CBSE Class 12 Chemical Kinetics PYQ | Online Mock TestBy admin / January 12, 2025 Uncategorized 5/5 - (2 votes) 🧪Chemical Kinetics – PYQs MCQ Quiz 🔢 Number of Questions: 20 MCQs ⏱️ Time Required: 20 minutes 📚 Subject: Chemistry (Chemical Kinetics) 🎯 Difficulty Level: Standard PYQs (10+2 Level) 👨🎓 Ideal For: CBSE Class XII aspirants 10+2 Board/NCERT/17+Age Chemistry students Competitive exam learners 📌 Key Features: Concept-based questions Covers important periodic trends Helps in quick revision Useful for mock test practice 🚀 Outcome: After completing this quiz, you will be able to confidently understand and apply Chemical Kinetics in exams. 1 / 20 Identify the order of reaction from the following unit for its rate constant: Lmol⁻¹s⁻¹ [PYQ 2022] Zero order First order Second order Third order Explanation: Unit Lmol⁻¹s⁻¹ corresponds to second order reaction because rate constant unit for second order is concentration⁻¹ time⁻¹. 2 / 20 [CBSE PYQ 2020] In a chemical reaction X → Y, it is found that the rate of reaction doubles when the concentration of X is increased fourtimes. The order of the reaction with respect to X is (a) 1 (b) 0 (c) 2 (d) 1/2 Explanation: Rate ∝ [X]ⁿ, 2 = (4)ⁿ ⇒ n = 1/2. 3 / 20 A first order reaction is 25% complete in 40 minutes. In what time will the reaction be 80% completed? (a) 180 min (b) 200 min (c) 223 min (d) 250 min Explanation: For first order reaction, k = (2.303/t) log([A]₀/[A]). Using 25% completion in 40 min, k ≈ 7.2×10⁻³ min⁻¹. For 80% completion, t = (2.303/k) log(100/20) ≈ 223 min. 4 / 20 A bimolecular reaction follows first order kinetics under which condition? [CBSE 2020] (a) When both reactants are in equal concentration (b) When one reactant is present in large excess (c) When temperature is very low (d) When catalyst is absent Explanation: If one reactant is present in large excess, its concentration remains nearly constant, so the bimolecular reaction behaves like a pseudo-first order reaction. 5 / 20 A first order reaction is 50% complete in 30 minutes at 300 K and in 10 minutes at 320 K. Calculate activation energy (Ea) for the reaction. [CBSE 2020] (a) 52 kJ mol⁻¹ (b) 53 kJ mol⁻¹ (c) 54 kJ mol⁻¹ (d) 55 kJ mol⁻¹ 6 / 20 Write the two conditions for collisions to be effective collisions. [CBSE 2020] (a) Low energy and wrong orientation (b) Proper orientation and sufficient activation energy (c) High pressure and low temperature (d) Large volume and low concentration Explanation: Effective collisions occur only when molecules collide with proper orientation and energy equal to or greater than activation energy. 7 / 20 In the case of a complex reaction, which statement correctly explains the difference between order of reaction and molecularity? [CBSE 2020] (a) Both order and molecularity are always experimentally determined and may be fractional (b) Order of reaction is obtained from rate law and can be zero, Integers or fractional, while molecularity is theoretical and always a positive whole number (c) Molecularity depends on concentration of reactants, whereas order never depends on experiment (d) Molecularity can be zero for some reactions, but order can never be zero Explanation: Order of reaction is determined experimentally from the rate equation and may be zero, fractional, or integral. Molecularity refers to the number of molecules colliding in an elementary step, so it is always a fixed positive whole number and never zero or fractional. 8 / 20 The decomposition of NH₃ on platinum surface is a zero order reaction. If the rate constant (k) is 4 × 10⁻³ Ms⁻¹, how long will it take to reduce the initial concentration of NH₃ from 0.1 M to 0.064 M? [CBSE 2019] (a) 6 s (b) 9 s (c) 12 s (d) 15 s Explanation: For zero order reaction, [A] = [A]₀ − kt.t = ([A]₀ − [A])/k = (0.1 − 0.064)/(4 × 10⁻³) = 0.036/0.004 = 9 s. 9 / 20 For the reaction 2N₂O₅(g) → 4NO₂(g) + O₂(g), the rate of formation of NO₂(g) is 2.8 × 10⁻³ M s⁻¹. Calculate the rate of disappearance of N₂O₅(g). [CBSE 2018] (a) 1.4 × 10⁻³ M s⁻¹ (b) 2.8 × 10⁻³ M s⁻¹ (c) 5.6 × 10⁻³ M s⁻¹ (d) 7.0 × 10⁻³ M s⁻¹ Explanation: From stoichiometry,−(1/2)d[N₂O₅]/dt = (1/4)d[NO₂]/dtSo, rate of disappearance of N₂O₅ = (2/4) × 2.8 × 10⁻³ = 1.4 × 10⁻³ M s⁻¹. 10 / 20 A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. (Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK⁻¹ mol⁻¹) [CBSE 2019] (a) 52.8 kJ mol⁻¹ (b) 53.1 kJ mol⁻¹ (c) 54.2 kJ mol⁻¹ (d) 27.7 kJ mol⁻¹ 11 / 20 For a reaction R → P, the half-life (t₁/₂) is observed to be independent of the initial concentration of reactants. What is the order of the reaction? [CBSE 2017] (a) Zero order (b) First order (c) Second order (d) Third order Explanation: In a first order reaction, half-life is given by t₁/₂ = 0.693/k, which is independent of the initial concentration of reactant. 12 / 20 A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed. (Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021) [CBSE 2017] (a) 20 min (b) 30 min (c) 40 min (d) 60 min Explanation: For first order reaction,k = (2.303/t) log([A]₀/[A]) For 25% decomposition, 75% remains:k = (2.303/20) log(4/3) For 75% decomposition, 25% remains:t = (2.303/k) log(4) Using values, t = 40 min. 13 / 20 For the Arrhenius equation log k = 14.2 − (1.0 × 10⁴ / T), calculate the activation energy (Ea). If the half-life of the first order reaction is 200 minutes, calculate the rate constant (k). [CBSE 2016] (a) Ea = 1.91 × 10⁵ J mol⁻¹, k = 5.78 × 10⁻⁵ s⁻¹ (b) Ea = 1.91 × 10⁴ J mol⁻¹, k = 5.78 × 10⁻⁴ s⁻¹ (c) Ea = 2.30 × 10⁵ J mol⁻¹, k = 6.93 × 10⁻⁵ s⁻¹ (d) Ea = 1.50 × 10⁵ J mol⁻¹, k = 4.50 × 10⁻⁵ s⁻¹ Explanation: Comparing log k = log A − Ea/(2.303RT) with log k = 14.2 − (1.0 × 10⁴/T), Ea/(2.303 × 8.314) = 1.0 × 10⁴ Ea = 1.0 × 10⁴ × 2.303 × 8.314 ≈ 1.91 × 10⁵ J mol⁻¹. For first order reaction,k = 0.693/t₁/₂ t₁/₂ = 200 × 60 = 12000 s k = 0.693/12000 ≈ 5.78 × 10⁻⁵ s⁻¹. 14 / 20 The rate constant of a first-order reaction increases from 2 × 10⁻² to 4 × 10⁻² when the temperature changes from 300 K to 310 K. Calculate the activation energy (Ea). (Given: log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021) [CBSE 2015] (a) 52.8 kJ mol⁻¹ (b) 53.6 kJ mol⁻¹ (c) 54.8 kJ mol⁻¹ (d) 55.9 kJ mol⁻¹ Explanation: Using Arrhenius equation, log(k₂/k₁) = Ea/(2.303R) × (T₂−T₁)/(T₁T₂) Here, k₂/k₁ = (4 × 10⁻²)/(2 × 10⁻²) = 2 So, 0.301 = Ea/(2.303 × 8.314) × (10)/(300 × 310) Ea ≈ 5.36 × 10⁴ J mol⁻¹ = 53.6 kJ mol⁻¹. 15 / 20 A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction. (Given: log 2 = 0.3010) [CBSE 2015] (a) 60 min (b) 90 min (c) 100 min (d) 120 min Explanation: For first order reaction,t₁/₂ = 0.693/k Given t₁/₂ = 30 min,k = 0.693/30 For 90% completion, 10% reactant remains: t = (2.303/k) log(100/10)= (2.303/k) × 1≈ 100 min. 16 / 20 For a reaction A + B → P, the rate law is given by r = k [A]¹ᐟ² [B]². What is the order of this reaction? [CBSE 2014] (a) 1 (b) 2 (c) 2.5 (d) 3 Explanation: Order of reaction is the sum of powers of concentration terms in the rate law.Order = 1/2 + 2 = 2.5. 17 / 20 A first order reaction is found to have a rate constant k = 5.5 × 10⁻¹⁴ s⁻¹. Find the half-life of the reaction. [CBSE 2014] (a) 1.26 × 10¹³ s (b) 1.26 × 10¹² s (c) 1.50 × 10¹³ s (d) 2.00 × 10¹³ s Explanation: For a first order reaction,t₁/₂ = 0.693/k t₁/₂ = 0.693 / (5.5 × 10⁻¹⁴)≈ 1.26 × 10¹³ s. 18 / 20 The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (Ea) of the reaction assuming that it does not change with temperature. [Given: R = 8.314 J K⁻¹ mol⁻¹, log 4 = 0.6021] [CBSE 2014] (a) 52.8 kJ mol⁻¹ (b) 53.8 kJ mol⁻¹ (c) 54.9 kJ mol⁻¹ (d) 56.2 kJ mol⁻¹ Explanation: Using Arrhenius equation, log(k₂/k₁) = Ea/(2.303R) × (T₂−T₁)/(T₁T₂) Here,k₂/k₁ = 4,T₁ = 293 K,T₂ = 313 K 0.6021 = Ea/(2.303 × 8.314) × (20)/(293 × 313) Ea ≈ 5.28 × 10⁴ J mol⁻¹ = 52.8 kJ mol⁻¹. 19 / 20 Write two differences between 'order of reaction' and 'molecularity of reaction'. [CBSE 2014] (a) Order is determined experimentally, while molecularity is obtained from reaction mechanism (b) Order can be zero or fractional, while molecularity is always a whole number (c) Both (a) and (b) (d) None of these Explanation: Order of reaction is calculated from the rate law and may be zero, fractional, or integral. Molecularity is the number of reacting species in an elementary step and is always a fixed positive integer. 20 / 20 The following data were obtained during the first-order thermal decomposition of SO₂Cl₂ at constant volume:SO₂Cl₂(g) → SO₂(g) + Cl₂(g) At t = 0 s, total pressure = 0.4 atmAt t = 100 s, total pressure = 0.7 atm Calculate the rate constant.(Given: log 4 = 0.6021 and log 2 = 0.3010) [CBSE 2014] (a) 6.93 × 10⁻³ s⁻¹ (b) 13.86 × 10⁻³ s⁻¹ (c) 3.01 × 10⁻³ s⁻¹ (d) 2.30 × 10⁻³ s⁻¹ Your score isThe average score is 100% Restart quiz Download CBSE Class 12 Chemical Kinetics PYQ 2013-2022 – PDF Post Views: 39
[Bengali] Haloalkanes and Haloarenes Online Mock Test 20 MCQ-2 Uncategorized / By admin / May 24, 2024
