India's #1 AI Tutorimportant-questions · Chemistry · Chapter 3
Important Questions: CBSE Class 12 Chemistry Chapter 3 Chemical Kinetics
Chemical Kinetics is a scoring yet calculation-intensive chapter in CBSE Class 12 Chemistry, contributing 5 marks to your board exam. Questions test your understanding of rate, order, molecularity and the Arrhenius equation through a mix of MCQs, short-answer theory and multi-step numericals. This page organizes important exam-style questions by marks weightage, provides model answers grounded in NCERT terminology, and highlights the question patterns CBSE has favoured in recent years.
Your child's private AI tutor — trained on NCERT.
3-day free trial · ₹1 to start · Cancel anytime.
Key takeaways
- ✓Chemical Kinetics carries 5 marks in the CBSE Class 12 board exam, typically one 3-mark and one 2-mark question or one 5-mark case-based question.
- ✓Order of reaction and integrated rate laws are the most frequently asked 3-mark numericals in past CBSE papers.
- ✓Arrhenius equation problems appear almost every year, testing activation energy calculation and temperature dependence of rate constants.
- ✓Distinguish between order and molecularity clearly — a common error trap in 1-mark and 2-mark questions.
- ✓Case-based questions often link collision theory with graphs; practice interpreting ln k vs 1/T plots.
- ✓Show all calculation steps with units for numerical questions to avoid losing method marks even if the final answer is wrong.
- ✓CBSETUTOR.ai offers 24×7 doubt solving by uploading photos of tricky numericals at ₹999/month across all subjects for Class 12, with a 3-day free trial.
Chapter Overview and Marks Weightage in CBSE Exam
Chemical Kinetics sits in Unit 4 (Chemical Kinetics) of the CBSE Class 12 Chemistry syllabus and carries a total of 5 marks in the board examination. The chapter introduces the concept of reaction rates, factors affecting them (concentration, temperature, catalyst), and mathematical frameworks to describe how fast reactions proceed. NCERT Class 12 Chemistry Part I covers four core topics: rate of reaction and factors influencing it, integrated rate equations for zero-order and first-order reactions, temperature dependence of rate (Arrhenius equation, activation energy), and collision theory with the concept of molecularity. Typically, the board paper includes one 2-mark question (definition, derivation snippet or graph interpretation) and one 3-mark numerical, or sometimes a single 5-mark case-based question combining theory and calculation. In the 2024 and 2023 CBSE papers, order determination from experimental data and Arrhenius equation numericals were prominent. Mastering unit conversions and graph-reading is essential for full marks.
- Total weightage: 5 marks in the 70-mark theory paper
- Common pattern: one 2-mark + one 3-mark question, or one 5-mark case study
- High-frequency topics: integrated rate laws, half-life formulas, Arrhenius equation, collision theory
- Graph-based questions (concentration vs time, ln k vs 1/T) appear frequently in recent years
1-Mark Questions (MCQ and Very Short Answer)
One-mark questions test recall of definitions, units and basic distinctions between order and molecularity. CBSE often asks MCQs on half-life expressions, unit of rate constant or matching reaction types to their integrated rate laws. These are quick-win marks if you have conceptual clarity. Below are representative questions drawn from NCERT exemplar and past board trends, each with the correct answer or brief working to guide your preparation. Practice these regularly to build speed and accuracy in the objective section of your exam, where even a single mark can shift your overall grade.
- Q1. The unit of rate constant for a zero-order reaction is: (a) mol L⁻¹ s⁻¹ (b) s⁻¹ (c) mol⁻¹ L s⁻¹ (d) mol⁻² L² s⁻¹. Answer: (a) mol L⁻¹ s⁻¹
- Q2. For a first-order reaction, the time required for 99% completion is twice the time required for 90% completion. True or False? Answer: False; it is approximately ten times (from t = 2.303/k × log 100).
- Q3. What is the order of a reaction if doubling the concentration of reactant quadruples the rate? Answer: Order = 2 (rate ∝ [A]², 2² = 4).
- Q4. Molecularity of a reaction can be zero or fractional. True or False? Answer: False; molecularity is always a whole number and refers to the number of molecules participating in an elementary step.
- Q5. The slope of a plot of ln[A] vs time for a first-order reaction is: (a) +k (b) –k (c) +k/2.303 (d) –k/2.303. Answer: (b) –k.
2-Mark Questions with Model Answers
Two-mark questions in CBSE Class 12 Chemistry Chapter 3 typically ask for short derivations, graph sketches or conceptual explanations. You might be asked to write the integrated rate law for a zero-order reaction and state its graphical representation, or to define activation energy and explain its significance. Examiners look for precise NCERT terminology and correct units. Each answer below is written to match the marking scheme used in board exams, ensuring you capture both content points. Practicing these formats will help you frame concise yet complete answers under exam time pressure.
- Q6. Define order of reaction and write the unit of rate constant for a second-order reaction. Answer: Order of reaction is the sum of powers of concentration terms in the rate law equation. For a second-order reaction (rate = k[A]²), unit of k = (mol L⁻¹)¹⁻² s⁻¹ = mol⁻¹ L s⁻¹.
- Q7. What is the half-life of a zero-order reaction? Derive the expression. Answer: For zero-order, rate = k (constant). Integrated form: [A] = [A₀] – kt. At t = t₁/₂, [A] = [A₀]/2. Substituting, [A₀]/2 = [A₀] – kt₁/₂ ⇒ t₁/₂ = [A₀]/(2k). Hence half-life is directly proportional to initial concentration.
- Q8. Distinguish between molecularity and order of reaction with one example each. Answer: Molecularity is the number of molecules colliding in an elementary step (always a whole number, e.g. 2 for 2HI → H₂ + I₂). Order is the sum of exponents in the rate law (can be zero, fractional or integer, e.g. decomposition of NH₃ on Pt is zero order). Molecularity is theoretical; order is experimental.
- Q9. A first-order reaction is 20% complete in 10 minutes. Calculate the rate constant. Answer: For first-order, k = (2.303/t) log([A₀]/[A]). 20% complete means 80% remains. k = (2.303/10) log(100/80) = 0.2303 × 0.0969 ≈ 0.0223 min⁻¹.
- Q10. What is meant by activation energy? How does a catalyst affect it? Answer: Activation energy (Eₐ) is the minimum energy required for reactant molecules to form the activated complex. A catalyst lowers Eₐ by providing an alternative reaction pathway, thereby increasing the rate constant k as per Arrhenius equation k = A e^(–Eₐ/RT).
3-Mark Questions with Detailed Solutions
Three-mark questions form the backbone of Chemical Kinetics scoring in CBSE exams. They test your ability to apply integrated rate laws, derive half-life expressions and solve multi-step numericals involving Arrhenius equation or order determination. The marking scheme awards one mark for the correct formula, one for substitution with units and one for the final answer. Always write the formula first, show every substitution step and box or underline the final answer with units. Below are five representative 3-mark questions with complete worked solutions mirroring the board exam answer key format.
- Q11. For a first-order reaction, show that the time required for 99% completion is twice the time required for 90% completion. Answer: For first-order, t = (2.303/k) log([A₀]/[A]). For 90% completion, 10% remains: t₉₀ = (2.303/k) log(100/10) = (2.303/k) × 1 = 2.303/k. For 99% completion, 1% remains: t₉₉ = (2.303/k) log(100/1) = (2.303/k) × 2 = 4.606/k. Ratio t₉₉/t₉₀ = (4.606/k)/(2.303/k) = 2. Hence proved.
- Q12. The rate constant for a first-order reaction is 60 s⁻¹. Calculate the time required for 75% of the reaction to complete. Answer: k = 60 s⁻¹. For 75% completion, 25% remains. t = (2.303/k) log([A₀]/[A]) = (2.303/60) log(100/25) = (2.303/60) × log 4 = (2.303/60) × 0.602 ≈ 0.0231 s.
- Q13. A reaction is second order with respect to a reactant A. How is the rate affected if the concentration of A is (i) doubled and (ii) reduced to half? Answer: Rate = k[A]². (i) If [A] is doubled, rate = k(2[A])² = 4k[A]², i.e. rate becomes 4 times. (ii) If [A] is halved, rate = k([A]/2)² = (1/4)k[A]², i.e. rate becomes one-fourth.
- Q14. The rate of a reaction increases four times when temperature changes from 300 K to 320 K. Calculate the activation energy (R = 8.314 J K⁻¹ mol⁻¹, log 4 = 0.602). Answer: log(k₂/k₁) = (Eₐ/2.303R)[(T₂ – T₁)/(T₁T₂)]. 0.602 = (Eₐ/(2.303×8.314))×[(20)/(300×320)]. Eₐ = (0.602 × 2.303 × 8.314 × 96000)/20 ≈ 55,400 J mol⁻¹ or 55.4 kJ mol⁻¹.
- Q15. Derive the integrated rate equation for a zero-order reaction. Answer: For zero-order, rate = –d[A]/dt = k. Integrating, ∫d[A] = –k∫dt ⇒ [A] = –kt + C. At t = 0, [A] = [A₀], so C = [A₀]. Hence [A] = [A₀] – kt. This is the integrated rate law for zero-order reaction.
5-Mark Questions and Case-Based Problems
Five-mark questions in CBSE Class 12 Chemistry often appear as case studies or multi-part numericals combining theory and calculation. A typical case-based question presents a paragraph on collision theory or catalyst action, followed by sub-questions testing graph interpretation, order determination and Arrhenius equation. Alternatively, you may be asked to derive the integrated rate law, sketch the graph and solve a numerical all within one question. These carry significant weightage and require clear step-wise presentation. Below are two representative 5-mark questions with full model answers structured as CBSE expects.
- Q16. (a) Write the rate law for a reaction A + B → C if it is first order in A and second order in B. (b) What is the overall order? (c) How will the rate change if concentration of A is tripled and B is doubled? (d) Calculate the half-life for a first-order reaction with k = 0.0693 min⁻¹. Answer: (a) Rate = k[A]¹[B]². (b) Overall order = 1 + 2 = 3. (c) New rate = k(3[A])(2[B])² = k × 3 × 4[A][B]² = 12 times original rate. (d) t₁/₂ = 0.693/k = 0.693/0.0693 = 10 min.
- Q17. Case Study: The decomposition of N₂O₅ in CCl₄ at 318 K follows first-order kinetics. The rate constant is 3.46 × 10⁻⁵ s⁻¹. (i) Write the integrated rate equation for first-order reaction. (ii) Calculate the half-life. (iii) How long will it take for 90% decomposition? (iv) What fraction remains after 2 hours? Answer: (i) ln[A] = ln[A₀] – kt. (ii) t₁/₂ = 0.693/(3.46×10⁻⁵) ≈ 20,000 s ≈ 5.56 h. (iii) t = (2.303/k) log(100/10) = (2.303/(3.46×10⁻⁵)) × 1 ≈ 66,560 s ≈ 18.5 h. (iv) 2 h = 7200 s. ln([A]/[A₀]) = –3.46×10⁻⁵×7200 ≈ –0.249; [A]/[A₀] = e⁻⁰·²⁴⁹ ≈ 0.78, so 78% remains.
How CBSE Frames Questions from Chemical Kinetics
CBSE question-setters follow NCERT closely and favour numericals that integrate multiple concepts in a single problem. Over the past five years, the board has consistently asked one graph-based question (plotting concentration vs time or ln k vs 1/T and extracting rate constant or activation energy), one order-determination numerical using initial rate data or integrated rate law, and one Arrhenius equation problem linking rate constants at two temperatures. Case-based questions introduced in 2021 often present real-world scenarios — drug stability, catalyst in industrial ammonia synthesis, or enzyme kinetics — followed by sub-parts testing definitions, order, half-life and temperature dependence. Important exam patterns include asking students to justify why a reaction is first-order by showing that a plot of ln[A] vs t is linear, or to calculate the percentage of reactant remaining after a given time. The marking scheme awards partial credit for correct formula and method even if arithmetic is wrong, so always write the starting equation and show dimensional analysis. CBSE also tests conceptual traps: stating that molecularity can be fractional (false), or confusing pseudo-first-order with true first-order kinetics.
- Graphical analysis: ln[A] vs t (first-order slope = –k), 1/[A] vs t (second-order slope = +k), ln k vs 1/T (Arrhenius slope = –Eₐ/R)
- Order determination: using method of initial rates or by checking which integrated rate law gives a straight-line plot
- Half-life questions: deriving or applying t₁/₂ = 0.693/k (first-order) or t₁/₂ = [A₀]/(2k) (zero-order)
- Arrhenius two-temperature problems: log(k₂/k₁) = (Eₐ/2.303R)[(T₂–T₁)/(T₁T₂)]
- Case studies linking collision theory, activation energy and catalyst effect in industrial or biological contexts
Common Mistakes Students Make in Chemical Kinetics Questions
Several recurring errors cost students easy marks in Chemical Kinetics. The most frequent mistake is confusing order and molecularity — writing that molecularity can be zero or fractional, or stating that order is always equal to molecularity. Remember, molecularity applies only to elementary reactions and is always a whole number, while order is determined experimentally and can be zero, fractional or integer for overall reactions. Another common pitfall is unit mismatch: forgetting to convert minutes to seconds when k is given in s⁻¹, or mixing concentration units (mol L⁻¹ vs mol dm⁻³). In Arrhenius equation problems, students often use the wrong form of the gas constant R (use 8.314 J K⁻¹ mol⁻¹, not 0.0821 L atm, for energy calculations) or forget to convert activation energy from kJ to J. Graph interpretation errors include reading the slope sign incorrectly — for a first-order reaction ln[A] vs t has a negative slope equal to –k, not +k. During half-life calculations, many assume the formula t₁/₂ = 0.693/k applies to all orders, when in fact it is valid only for first-order; for zero-order, t₁/₂ = [A₀]/(2k) and depends on initial concentration. Finally, in multi-step case-based questions, students skip showing intermediate steps or omit units in the final answer, losing method marks even when the numerical value is correct.
- Confusing order with molecularity — always state which is theoretical and which is experimental
- Unit errors: ensure consistent time units (s or min) and energy units (J or kJ) throughout the problem
- Using wrong R value in Arrhenius equation (use 8.314 J K⁻¹ mol⁻¹ for Eₐ in J, or 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹ for Eₐ in kJ)
- Forgetting to apply log or ln correctly: log is base-10 (2.303 factor), ln is base-e (natural log)
- Assuming all half-life formulas are the same — know the correct expression for each order
- Not showing working steps: CBSE awards partial marks for method, so write formula, substitution and units
Preparation Strategy and Resources for Chemical Kinetics
To master Chemical Kinetics, start with NCERT Class 12 Chemistry Part I Chapter 4. Read the theory sections on rate laws, integrated rate equations and Arrhenius theory carefully, noting all bold-faced definitions and equations. Work through every in-text example and end-of-chapter exercise; CBSE often adapts these directly into board questions. Next, create a formula sheet covering integrated rate laws for zero, first and second order, half-life expressions, Arrhenius equation in both forms (k = Ae^(–Eₐ/RT) and log form), and unit tables for rate constants of each order. Practice plotting graphs on graph paper — ln[A] vs t, log k vs 1/T — to build confidence in slope and intercept interpretation. Solve previous years' CBSE board papers from 2015 onwards; you will notice recurring patterns in order determination and Arrhenius numericals. Use NCERT Exemplar for challenging MCQs and assertion-reason questions. Time yourself on 3-mark and 5-mark questions to improve speed. For doubt-clearing, especially tricky derivations or multi-step numericals, CBSETUTOR.ai provides instant AI-powered solutions — upload a photo of your problem and get step-by-step working at any hour. At ₹999 per month for all subjects across Class 12, with a 3-day free trial, it is a cost-effective alternative to expensive weekend coaching. Finally, revise common mistakes and graph interpretations a week before the exam; Chemical Kinetics is a high-scoring chapter if you practice systematically.
- Read NCERT Chapter 4 thoroughly; underline all equations and mark example problems
- Prepare a one-page formula sheet with integrated rate laws, half-life formulas and Arrhenius equation variants
- Practice graph plotting and slope calculation on paper, not just digitally
- Solve CBSE previous years' papers (2015–2024) to identify recurring question types
- Use NCERT Exemplar for MCQs and assertion-reason practice
- Time your answers: aim for 2-mark in 3 minutes, 3-mark in 5 minutes, 5-mark in 8 minutes
- Clarify doubts instantly with CBSETUTOR.ai photo-upload solving — ₹999/month, 3-day free trial
How CBSETUTOR.ai Supports Your Chemical Kinetics Preparation
Chemical Kinetics numericals can be daunting — multi-step Arrhenius calculations, logarithm manipulations and unit conversions trip up even diligent students. CBSETUTOR.ai offers a 24×7 AI tutor that you can access from your phone: snap a photo of any problem from NCERT, Exemplar or your school worksheet, upload it and receive a detailed step-by-step solution within seconds. The AI breaks down each calculation, highlights where to apply 2.303 for base-10 log, shows unit conversions explicitly and explains conceptual steps like why the slope of ln[A] vs t is negative. Beyond question-solving, the platform provides chapter-wise notes, formula sheets and video explanations aligned with the latest CBSE syllabus. Parents appreciate the flat ₹999 per month pricing for all subjects (Physics, Chemistry, Maths, Biology) across Class 12, making it far more affordable than traditional coaching where Chemistry alone can cost ₹3,000–5,000 monthly. A 3-day free trial lets your child test the service before committing. Students in Delhi NCR coaching hubs and Tier-2 cities alike rely on CBSETUTOR.ai for instant doubt resolution, especially late at night when conventional tutors are unavailable. For Chemical Kinetics, where a single sign error or wrong R value can cost 3 marks, having on-demand expert guidance ensures you practise correctly and build confidence heading into the board exam.
- Upload photos of Chemical Kinetics numericals and get step-by-step solutions anytime
- AI tutor explains each formula, substitution and unit conversion in detail
- Access chapter notes, formula sheets and video walkthroughs for quick revision
- Flat ₹999/month for all Class 12 subjects — Chemistry, Physics, Maths, Biology
- 3-day free trial to explore the platform risk-free
- Ideal for students in areas with limited coaching access or those seeking flexible, self-paced learning
Frequently asked questions
How many marks does Chemical Kinetics carry in CBSE Class 12 board exam?+
Chemical Kinetics (Chapter 3 in NCERT Part I) carries 5 marks in the CBSE Class 12 Chemistry board exam, typically split as one 2-mark and one 3-mark question or one 5-mark case-based question.
What is the difference between order and molecularity of 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 participating in an elementary step, always a whole number (1, 2, 3) and is a theoretical concept.
Which formula should I use for half-life in a first-order reaction?+
For a first-order reaction, half-life t₁/₂ = 0.693/k, where k is the rate constant. This value is independent of initial concentration and depends only on k.
How do I determine the order of a reaction from experimental data?+
Use the method of initial rates: compare how the rate changes when concentration of one reactant is varied while others are kept constant. If doubling [A] doubles the rate, order in A is 1; if it quadruples the rate, order is 2.
What is the Arrhenius equation and when is it used?+
The Arrhenius equation k = A e^(–Eₐ/RT) relates the rate constant k to temperature T and activation energy Eₐ. Use the log form log(k₂/k₁) = (Eₐ/2.303R)[(T₂–T₁)/(T₁T₂)] to calculate Eₐ when k is known at two temperatures.
Why does a plot of ln[A] vs time give a straight line for first-order reactions?+
The integrated rate law for first-order is ln[A] = ln[A₀] – kt, which is of the form y = c + mx. Plotting ln[A] (y-axis) vs t (x-axis) yields a straight line with slope –k and intercept ln[A₀].
What are common mistakes students make in Chemical Kinetics numericals?+
Common errors include confusing order with molecularity, using wrong units for k or R, forgetting to convert kJ to J in Arrhenius problems, applying the wrong half-life formula and not showing working steps clearly.
How should I prepare for case-based questions in Chemical Kinetics?+
Read the case paragraph carefully, identify which concepts (collision theory, catalyst, Arrhenius equation) are involved, then solve sub-parts methodically. Practice CBSE sample papers and previous years' case studies to familiarize yourself with the format.
Which topics in Chemical Kinetics are most frequently asked in board exams?+
Integrated rate laws for zero and first order, half-life derivations and calculations, order determination from initial rate data, Arrhenius equation problems and graph interpretation (ln[A] vs t, ln k vs 1/T) appear almost every year.
Can CBSETUTOR.ai help me solve Chemical Kinetics numericals step-by-step?+
Yes, CBSETUTOR.ai lets you upload a photo of any Chemical Kinetics problem and provides detailed step-by-step solutions instantly. It covers all question types — MCQs, numericals, derivations and case-based — at ₹999/month for all Class 12 subjects with a 3-day free trial.
Related resources
CBSE Class 12 Chemistry Chapter 3 Chemical Kinetics Worksheet with AnswersClass 12 Chemistry Chapter 3 Chemical Kinetics — Formulas & Key PointsClass 12 Chemistry Chapter 2 Electrochemistry — Formulas & Key PointsCBSE Class 12 Chemistry Chapter 2 Electrochemistry Worksheet with AnswersAI Tutor for Class 12: The Smart Alternative to TuitionAI Tutor for Class 12 Accountancy: Learn Faster with Instant HelpOnline Class 9 Chemistry Tutor in Mumbai — NCERT-Aligned Board Exam CoachingImportant Questions: CBSE Class 9 Mathematics Chapter 2 Polynomials
Keep learning — related guides
Class 12Chemistry
Class 12 Chemistry Tutor in Rushikonda, Visakhapatnam
Class 12Chemistry
Class 12 Chemistry Chapter 9 Amines — Formulas & Key Points
Class 12Chemistry
Class 12 Chemistry Chapter 10 Biomolecules — Formulas & Key Points
Class 12Chemistry
CBSE Class 12 Chemistry Chapter 10 Biomolecules Worksheet with Answers
Class 12Chemistry
Important Questions: CBSE Class 12 Chemistry Chapter 8 Aldehydes, Ketones and Carboxylic Acids
Class 12Chemistry
Important Questions: CBSE Class 12 Chemistry Chapter 7 Alcohols, Phenols and Ethers
Ready to give your Class 12 child the tutor that never sleeps?
CBSETUTOR.ai covers every chapter in the Class 12 NCERT syllabus — Maths, Science, Social Science, English, Hindi and more. 24×7. Patient. Unlimited. 3-day free trial.
Start your child's 3-day free trial →