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CBSE Class 12 Chemistry Chapter 3 Chemical Kinetics — 20 MCQs with Answers

Chemical Kinetics is Chapter 3 in the NCERT Class 12 Chemistry textbook and carries significant weight in the CBSE board exam—expect 3–5 marks in Section A (MCQs) and at least one numerical in Section B. The chapter covers rate of reaction, rate laws, order and molecularity, integrated rate equations, temperature dependence via the Arrhenius equation, and collision theory. Mastering MCQs on these topics builds speed and confidence. Below are 20 carefully crafted multiple-choice questions mirroring the style and difficulty of recent CBSE papers, complete with answers and explanations.

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Key takeaways

  • Rate of reaction is measured in mol L⁻¹ s⁻¹ and can be expressed as −d[R]/dt or +d[P]/dt depending on stoichiometry.
  • Order of reaction (sum of exponents in rate law) is experimentally determined and may differ from stoichiometric coefficients.
  • Molecularity is always a whole number (1, 2, or 3) and refers to the number of molecules participating in an elementary step.
  • Integrated rate equations link concentration and time: zero-order ([A]=[A]₀−kt), first-order (ln[A]=ln[A]₀−kt), second-order (1/[A]=1/[A]₀+kt).
  • The Arrhenius equation k=Ae^(−Ea/RT) shows reaction rate constant increases exponentially with temperature and decreases with higher activation energy.
  • Half-life for first-order reactions is independent of initial concentration (t₁/₂=0.693/k), a key exam concept.
  • Collision theory explains that only effective collisions—with proper orientation and energy ≥Ea—lead to product formation.

Rate of Reaction and Factors Affecting It — MCQs 1–4

The rate of a chemical reaction quantifies how quickly reactants transform into products. For a reaction aA + bB → cC + dD, the average rate is −(1/a)(Δ[A]/Δt) or +(1/c)(Δ[C]/Δt). NCERT emphasizes that rate depends on concentration, temperature, catalyst presence, and surface area (for heterogeneous reactions). Understanding units—mol L⁻¹ s⁻¹ for concentration-time rate and bar s⁻¹ for gaseous reactions—is crucial. The following MCQs test basic definitions, unit conversions, and conceptual understanding of instantaneous versus average rate. These are commonly the first 1–2 MCQs in Section A of the CBSE paper.
  • Q1. For the reaction 2N₂O₅ → 4NO₂ + O₂, if the rate of formation of NO₂ is 0.04 mol L⁻¹ s⁻¹, the rate of disappearance of N₂O₅ is: (A) 0.01 mol L⁻¹ s⁻¹ (B) 0.02 mol L⁻¹ s⁻¹ (C) 0.04 mol L⁻¹ s⁻¹ (D) 0.08 mol L⁻¹ s⁻¹
  • Answer: (B) 0.02 mol L⁻¹ s⁻¹. Rate of disappearance = (2/4) × rate of formation of NO₂ = 0.02 mol L⁻¹ s⁻¹ by stoichiometry.
  • Q2. Which factor does NOT affect the rate of reaction between two gases? (A) Temperature (B) Catalyst (C) Pressure (D) Molar mass of the container
  • Answer: (D) Molar mass of the container. Reaction rate depends on reactant properties and conditions, not the container material or its molar mass.
  • Q3. The instantaneous rate of reaction at time t is given by: (A) Δ[A]/Δt (B) d[A]/dt (C) [A]₀−[A] (D) k[A]
  • Answer: (B) d[A]/dt. Instantaneous rate is the derivative (slope of tangent) on a concentration-time graph at a specific moment.
  • Q4. Units of rate constant for a zero-order reaction are: (A) s⁻¹ (B) mol L⁻¹ s⁻¹ (C) L mol⁻¹ s⁻¹ (D) L² mol⁻² s⁻¹
  • Answer: (B) mol L⁻¹ s⁻¹. For zero-order, rate = k, so units of k match units of rate.

Rate Law, Order, and Molecularity — MCQs 5–8

Rate law expresses rate as k[A]ˣ[B]ʸ, where x and y are determined experimentally and their sum is the overall order. Order can be zero, fractional, or integer, but molecularity (number of molecules in an elementary step) is always a positive integer—1 (unimolecular), 2 (bimolecular), or rarely 3 (termolecular). NCERT stresses that order ≠ stoichiometric coefficient unless the reaction is elementary. These distinctions are favorite traps in CBSE MCQs. The questions below test your ability to deduce order from experimental data, recognize elementary steps, and avoid common pitfalls. Expect at least one such MCQ in every CBSE Class 12 Chemistry paper, often worth 1 mark in Section A.
  • Q5. For the reaction A + 2B → C, if doubling [A] doubles the rate and doubling [B] quadruples the rate, the order is: (A) 1 (B) 2 (C) 3 (D) 4
  • Answer: (C) 3. Rate ∝ [A]¹[B]², so overall order = 1+2 = 3.
  • Q6. Which statement is correct? (A) Order can be fractional; molecularity cannot. (B) Molecularity can be zero; order cannot. (C) Order is always an integer. (D) Molecularity is determined from the balanced equation.
  • Answer: (A) Order can be fractional; molecularity cannot. Molecularity is a count of molecules in an elementary step, always a whole number.
  • Q7. A reaction occurs in two steps: (i) A → B (slow), (ii) B + C → D (fast). The molecularity of the rate-determining step is: (A) 1 (B) 2 (C) 3 (D) cannot be determined
  • Answer: (A) 1. The slow step (i) involves one molecule of A, so molecularity = 1 (unimolecular).
  • Q8. For a reaction with rate = k[A]⁰[B]¹, if [A] is tripled, the rate: (A) triples (B) remains unchanged (C) becomes nine times (D) becomes one-third
  • Answer: (B) remains unchanged. Zero order in A means [A] has no effect on rate; rate depends only on [B].

Integrated Rate Equations and Half-Life — MCQs 9–12

Integrated rate equations relate concentration to time. For zero-order: [A]=[A]₀−kt (linear with time); first-order: ln[A]=ln[A]₀−kt (exponential decay), and t₁/₂=0.693/k independent of [A]₀; second-order: 1/[A]=1/[A]₀+kt. NCERT devotes significant space to deriving these and showing graphical tests (plot [A] vs t for zero-order gives a straight line; ln[A] vs t for first-order; 1/[A] vs t for second-order). Half-life formulas are high-yield: zero-order t₁/₂=[A]₀/(2k), first-order t₁/₂=0.693/k, second-order t₁/₂=1/(k[A]₀). Recognizing which plot is linear for a given order is a classic CBSE trap. Practice these four MCQs to cement the distinctions and avoid mixing up formulas under exam pressure.
  • Q9. A first-order reaction has k=0.02 s⁻¹. The half-life is approximately: (A) 17.3 s (B) 34.6 s (C) 50 s (D) 69.3 s
  • Answer: (B) 34.6 s. t₁/₂=0.693/k=0.693/0.02≈34.6 s.
  • Q10. For a zero-order reaction, a plot of [A] versus time is: (A) linear with negative slope (B) exponential curve (C) hyperbola (D) parabola
  • Answer: (A) linear with negative slope. [A]=[A]₀−kt is the equation of a straight line with slope −k.
  • Q11. Which statement about half-life is FALSE? (A) For first-order, t₁/₂ is independent of [A]₀. (B) For zero-order, t₁/₂ ∝ [A]₀. (C) For second-order, t₁/₂ ∝ 1/[A]₀. (D) For all orders, t₁/₂ ∝ 1/k.
  • Answer: (D) For all orders, t₁/₂ ∝ 1/k. True for first-order, but zero-order t₁/₂ also depends on [A]₀, and second-order t₁/₂ on [A]₀.
  • Q12. A reaction A→B follows first-order kinetics. If 75% of A decomposes in 60 minutes, t₁/₂ is: (A) 15 min (B) 20 min (C) 30 min (D) 40 min
  • Answer: (C) 30 min. 75% decomposed means two half-lives elapsed (50% + 25% of remaining 50%). So 2×t₁/₂=60 min ⇒ t₁/₂=30 min.

Arrhenius Equation and Temperature Dependence — MCQs 13–16

The Arrhenius equation k=Ae^(−Ea/RT) links the rate constant k to absolute temperature T and activation energy Ea. Taking natural logarithm: ln k = ln A − Ea/(RT), which is linear in 1/T with slope −Ea/R. NCERT explains that A is the pre-exponential (frequency) factor, and Ea is the minimum energy colliding molecules need to react. A 10°C rise typically doubles or triples the reaction rate for many reactions. The equation also appears in the two-temperature form: ln(k₂/k₁)=(Ea/R)×(1/T₁−1/T₂). CBSE loves numerical-based MCQs here: given k at two temperatures, find Ea, or vice versa. Understanding units (Ea in J mol⁻¹, R=8.314 J K⁻¹ mol⁻¹) and the inverse relationship between Ea and rate is critical. These four questions drill the concepts and common calculation traps that appear in Section A or as assertion-reason pairs.
  • Q13. According to the Arrhenius equation, increasing temperature: (A) increases Ea (B) decreases Ea (C) increases k (D) has no effect on k
  • Answer: (C) increases k. Higher T makes the exponent −Ea/RT less negative, so e^(−Ea/RT) and hence k increase.
  • Q14. A plot of ln k versus 1/T for a reaction gives a straight line with slope −5000 K. The activation energy Ea (R=8.314 J K⁻¹ mol⁻¹) is: (A) 5 kJ mol⁻¹ (B) 41.57 kJ mol⁻¹ (C) 8.314 kJ mol⁻¹ (D) 5000 J mol⁻¹
  • Answer: (B) 41.57 kJ mol⁻¹. Slope = −Ea/R ⇒ Ea = 5000×8.314 = 41,570 J = 41.57 kJ mol⁻¹.
  • Q15. For a reaction, k doubles when temperature increases from 300 K to 310 K. The activation energy is approximately (R=8.314 J K⁻¹ mol⁻¹): (A) 26 kJ mol⁻¹ (B) 53 kJ mol⁻¹ (C) 13 kJ mol⁻¹ (D) 100 kJ mol⁻¹
  • Answer: (B) 53 kJ mol⁻¹. Using ln(k₂/k₁)=ln2=0.693=(Ea/R)×(10/(300×310)), solve Ea ≈ 53 kJ mol⁻¹.
  • Q16. Which factor does NOT appear in the Arrhenius equation? (A) Activation energy (B) Gas constant (C) Temperature (D) Enthalpy change of reaction
  • Answer: (D) Enthalpy change of reaction. k=Ae^(−Ea/RT) involves Ea (activation energy), R, T, and A, not ΔH.

Collision Theory and Activation Energy — MCQs 17–18

Collision theory posits that molecules must collide with energy ≥ Ea and proper orientation to react. NCERT defines the fraction of molecules with energy ≥ Ea as e^(−Ea/RT), which explains the exponential term in the Arrhenius equation. Not all collisions are effective; steric factors (orientation) reduce the number of successful collisions, captured by the pre-exponential factor A. Higher Ea means fewer molecules cross the energy barrier at a given T, slowing the reaction. Catalysts lower Ea without being consumed, increasing k. These MCQs test conceptual clarity on effective collisions, the role of Ea, and the distinction between thermodynamic stability (ΔG) and kinetic stability (Ea). Expect one question on collision theory or catalysts in most CBSE papers, sometimes as an assertion-reason type, particularly valuable because assertion-reason MCQs carry specific marking in the new CBSE pattern.
  • Q17. According to collision theory, the rate of reaction depends on: (A) number of collisions only (B) number of effective collisions with E ≥ Ea and proper orientation (C) temperature only (D) concentration of products
  • Answer: (B) number of effective collisions with E ≥ Ea and proper orientation. Only collisions meeting both criteria lead to product formation.
  • Q18. A catalyst increases the rate of reaction by: (A) increasing ΔH (B) increasing Ea (C) decreasing Ea (D) increasing the concentration of reactants
  • Answer: (C) decreasing Ea. A catalyst provides an alternate pathway with lower activation energy, raising the fraction of effective collisions.

HOTS and Assertion-Reason MCQs — MCQs 19–20

CBSE papers increasingly include higher-order thinking (HOTS) and assertion-reason (A-R) MCQs in Section A. HOTS questions require application of two or more concepts—for example, combining integrated rate laws with Arrhenius equation or deducing mechanism from experimental data. Assertion-reason MCQs present two statements: you must decide if both are true, if one is true, and if the reason correctly explains the assertion. NCERT examples show that many students lose marks here by not reading carefully. The two MCQs below are modeled on the 2024–25 CBSE pattern. Practice dissecting each statement independently, then checking logical linkage. These questions often distinguish top scorers from the rest because they test depth, not rote memory. Spending an extra 30 seconds on A-R questions can save a mark, which can matter when cutoffs are 90+.
  • Q19. Assertion (A): The half-life of a first-order reaction is independent of initial concentration. Reason (R): The rate constant k for a first-order reaction has units of s⁻¹. (A) Both A and R true; R is correct explanation of A. (B) Both A and R true; R is NOT correct explanation of A. (C) A true, R false. (D) A false, R true.
  • Answer: (B) Both A and R true; R is NOT correct explanation of A. A is true (t₁/₂=0.693/k, no [A]₀ term). R is true (units of k for first-order are indeed s⁻¹). But R does not explain why t₁/₂ is independent of [A]₀; that follows from the integrated rate law form.
  • Q20. A reaction has rate = k[A]²[B]. If [A] is halved and [B] is doubled, the new rate relative to the original is: (A) unchanged (B) halved (C) doubled (D) quartered
  • Answer: (B) halved. New rate = k(½[A])²(2[B]) = k × ¼[A]² × 2[B] = ½ × (k[A]²[B]) = ½ original rate.

How to Attempt MCQs in the CBSE Chemistry Paper

Section A of the 2025 CBSE Class 12 Chemistry paper carries 20 MCQs (16×1 mark + 4×1 mark assertion-reason), total 20 marks. Timing is tight: aim for 20–25 minutes for the entire section, leaving buffer for numericals and long answers. First, scan all 20 questions and mark the 5–7 you can answer in under 15 seconds (definitions, direct formula recall). Answer those immediately to bank easy marks and build confidence. Next, tackle numerical MCQs (rate law, Arrhenius, half-life calculations) because they have one correct answer—no ambiguity. Show quick working in the margin if allowed; even if you can't finish, you might spot the right option by eliminating impossible values. For assertion-reason, read the assertion first, decide if it's true or false, then read the reason independently. Only then check if R explains A. Beware of statements that are individually true but logically unconnected—CBSE designs these to catch hasty readers. If stuck between two options, use elimination: cross out clearly wrong choices, then guess intelligently from the rest. Never leave an MCQ blank; there is no negative marking in CBSE board exams. Finally, if time remains, revisit any MCQ you flagged. A second read often reveals a missed keyword like 'NOT' or 'EXCEPT'. Practice 20 MCQs under 20-minute time limits weekly in the two months before boards. Use NCERT Exemplar and previous years' CBSE question papers as your primary sources—third-party guides often include non-syllabus or poorly worded questions. Remember, Chemical Kinetics MCQs are usually straightforward if you have clarity on order versus molecularity, half-life formulas, and the Arrhenius equation. Drill these core concepts until they are reflexive, and you will comfortably score full marks in this section on exam day.
  • Read the question stem completely before looking at options; keywords like 'NOT', 'EXCEPT', 'FALSE' flip the logic.
  • For numerical MCQs, plug in values quickly and check units—mismatched units often signal a wrong option.
  • In assertion-reason, evaluate A and R separately first, then check causation; don't assume R explains A just because both are true.
  • If two options seem close, reread the question to spot which concept is actually being tested (order vs molecularity is a common trap).
  • Use the 'strike-through' method: lightly cross out impossible options in the question paper to narrow choices.
  • Budget roughly 1 minute per MCQ; if one takes longer, flag it and return after finishing easier questions.
  • CBSE board papers have no negative marking, so never leave a question unattempted—guess intelligently if needed.

Common Mistakes Students Make in Chemical Kinetics MCQs

Year after year, CBSE examiners report recurring errors in Chemical Kinetics MCQs. First, students confuse order with molecularity. Remember: order is experimental (can be 0, fractional, or integer), while molecularity is theoretical (1, 2, or 3 for elementary steps). Second, half-life formula mix-ups cost marks. Write down the three key formulas on your rough sheet at the start: zero-order [A]₀/(2k), first-order 0.693/k, second-order 1/(k[A]₀). Third, in Arrhenius calculations, unit errors are rampant—Ea must be in J mol⁻¹ if R=8.314 J K⁻¹ mol⁻¹, not kJ. Fourth, many students ignore stoichiometry when relating rates of different species; always divide the rate by the stoichiometric coefficient. Fifth, assertion-reason MCQs trip students who read only the assertion and assume the reason is correct. Treat A and R as independent true/false statements first. Sixth, time pressure leads to misreading 'increases' as 'decreases' or missing a negative sign in −Ea/RT. Underlining key terms as you read can prevent this. Seventh, students apply integrated rate laws without checking which order the reaction is—always identify order from given data before picking a formula. Lastly, neglecting NCERT in-text questions and exercise numericals is a mistake. At least 3–4 MCQs in every CBSE paper are direct adaptations of NCERT exercise problems. Work through all 35 NCERT exercise questions for Chapter 3, especially the ones involving graphical interpretation and two-temperature Arrhenius problems. Avoiding these pitfalls can easily lift your MCQ score from 14–15/20 to 18–19/20. Use CBSETUTOR.ai 24×7 to upload a photo of any MCQ you got wrong and get an instant step-by-step video-style explanation—this rapid feedback loop fixes misconceptions before they cost you marks in the actual exam. Available at a flat ₹999/month for all subjects and classes (6–12), with a 3-day free trial so you can test the AI tutor risk-free.
  • Confusing order (experimental, can be fractional) with molecularity (integer, elementary step only).
  • Using the wrong half-life formula—write all three down at the start of the exam to avoid mid-question panic.
  • Unit mismatches in Arrhenius calculations: always convert Ea to J mol⁻¹ when using R=8.314 J K⁻¹ mol⁻¹.
  • Forgetting to divide rate by stoichiometric coefficient when relating rates of different species in the same reaction.
  • Rushing through assertion-reason MCQs and assuming R explains A without independent verification.
  • Misreading question stems under time pressure—circle keywords like NOT, EXCEPT, INCREASES, DECREASES.
  • Skipping NCERT exercise questions—at least 3–4 MCQs per paper are near-replicas of textbook problems.

Frequently asked questions

How many MCQs on Chemical Kinetics appear in the CBSE Class 12 Chemistry board exam?+
Typically 2–3 MCQs out of 20 in Section A come from Chapter 3 Chemical Kinetics, worth 2–3 marks. Additionally, you may see one assertion-reason MCQ on topics like half-life or Arrhenius equation, bringing the total to 3–4 marks from this chapter in Section A alone.
What is the difference between order and molecularity in a reaction?+
Order is the sum of exponents in the experimentally determined rate law and can be zero, fractional, or integer. Molecularity is the number of molecules that participate in a single elementary step and is always a positive integer (1, 2, or 3). Order is overall; molecularity applies only to elementary reactions.
Which half-life formula should I memorize for the CBSE exam?+
Memorize three: zero-order t₁/₂=[A]₀/(2k); first-order t₁/₂=0.693/k (independent of [A]₀); second-order t₁/₂=1/(k[A]₀). The first-order formula is most frequently tested because of its unique property of concentration independence.
How do I solve Arrhenius equation problems quickly in MCQs?+
Use the two-temperature form: ln(k₂/k₁)=(Ea/R)(1/T₁−1/T₂). Write down R=8.314 J K⁻¹ mol⁻¹ on your rough sheet. If Ea is given in kJ, convert to J by multiplying by 1000 before plugging in. Pre-calculate common terms like 1/300−1/310 mentally or with a calculator if allowed.
Are graphical questions on rate laws asked as MCQs?+
Yes. CBSE often shows a plot—[A] vs t, ln[A] vs t, or 1/[A] vs t—and asks you to identify the order. A straight line for [A] vs t indicates zero-order; straight line for ln[A] vs t indicates first-order; straight line for 1/[A] vs t indicates second-order. Know these cold.
What are effective collisions in collision theory?+
Effective collisions are those in which colliding molecules have kinetic energy ≥ activation energy (Ea) and proper spatial orientation to form products. Only a fraction e^(−Ea/RT) of all collisions are effective, which explains why reaction rate increases exponentially with temperature.
Can the order of a reaction be determined from the balanced chemical equation?+
No, not unless the reaction is elementary. For complex (multi-step) reactions, the overall balanced equation does not reveal the rate law. Order must be determined experimentally by measuring how rate changes with concentration. Only for elementary steps does order match stoichiometry.
How does a catalyst affect the rate constant k and activation energy Ea?+
A catalyst lowers the activation energy Ea by providing an alternate reaction pathway. According to k=Ae^(−Ea/RT), a smaller Ea increases k, thus accelerating the reaction. The catalyst is unchanged at the end and does not alter ΔH or equilibrium constant.
Why is the first-order half-life independent of initial concentration?+
For first-order kinetics, t₁/₂=0.693/k. Notice [A]₀ does not appear in this formula. This arises because the integrated rate law ln([A]/[A]₀)=−kt becomes ln(1/2)=−kt₁/₂, which simplifies to t₁/₂=ln2/k. It is a unique property of first-order reactions, frequently tested in MCQs.
How can I prepare Chemical Kinetics MCQs without coaching in a Tier-2 city?+
Focus on NCERT textbook exercises (all 35 questions in Chapter 3), NCERT Exemplar MCQs, and past 5 years' CBSE board papers. Use CBSETUTOR.ai to upload photos of any confusing MCQ or numerical—get instant step-by-step solutions and video-style explanations 24×7, at ₹999/month for all classes 6–12. A 3-day free trial lets you try before committing.

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