Chemical Kinetics: Class 12 Important Questions with Answers
72 important questions for Class 12 Chemistry Chapter 3, Chemical Kinetics, the way they come in the board exam: MCQs, assertion–reason, case-based and short and long answer questions, with answers. Taken from our Class 12 Chemistry notes for 2026-27.
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Jump to: Practice set: test yourself20 MCQsAssertion–ReasonAnswers to the practice set
Practice set: test yourself
Try every question before looking at the answers below. 64 or more out of 80 means you are ready for the board.
A. Fill in the blanks (1 mark each)
- The unit of k of a first order reaction is ______.
- The slowest step of a reaction is called the ______ step.
- For zero order, t½ = ______.
- Molecularity can never be ______ or a fraction.
- A catalyst lowers the ______.
- Rate constant nearly ______ for a 10 K rise.
- The plot of ln k against 1/T has slope ______.
- Radioactive decay is ______ order.
B. True or false? (1 mark each)
- Order can be found from the balanced equation.
- t½ of a first order reaction depends on [R]₀.
- Hydrolysis of an ester in acid is pseudo first order.
- A catalyst changes ΔG of a reaction.
- k depends on temperature.
- Order of a complex reaction can be fractional.
- Rate of a zero order reaction falls with time.
- e^(−Ea/RT) is the fraction of molecules with E ≥ Ea.
C. Very short answer (1 mark each)
- Define rate of reaction.
- Define rate constant.
- Give an example of a zero order reaction.
- Write the unit of k for a second order reaction.
- What is an activated complex?
- Why is molecularity not defined for complex reactions?
- Write the Arrhenius equation.
- What is the effect of a catalyst on K?
- What is an effective collision?
- Give the relation between t½ and k (first order).
D. Short answer I (2 marks each)
- Distinguish order and molecularity (two points).
- A zero order reaction has k = 0.02 mol L⁻¹ s⁻¹ and [R]₀ = 0.5 M. Find t½.
- Find t½ of a first order reaction with k = 0.0693 min⁻¹.
- Rate = k[A]²[B]: give the order and the unit of k.
- For 2N₂O₅ → 4NO₂ + O₂, relate the rates of the three species.
- Draw the energy profile showing Ea with and without a catalyst.
- Why does a 10 K rise nearly double the rate?
- What is pseudo first order? Give an example.
| Section | A /8 | B /8 | C /10 | D /16 | E /18 | F /20 | G (bonus) | Total /80 |
|---|---|---|---|---|---|---|---|---|
| My marks |
Rough work
E. Short answer II (3 marks each)
- A first order reaction is 20% complete in 10 min. Find k and t½.
- k doubles from 300 K to 310 K. Calculate Ea.
- Initial rates: [A] 0.1, 0.2, 0.1; [B] 0.1, 0.1, 0.2; rate 2 × 10⁻³, 8 × 10⁻³, 4 × 10⁻³ mol L⁻¹ s⁻¹. Find the rate law and k.
- Derive the integrated rate equation of a zero order reaction.
- A first order reaction has t½ = 30 min. How long for 75% completion?
- Explain the effect of temperature using the Maxwell–Boltzmann curve.
F. Long answer (5 marks each)
- (a) Derive k = (2.303/t) log([R]₀/[R]) for a first order reaction. (b) Show that t½ is independent of [R]₀. (c) A first order reaction has k = 1.0 × 10⁻³ s⁻¹. Find t½ and the time for 90% completion.
- (a) State the Arrhenius equation and explain each term. (b) Ea = 75 kJ mol⁻¹: by what factor does k rise from 300 K to 310 K? (c) Give two effects of a catalyst.
G. HOTS (higher-order thinking)
- Why can't a reaction of order 5/2 be elementary?
- Why does the rate of a reaction fall with time even though k is constant?
- Why is the rate of NH₃ decomposition on hot Pt independent of [NH₃] at high pressure?
- Why do food items keep longer in a freezer?
20 MCQs
- Unit of k for a zero order reaction(a) s⁻¹ (b) L mol⁻¹ s⁻¹ (c) mol L⁻¹ s⁻¹ (d) L² mol⁻² s⁻¹
- For a first order reaction t½ is(a) 0.693/k (b) [R]₀/2k (c) 2.303/k (d) 1/k[R]₀
- Order of Rate = k[A]^½[B]²(a) 2 (b) 3 (c) 1.5 (d) 2.5
- Molecularity can be(a) 0 (b) 2 (c) 1.5 (d) −1
- A catalyst changes(a) ΔH (b) Ea (c) K (d) ΔG
- 99.9% of a first order reaction takes(a) 2 t½ (b) 3 t½ (c) 100 t½ (d) 10 t½
- Slope of log k vs 1/T(a) −Ea/2.303R (b) −Ea/R (c) Ea/R (d) log A
- Rate × 9 when [A] × 3: order(a) 1 (b) 3 (c) 2 (d) 0
- Acid hydrolysis of ethyl acetate is(a) zero order (b) third order (c) second order (d) pseudo first order
- For zero order the straight-line plot is(a) ln[R] vs t (b) [R] vs t (c) 1/[R] vs t (d) t½ vs t
- k = 3 × 10⁻⁴ s⁻¹ means order(a) 0 (b) 2 (c) 1 (d) 3
- Activation energy is the energy to form(a) the activated complex (b) products (c) reactants (d) a catalyst
- Fraction of molecules with E ≥ Ea(a) e^(Ea/RT) (b) e^(−Ea/RT) (c) Ea/RT (d) −Ea/RT
- Rate of 2HI → H₂ + I₂(a) −d[HI]/dt (b) −2 d[HI]/dt (c) −½ d[HI]/dt (d) d[HI]/dt
- The rate constant depends on(a) concentration (b) volume (c) time (d) temperature
- t₉₉/t₉₀ for first order(a) 2 (b) 1 (c) 10 (d) 0.5
- Decomposition of NH₃ on hot Pt is(a) zero order (b) first order (c) second order (d) third order
- Rate-determining step is the(a) fastest (b) last (c) first (d) slowest
- ¼ of a first order reactant left after(a) 1 t½ (b) 2 t½ (c) 3 t½ (d) 4 t½
- Arrhenius factor A is the(a) activation energy (b) rate (c) frequency factor (d) order
1 (c) · 2 (a) · 3 (d) · 4 (b) · 5 (b) · 6 (d) · 7 (a) · 8 (c) · 9 (d) · 10 (b) · 11 (c) · 12 (a) · 13 (b) · 14 (c) · 15 (d) · 16 (a) · 17 (a) · 18 (d) · 19 (b) · 20 (c)
Assertion–Reason (6)
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
- A: t½ of a first order reaction does not depend on [R]₀. R: t½ = 0.693/k.
- A: Molecularity of a reaction cannot be zero. R: At least one species must take part in an elementary step.
- A: A catalyst increases the rate. R: A catalyst increases ΔG of the reaction.
- A: Rate of reaction increases with temperature. R: More molecules have E ≥ Ea at higher temperature.
- A: Order can be found from the balanced equation. R: Order is an experimental quantity.
- A: Units of k and rate are the same for zero order. R: For zero order, Rate = k.
Key: 1 (a) · 2 (a) · 3 (c) · 4 (a) · 5 (d) · 6 (a)
Answers to the practice set
A. 1. s⁻¹ 2. rate-determining 3. [R]₀/2k 4. zero 5. activation energy 6. doubles 7. −Ea/R 8. first
B. 1. F 2. F 3. T 4. F 5. T 6. T 7. F (it stays constant until the reactant is used up) 8. T
C. 1. change in concentration per unit time 2. rate when all concentrations are unity 3. 2NH₃ → N₂ + 3H₂ on hot Pt 4. L mol⁻¹ s⁻¹ 5. the unstable high-energy intermediate at the top of the barrier 6. it has several steps, each with its own molecularity 7. k = Ae^(−Ea/RT) 8. none; K unchanged 9. a collision with E ≥ Ea and proper orientation 10. t½ = 0.693/k
D. 1. p. 4 table 2. 0.5/(2 × 0.02) = 12.5 s 3. 10 min 4. order 3, L² mol⁻² s⁻¹ 5. −½ d[N₂O₅]/dt = ¼ d[NO₂]/dt = d[O₂]/dt 6. p. 7 figure 7. the fraction with E ≥ Ea roughly doubles 8. p. 6
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