CBSE Class 12 Chemistry Chemical Kinetics MCQs Set 15

Answer: (c) 2 : 1

Explanation:
1. t_3/4 = \( \frac{2.303}{k} \) log \( \frac{a}{a/4} \) = \( \frac{2.303}{k} \) log 4 = 2 × \( \frac{2.303}{k} \) log 2 = 2t_1/2
2. Hence t_3/4 : t_1/2 = 2 : 1.

Teacher's Note:
1. 75% completion = two half-lives (100 → 50 → 25) for any first order reaction.

Answer: (c) 12.5 g

Explanation:
1. Number of half-lives in 15 days, n = \( \frac{15}{5} \) = 3
2. Amount left = \( \frac{100}{2^3} \) = 12.5 g

Teacher's Note:
1. Convert time to number of half-lives first, then halve that many times: 100 → 50 → 25 → 12.5 g.

Answer: (a) 3

Explanation:
1. t_1/8 is the time in which the reactant falls to 1/8 of its initial amount: t_1/8 = \( \frac{2.303}{k} \) log 8 = 3 × \( \frac{2.303}{k} \) log 2 = 3t_1/2
2. Hence t_1/8 : t_1/2 = 3.

Teacher's Note:
1. 1/8 = (1/2)³, so three half-lives are needed.

Answer: (d) 2 × 10⁻² sec⁻¹

Explanation:
1. k = \( \frac{0.693}{t_{1/2}} \) = \( \frac{0.693}{34.65\ s} \) = 0.02 s⁻¹ = 2 × 10⁻² s⁻¹

Teacher's Note:
1. 0.693 ÷ 34.65 is exactly 0.02 – such neat numbers are a good check that you used the right formula.

Answer: (c) The fraction of the reacting molecules possessing an energy equal to the activation energy is more

Explanation:
1. With rise in temperature, the fraction of molecules having energy equal to or greater than the activation energy increases (the Boltzmann factor e^−E_a/RT increases), so more collisions are effective.

Teacher's Note:
1. Temperature does not change E_a; it changes how many molecules can cross it.

Answer: (a) It changes the equilibrium constant

Explanation:
1. A catalyst speeds up the forward and backward reactions to the same extent; it does not change the equilibrium constant, it only helps the equilibrium to be attained quickly.

Teacher's Note:
1. A catalyst changes the rate (via a new path of lower E_a) but never ΔG, ΔH or K.

Answer: (d) W → Z, E_a = 80 kJ

Explanation:
1. log \( \frac{k_2}{k_1} \) = \( \frac{E_a}{2.303R} \)\( \left(\frac{T_2 - T_1}{T_1 T_2}\right) \); for the same rise in temperature, the ratio k₂/k₁ is larger when E_a is larger.
2. Hence the reaction with E_a = 80 kJ has the maximum temperature coefficient.

Teacher's Note:
1. Reactions with higher activation energy are more sensitive to temperature.

Answer: (d) Energy of activation as well as frequency factor

Explanation:
1. log k = log A − \( \frac{E_a}{2.303RT} \). The plot of log k vs 1/T is a straight line.
2. Slope = −\( \frac{E_a}{2.303R} \) gives E_a and the intercept = log A gives the frequency factor A.

Teacher's Note:
1. From one straight line you get two quantities: E_a from the slope and A from the intercept.

Answer: (a) ΔH = 0

Explanation:
1. ΔH = E_a(forward) − E_a(backward) = 0

Teacher's Note:
1. On an energy profile, ΔH is the gap between the two activation energies; equal barriers mean reactants and products are at the same energy.

Answer: (c) 32 times

Explanation:
1. A rise of 50°C = 5 steps of 10°. Rate increases 2⁵ = 32 times.

Teacher's Note:
1. The increase is multiplicative (2 × 2 × 2 × 2 × 2), not additive (2 × 5 = 10).

Answer: (b) 53.6 kJ mol⁻¹

Explanation:
1. log \( \frac{k_2}{k_1} \) = \( \frac{E_a}{2.303R} \)\( \left(\frac{T_2 - T_1}{T_1 T_2}\right) \)
2. log 2 = \( \frac{E_a}{2.303\times8.314} \) × \( \frac{310-300}{300\times310} \)
3. 0.3010 = \( \frac{E_a}{2.303\times8.314} \) × \( \frac{10}{300\times310} \) or E_a = 53.6 kJ mol⁻¹

Teacher's Note:
1. Use R = 8.314 J K⁻¹ mol⁻¹ and convert the answer from J to kJ at the end.

Answer: (c) more than ΔH

Explanation:
1. For an endothermic reaction, the products lie ΔH above the reactants and the activated complex lies still higher, so E_a > ΔH.

Teacher's Note:
1. Sketch the energy profile: the peak must be above the product level, so for an endothermic reaction E_a always exceeds ΔH.

Answer: (b) B + 3D → 4A + 2C

Explanation:
1. Minus signs are for reactants and positive signs for products. Dividing numbers are the coefficients.
2. Hence, B and D are the reactants and A and C are the products: B + 3D → 4A + 2C.

Teacher's Note:
1. Read the sign first (reactant or product) and the divisor second (coefficient).

Answer: (c) 1.25 × 10⁻² mol L⁻¹ s⁻¹ and 3.125 × 10⁻³ mol L⁻¹ s⁻¹

Explanation:
1. Rate = −\( \frac{d[N_2O_5]}{dt} \) = +\( \frac{1}{2} \) \( \frac{d[NO_2]}{dt} \) = 2 \( \frac{d[O_2]}{dt} \)
2. Given −\( \frac{d[N_2O_5]}{dt} \) = 6.25 × 10⁻³ mol L⁻¹ s⁻¹
3. Rate of formation of NO₂ = \( \frac{d[NO_2]}{dt} \) = −2 \( \frac{d[N_2O_5]}{dt} \) = 2 × 6.25 × 10⁻³ = 12.50 × 10⁻³ = 1.25 × 10⁻² mol L⁻¹ s⁻¹
4. Rate of formation of O₂ = \( \frac{d[O_2]}{dt} \) = −\( \frac{1}{2} \) \( \frac{d[N_2O_5]}{dt} \) = \( \frac{1}{2} \) × 6.25 × 10⁻³ = 3.125 × 10⁻³ mol L⁻¹ s⁻¹

Teacher's Note:
1. With a fractional coefficient (½ O₂), dividing by ½ means multiplying by 2 – a common place for slips.

Answer: (b) 0.25 times

Explanation:
1. CH₃COOH (1 M) + CH₃OH (2 M) → CH₃COOCH₃ + H₂O
2. Rate (r₁) = k [CH₃COOH] [CH₃OH] = k (1) (2) = 2k
3. When each solution is diluted with equal volume of water, concentration of each is halved. Now, rate r₂ = k \( \left(\frac{1}{2}\right)\left(\frac{2}{2}\right) \) = \( \frac{k}{2} \)
4. ∴ \( \frac{r_2}{r_1} \) = \( \frac{k/2}{2k} \) = \( \frac{1}{4} \) = 0.25 or r₂ = 0.25 r₁

Teacher's Note:
1. Halving both concentrations in a reaction that is first order in each reactant reduces the rate to (½)² = ¼.

Answer: (a) 1

Explanation:
1. Rate = kC_A^3/2 C_B^−1/2
2. Overall order = \( \frac{3}{2} \) + \( \left(-\frac{1}{2}\right) \) = 1

Teacher's Note:
1. Overall order is the algebraic sum of the powers, including negative ones.

Answer: (c) two

Explanation:
1. Based on the units of rate constant (L mol⁻¹ s⁻¹ = (mol L⁻¹)¹⁻ⁿ s⁻¹ with n = 2), order of reaction = 2.

Teacher's Note:
1. Units of k: zero order mol L⁻¹ s⁻¹, first order s⁻¹, second order L mol⁻¹ s⁻¹.

Answer: (b) increase to eight times of its initial value

Explanation:
1. Rate = k [NO]² [O₂]. Initially, rate = ka²b
2. If volume is reduced to half, concentrations are doubled so that new rate = k (2a)² (2b) = 8ka²b, i.e., 8 times.

Teacher's Note:
1. Halving the volume doubles every gaseous concentration; then apply 2^{overall order} = 2³.

Answer: (b) 2.8 × 10⁻⁴ s⁻¹

Explanation:
1. k = \( \frac{2.303}{1.386\times60\times60} \) log \( \frac{a}{a-0.75a} \)
2. = \( \frac{2.303}{1.386\times60\times60} \) log 4 = 2.8 × 10⁻⁴ s⁻¹

Teacher's Note:
1. The options are in s⁻¹, so convert 1.386 h into seconds (× 3600) before substituting.

Answer: (c) 7\( \frac{1}{2} \) hours

Explanation:
1. k = \( \frac{0.693}{45} \) min⁻¹ = \( \frac{2.303}{t_{99.9\%}} \) log \( \frac{a}{a-0.999a} \)
2. or t_99.9% = \( \frac{2.303\times45}{0.693} \) log 10³ = 448 min ≈ 7\( \frac{1}{2} \) hours.

Teacher's Note:
1. 99.9% completion of a first order reaction always takes about 10 half-lives (log 1000 = 3).

Answer: (c) 4.8

Explanation:
1. For 75% completion, t_75% = \( \frac{2.303}{k} \) log \( \frac{a}{a-0.75a} \) = \( \frac{2.303}{k} \) log 4
2. For 25% completion, t_25% = \( \frac{2.303}{k} \) log \( \frac{a}{a-0.25a} \) = \( \frac{2.303}{k} \) log \( \frac{4}{3} \)
3. ∴ \( \frac{t_{75\%}}{t_{25\%}} \) = \( \frac{\log 4}{\log (4/3)} \) = \( \frac{0.6021}{0.6021-0.4771} \) = \( \frac{0.6021}{0.125} \) = 4.8

Teacher's Note:
1. The temperature (27°C) is a distractor here – k cancels out in the ratio.

Answer: (a) \( \frac{\ln 2}{k} \)

Explanation:
1. Half-life of a first order reaction does not depend upon initial concentration. It is equal to ln 2/k.

Teacher's Note:
1. ln 2 = 0.693 (natural log); log 2 = 0.301 (base 10) – do not mix them up.

Answer: (a) 0

Explanation:
1. t_1/2 ∝ \( \frac{1}{[A_0]^{n-1}} \); when n = 0, t_1/2 ∝ [A₀].

Teacher's Note:
1. For zero order, t_1/2 = [A]₀/2k, which is directly proportional to [A]₀.

Answer: (b) \( \frac{1}{3} \)

Explanation:
1. (i) r = kaⁿ (ii) 2r = k(8a)ⁿ. Dividing (ii) by (i), 2 = 8ⁿ or 2³ⁿ = 2 or 3n = 1 or n = 1/3.

Teacher's Note:
1. Orders need not be whole numbers; write both sides as powers of the same base (here 2).

Answer: (d) 5

Explanation:
1. From expt. 1 and 2, [Y₀] is kept constant, [X₀] is doubled, rate becomes 4 times. Hence, rate ∝ [X₀]².
2. From expt. 1 and 3, [X₀] is kept constant, [Y₀] is doubled, rate becomes 8 times. Hence, rate ∝ [Y₀]³.
3. Hence, overall rate is: Rate = k [X₀]² [Y₀]³. ∴ Overall order = 2 + 3 = 5.

Teacher's Note:
1. Compare two experiments in which only one concentration changes; the factor by which the rate changes, written as 2ⁿ, gives that order.

Answer: (c) 10 kcal/mole

Explanation:
1. For A → B, E_a = 15 kcal/mol and the product B lies 5 kcal/mol above A (endothermic).
2. E_a (B → A) = E_a (A → B) − ΔH = 15 − 5 = 10 kcal/mol.

Teacher's Note:
1. For the reverse of an endothermic reaction, subtract ΔH from the forward activation energy; for the reverse of an exothermic reaction you would add it.

Answer: (b) 512

Explanation:
1. Increase in steps of 10° has been made 9 times. Hence, rate of reaction should increase 2⁹ times, i.e., 512 times.

Teacher's Note:
1. This assumes the rule of thumb that the rate doubles for every 10° rise; count the steps: (100 − 10)/10 = 9.

Answer: (d) k = Ae^−E_a/RT

Explanation:
1. Arrhenius equation is k = Ae^−E_a/RT, where A is the frequency (pre-exponential) factor.

Teacher's Note:
1. Option (a) misses the pre-exponential factor A, which carries the units of k.

Answer: (d) Catalyst provides an alternative path of lower activation energy.

Explanation:
1. Catalyst provides an alternative path of lower activation energy.

Teacher's Note:
1. A catalyst does not supply energy; it lowers the barrier, so more molecules already have enough energy to react.

Answer: (c) −k

Explanation:
1. For 1st order reaction, k = \( \frac{1}{t} \) ln \( \frac{[A]_0}{[A]} \)
2. or kt = ln [A]₀ − ln [A] or ln [A] = ln [A]₀ − kt
3. Hence, plot of ln [A] vs t will be linear with slope = −k.

Teacher's Note:
1. With ln the slope is −k; with log₁₀ it would be −k/2.303.

Answer: (d) \( \frac{[R_0]}{2k} \)

Explanation:
1. For a zero order reaction, [R] = [R]₀ − kt. At t_1/2, [R] = [R]₀/2.
2. ∴ t_1/2 = \( \frac{[R]_0}{2k} \)

Teacher's Note:
1. 0.693/k is for first order only; for zero order the half-life depends on [R]₀.

Answer: (b) −k

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

Teacher's Note:
1. Compare with y = mx + c: [R] = −kt + [R]₀ gives slope −k and intercept [R]₀.

Answer: (b) Zero order reaction: Units of k = mol L⁻¹ s⁻¹ :: Second order reaction: Units of k = L mol⁻¹ s⁻¹

Explanation:
1. (a) is not correct because for zero order reaction, t_1/2 = \( \frac{[A]_0}{2k} \).
2. (c) is not correct because acid catalysed hydrolysis of ethyl acetate is a first order reaction whereas base catalysed hydrolysis of ethyl acetate is a second order reaction.
3. (d) is not correct because 2HI → H₂ + I₂ is a reaction of second order.
4. Hence (b) is correct.

Teacher's Note:
1. Acid hydrolysis of an ester is pseudo first order (water in excess); with NaOH, both ester and OH⁻ concentrations change, so it is second order.

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

Explanation:
1. Greater the half-life period, slower is the reaction.
2. R is true: t_1/2 ∝ \( \frac{1}{[A]_0^{n-1}} \).

Teacher's Note:
1. A long half-life means the reactant takes longer to fall to half – i.e., a slow reaction.

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

Explanation:
1. The time taken for half of the substance to decay is the same, as radioactive decay is a reaction of 1st order.
2. So both times are equal and A is false; R is true.

Teacher's Note:
1. Half-life of a radioactive isotope does not depend on how much of it is present.

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

Explanation:
1. In the presence of catalyst, there is no change in the value of enthalpy change (ΔH).
2. A catalyst lowers the activation energy, so R is true and A is false.

Teacher's Note:
1. A catalyst lowers the hump, not the levels of reactants and products, so ΔH stays the same.

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

Explanation:
1. Only effective collisions, i.e., collisions in which molecules have energy equal to or more than the threshold energy and proper orientation, lead to reaction.
2. So the rate is decided by the number of effective collisions and not by the total number of collisions; R correctly explains A.

Teacher's Note:
1. Effective collision = sufficient energy + proper orientation; both conditions must be met.

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

Explanation:
1. Rate of reaction is always positive.
2. The minus sign with d[A]/dt only makes the rate positive because [A] decreases; so A is false and R is true.

Teacher's Note:
1. The minus sign is a convention to keep the rate positive, not a sign that the rate itself is negative.

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

Explanation:
1. If equilibrium constant of the reaction 2N₂ + 6H₂ ⇄ 4NH₃ is K, then equilibrium constant for the reaction N₂ + 3H₂ ⇄ 2NH₃ is √K.
2. So R is false, while A is true.

Teacher's Note:
1. Multiplying an equation by 2 squares its equilibrium constant; it does not double it.

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

Explanation:
1. Both A and R are correct and R is the correct explanation of A.

Teacher's Note:
1. Molecularity is written from the balanced equation of an elementary step, so it is always a whole number; order comes from experiment and can be any value.

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

Explanation:
1. For zero order, t_1/2 = \( \frac{[A]_0}{2k} \), so A is true.
2. Rate of a zero order reaction is independent of initial concentration, so R is false.

Teacher's Note:
1. For zero order, half-life changes with [A]₀ but the rate does not.

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

Explanation:
1. Both A and R are true and R is the correct explanation of A: the catalyst forms an intermediate complex with the reactants, which provides a new path of lower activation energy.

Teacher's Note:
1. The intermediate complex later breaks up to give products and the catalyst is regenerated unchanged.

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

Explanation:
1. Correct explanation: Arrhenius equation is k = Ae^−E_a/RT so that ln k = ln A − \( \frac{E_a}{RT} \).
2. Hence, plot of ln k vs \( \frac{1}{T} \) has slope = −\( \frac{E_a}{R} \). R is true but does not explain A.

Teacher's Note:
1. It is the logarithmic form of the Arrhenius equation (y = c + mx) that makes the plot linear, with slope −E_a/2.303R for log₁₀.