Class 9 Chemistry Chapter 3 Chemical Kinetics: 18 Important Questions with Complete Answers

Chemical Kinetics (Chapter 3) carries significant weight in CBSE Class 9 Chemistry because it bridges qualitative observations with quantitative analysis. The 2024-25 rationalized syllabus emphasizes core concepts: rate of reaction, factors affecting rate, order of reaction, molecularity, and the Arrhenius equation. Board exams typically allocate 8–12 marks to this chapter across 1-mark, 2-mark, and 3-mark questions, with occasional 5-mark applications. Students often struggle with three key areas: (1) distinguishing between order and molecularity, (2) interpreting rate-concentration graphs, and (3) applying the Arrhenius equation in word problems. The important questions below target these exact pain points. By practising these, you'll recognize question patterns instantly, avoid common misconceptions (like confusing rate constant with rate of reaction), and earn full marks on definition-based and calculation-based questions. This chapter also builds conceptual foundations for Class 11 chemical kinetics, making mastery now incredibly valuable.

Multiple-choice questions test quick recall and definition accuracy. These are typically worth 1 mark each and form 20–25% of the chemistry paper. Below are 5 representative MCQs from Chemical Kinetics: **Q1.** Rate of reaction is expressed in units of: (A) mol·L⁻¹ (B) mol·L⁻¹·s⁻¹ (C) mol·L⁻¹·s (D) mol·s⁻¹ **Answer: (B) mol·L⁻¹·s⁻¹** Explanation: Rate is the change in concentration per unit time. Concentration is in mol·L⁻¹ and time in seconds, so rate has units mol·L⁻¹·s⁻¹. **Q2.** Molecularity of a reaction is defined as: (A) Number of moles of reactants (B) Number of atoms in the reaction (C) Number of molecules participating in an elementary step (D) Order of the overall reaction **Answer: (C) Number of molecules participating in an elementary step** Explanation: Molecularity refers to the number of molecules (reactant particles) that participate in a single elementary reaction step, not the overall reaction. **Q3.** Which factor does NOT affect the rate of a chemical reaction? (A) Catalyst (B) Temperature (C) Pressure (D) Colour of reactants **Answer: (D) Colour of reactants** Explanation: Rate is affected by concentration, temperature, pressure (for gases), surface area, and catalyst. Colour is a physical property unrelated to kinetics. **Q4.** The Arrhenius equation is: k = Ae^(−Eₐ/RT). Here, Eₐ represents: (A) Energy of products (B) Activation energy (C) Total energy released (D) Average kinetic energy **Answer: (B) Activation energy** Explanation: Eₐ is the minimum energy required for reactants to form products. The exponential term e^(−Eₐ/RT) gives the fraction of molecules with sufficient energy. **Q5.** Order of reaction is determined by: (A) Stoichiometric coefficients only (B) Experimental rate law data (C) Molecular weight of products (D) Temperature of reaction **Answer: (B) Experimental rate law data** Explanation: Order cannot be predicted from balanced equations alone; it must be found experimentally by studying how rate varies with concentration.

Two-mark questions require brief explanations, definitions, or simple calculations. These test both knowledge and clarity of expression. **Q1.** Distinguish between rate of reaction and rate constant. **Answer:** Rate of reaction is the speed at which reactants are consumed or products are formed, expressed as the change in concentration per unit time (Δ[conc]/Δt). It has units mol·L⁻¹·s⁻¹ and varies with concentration. Rate constant (k) is a fixed value for a given reaction at a fixed temperature. It does not depend on concentration but only on temperature. Its units depend on the order of the reaction. For example, for a first-order reaction, k has units s⁻¹. **Q2.** Why does a catalyst increase the rate of reaction without being consumed? **Answer:** A catalyst provides an alternative reaction pathway with a lower activation energy (Eₐ). By lowering Eₐ, more reactant molecules possess sufficient energy to undergo reaction at the same temperature. Since the catalyst is regenerated at the end of the reaction, it is not consumed. The overall energy change (ΔH) remains unchanged. **Q3.** A reaction has order 2 with respect to reactant A and order 1 with respect to reactant B. Write the rate law expression. **Answer:** Rate law: Rate = k[A]²[B]¹ or Rate = k[A]²[B] Here, the overall order = 2 + 1 = 3 (third-order reaction). The rate constant k must be determined experimentally by measuring the rate at known concentrations of A and B. **Q4.** The activation energy of a reaction is 50 kJ·mol⁻¹. How would doubling the temperature affect the rate constant according to the Arrhenius equation? **Answer:** Using the Arrhenius equation, k = Ae^(−Eₐ/RT). When temperature doubles, the exponent −Eₐ/RT becomes less negative (because T is larger), so e^(−Eₐ/RT) increases significantly. For most reactions with Eₐ ≈ 50 kJ·mol⁻¹, doubling absolute temperature (e.g., 300 K to 600 K) increases k by a factor of roughly 10–100 (the exact factor depends on the original temperature). Lower Eₐ values show larger increases. **Q5.** State the relationship between order and molecularity for elementary reactions. **Answer:** For elementary reactions (single-step reactions), the order equals the molecularity. Both are determined by the stoichiometric coefficients of the elementary step. For example, if an elementary step is 2A → B, the molecularity is 2 and the order is also 2. However, for multi-step reactions, the order is NOT equal to molecularity; order is determined experimentally from the overall rate law.

Three-mark questions demand deeper understanding, short derivations, or problem-solving that combines multiple concepts. **Q1.** Define activation energy and explain why reactions with lower activation energy occur faster at the same temperature. **Answer:** Activation energy (Eₐ) is the minimum amount of energy that reactant molecules must possess to transform into products, measured from the energy level of reactants to the transition state. At a given temperature T, the fraction of molecules with energy ≥ Eₐ is proportional to e^(−Eₐ/RT) (from the Arrhenius equation). When Eₐ is lower, the exponent −Eₐ/RT is less negative, making e^(−Eₐ/RT) larger. This means a larger fraction of reactant molecules have sufficient energy to react. Consequently, more collisions are successful, and the reaction rate increases. For example, a reaction with Eₐ = 30 kJ·mol⁻¹ will be faster than one with Eₐ = 100 kJ·mol⁻¹ at the same temperature because more molecules exceed the lower energy barrier. **Q2.** For the reaction 2NO + O₂ → 2NO₂, the rate law is found to be Rate = k[NO]²[O₂]. Explain why the order with respect to NO is 2 even though its stoichiometric coefficient is 2. **Answer:** The stoichiometric coefficient (2 for NO in the overall equation) does NOT directly determine the order of reaction. Order is an experimental property determined from the rate law, which must be found by performing experiments at different concentrations. In this case, the experimentally determined rate law is Rate = k[NO]²[O₂], meaning the order is 2 with respect to NO and 1 with respect to O₂ (overall order = 3). This indicates the reaction likely occurs via a multi-step mechanism (not a single elementary step). The rate-determining step probably involves collision of two NO molecules with one O₂ molecule, or the reaction proceeds through intermediate steps. The rate law reflects the actual molecular-level process, not the overall stoichiometry. **Q3.** A first-order reaction has a rate constant k = 0.0693 s⁻¹ at 25 °C. If the reaction is heated to 35 °C, use the integrated form of the Arrhenius equation to estimate the new rate constant (assume Eₐ = 50 kJ·mol⁻¹, R = 8.314 J·mol⁻¹·K⁻¹). **Answer:** Using: ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂) Given: k₁ = 0.0693 s⁻¹, T₁ = 25 + 273 = 298 K, T₂ = 35 + 273 = 308 K, Eₐ = 50,000 J·mol⁻¹ ln(k₂/0.0693) = (50,000/8.314)(1/298 − 1/308) ln(k₂/0.0693) = 6,014.6 × (0.003356 − 0.003247) ln(k₂/0.0693) = 6,014.6 × 0.000109 ln(k₂/0.0693) = 0.656 k₂/0.0693 = e^0.656 ≈ 1.928 k₂ ≈ 0.134 s⁻¹ The rate constant nearly doubles with a 10 K temperature increase, confirming that rate is highly temperature-dependent. **Q4.** Explain the role of a reaction intermediate in a multi-step reaction mechanism. Use an example. **Answer:** A reaction intermediate is a substance produced in one elementary step and consumed in a subsequent elementary step. It does not appear in the overall balanced equation. Example: The reaction O₃ + NO → O₂ + NO₂ is believed to occur via two steps: Step 1 (slow): O₃ + NO → NO₂ + O₂ Step 2 (fast): O + NO₂ → NO₃ Actually, a corrected example: O₃ + NO → NO₂ + O₂ Mechanism: Step 1 (slow): O₃ + NO → O₂ + NO₂ Step 2 (fast): NO₂ → NO + O₂ (this doesn't work either). Better example: H₂ + I₂ → 2HI occurs via: Step 1 (fast equilibrium): I₂ ⇌ 2I Step 2 (slow): H₂ + 2I → 2HI Here, atomic iodine (I) is the intermediate—formed in Step 1 and consumed in Step 2. The rate law depends on the slow step and the equilibrium from Step 1. Intermediates are crucial for explaining why the experimental rate law does not match the overall stoichiometry.

Five-mark questions test comprehensive understanding, derivations, and applications across multiple subtopics. **Q1.** Derive the integrated rate law for a first-order reaction and show how it is used to find the age of archaeological samples using carbon-14 dating. **Full Solution:** *Derivation of integrated rate law:* For a first-order reaction: Rate = −d[A]/dt = k[A] Rearranging: d[A]/[A] = −k·dt Integrating both sides: ∫d[A]/[A] = −k∫dt ln[A] = −kt + C At t = 0, [A] = [A]₀, so C = ln[A]₀ Therefore: ln[A] = −kt + ln[A]₀ Or: ln([A]₀/[A]) = kt Alternatively: [A] = [A]₀e^(−kt) This equation shows that for a first-order reaction, the concentration of reactant decreases exponentially with time. *Application to carbon-14 dating:* Radioactive decay of ¹⁴C follows first-order kinetics. The rate constant for ¹⁴C decay is k = 1.21 × 10⁻⁴ year⁻¹ (corresponding to a half-life of 5,730 years). When a living organism dies, it stops exchanging ¹⁴C with the atmosphere. The ¹⁴C already present decays according to: ln([C₀]/[C]) = kt Where [C₀] is the initial ¹⁴C content and [C] is the current amount. By measuring the current radioactivity of a sample and comparing it to the known initial activity, we can calculate t (age). Example: An archaeological sample shows 25% of its original ¹⁴C remains. ln(100/25) = 1.21 × 10⁻⁴ × t ln(4) = 1.21 × 10⁻⁴ × t 1.386 = 1.21 × 10⁻⁴ × t t ≈ 11,450 years **Q2.** The reaction 2H₂ + 2NO → N₂ + 2H₂O is found to have the rate law: Rate = k[H₂][NO]². (a) What is the order of the reaction? (b) How would the rate change if [H₂] is doubled and [NO] is halved? (c) Suggest a plausible mechanism. **Full Solution:** (a) *Order of reaction:* Order with respect to H₂ = 1 Order with respect to NO = 2 Overall order = 1 + 2 = 3 (third-order reaction) (b) *Effect of concentration changes:* Initial rate: Rate₁ = k[H₂]₁[NO]₁² New rate: Rate₂ = k[H₂]₂[NO]₂² = k(2[H₂]₁)(0.5[NO]₁)² Rate₂ = k(2[H₂]₁)(0.25[NO]₁²) = 0.5 × k[H₂]₁[NO]₁² Rate₂ = 0.5 × Rate₁ The rate decreases to 50% of the original rate. Doubling [H₂] would double the rate, but halving [NO] reduces the rate by a factor of 4 (since order is 2). The net effect is 2 × (1/4) = 1/2. (c) *Plausible mechanism:* Step 1 (fast equilibrium): 2NO ⇌ N₂O₂ Step 2 (slow): N₂O₂ + H₂ → N₂O + H₂O Step 3 (fast): N₂O + H₂ → N₂ + H₂O From the slow step, Rate = k₂[N₂O₂][H₂]. But [N₂O₂] is in equilibrium with NO, so: K = [N₂O₂]/[NO]², giving [N₂O₂] = K[NO]². Substituting: Rate = k₂K[NO]²[H₂] = k[NO]²[H₂] ✓ (matches the experimental rate law) **Q3.** The rate constant of a reaction increases from 2.5 × 10⁻⁴ s⁻¹ at 20 °C to 1.0 × 10⁻³ s⁻¹ at 40 °C. Calculate the activation energy for this reaction. (R = 8.314 J·mol⁻¹·K⁻¹) **Full Solution:** Using the two-temperature Arrhenius form: ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂) Given: k₁ = 2.5 × 10⁻⁴ s⁻¹ at T₁ = 20 + 273 = 293 K k₂ = 1.0 × 10⁻³ s⁻¹ at T₂ = 40 + 273 = 313 K R = 8.314 J·mol⁻¹·K⁻¹ ln(1.0 × 10⁻³ / 2.5 × 10⁻⁴) = (Eₐ / 8.314) × (1/293 − 1/313) ln(4) = (Eₐ / 8.314) × (0.003413 − 0.003195) 1.386 = (Eₐ / 8.314) × 0.000218 1.386 = Eₐ × 2.622 × 10⁻⁵ Eₐ = 1.386 / 2.622 × 10⁻⁵ Eₐ ≈ 52,850 J·mol⁻¹ ≈ 52.9 kJ·mol⁻¹ The activation energy is approximately 53 kJ·mol⁻¹. This moderate value explains why the rate constant changes significantly over a 20 K temperature range.

**Case Study: Industrial Synthesis of Ammonia** The Haber process synthesizes ammonia from nitrogen and hydrogen: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ·mol⁻¹ Industrially, this reaction is conducted at 400–500 °C under 150–300 atm pressure with an iron catalyst. The uncatalysed reaction is extremely slow at room temperature. **Case-based questions:** **(i) Explain why the Haber process uses high temperature despite the reaction being exothermic.** At room temperature, the activation energy is so high that the reaction is kinetically slow, even though it is thermodynamically favourable (ΔG < 0). High temperature increases the fraction of molecules with energy ≥ Eₐ, speeding up the forward reaction. Although high temperature shifts equilibrium leftward (Le Chatelier), the dramatic increase in forward rate at high temperature more than compensates for the equilibrium shift. **(ii) What is the role of the iron catalyst in this process? Why doesn't it affect the equilibrium position?** The iron catalyst provides an alternative reaction pathway with lower activation energy. This speeds up both forward and reverse reactions equally by the same factor. Since the catalyst affects k equally for forward and reverse steps, the equilibrium constant K = k_forward / k_reverse remains unchanged. The catalyst reduces Eₐ, allowing the system to reach equilibrium faster without altering the final ratio of products to reactants. **(iii) Pressure is maintained at 150–300 atm. Using Le Chatelier's principle and kinetic considerations, explain this choice.** High pressure shifts equilibrium rightward (toward NH₃, the side with fewer moles: 2 moles vs. 4 moles). From kinetics, high pressure increases collision frequency and effective concentration, raising the rate constant and forward rate. However, extremely high pressure (>300 atm) increases equipment cost and energy input without proportional gains in conversion. 150–300 atm represents an industrial optimum balancing conversion, rate, and economics. **(iv) If the rate constant at 400 °C is 8.4 × 10⁻³ L²·mol⁻²·s⁻¹, calculate it at 500 °C (assume Eₐ = 190 kJ·mol⁻¹, R = 8.314 J·mol⁻¹·K⁻¹).** Using: ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂) T₁ = 400 + 273 = 673 K, T₂ = 500 + 273 = 773 K k₁ = 8.4 × 10⁻³ L²·mol⁻²·s⁻¹ ln(k₂ / 8.4 × 10⁻³) = (190,000 / 8.314) × (1/673 − 1/773) ln(k₂ / 8.4 × 10⁻³) = 22,837 × (0.001486 − 0.001294) ln(k₂ / 8.4 × 10⁻³) = 22,837 × 0.000192 ln(k₂ / 8.4 × 10⁻³) = 4.384 k₂ / 8.4 × 10⁻³ = e^4.384 ≈ 80.2 k₂ ≈ 0.673 L²·mol⁻²·s⁻¹ The rate constant increases roughly 80-fold with a 100 K rise, showing the extreme temperature sensitivity of the Haber process due to its high activation energy.

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