Velocity constant plays a central role in understanding chemical kinetics for senior secondary aspirants. When learners prepare for competitive exams, they repeatedly encounter velocity constant calculation MCQs Class 12 because these problems test both conceptual clarity and numerical accuracy. A strong foundation in the definition and application of the rate constant is essential to solve velocity constant calculation MCQs Class 12 efficiently and confidently.

In chemical kinetics, the velocity constant, commonly denoted by k, appears in the rate law expression. For a general reaction, rate = k[A]ⁿ[B]ᵐ, the value of k depends only on temperature and the nature of the reaction. Aspirants preparing for velocity constant calculation MCQs Class 12 must remember that k is independent of concentration for a given temperature. This principle is frequently used while solving velocity constant calculation MCQs in board exams and entrance tests.

For a first order reaction, the integrated rate equation is:

k = (2.303/t) log (a/(a – x))

Here, a is the initial concentration and x is the amount reacted in time t. Most velocity constant calculation MCQs Class 12 are based on this logarithmic form. When half-life is given, students can use the relation:

k = 0.693 / t₁/₂

Understanding this formula simplifies many velocity constant calculation MCQs Class 12 where half-life data is provided directly.

For zero order reactions, the integrated form is:

k = (a – x)/t

In such cases, graphical interpretation is also common. Many velocity constant calculation MCQs Class 12 involve plotting concentration versus time and determining k from the slope. Recognizing the linear nature of zero order plots is extremely helpful for velocity constant calculation.

Second order reactions have a different integrated form:

k = (1/t) (1/(a – x) – 1/a)

Because the units of k change with reaction order, aspirants must be careful with dimensional analysis. Units are frequently tested in velocity constant calculation MCQs Class 12, especially to identify reaction order from the unit of k.

Temperature dependence is another critical concept. According to the Arrhenius equation:

k = A e^(–Ea/RT)

This shows that velocity constant increases with temperature. Many velocity constant calculation MCQs Class 12 require comparing k at two temperatures or calculating activation energy. Using the logarithmic Arrhenius form:

log(k₂/k₁) = Ea / (2.303R) (1/T₁ – 1/T₂)

Aspirants can solve temperature-based velocity constant calculation MCQs Class 12 effectively.

Another common situation involves percentage decomposition. For example, if 75% of a reactant decomposes in a given time, the remaining concentration can be substituted into the integrated equation. Such percentage-based numericals are classic velocity constant calculation MCQs Class 12 patterns.

Graph-based problems are also important. A straight line in a plot of ln[A] versus time indicates first order kinetics, and the slope equals –k. Identifying slope values quickly helps in velocity constant calculation MCQs Class 12 that involve experimental data interpretation.

It is also essential to understand that the magnitude of the velocity constant indicates reaction speed. A larger k implies a faster reaction. Many conceptual velocity constant calculation MCQs Class 12 test this qualitative interpretation without heavy numerical work.

Dimensional consistency should always be checked. For first order reactions, the unit of k is time⁻¹. For second order reactions, it is L mol⁻¹ time⁻¹. Recognizing these units can immediately solve several velocity constant calculation MCQs Class 12 without lengthy calculations.

Careful substitution of values, proper use of logarithms, and clarity in units are the keys to mastering velocity constant calculation MCQs Class 12. ASpirants should practice converting minutes to seconds and Kelvin temperature consistently, since unit errors often lead to mistakes in velocity constant calculation MCQs Class 12.

Velocity Constant Calculation MCQs Class 12:

1. With respect to the velocity constant of a reaction, which one of the following statements is NOT correct?

A. It is a measure of the velocity of the chemical reaction.
B. It is dependent on initial concentration of reactants.
C. It depends on temperature.
D. During the progress of the reaction even though the velocity decreases, the velocity constant remains constant.

Answer: B

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2. A(g) → B(g) is a first order reaction. The initial concentration of A is 0.2 mol L⁻¹. After 10 min, the concentration of B is found to be 0.18 mol L⁻¹. The rate constant (in min⁻¹) for the reaction is:

A. 2.303
B. 0.2303
C. 0.693
D. 0.01

Answer: B

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3. Unit of k for third order reaction is:

A. (Litre/Mole).sec
B. (Mole/Litre).sec
C. (Litre/Mole)⁻¹.sec⁻¹
D. (Mole/Litre)⁻².sec⁻¹

Answer: D

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4. Assertion: A catalyst does not alter the equilibrium constant of a reaction.

Reason: The catalyst forms a complex with the reactants and provides an alternate path with lower energy of activation; the forward and backward reactions are affected to the same extent.

A. Both Assertion and Reason are correct and Reason is the correct explanation.
B. Both correct but Reason is not correct explanation.
C. Assertion correct but Reason incorrect.
D. Both incorrect.

Answer: A

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5. Assertion: The rate of reaction is the rate of change of concentration of a reactant or product.

Reason: Rate of reaction remains constant during the course of reaction.

A. Both correct and explanation correct
B. Both correct but explanation incorrect
C. Assertion correct but Reason incorrect
D. Both incorrect

Answer: C

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6. Assertion: In rate law, exponents do not necessarily match stoichiometric coefficients.

Reason: Reaction mechanism governs rate.

A. Both correct and explanation correct
B. Both correct but explanation incorrect
C. Assertion correct but Reason incorrect
D. Both incorrect

Answer: A

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7. For the reaction 2N₂O₅ → 4NO₂ + O₂, k = 3.0 × 10⁻⁴ s⁻¹. If started with 1.0 mol L⁻¹ N₂O₅, calculate rate of formation of NO₂ when [O₂] = 0.1 mol L⁻¹.

A. 2.7 × 10⁻⁴
B. 2.4 × 10⁻⁴
C. 4.8 × 10⁻⁴
D. 9.6 × 10⁻⁴

Answer: D

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8. For the reaction 2N₂O₅ → 4NO₂ + O₂, the rate of reaction is:

A. (1/2) d[N₂O₅]/dt
B. 2 d[N₂O₅]/dt
C. (1/4) d[NO₂]/dt
D. 4 d[NO₂]/dt

Answer: C

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9. Decomposition of N₂O₅ follows first order kinetics. Hence:

A. Reaction is unimolecular
B. Reaction is bimolecular
C. t₁/₂ ∝ a⁰
D. None

Answer: C

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10. The rate constant is 1 × 10⁻² mol⁻² L² s⁻¹. The order is:

A. 0
B. 1
C. 2
D. 3

Answer: D

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11. For mechanism:

O₃ ⇌ O₂ + O (fast)
O + O₃ → 2O₂ (slow)

Rate law is:

A. r = k[O₃]²
B. r = k[O₃]²[O₂]⁻¹
C. r = k³[O₃]²[O₂]⁻¹
D. r = [O₃][O₂]

Answer: B

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12. Velocity constant at 290 K is 3.2 × 10⁻³. At 300 K it will be:

A. 1.28 × 10⁻²
B. 9.6 × 10⁻⁴
C. 6.4 × 10⁻³
D. 3.2 × 10⁻⁴

Answer: C

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13. 2N₂O₅ ⇌ 4NO₂ + O₂

Rate = 2.40 × 10⁻⁵ mol L⁻¹ s⁻¹
k = 3 × 10⁻⁵ s⁻¹

Concentration of N₂O₅:

A. 1.4
B. 1.2
C. 0.4
D. 0.8

Answer: D

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14. X₂(g) → Y + (1/2)Z(g)

Pressure increases from 100 mm to 120 mm in 5 min. Rate of disappearance of X₂:

A. 8 mm min⁻¹
B. 2 mm min⁻¹
C. 16 mm min⁻¹
D. 4 mm min⁻¹

Answer: A

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15. A + B → Product

Doubling A → rate ×4
Doubling both → rate ×8

Differential rate equation:

A. dC/dt = kC_A C_B
B. dC/dt = kC_A² C_B³
C. dC/dt = kC_A² C_B
D. dC/dt = kC_A² C_B²

Answer: C

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16. First order reaction k = 2.303 × 10⁻² s⁻¹. Time to reduce concentration to 1/10:

A. 10 s
B. 100 s
C. 2303 s
D. 230.3 s

Answer: B

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17. Rate doubles for every 10°C rise. If temperature raised by 50°C, rate increases by:

A. 10 times
B. 24 times
C. 32 times
D. 64 times

Answer: C

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18. For 2A + B → product

Doubling B → no change in t₁/₂
Doubling A → rate doubles

Unit of k:

A. L mol⁻¹ s⁻¹
B. No unit
C. mol L⁻¹ s⁻¹
D. s⁻¹

Answer: A

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19. Reaction order determination from data:

Rate law:
A. dc/dt = k[A][B]
B. dc/dt = k[A]²[B]
C. dc/dt = k[A][B]²
D. dc/dt = k[A]

Answer: D

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20. For 2A + B → C

Rate of formation of C = 2.2 × 10⁻³ mol L⁻¹ min⁻¹
Find -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

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21. First order reaction, 75% disappears in 1.386 hr. Rate constant:

A. 3.0 × 10⁻³ s⁻¹
B. 2.8 × 10⁻⁴ s⁻¹
C. 2.7 × 10⁻³ s⁻¹
D. 1.8 × 10⁻⁴ s⁻¹

Answer: C

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22. From given data:

Rate law:

A. r = k[A][B]
B. r = k[A]²[B]
C. r = k[A]³
D. r = k[A]

Answer: B

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23. Time for 100% completion of zero order reaction:

A. ak
B. a/2k
C. a/k
D. 2k/a

Answer: C

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24. From data for 2A → C + D, value of k:

A. 0.01 s⁻¹
B. 5 × 10⁻² mol L⁻¹ s⁻¹
C. 5.0 × 10⁻¹ mol⁻¹ L s⁻¹
D. 5.0 × 10⁻⁴ mol L⁻¹ s⁻¹

Answer: C

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25. 2A + B → 3C

[A] decreases 0.5 to 0.3 in 10 min. Rate of production of C:

A. 0.01
B. 0.03
C. 0.05
D. 0.02

Answer: B

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26. Rate constant doubles for 10°C rise:

A. Nearly doubled
B. Nearly tripled
C. Increases 5 times
D. Increases 4 times

Answer: A

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27. For 2N₂O₅ → 4NO₂ + O₂

Order and molecularity:

A. Order 1, Molecularity 1
B. Order 1, Molecularity 2
C. Order 2, Molecularity 2
D. Order 2, Molecularity 1

Answer: A

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28. In reaction 2A + B → 3A

B disappears at:

A. Half the rate as A decreases
B. Same rate as A decreases
C. Twice the rate
D. Half rate as AB forms

Answer: A

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29. Rate constant independent of concentration:

A. It is independent of concentration
B. Arrhenius constant
C. Dimensionless
D. Independent of temperature

Answer: A

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30. Ammonium + Nitrite reaction rate law:

A. Rate = k[NH₄⁺][NO₂⁻]⁴
B. Rate = k[NH₄⁺][NO₂⁻]
C. Rate = k[NH₄⁺][NO₂⁻]²
D. Rate = k[NH₄⁺]⁴[NO₂⁻]

Answer: C