- iamal
- February 17, 2026
Smart Kinetics Mechanism Consistency Problem MCQs Solutions for NEET, JEE and CUET
Kinetics mechanism consistency problem mcqs solutions are an important part of chemical kinetics in Class 12 and competitive examinations. These problems test whether a proposed reaction mechanism agrees with the experimentally observed rate law. While aspirants often memorize rate equations, they sometimes struggle to check if a mechanism is actually consistent with experimental data. That is why practicing kinetics mechanism consistency problem mcqs solutions helps build conceptual clarity and analytical thinking.
In chemical kinetics, a reaction mechanism consists of elementary steps that together give the overall reaction. Each elementary step has its own molecularity and rate expression. However, the experimentally determined rate law may not match the stoichiometric equation of the overall reaction. In such cases, kinetics mechanism consistency problem mcqs solutions focus on verifying whether the slow (rate-determining) step produces the observed rate law.
The key principle behind kinetics mechanism consistency problem mcqs solutions is that the rate law must be derived from the slowest step of the mechanism. If the slow step contains only reactants that appear in the final rate law, the mechanism may be consistent. If intermediates appear in the rate expression, they must be eliminated using equilibrium or steady-state approximations. Understanding this logic is essential for solving kinetics mechanism consistency problem mcqs solutions accurately.
For example, consider a reaction where the experimental rate law is Rate = k[A][B]. If a proposed mechanism suggests that the slow step involves only A and B, then the mechanism is likely valid. But if the slow step involves an intermediate species not present in the observed rate law, then further analysis is required. Many kinetics mechanism consistency problem mcqs solutions revolve around identifying such inconsistencies.
Another important concept in kinetics mechanism consistency problem mcqs solutions is the pre-equilibrium approximation. In this approach, a fast reversible step establishes equilibrium before the slow step occurs. The concentration of the intermediate is expressed in terms of reactants using the equilibrium constant. Substituting this into the slow-step rate expression should reproduce the experimental rate law. If it does, the mechanism is consistent. This logical verification is central to kinetics mechanism consistency problem mcqs solutions.
The steady-state approximation is also frequently tested in kinetics mechanism consistency problem mcqs solutions. According to this assumption, the concentration of the intermediate remains nearly constant during the reaction. By setting the rate of formation of the intermediate equal to its rate of consumption, one can derive a rate expression that must match the experimental rate law. If the derived expression matches, the mechanism is acceptable. If not, it must be rejected.
Aspirants often confuse molecularity with order of reaction. Kinetics mechanism consistency problem mcqs solutions clearly demonstrate that molecularity applies only to elementary steps, while order is determined experimentally. A complex reaction may show fractional order even though individual elementary steps have whole-number molecularity. Recognizing this difference is crucial when analyzing kinetics mechanism consistency problem mcqs solutions.
Temperature effects, catalysts, and reaction intermediates also appear in kinetics mechanism consistency problem mcqs solutions. A catalyst may introduce an alternative pathway with lower activation energy but must still produce a rate law consistent with experimental findings. Similarly, intermediates must not appear in the final rate equation unless properly substituted.
To master kinetics mechanism consistency problem mcqs solutions, aspirants should follow a systematic approach. First, identify the slow step. Second, write its rate expression. Third, eliminate intermediates using equilibrium or steady-state conditions. Finally, compare the derived rate law with the experimental one. If they match exactly, the mechanism is consistent. If not, it is incorrect.
Repeated practice of kinetics mechanism consistency problem mcqs solutions improves logical reasoning and strengthens understanding of reaction mechanisms. These problems are highly scoring in board exams, NEET, and JEE because they test conceptual clarity rather than memorization. With careful study and stepwise analysis, aspirants can confidently solve kinetics mechanism consistency problem mcqs with solutions and avoid common mistakes.
Kinetics Mechanism Consistency Problem MCQs Solutions
1.
Total order of reaction X + Y → XY is 3. The order of reaction with respect to X is 2. State the differential rate equation for the reaction.
A. −d[X]dt=k[X]3[Y]0-\frac {d[X]}{dt} = k[X]^{3}[Y]^{0}
B. −d[X]dt=k[X]0[Y]3-\frac {d[X]}{dt} = k[X]^{0}[Y]^{3}
C. −d[X]dt=k[X]2[Y]-\frac {d[X]}{dt} = k[X]^{2}[Y]
D. −d[X]dt=k[X][Y]2-\frac {d[X]}{dt} = k[X][Y]^{2}
Answer: C
2.
For a reaction 2A → 3B, if the rate of formation of B is x mol L⁻¹ s⁻¹, the rate of consumption of A is:
A. 3x/2
B. 3x
C. 2x/3
D. x
Answer: C
3.
Which steps detetrmines the rate of reaction?
A. Fast step
B.Slow step
C. Intermediate step
D. Final step
Answer: B
4.
For the reaction 2A + B → C, the rate of formation of C is 2.2 × 10⁻³ mol L⁻¹ min⁻¹. What is the value of −d[A]dt-\frac {d[A]}{dt}?
A. 2.2 × 10⁻³
B. 1.1 × 10⁻³
C. 4.4 × 10⁻³
D. 5.5 × 10⁻³
Answer: C
5.
The concentrations of reactant A in reaction A → B decrease linearly with time. The rate constant according to correct order is:
A. 0.001 M min⁻¹
B. 0.001 M⁻¹ min⁻¹
C. 0.001 M⁻² min
D. 0.001 M⁻² min⁻¹
Answer: A
6.
The most probable velocity of a gas molecule at 298 K is 300 m s⁻¹. Its RMS velocity is:
A. 420
B. 245
C. 402
D. 367
Answer: D
7.
For Cl₂ + 2I⁻ → 2Cl⁻ + I₂, concentration of I⁻ changes from 0.60 to 0.56 mol L⁻¹ in 10 minutes. Rate of disappearance of I⁻ and rate of appearance of I₂ respectively are:
A. 0.004 and 0.002
B. 0.002 and 0.004
C. 0.004 and 0.004
D. 0.002 and 0.002
Answer: A
8.
For gaseous reaction r = k[A][B], if volume is halved, rate becomes:
A. 1/4
B. 4
C. 1/2
D. 2
Answer: B
9.
If order with respect to A is 2 and B is 3, doubling both concentrations increases rate by factor:
A. 12
B. 16
C. 32
D. 10
Answer: C
10.
Rate doubles for every 10°C rise. For 60°C increase, rate increases by:
A. 20 times
B. 32 times
C. 64 times
D. 128 times
Answer: C
11.
If temperature increases from 30°C to 80°C (50°C rise), rate increases by:
A. 16
B. 32
C. 64
D. 4
Answer: B
12.
For A + 3B → 2C, if −d[A]dt=3×10−3-\frac{d[A]}{dt} = 3 × 10⁻³, then −d[B]dt-\frac{d[B]}{dt} equals:
A. 3 × 10⁻³
B. 9 × 10⁻³
C. 10⁻³
D. 1.5 × 10⁻³
Answer: B
13.
For 5Br⁻ + BrO₃⁻ → 3Br₂ + 3H₂O, the relation between rates is:
A. d[Br2]dt=−53d[Br−]dt\frac {d[Br_2]}{dt} = -\frac{5}{3} \frac {d[Br^-]}{dt}
B. Equal
C. 5/3 times
D. d[Br2]dt=35d[Br−]dt\frac {d[Br_2]}{dt} = \frac{3}{5} \frac {d[Br^-]}{dt}
Answer: D
14.
For A + 2B → Products, if initial rate of A is 2.6 × 10⁻², initial rate of B is:
A. 0.10
B. 2.6 × 10⁻²
C. 5.2 × 10⁻²
D. 6.5 × 10⁻²
Answer: C
15.
Increasing temperature increases rate due to:
A. More number of collisions
B. Decrease in mean free path
C. More energetic electrons
D. Less energetic electrons
Answer: A
16.
For 2NO + O₂ → 2NO₂, halving volume increases rate by:
A. 1/4
B. 1/8
C. 8 times
D. 4 times
Answer: C
17.
For S₂O₈²⁻ + 3I⁻ → 2SO₄²⁻ + I₃⁻, if rate of I⁻ disappearance is (9/2) × 10⁻³, rate of SO₄²⁻ formation is:
A. 3 × 10⁻³
B. 2 × 10⁻³
C. 10⁻³
D. 4 × 10⁻³
Answer: A
18.
Dilution effect on esterification reaction reduces rate by:
A. 0.5 times
B. 2 times
C. 4 times
D. 0.25 times
Answer: D
19.
For 2N₂O₅ ⇌ 4NO₂ + O₂, incorrect rate relation is:
A. −d[N2O5]dt=2d[O2]dt-\frac {d[N_2O_5]}{dt} = 2 \frac {d[O_2]}{dt}
B. −2d[N2O5]dt=d[NO2]dt-2\frac {d[N_2O_5]}{dt} = \frac {d[NO_2]}{dt}
C. d[NO2]dt=4d[O2]dt\frac {d[NO_2]}{dt} = 4 \frac {d[O_2]}{dt}
D. Long incorrect relation
Answer: B
20.
Enzyme lowers activation energy; value is:
A. 6/RT
B. P required
C. Different from lab Ea
D. Can’t say
Answer: C
21.
For 3A → 2B, relation is:
A. -3/2 d[A]/dt
B. -2/3 d[A]/dt
C. -1/3 d[A]/dt
D. +2 d[A]/dt
Answer: B
22.
If doubling A makes rate four times and B has no effect, rate law is:
A. k[A][B]
B. k[A]²
C. k[A][B]²
D. k[A]²[B]²
Answer: B
23.
First order reaction: rate when [A] = 0.01 M is:
A. 1.735×10⁻²
B. 3.47×10⁻³
C. 3.47×10⁻⁴
D. 1.735×10⁻⁴
Answer: C
24.
Correct mechanism consistent with rate = k[Cl₂][H₂S] is:
A. II only
B. Neither
C. Only I
D. Both
Answer: C
25.
For ½A → 2B, relation is:
A. -d[A]/dt = ½ d[B]/dt
B. -d[A]/dt = ¼ d[B]/dt
C. Equal
D. -d[A]/dt = 4 d[B]/dt
Answer: B
26.
Factor not influencing rate:
A. Molecularity
B. Temperature
C. Concentration
D. Nature
Answer: A
27.
For Haber process, rate in terms of N₂ is:
A. 1 × 10⁻⁴
B. 4 × 10⁻⁴
C. 5 × 10⁻⁴
D. 10 × 10⁻⁴
Answer: A
28.
For r = k[A][B], volume reduced to ¼, rate becomes:
A. 4 times
B. 8 times
C. 16 times
D. 1/8 times
Answer: C
29.
For 2N₂O₅ → 4NO₂ + O₂, if [NO₂] increases by 5.2×10⁻³ in 100 s, rate is:
A. 1.3 × 10⁻⁵
B. 0.5 × 10⁻⁴
C. 7.5 × 10⁻⁴
D. 2 × 10⁻⁵
Answer: A
30.
In multistep reaction, overall rate equals:
A. Rate of slowest step
B. Rate of fastest step
C. Average rate
D. Last step
Answer: A
Conclusion on Kinetics Mechanism Consistency Problem MCQs Solutions
In conclusion, kinetics mechanism consistency problem mcqs solutions form a vital component of chemical kinetics preparation. They train aspirants to connect experimental rate laws with theoretical mechanisms, apply approximations correctly, and think critically about reaction pathways. Mastery of kinetics mechanism consistency problem mcqs solutions ensures a deeper understanding of how reactions actually proceed at the molecular level.