\frac{1}{2} {O}_2. After 15 minutes the volume of O2{O}_2 produced is 9 ml9 \, \mathrm{ml} and at the end of the reaction 35 mL35 \mathrm{~mL}. The rate constant it equal to
View answer and explanation
Correct option: C
115In⁡4435\frac{1}{15} \operatorname{In} \frac{44}{35}
Option C (115In⁡4435\frac{1}{15} \operatorname{In} \frac{44}{35}) is correct because it is derived from the integrated rate law for a first-order reaction, considering the volume of oxygen produced at different times.
282 Medium
282 of 301
Application +4 | -1
Which of the following represents the expression for 3/43 / 4th the life of a first-order reaction?
View answer and explanation
Correct option: C
2.303klog⁡4\frac{2.303}{k} \log 4
Option C (2.303klog⁡4\frac{2.303}{k} \log 4) is correct because it represents the time taken for 75% completion of a first-order reaction.
283 Medium
283 of 301
Application +4 | -1
Timetaken for 12.8 g12.8 {~g} of a radioactive substance to decay to 0.4 g0.4 {~g} is: (half life is 138 s138 {~s} )
View answer and explanation
Correct option: B
690 s690 {~s}
Option B (690 s690 {~s}) is correct because it is calculated by determining the number of half-lives required for the decay and multiplying by the half-life duration.
284 Medium
284 of 301
Application +4 | -1
Which of the following rate law has an overall order of 0.5 for reaction involving substances X,Y\mathrm{X}, \mathrm{Y} and Z\mathrm{Z} ?
View answer and explanation
Correct option: C
rate =k(CX)1.5(CY)−1(CZ)0=k\left(C_{\mathrm{X}}\right)^{1.5}\left(C_{\mathrm{Y}}\right)^{-1}\left(C_{\mathrm{Z}}\right)^0
The correct answer is C (rate =k(CX)1.5(CY)−1(CZ)0=k(C_{\mathrm{X}})^{1.5}(C_{\mathrm{Y}})^{-1}(C_{\mathrm{Z}})^0) because the sum of the exponents in the rate law equals 0.5.
285 Medium
285 of 301
Application +4 | -1
t99.9%{t_{99.9\% }} of a first order reaction is 100 minutes. Rate constant of the reaction is
View answer and explanation
Correct option: C
0.0693 minutes
Option C (0.0693 minutes) is correct because the rate constant is calculated using the relationship between t99.9%t_{99.9\%} and half-life for a first-order reaction.
286 Easy
286 of 301
Analyzing +4 | -1
Which of the following statement is not correct for the reaction: 4A+B→2C+2D4A + B \to 2C + 2D
View answer and explanation
Correct option: A
The rate of disappearance of ‘B’ is twice the rate of appearance of ‘C’
Option A is the incorrect statement because the rate of disappearance of B is one-half the rate of appearance of C, not twice.
287 Medium
287 of 301
Analyzing +4 | -1
Consider the following statements,

The rate law for the acid catalysed hydrolysis of an ester being given as

Rate=k  [H+]  [ester]=k′  [ester]Rate = k\;\left[ {{H^ + }} \right]\;\left[ {ester} \right] = k'\;\left[ {ester} \right]

If the acid concentration is doubled at constant ester concentration

1. The second order rate constant, kk is doubled.

2. The pseudo first order rate constant, kk is double.

3. The rate of the reaction is doubled.

Which of the above statements are correct?
View answer and explanation
Correct option: B
2 and 3
Option B (2 and 3) is correct because doubling the acid concentration doubles both the pseudo-first-order rate constant and the reaction rate, while the second-order rate constant remains unchanged.
288 Easy
288 of 301
Application +4 | -1
How much faster would a reaction proceed at 25∘C25^\circ C than at 0∘C0^\circ C if the activation energy is 65  kJ?65\;kJ?
View answer and explanation
Correct option: C
11 times
The correct answer is C (11 times) because the reaction proceeds approximately 11 times faster at 25°C than at 0°C, based on the Arrhenius equation and the given activation energy.
289 Easy
289 of 301
Analyzing +4 | -1
For the reaction system

2NO(g)+O2(g)→2NO2(g)2\mathrm{NO}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right) \to 2\mathrm{N}{\mathrm{O}_2}\left( \mathrm{g} \right) if the volume of the reaction vessel is reduced to one-third of its original volume, what will be the order of the reaction?
View answer and explanation
Correct option: C
Increase to twenty seven times of its initial value
Option C (0.01 M(NH4)3PO40.01 \mathrm{~M}\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4) is correct because it produces the highest concentration of NH4+NH_4^+ ions, leading to the lowest pH (most acidic) among the given options, and the question is asking for the maximum pH, so the least acidic has the highest pH.
290 Easy
290 of 301
Application +4 | -1
DDT on exposure to water decomposes. Half-life is 10 yr. How much time it will take for its decomposition to 99%?
View answer and explanation
Correct option: B
70 Yr
Option B (70 Yr) is correct because it is the approximate time calculated using the first-order kinetics equation for 99% decomposition.
291 Medium
291 of 301
Analyzing +4 | -1
For the reaction A→K1B→K2C→K3D,\mathrm{A} \xrightarrow{K_1} \mathrm{B} \xrightarrow{K_2} \mathrm{C} \xrightarrow{K_3} \mathrm{D}, where K3>K2>K1;K_3>K_2>K_1; the rate of formation of D\mathrm{D} is determined by
View answer and explanation
Correct option: A
The rate of reaction: A→B\mathrm{A} \rightarrow \mathrm{B}
The rate of formation of D is determined by the rate of reaction A → B because K1 is the smallest rate constant, making it the slowest and rate-determining step.
292 Easy
292 of 301
Remembering +4 | -1
The minimum energy required for the reacting molecules to undergo reaction is :
View answer and explanation
Correct option: D
Activation energy
Option D (Activation energy) is correct because activation energy is the minimum energy required for a chemical reaction to occur.
293 Medium
293 of 301
Application +4 | -1
For the reaction 2N2O5→4NO2+O22{N_2}O5 \to 4N{O_2} + {O_2} rate and rate constant are 1.02×10−4  mol.L−1.sec⁡−11.02 \times {10^{ - 4}}\,\,mol.{L^{ - 1}}.{\sec ^{ - 1}} and 3.4×10−5sec⁡−13.4 \times {10^{ - 5}}{\sec ^{ - 1}} respectively then concentration of N2O5{N_2}{O_5} at that time will be
View answer and explanation
Correct option: B
3 M3\,M
Option B (3 M) is correct because the concentration of N2O5 is calculated using the rate law for the reaction.
294 Medium
294 of 301
Application +4 | -1
Half life of a first order reaction is 4 s4 {~s} and the initial concentration of the reactant is 0.12M0.12 {M}. The concentration of the reactant left after 16 s16 {~s} is
View answer and explanation
Correct option: A
0.0075M0.0075 {M}
Option D (-x) is correct because the energy of an electron depends on the atomic number squared divided by the principal quantum number squared.
295 Medium
295 of 301
Analyzing +4 | -1
Consider the following statements

a) Increase in concentration of reactant increases the rate of zero order reaction

b) Rate constant K is equal to collision frequency A if Ea=0{E_a} = 0

c) Rate constant K is equal to collision frequency A if Ea= ∝{E_a} = \, \propto

d) ln K Vs T is a straight line

e) ln K Vs 1/T is a straight line

Correct statements are
View answer and explanation
Correct option: B
b and e
Option B (b and e) is correct because the rate constant equals the collision frequency when activation energy is zero, and ln K vs 1/T yields a straight line according to the Arrhenius equation.
296 Hard
296 of 301
Application +4 | -1
The rate constant of a first order reaction at 27∘ C27^\circ \,C is 10−3 min−1{10^{ - 3}}\,\mathrm{mi}{\mathrm{n}^{ - 1}}. The temperature coefficient of this reaction is 2. What is the rate constant (in min−1\mathrm{mi}{\mathrm{n}^{ - 1}}) at 17∘ C17^\circ \,C for this reaction?
View answer and explanation
Correct option: B
5×10−45 \times {10^{ - 4}}
Option B (5×10−45 \times {10^{ - 4}}) is correct because the rate constant at the lower temperature is half of the rate constant at the higher temperature, given a temperature coefficient of 2.
297 Medium
297 of 301
Application +4 | -1
If for a radio-isotope, at the end of 4.2 days, N=0.798 N0{N}=0.798 {~N}_0. What will be half life period of its nucleus?
View answer and explanation
Correct option: B
12.83 days
Option B (12.83 days) is correct because it is the half-life period calculated using the radioactive decay formula.
298 Medium
298 of 301
Application +4 | -1
The ratio of 14C12C\frac{{ }^{14} \mathrm{C}}{{ }^{12} \mathrm{C}} in a piece of wood is 18\frac{1}{8} part that of atmosphere. If half life of 14C{ }^{14} \mathrm{C} is 5730 years, the age of wood sample is …..years.
View answer and explanation
Correct option: A
17190
Option A (17190) is correct because it accurately calculates the age of the wood sample using the half-life of Carbon-14 and the given ratio.
299 Medium
299 of 301
Analyzing +4 | -1
Statement I A high activation energy corresponds to a
reaction rate that is very sensitive to temperature.
Statement II In following gaseous phase reaction CH3+CH3⟶CH3−CH3\mathrm{CH}_3+\mathrm{CH}_3 \longrightarrow \mathrm{CH}_3-\mathrm{CH}_3 rate is independent of temperature.
View answer and explanation
Correct option: B
Both Statement I and Statement II are correct but
Statement II is not the correct explanation of Statement I
Option B (Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I) is correct because both statements are factually true, but the second statement doesn't explain the first.
300 Easy
300 of 301
Application +4 | -1
In the reaction
BrO3−(aq)+5 Br−(aq)+6H+(aq)→3Br2(l)+3H2O(l)BrO_3^ - \left( {aq} \right) + 5\,B{r^ - }\left( {aq} \right) + 6{H^ + }\left( {aq} \right) \to 3B{r_2}\left( l \right) + 3{H_2}O\left( l \right)
The rate of appearance of bromine (Br2)\left( {B{r_2}} \right) is related to rate of disappearance of Bromide ions is
View answer and explanation
Correct option: D
d[Br2]dt=35  d[Br−]dt\frac{{d\left[ {B{r_2}} \right]}}{{dt}} = \frac{3}{5}\,\,\frac{{d\left[ {B{r^ - }} \right]}}{{dt}}
Option D (d[Br2]dt=35d[Br−]dt\frac{{d[B{r_2}]}}{{dt}} = \frac{3}{5}\frac{{d[B{r^ - }]}}{{dt}}) is correct because it correctly relates the rate of appearance of bromine to the rate of disappearance of bromide ions based on the stoichiometric coefficients.
Revision notes and key takeaways

Review CHEMICAL KINETICS in your textbook or class notes, then use this set to check your understanding.

Showing 271-300 of 301 questions
0 /301
0✓ 0✗
0% Accuracy
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.