JEE Chemistry Chemical Kinetics MCQs Set 20

Answer: (b) Halved on reducing the concentration of alkyl halide to one half

Explanation:
1. The rate depends only on [RCl] (first order), so halving [RCl] halves the rate.
2. NaOH does not appear in the rate law, so changing [NaOH] has no effect. Raising the temperature increases k and hence the rate.

Teacher's Note:
1. The rate law is determined experimentally and need not match the stoichiometry.
2. Rate constants increase with temperature.

Answer: (b) Increase 4 times

Explanation:
1. For an elementary reaction the rate law follows the stoichiometry: rate = k[A]²[B].
2. Doubling [A] with [B] constant multiplies the rate by 2² = 4.

Teacher's Note:
1. Only for elementary reactions can the order be read from the balanced equation.
2. Active mass means molar concentration.

Answer: (d) 16

Explanation:
1. Reducing the volume to 1/4 increases each gas concentration by a factor of 4.
2. New rate = k(4[A])(4[B]) = 16 k[A][B], that is 16 times the original rate.

Teacher's Note:
1. Concentration of a gas is inversely proportional to volume at constant temperature and moles.
2. Multiply the factors for each concentration term.

Answer: (a) \( -\frac13\frac{d[A]}{dt}=\frac{-d[B]}{dt}=\frac{+d[C]}{dt}=\frac{+d[D]}{dt} \)

Explanation:
1. The rate of reaction is the rate of change of concentration of each species divided by its stoichiometric coefficient.
2. Reactants decrease (negative sign) and products increase (positive sign), so −(1/3)d[A]/dt = −d[B]/dt = d[C]/dt = d[D]/dt.

Teacher's Note:
1. Divide each rate by its stoichiometric coefficient to get the rate of reaction.
2. Use a negative sign for reactants.

Answer: (a) \( 5\times10^{-4}\ \text{mol L}^{-1}\text{s}^{-1} \)

Explanation:
1. Rate of appearance of B = Δ[B]/Δt = 5 × 10⁻³/10 = 5 × 10⁻⁴ mol L⁻¹ s⁻¹.
2. The stoichiometric coefficient is not needed for the rate of appearance of one species.

Teacher's Note:
1. Do not divide by the coefficient 4 when asked for the rate of appearance of B itself.
2. Rate of reaction would be 1/4 of this.

Answer: (c) Rate = k(Cx)^1.5(Cy)^−1(Cz)^0

Explanation:
1. Overall order is the sum of the exponents.
2. (a) 1 + 1 + 1 = 3. (b) 0.5 × 3 = 1.5. (c) 1.5 − 1 + 0 = 0.5. (d) 1 + n − 2 = n − 1, which is 0.5 only for a specific n.
3. Only (c) gives 0.5.

Teacher's Note:
1. Negative exponents are allowed in empirical rate laws; they correspond to inhibition.
2. Add all the exponents.

Answer: (d) 2 d[B]/dt = d[A]/dt

Explanation:
1. Rate r = −d[A]/dt = −(1/2)d[B]/dt = (1/3)d[C]/dt = d[D]/dt.
2. So d[C]/dt = 3r = −3d[A]/dt (a is true), d[C]/dt = 3d[D]/dt (b is true), and 3d[B]/dt = −6r = −2d[C]/dt (c is true).
3. But 2d[B]/dt = −4r, while d[A]/dt = −r, so (d) is false.

Teacher's Note:
1. Write the rate of reaction first, then each rate of change is its coefficient times r.
2. Signs: negative for reactants, positive for products.

Answer: (c) Not defined, 3/2

Explanation:
1. Order = 1 + 1/2 = 3/2.
2. The fractional order shows that the reaction occurs by a multistep (chain) mechanism, so the overall reaction has no single molecularity; molecularity is defined only for elementary steps and is always a whole number.

Teacher's Note:
1. Order can be fractional; molecularity cannot.
2. Molecularity applies only to an elementary step.

Answer: (d) 4.65

Explanation:
1. For first order reactions k = (2.303/t) log[a/(a − x)] with the same time t.
2. k₁ = (2.303/t) log 2 and k₂ = (2.303/t) log(100/4) = (2.303/t) log 25.
3. k₂/k₁ = log 25/log 2 = 1.398/0.301 = 4.65.

Teacher's Note:
1. The same time t cancels in the ratio.
2. 96% completed means 4% remains, so a/(a − x) = 100/4 = 25.

Answer: (c) \( 1.386\times10^{-2} \)

Explanation:
1. \( k=\frac{2.303}{t}\log\frac{a}{a-x}=\frac{2.303}{200}\log\frac{800}{50} \).
2. log 16 = 1.204, so k = 2.303 × 1.204/200 = 1.386 × 10⁻² s⁻¹.

Teacher's Note:
1. 800 to 50 is four half-lives (800 → 400 → 200 → 100 → 50), so t₁/₂ = 50 s and k = 0.693/50 = 0.01386.
2. Use natural or common logs consistently.

Answer: (b) 30 minutes

Explanation:
1. 0.08 → 0.04 → 0.02 → 0.01 takes 3 half-lives.
2. t = 3 × 10 = 30 minutes.

Teacher's Note:
1. Ratio 0.08/0.01 = 8 = 2³, so three half-lives.
2. For a first order reaction the half-life is independent of concentration.

Answer: (b) 4.0 g

Explanation:
1. Number of half-lives = 18/3 = 6.
2. Mass left = 256/2⁶ = 256/64 = 4.0 g.

Teacher's Note:
1. Radioactive decay follows first order kinetics.
2. N = N₀/2ⁿ with n the number of half-lives.

Answer: (a) CH₃COOCH₃ + H₂O → CH₃COOH + CH₃OH (H⁺ catalyst)

Explanation:
1. In acid-catalysed hydrolysis of an ester, water is present in large excess, so [H₂O] is practically constant and the rate = k[ester], which is first order, though two molecules take part.
2. Hydrolysis in base is really second order (OH⁻ is used up). The other two reactions are genuinely bimolecular or higher order.

Teacher's Note:
1. Pseudo first order: a bimolecular reaction in which one reactant is in large excess.
2. Inversion of cane sugar is another example.

Answer: (b) 2 atm

Explanation:
1. Initial pressures: NO₂ = 2 atm and F₂ = 1 atm.
2. They react exactly in the stoichiometric ratio 2 : 1 to give 2 atm of NO₂F.
3. Total pressure after completion = 2 atm.

Teacher's Note:
1. At constant T and V, pressure is proportional to the number of moles.
2. 3 moles of gas become 2 moles.

Answer: (b) 8

Explanation:
1. Let x mm of A₂ be consumed. Pressure of A₂ = 100 − x; B = x; C = x/2.
2. Total = 100 + x/2 = 120 gives x = 40 mm.
3. Rate of disappearance of A₂ = 40/5 = 8 mm min⁻¹.

Teacher's Note:
1. Use partial pressures in place of concentrations for gases at constant volume.
2. One mole of A₂ gives 1.5 mol of products, so the net increase is half the amount reacted.

Answer: (b) \( \frac{2.303}{1200}\log\frac73\ \text{s}^{-1} \)

Explanation:
1. Volume at infinite time V∞ = 70 mL measures the initial concentration; V∞ − Vₜ = 30 mL measures the amount left.
2. \( k=\frac{2.303}{t}\log\frac{V_\infty}{V_\infty-V_t}=\frac{2.303}{20\ \text{min}}\log\frac{70}{30} \).
3. With t = 20 min = 1200 s: k = (2.303/1200) log(7/3) s⁻¹.

Teacher's Note:
1. For gas evolution reactions use V∞ for the initial amount and (V∞ − Vₜ) for the amount remaining.
2. Note that the unit in option (b) is seconds, matching 1200 s.

Answer: (b) 76.83% B and 23.17% C

Explanation:
1. For parallel first order reactions the product ratio equals the ratio of rate constants: [B]/[C] = k₁/k₂.
2. Fraction of B = k₁/(k₁ + k₂) = 1.26/(1.26 + 0.38) = 0.7683.
3. So 76.83% B and 23.17% C.

Teacher's Note:
1. Total rate constant k = k₁ + k₂ for the disappearance of A.
2. Both paths share the same time, so the yield depends only on k.

Answer: (b) E_f < E_r

Explanation:
1. ΔH = E_f − E_r.
2. For an exothermic reaction ΔH is negative, so E_f < E_r.

Teacher's Note:
1. In an energy profile diagram, the products of an exothermic reaction lie below the reactants.
2. For an endothermic reaction E_f > E_r.

Answer: (a) \( \log\frac{K_2}{K_1}=\frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_1T_2}\right] \)

Explanation:
1. Arrhenius: log k = log A − Ea/(2.303RT).
2. Subtract the equations at T₁ and T₂: log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂) = (Ea/2.303R)(T₂ − T₁)/(T₁T₂).
3. Option (b) has the wrong sign. Option (c) is also wrong: the correct form is log(k₁/k₂) = (Ea/2.303R)(T₁ − T₂)/(T₁T₂), without the extra minus sign. So only (a) is correct.

Teacher's Note:
1. Rate constant increases with temperature, so log(k₂/k₁) must be positive when T₂ > T₁.
2. Check the sign: (T₂ − T₁) > 0 gives a positive value.

Answer: (d) r = k[A][B]

Explanation:
1. The rate of the overall reaction is determined by the slow step: A + B → C + D.
2. The slow step is elementary, so rate = k[A][B].

Teacher's Note:
1. The rate law is that of the rate determining (slowest) step.
2. The fast step does not appear in the rate law.