CBSE Class 12 Chemistry Chemical Kinetics MCQs Set 14

Answer: (a) first order

Explanation:
1. The rate of radioactive decay is proportional to the number of nuclei present, so it follows first order kinetics.

Teacher's Note:
1. This is why every radioactive isotope has a constant half-life.

Answer: (c) 100 times

Explanation:
1. log \( \frac{k_2}{k_1} \) = \( \frac{\Delta E_a}{2.303RT} \) = \( \frac{4606}{2.303 \times 2 \times 500} \) = 2 (T = 227 + 273 = 500 K, R ≈ 2 cal K⁻¹ mol⁻¹)
2. \( \frac{k_2}{k_1} \) = 10² = 100

Teacher's Note:
1. Since E_a is in calories here, use R = 2 cal K⁻¹ mol⁻¹ (not 8.314) and convert °C to K.

Answer: (c) t_1/2 ∝ a⁰

Explanation:
1. For a first order reaction, t_1/2 = 0.693/K, which is independent of the initial concentration a, i.e. t_1/2 ∝ a⁰.
2. The unit of K for a first order reaction is s⁻¹, and molecularity cannot be decided from the order.

Teacher's Note:
1. Order (from experiment) tells you nothing directly about molecularity of a complex reaction.

Answer: (a) of any reaction

Explanation:
1. Increase in temperature increases the number of molecules having energy above the activation energy, so the rate of any reaction (exothermic or endothermic) increases.

Teacher's Note:
1. Do not confuse rate with equilibrium: temperature shifts equilibrium differently for exothermic and endothermic reactions, but it raises the rate of both.

Answer: (a) 2K₁ = K₂ = 4K₃

Explanation:
1. Rate = −\( \frac{1}{2} \)\( \frac{d[N_2O_5]}{dt} \) = \( \frac{1}{4} \)\( \frac{d[NO_2]}{dt} \) = \( \frac{d[O_2]}{dt} \)
2. \( \frac{K_1}{2} \) = \( \frac{K_2}{4} \) = K₃; multiplying by 4: 2K₁ = K₂ = 4K₃

Teacher's Note:
1. Divide each rate of change by its stoichiometric coefficient to get the single rate of reaction, then compare.

Answer: (b) order of reaction can not have fractional value

Explanation:
1. Order of reaction can be zero, a whole number or even fractional (e.g. decomposition of CH₃CHO has order 1.5). Statements (a), (c) and (d) are correct.

Teacher's Note:
1. It is molecularity, not order, that must always be a whole number.

Answer: (d) \( \frac{t_{87.5\%}}{t_{50\%}} \) = 3

Explanation:
1. For a first order reaction, 87.5% completion means the reactant is reduced to 1/8 = (1/2)³, i.e. three half-lives.
2. Hence t_87.5% = 3 t_50%; also t_75% = 2 t_50% and t_99.9% ≈ 10 t_50%.

Teacher's Note:
1. Count half-lives: 50% → 1, 75% → 2, 87.5% → 3, 93.75% → 4.

Answer: (c) A–Q, B–R, C–P

Explanation:
1. Unit of K = (mol L⁻¹)¹⁻ⁿ s⁻¹.
2. n = 0: mol L⁻¹ s⁻¹ (Q); n = 1: s⁻¹ (R); n = 2: L mol⁻¹ s⁻¹ (P).

Teacher's Note:
1. Use the general formula (mol L⁻¹)¹⁻ⁿ s⁻¹ rather than memorising units separately.

Answer: (d) 10

Explanation:
1. t_99.9% = \( \frac{2.303}{k} \) log \( \frac{100}{0.1} \) = \( \frac{2.303 \times 3}{k} \); t_50% = \( \frac{2.303 \times 0.3010}{k} \)
2. \( \frac{t_{99.9\%}}{t_{50\%}} \) = \( \frac{3}{0.3010} \) ≈ 10

Teacher's Note:
1. 99.9% completion takes about 10 half-lives – a quick check worth remembering.

Answer: (a) Rate = −\( \frac{d[NO_2]}{dt} \)

Explanation:
1. Rate of reaction = −\( \frac{d[NO_2]}{dt} \) = −\( \frac{d[CO]}{dt} \) = +\( \frac{d[NO]}{dt} \) = +\( \frac{d[CO_2]}{dt} \)
2. A reactant is used up, so its rate of disappearance carries a minus sign; only option (a) is written correctly.

Teacher's Note:
1. Minus sign for a reactant, plus sign for a product – this keeps the rate of reaction a positive quantity.

Answer: (b) The rate of disappearance of O₂ is 50 mol min⁻¹

Explanation:
1. −\( \frac{1}{2} \) \( \frac{d[SO_2]}{dt} \) = −\( \frac{d[O_2]}{dt} \) = \( \frac{1}{2} \) \( \frac{d[SO_3]}{dt} \)
2. Rate of disappearance of O₂ = \( \frac{1}{2} \) × 100 = 50 mol min⁻¹; rate of disappearance of SO₂ = 100 mol min⁻¹.
3. A catalyst speeds up both forward and backward reactions, and order cannot be found from the balanced equation, so (c) and (d) are wrong.

Teacher's Note:
1. Stoichiometric ratios apply to moles, not to masses – convert kg to moles before using the coefficients.

Answer: (a) 1 × 10⁻⁴ M s⁻¹

Explanation:
1. Rate = \( \frac{1}{2} \) \( \frac{d[NH_3]}{dt} \) = −\( \frac{1}{3} \) \( \frac{d[H_2]}{dt} \) = −\( \frac{d[N_2]}{dt} \)
2. ∴ −\( \frac{d[N_2]}{dt} \) = \( \frac{1}{2} \) \( \frac{d[NH_3]}{dt} \) = \( \frac{1}{2} \) × (2 × 10⁻⁴ M s⁻¹) = 10⁻⁴ M s⁻¹

Teacher's Note:
1. N₂ is consumed at half the rate at which NH₃ is formed, because 1 N₂ gives 2 NH₃.

Answer: (c) 2^{(n − m)}

Explanation:
1. Earlier rate = k aⁿ b^m
2. New rate = k (2a)ⁿ \( \left(\frac{b}{2}\right)^m \)
3. \( \frac{New\ rate}{Earlier\ rate} \) = \( \frac{2^n a^n b^m 2^{-m}}{a^n b^m} \) = 2ⁿ . 2^−m = 2^{n − m}

Teacher's Note:
1. Doubling multiplies the rate by 2^order; halving multiplies it by 2^−order.

Answer: (b) zero

Explanation:
1. The unit of k is (mol L⁻¹)¹⁻ⁿ s⁻¹. With mol L⁻¹ s⁻¹, 1 − n = 1, so n = 0.
2. Hence the reaction is of zero order.

Teacher's Note:
1. A rate constant with the same units as rate (mol L⁻¹ s⁻¹) always means zero order.

Answer: (c) NO₂Cl → NO₂ + Cl

Explanation:
1. The rate law contains only one molecule of NO₂Cl, so the slow (rate controlling) step must involve a single NO₂Cl molecule:
2. NO₂Cl → NO₂ + Cl (slow)
3. NO₂Cl + Cl → NO₂ + Cl₂ (fast)

Teacher's Note:
1. The rate law reflects the slowest step; species and powers in the rate law tell you which molecules take part in that step.

Answer: (a) Low probability of simultaneous collision of all the reacting species

Explanation:
1. Higher order (> 3) reactions are rare because there is low probability of molecules to come together simultaneously and collide.

Teacher's Note:
1. The chance of three or more particles colliding at the same instant with proper orientation is extremely small.

Answer: (c) first

Explanation:
1. t_1/2 ∝ \( \frac{1}{[A_0]^{n-1}} \). t_1/2 is independent of [A₀] when n = 1.

Teacher's Note:
1. Remember: zero order t_1/2 ∝ [A]₀, first order independent, second order ∝ 1/[A]₀.

Answer: (b) The rate of reaction remains the same at any concentration of A

Explanation:
1. For a zero order reaction, rate = k[A]⁰ = k, i.e., the rate does not depend on the concentration of the reactant A.

Teacher's Note:
1. Order is about the reactant concentrations; products B and C do not appear in this rate law at all.

Answer: (b) is halved by reducing concentration of RCl by one half

Explanation:
1. Rate = k[RCl] is first order in RCl and independent of [NaOH]; halving [RCl] halves the rate.
2. Rate always increases with rise in temperature, so (c) and (d) are wrong.

Teacher's Note:
1. A reactant that does not appear in the rate law (here NaOH) has no effect on the rate, however much it is changed.

Answer: (b) 3

Explanation:
1. The given reaction is a reaction of 1st order (unit of k is s⁻¹): Rate = k[N₂O₅]
2. [N₂O₅] = \( \frac{Rate}{k} \) = \( \frac{1.02\times10^{-4}}{3.4\times10^{-5}} \) = 3 mol L⁻¹

Teacher's Note:
1. The unit s⁻¹ of k tells you the order is one, so rate = k[N₂O₅] with power 1, not 2.

Answer: (b) 0.72 M

Explanation:
1. The units mol L⁻¹ s⁻¹ of the rate constant show that it is a reaction of zero order. For a zero order reaction, x = kt.
2. ∴ Amount of A reacted (x) = (0.6 × 10⁻³ mol L⁻¹ s⁻¹) (20 × 60 s) = 0.72 mol L⁻¹
3. [B] formed = [A] reacted = 0.72 mol L⁻¹

Teacher's Note:
1. Convert minutes to seconds before multiplying – the units of k are per second.

Answer: (c) 690 s

Explanation:
1. Amount left after n half lives = \( \frac{[A]_0}{2^n} \)
2. 0.04 = \( \frac{1.28}{2^n} \) or 2ⁿ = 32 = 2⁵ ∴ n = 5
3. Time taken = 5 half lives = 5 × 138 s = 690 s.

Teacher's Note:
1. Count half-lives by repeated halving: 1.28 → 0.64 → 0.32 → 0.16 → 0.08 → 0.04 is five steps.

Answer: (b) Graph (b)

Explanation:
1. For zero order reaction, k = \( \frac{1}{t} \) {[A]₀ − [A]}
2. or [A] = −kt + [A]₀. Thus, plot of [A] vs t is linear with −ve slope (= −k).

Teacher's Note:
1. Concentration of a reactant can only fall with time, so rising graphs (c) and (d) are ruled out at once; the straight falling line is zero order, the curve is first order.

Answer: (d) 1

Explanation:
1. With respect to P, as t_3/4 is twice that of t_1/2, therefore, order = 1.
2. With respect to Q, from the graph (straight line, [Q] falling linearly with time), order = 0.
3. Hence, overall order = 1 + 0 = 1.

Teacher's Note:
1. t_75% = 2t_50% is a quick test for first order; a linear fall of concentration with time means zero order.

Answer: (a) 40%

Explanation:
1. For a first order reaction, time taken for a definite fraction to complete is independent of initial concentration.
2. In other words, in same time, same % age of reaction will be completed, i.e., 40%.

Teacher's Note:
1. For first order, k = (2.303/t) log(a/(a − x)) depends only on the ratio a/(a − x), so the percentage completed in a given time is the same for any starting amount.

Answer: (a) 40 mins

Explanation:
1. t = \( \frac{2.303}{k} \) log \( \frac{a}{a-x} \) ∴ 20 = \( \frac{2.303}{k} \) log \( \frac{a}{a-0.10a} \) = \( \frac{2.303}{k} \) log \( \frac{100}{90} \) ...(i)
2. In 2nd case, t = \( \frac{2.303}{k} \) log \( \frac{a}{a-0.19a} \) = \( \frac{2.303}{k} \) log \( \frac{100}{81} \) = \( \frac{2.303}{k} \) log \( \left(\frac{10}{9}\right)^2 \) ...(ii)
3. Eqn. (ii)/Eqn. (i) gives \( \frac{t}{20} \) = 2 or t = 40 min.

Teacher's Note:
1. 19% completion means 81% left, which is (90%)² – i.e., two successive 10% stages, so twice the time.

Answer: (c) 1

Explanation:
1. Rate, r = k [A]^a [B]^b ...(i); 2r = k [2A]^a [B]^b ...(ii); 4r = k [4A]^a [4B]^b ...(iii)
2. Dividing eqn. (ii) by eqn. (i), 2 = 2^a ∴ a = 1
3. Dividing eqn. (iii) by eqn. (i), 4 = 4^a 4^b = 4 × 4^b = 4^b+1 ∴ b + 1 = 1 or b = 0
4. Hence, overall order = 1.

Teacher's Note:
1. The coefficient 2 of B in the equation does not decide its order – the experimental data show the rate is independent of [B].

Answer: (c) 2

Explanation:
1. t_1/2 ∝ \( \frac{1}{a^{n-1}} \); For n = 2, t_1/2 ∝ \( \frac{1}{a} \).
2. The graph of T_1/2 vs 1/a₀ is a straight line through the origin, so the order is 2.

Teacher's Note:
1. Look at what is on the x-axis: a straight line against 1/a₀ (not a₀) means t_1/2 ∝ 1/a₀, i.e., second order.

Answer: (b) 3

Explanation:
1. Initially, rate r = k [M]^α = ka^α ...(i)
2. On doubling the concentration of M, 8r = k (2a)^α ...(ii)
3. Dividing eqn. (ii) by eqn. (i), 8 = 2^α or 2^α = 2³ or α = 3

Teacher's Note:
1. Write the rate change as 2ⁿ: a factor of 8 = 2³ gives order 3.

Answer: (b) is doubled

Explanation:
1. For a zero order reaction, t_1/2 = \( \frac{[A]_0}{2k} \). Thus, if [A]₀ is doubled, t_1/2 is also doubled.

Teacher's Note:
1. For zero order, t_1/2 is directly proportional to [A]₀ – the opposite trend to second order.

Answer: (c) pseudo first order reaction

Explanation:
1. When B is in large excess, its concentration remains practically constant, so rate = k[A][B] = k′[A].
2. The reaction behaves as first order though it is actually second order; it is called a pseudo first order reaction.

Teacher's Note:
1. Pseudo first order: one reactant in large excess (often the solvent water), so only the other reactant’s concentration changes appreciably.

Answer: (a) \( \frac{2.303}{t} \) log \( \frac{100}{80} \)

Explanation:
1. Radioactive decay follows first order kinetics: k = \( \frac{2.303}{t} \) log \( \frac{[A]_0}{[A]} \)
2. [A]₀ = 100, [A] = 80 (80% of ¹⁴C left), so k = \( \frac{2.303}{t} \) log \( \frac{100}{80} \)
3. The same k is also given by the half-life: k = \( \frac{0.693}{5730} \) year⁻¹.

Teacher's Note:
1. Put the amount LEFT (80), not the amount decayed (20), in the denominator of the log term.

Answer: (d) A is false, but R is true.

Explanation:
1. Only effective collisions (proper orientation and energy equal to or greater than the threshold energy) lead to products, so A is false and R is true.

Teacher's Note:
1. If every collision were effective, most reactions would be over almost instantly – that is why only a small fraction of collisions count.

Answer: (c) A is true, but R is false.

Explanation:
1. A catalyst increases the rate by providing an alternative path of lower activation energy.
2. A catalyst does not change ΔG° of a reaction, so the Reason is false.

Teacher's Note:
1. A catalyst changes only the kinetics (rate), never the thermodynamics (ΔG, ΔH, K) of a reaction.

Answer: (c) A is true, but R is false.

Explanation:
1. With lower activation energy, a larger fraction of molecules can cross the energy barrier, so the reaction is faster.
2. A catalyst lowers the activation energy, so the Reason is false.

Teacher's Note:
1. Lowering E_a is exactly how a catalyst works, so any statement that a catalyst does not affect E_a is false.

Answer: (c) A is true, but R is false.

Explanation:
1. Rate law expression for the reaction NO₂ + CO → CO₂ + NO is: Rate = k [NO₂]².
2. The reaction is second order (in NO₂ alone), so the assertion is true but the reason is false.

Teacher's Note:
1. Rate laws are found by experiment; this reaction is not elementary, so its rate law does not follow the balanced equation.

Answer: (d) A is false, but R is true.

Explanation:
1. Order of reaction can be negative with respect to a product (e.g., in case of 2O₃ → 3O₂, order w.r.t. O₂ is −1).
2. The rate of reaction is always taken as positive. So A is false and R is true.

Teacher's Note:
1. A negative order means that species slows the reaction down as its concentration rises (rate = k[O₃]²[O₂]⁻¹).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. R is the correct explanation of A: order is found experimentally and is not tied to the coefficients, so it can be fractional, whereas molecularity is the number of molecules in an elementary step and must be a whole number.

Teacher's Note:
1. Molecularity counts particles colliding in one elementary step, so it can only be 1, 2 or 3.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. R is the correct explanation of A because for a zero order reaction, rate = k [conc]⁰ = k.

Teacher's Note:
1. If rate equals k, they must have the same units: mol L⁻¹ s⁻¹.

Answer: (c) A is true, but R is false.

Explanation:
1. Rate constant of any reaction is constant at constant temperature and does not change with time or concentration. However, Assertion is correct because
2. t = \( \frac{1}{k} \) {[A]₀ − [A]}
3. For 50%, t_50% = \( \frac{1}{k} \)\( \left\{[A]_0 - \frac{[A]_0}{2}\right\} \) = \( \frac{[A]_0}{2k} \) = 50 sec
4. For 75%, t_75% = \( \frac{1}{k} \)\( \left\{[A]_0 - \frac{[A]_0}{4}\right\} \) = \( \frac{[A]_0}{k} \) × \( \frac{3}{4} \) = \( \frac{[A]_0}{2k} \) × \( \frac{3}{2} \) = 75 sec

Teacher's Note:
1. The equation is fine as an expression for k; the wording “depends upon time” is what makes R false – k is constant at a given temperature.

Answer: (c) A is true, but R is false.

Explanation:
1. The rate of reaction is represented by k [RCl] [NaOH], i.e., it is a second order reaction.
2. Since the rate is first order in RCl, halving [RCl] halves the rate, so A is true; R is false.

Teacher's Note:
1. Being first order in one reactant does not make the whole reaction first order – add the orders of all reactants.