CBSE Class 12 Chemistry Chemical Kinetics MCQs Set 11

Answer: (b) bar min⁻¹, bar^−1/2 min⁻¹

Explanation:
1. In terms of pressure, Rate = k(p_CH3OCH3)^3/2
2. Units of rate = bar min⁻¹
3. Units of rate constant = \( \frac{\text{rate}}{(p_{CH_3OCH_3})^{3/2}} \) = \( \frac{\text{bar min}^{-1}}{\text{bar}^{3/2}} \) = bar^−1/2 min⁻¹

Teacher's Note:
1. Rate always has units of (pressure or concentration) per time; only the units of k change with order.
2. For a gas-phase rate in pressure terms, k has units bar¹⁻ⁿ time⁻¹; with n = 3/2 this gives bar^−1/2 min⁻¹.

Answer: (d) 32a

Explanation:
1. Rate = k[A]²[B]³ = a
2. When volume is reduced to one half then conc. of reactants will be doubled.
3. Rate = k[2A]²[2B]³ = 32 k[A]²[B]³ = 32a

Teacher's Note:
1. Halving the volume doubles every gaseous concentration, so the rate changes by 2 raised to the overall order (here 2 + 3 = 5).

Answer: (b) Rate of disappearance of Q = 2 × Rate of appearance of R

Explanation:
1. Rate = −d[P]/dt = −d[Q]/dt = \( \frac{1}{2} \) d[R]/dt = d[S]/dt
2. Rate of disappearance of Q = \( \frac{1}{2} \) × rate of appearance of R, so statement (b) is incorrect.

Teacher's Note:
1. Divide each rate by its stoichiometric coefficient to get the single rate of reaction; R is formed twice as fast as Q is used up.

Answer: (c) Order cannot be fractional.

Explanation:
1. Order of reaction can be zero, fractional or negative.

Teacher's Note:
1. Only molecularity must be a whole number; order is an experimental quantity and can be 0, 1/2, 3/2 or even negative.

Answer: (b) 1.969 min

Explanation:
1. \( \frac{dx}{dt} \) = k[X] (For a first order reaction)
2. 0.00352 = k × 0.01 ⇒ k = 0.352 min⁻¹
3. t_1/2 = \( \frac{0.693}{k} \) = \( \frac{0.693}{0.352} \) = 1.969 min

Teacher's Note:
1. Get k from rate = k[X] first; the half-life then depends only on k, not on the starting concentration.

Answer: (c) rate of reaction remains same at any concentration of X

Explanation:
1. Rate of a zero order reaction is independent of the concentration of reactants, so the rate stays the same at any concentration of X.

Teacher's Note:
1. Zero order refers to the reactant X: rate = k[X]⁰ = k. Do not confuse the 2 in 2X with the order.

Answer: (a) 34.07 s

Explanation:
1. t = \( \frac{2.303}{k} \) log \( \frac{a}{a-x} \) or t = \( \frac{2.303}{15\times10^{-3}} \) log \( \frac{5}{3} \) = 34.07 s

Teacher's Note:
1. Masses can be used directly in the log ratio because only the ratio of amounts matters in a first order equation.

Answer: (b) \( \left(\frac{2.303}{12}\log 8\right) \) min⁻¹

Explanation:
1. k = \( \frac{2.303}{t} \) log \( \frac{a}{a-x} \) (for first order)
2. k = \( \frac{2.303}{12} \) log \( \frac{1}{1/8} \) = \( \left(\frac{2.303}{12}\log 8\right) \) min⁻¹

Teacher's Note:
1. Keep the ratio as a/(a − x), initial over remaining; writing log(1/8) gives a negative k, which is a quick sign that the ratio is upside down.
2. Reducing to 1/8 means three half-lives, so 0.693/4 min⁻¹ is the same value.

Answer: (d) The unit of rate of reaction changes with order of reaction.

Explanation:
1. The unit of rate of reaction is mol L⁻¹ s⁻¹. It does not change with order.

Teacher's Note:
1. Only the units of the rate constant depend on order; the rate itself is always concentration per unit time.

Answer: (b) 0.0031 mol L⁻¹ s⁻¹

Explanation:
1. Rate = \( \frac{1}{4} \) \( \frac{d[NO_2]}{dt} \) = \( \frac{1}{4} \) × 0.0125 = 0.0031 mol L⁻¹ s⁻¹

Teacher's Note:
1. The rate of reaction is the rate of change of any species divided by its coefficient in the balanced equation.

Answer: (b) Figure (b)

Explanation:
1. In a reaction P → Q, concentration of reactant decreases as concentration of product increases during the course of a reaction.
2. In figure (b), [P] falls exponentially while [Q] rises exponentially and levels off.

Teacher's Note:
1. Check the labels as well as the shapes: in (d) the curves are right but [P] and [Q] are swapped.
2. An exponential rise of product always flattens out as the reactant is used up; it cannot keep curving upward as in (a).

Answer: (b) molecularity of a reaction

Explanation:
1. The number of molecules of the reactants taking part in a single step of the reaction tells about molecularity of the reaction.

Teacher's Note:
1. Molecularity is defined for an elementary step; order is found experimentally from the rate law.

Answer: (b) (p) - rate of forward reaction, (q) - rate of backward reaction, (r) - equilibrium

Explanation:
1. Rate of forward reaction decreases and rate of backward reaction increases with passage of time. At equilibrium both the rates become equal.

Teacher's Note:
1. The forward rate starts high because only reactants are present; the backward rate starts at zero because no products exist yet.

Answer: (b) Rate = −\( \frac{1}{2} \)\( \frac{d[N_2O_5]}{dt} \) = \( \frac{1}{4} \)\( \frac{d[NO_2]}{dt} \) = \( \frac{d[O_2]}{dt} \)

Explanation:
1. For 2N₂O₅ → 4NO₂ + O₂, the rate of reaction can be expressed as
2. −\( \frac{1}{2} \)\( \frac{d[N_2O_5]}{dt} \) = \( \frac{1}{4} \)\( \frac{d[NO_2]}{dt} \) = \( \frac{d[O_2]}{dt} \)

Teacher's Note:
1. Reactant terms carry a minus sign and every term is divided by its own coefficient: 2 for N₂O₅, 4 for NO₂, 1 for O₂.

Answer: (a) 5 × 10⁻³ M min⁻¹

Explanation:
1. Average rate = −\( \frac{1}{2} \) \( \frac{\Delta[R]}{\Delta t} \) = −\( \frac{1}{2} \) × \( \frac{0.4 - 0.5}{10} \)
2. = \( \frac{1}{2} \) × \( \frac{0.1}{10} \) = 5 × 10⁻³ M min⁻¹

Teacher's Note:
1. Do not forget the factor 1/2 for HI; the rate of disappearance of HI itself is 0.01 M min⁻¹, which is twice the rate of reaction.

Answer: (a) zero order reaction

Explanation:
1. For a zero order reaction, rate = k[A]⁰ = k
2. Units = mol L⁻¹ s⁻¹

Teacher's Note:
1. Use the general unit of k, (mol L⁻¹)¹⁻ⁿ s⁻¹: only n = 0 gives mol L⁻¹ s⁻¹, the same as rate.

Answer: (c) one of the reactants is present in excess

Explanation:
1. When one of the reactants is in excess the reaction behaves as a first order reaction.

Teacher's Note:
1. Acid hydrolysis of an ester and inversion of cane sugar are the standard examples: water is in such excess that its concentration hardly changes.

Answer: (c) 3.3 × 10⁻⁴ M min⁻¹

Explanation:
1. Average rate = −\( \frac{\Delta[R]}{\Delta t} \) = −\( \frac{([R]_2 - [R]_1)}{t_2 - t_1} \)
2. = −\( \frac{(0.04 - 0.05)}{30} \) = \( \frac{0.01}{30} \) = 3.3 × 10⁻⁴ M min⁻¹

Teacher's Note:
1. The minus sign in front of Δ[R] makes the rate positive, since [R] decreases.

Answer: (a) Zero, mol L⁻¹ s⁻¹

Explanation:
1. A straight line with negative slope for concentration vs time shows a zero order reaction.
2. For a zero order reaction, rate = k = \( \frac{dx}{dt} \); units of k = mol L⁻¹ s⁻¹

Teacher's Note:
1. Straight line for [R] vs t means zero order; straight line for ln[R] vs t means first order. Always read the y-axis label.

Answer: (d) CH₃CHO → CH₄ + CO

Explanation:
1. CH₃CHO → CH₄ + CO
2. Rate = k[CH₃CHO]^3/2

Teacher's Note:
1. Thermal decomposition of acetaldehyde (order 3/2) is the standard textbook example of a fractional order reaction.

Answer: (b) the concentration of the reactant present in small amount

Explanation:
1. If one of the reactant is present in excess, the reaction becomes independent of the concentration of that reactant hence it becomes pseudo first order reaction.
2. Thus, rate of pseudo first order reaction depends upon the concentration of the reactant present in small amount.

Teacher's Note:
1. The reactant in excess hardly changes in concentration, so its term gets absorbed into the rate constant.

Answer: (b) halved on reducing the concentration of A to half

Explanation:
1. The rate of reaction depends upon concentration of only A.
2. Since rate = k[A], halving [A] halves the rate.

Teacher's Note:
1. The rate law, not the balanced equation, decides which concentrations matter; B does not appear in it, so changing [B] has no effect.

Answer: (a) 3 × 10⁻¹² mol L⁻¹ s⁻¹

Explanation:
1. For the reaction, 2NO₂ → 2NO + O₂
2. −\( \frac{1}{2} \)\( \frac{d[NO_2]}{dt} \) = \( \frac{1}{2} \)\( \frac{d[NO]}{dt} \) = \( \frac{d[O_2]}{dt} \)
3. −\( \frac{d[NO_2]}{dt} \) = 6 × 10⁻¹² mol L⁻¹ s⁻¹
4. \( \frac{d[O_2]}{dt} \) = 3 × 10⁻¹² mol L⁻¹ s⁻¹

Teacher's Note:
1. One O₂ forms for every two NO₂ used, so O₂ appears at half the rate at which NO₂ disappears.

Answer: (a) 5.4 × 10⁻³ mol L⁻¹ s⁻¹

Explanation:
1. −\( \frac{1}{4} \)\( \frac{d[NH_3]}{dt} \) = +\( \frac{1}{6} \)\( \frac{d[H_2O]}{dt} \)
2. \( \frac{d[H_2O]}{dt} \) = \( \frac{6}{4} \) × 3.6 × 10⁻³ = 5.4 × 10⁻³ mol L⁻¹ s⁻¹

Teacher's Note:
1. Ratio of rates = ratio of coefficients: H₂O forms 6/4 = 1.5 times as fast as NH₃ is used.

Answer: (b) 0.7

Explanation:
1. Rate = k[N₂O₅] (first order as unit of rate constant is s⁻¹)
2. [N₂O₅] = \( \frac{\text{Rate}}{k} \) = \( \frac{1.4\times10^{-5}\ \text{mol L}^{-1}\text{s}^{-1}}{2\times10^{-5}\ \text{s}^{-1}} \) = 0.7 mol L⁻¹

Teacher's Note:
1. The unit of k tells you the order: s⁻¹ means first order, so rate = k[N₂O₅].

Answer: (c) keeps on decreasing with time

Explanation:
1. Rate of reaction ∝ conc. of reactants
2. As the reaction proceeds, concentration of the reactants decreases hence the rate also keeps on decreasing with time.

Teacher's Note:
1. This is why instantaneous rate is found from the slope of the tangent at a given time; the slope keeps getting smaller.

Answer: (a) the order and molecularity of the slowest step are equal to one

Explanation:
1. For a unimolecular reaction, both order and molecularity are one in rate determining step.

Teacher's Note:
1. Molecularity can never be zero and is a theoretical count, not an experimental result.

Answer: (a) 0.1 mol L⁻¹ min⁻¹

Explanation:
1. Rate = −\( \frac{1}{2} \) \( \frac{\Delta[X]}{\Delta t} \) = −\( \frac{1}{2} \) \( \frac{(2-3)}{5} \) = 0.1 mol L⁻¹ min⁻¹

Teacher's Note:
1. Rate of reaction is always positive; the minus sign in front cancels the negative change in [X].

Answer: (b) Rate = k[O₃]²[O₂]⁻¹

Explanation:
1. O₃ ⇌ O₂ + [O] (fast; forward rate constant k₁, backward k₋₁)
2. [O] + O₃ → 2O₂ (slow; rate constant k₂)
3. Rate of reaction is determined by slow step hence, Rate = k₂[O][O₃]
4. [O] is unstable intermediate so substitute the value of [O] in above equation.
5. Rate of forward reaction = k₁[O₃]; Rate of backward reaction = k₋₁[O₂][O]
6. At equilibrium, rate of forward reaction = rate of backward reaction: k₁[O₃] = k₋₁[O₂][O]
7. [O] = \( \frac{k_1[O_3]}{k_{-1}[O_2]} \)
8. Rate = k₂ \( \left(\frac{k_1[O_3]}{k_{-1}[O_2]}\right) \)[O₃]; Rate = \( \frac{k[O_3]^2}{[O_2]} \)

Teacher's Note:
1. An intermediate such as [O] must never appear in the final rate law; replace it using the fast pre-equilibrium.

Answer: (b) first order reaction

Explanation:
1. Radioactive disintegration is an example of first order reaction.
2. k = \( \frac{2.303}{t} \) log \( \frac{N_0}{N_t} \)

Teacher's Note:
1. Every radioactive isotope has a constant half-life whatever the amount present, which is the hallmark of first order kinetics.

Answer: (c) 1.386 × 10⁻² s⁻¹

Explanation:
1. For first order reaction, k = \( \frac{2.303}{t} \) log \( \frac{a}{a-x} \) = \( \frac{2.303}{200} \) log \( \frac{400}{25} \)
2. = \( \frac{2.303}{200} \) × 1.204 = 0.01386 s⁻¹ = 1.386 × 10⁻² s⁻¹

Teacher's Note:
1. 400 → 25 is a 16-fold fall, i.e., four half-lives in 200 s, so t_1/2 = 50 s and k = 0.693/50, a quick check.

Answer: (b) Rate = k

Explanation:
1. Rate = k[X]⁰[Y]⁰ or rate = k

Teacher's Note:
1. Zero overall order means every concentration term carries power zero, so the rate equals k.

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

Explanation:
1. Rate of reaction does not remain constant during the complete reaction because rate depends upon the concentration of reactants which decreases with time.

Teacher's Note:
1. Only a zero order reaction has a constant rate, and even that only until the reactant runs out.

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

Explanation:
1. The rate law equation cannot be written from stoichiometric equation; it is determined experimentally.

Teacher's Note:
1. Coefficients in the balanced equation tell you nothing reliable about the powers in the rate law, except for an elementary step.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. During hydrolysis of ester, water is always present in high concentration, thus there is very little change in its concentration and it practically remains constant. Thus, the order of reaction is 1 (pseudo first order reaction).
2. Two molecules react, so the molecularity is 2, but this does not explain why the order is 1.

Teacher's Note:
1. This is the classic pseudo first order case: molecularity 2, order 1, because water is in large excess.

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

Explanation:
1. Rate of any reaction is equal to the rate of its slowest step and here rate of given reaction = k[N₂O₅] because the slowest step has only N₂O₅ molecule involved.

Teacher's Note:
1. The rate law reflects the slow step, not the overall equation, which is why the coefficient 2 does not appear as a power.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both statements are true. t_1/2 = \( \frac{0.693}{k} \) contains no concentration term, and 75% completion (100 → 50 → 25) takes two half-lives.
2. The reason is a consequence of constant half-life, not the explanation of why it is independent of concentration.

Teacher's Note:
1. t_75% = 2t_1/2 and t_87.5% = 3t_1/2 hold only because first order half-life is constant.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both statements are true. The assertion defines the scope of chemical kinetics, while the reason lists some factors that affect rate; it does not explain the definition.

Teacher's Note:
1. Remember the factors affecting rate: nature and concentration of reactants, temperature, catalyst, surface area and radiation.

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

Explanation:
1. A bimolecular reaction may involve combination of two molecules or exchange of atoms or groups of atoms between the two reactant molecules.
2. H₂ + I₂ → 2HI involves two reactant molecules, so it is bimolecular.

Teacher's Note:
1. Molecularity is the count of reacting species in an elementary step: one H₂ + one I₂ = 2.

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

Explanation:
1. Hydrolysis of cane sugar is pseudo first order reaction. Since, water is always in excess, rate of reaction does not depend appreciably on its concentration, thus it is an example of pseudo unimolecular reactions.

Teacher's Note:
1. Rate = k[C₁₂H₂₂O₁₁][H₂O] becomes k′[C₁₂H₂₂O₁₁] because [H₂O] is practically constant.

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

Explanation:
1. For a gaseous reaction at constant T, concentration is directly proportional to the partial pressure of the species (p = cRT). Thus, the rate of reaction can be expressed in terms of partial pressure for gaseous reactants or products.
2. Partial pressure is proportional to, not equal to, concentration, so the reason is false.

Teacher's Note:
1. Watch for "equal" versus "proportional" in A-R statements; p = cRT, so p ∝ c only at constant temperature.