CBSE Class 12 Chemistry Chemical Kinetics MCQs Set 17

Answer: (b) The rate of a first order reaction remains constant as the reaction proceeds

Explanation:
1. For a first order reaction rate = k[A], and [A] falls with time, so the rate falls too.
2. The half-life t½ = 0.693/k does not involve concentration, and the units of k for first and second order reactions are s⁻¹ and mol⁻¹ L s⁻¹.
3. Only the rate of a zero order reaction stays constant.

Teacher's Note:
1. Only a zero order reaction has a constant rate throughout.
2. Learn the units of k for orders 0, 1 and 2.

Answer: (c) 4.8

Explanation:
1. t(75%) = (2.303/k) log 4 and t(25%) = (2.303/k) log (4/3).
2. Ratio = log 4 ÷ log (4/3) = 0.6021 ÷ 0.1249 = 4.8.

Teacher's Note:
1. Use t = (2.303/k) log (initial ÷ remaining).
2. The remaining fractions are 1/4 and 3/4.

Answer: (d) 10 times the half-life

Explanation:
1. t(99.9%) = (2.303/k) log 1000 = 6.909/k, while t½ = 0.693/k.
2. The ratio is 6.909 ÷ 0.693 ≈ 10.

Teacher's Note:
1. Compare the two times using the same k.
2. 1000 = 2¹⁰ roughly, so about ten half-lives are needed.

Answer: (a) 60 minutes

Explanation:
1. 0.08 → 0.04 → 0.02 → 0.01 takes three half-lives.
2. Time = 3 × 20 = 60 minutes.

Teacher's Note:
1. Count how many times the concentration must be halved.
2. Each halving takes the same time in a first order reaction.

Answer: (b) 399 minutes

Explanation:
1. k = 0.693/120 = 5.78 × 10⁻³ min⁻¹.
2. t = (2.303/k) log (100/10) = (2.303 ÷ 5.78 × 10⁻³) × 1 ≈ 399 minutes.

Teacher's Note:
1. For 90% completion the remaining fraction is 10%.
2. log 10 = 1 simplifies the calculation.

Answer: (d) 1.386 × 10⁻² s⁻¹

Explanation:
1. 0.40 → 0.025 is a 16-fold fall, so [A]₀/[A] = 16.
2. k = (2.303/200) log 16 = (2.303 ÷ 200) × 1.2041 = 1.386 × 10⁻² s⁻¹.

Teacher's Note:
1. Write the ratio [A]₀/[A] first.
2. log 16 = 4 log 2 = 1.204.

Answer: (a) 3.47 × 10⁻⁴ M min⁻¹

Explanation:
1. 0.1 M → 0.025 M is two half-lives, so t½ = 20 min and k = 0.693/20 = 3.47 × 10⁻² min⁻¹.
2. Rate = k[A] = 3.47 × 10⁻² × 0.01 = 3.47 × 10⁻⁴ M min⁻¹.

Teacher's Note:
1. Find k first; then use rate = k[A].
2. Two halvings take 40 minutes.

Answer: (c) 1118 s

Explanation:
1. t½ = 0.693/k = 0.693 ÷ (6.2 × 10⁻⁴) s.
2. t½ ≈ 1118 s (about 18.6 minutes).

Teacher's Note:
1. Use t½ = 0.693/k for first order only.
2. Check the power of ten when dividing.

Answer: (b) 100 minutes

Explanation:
1. In the first case 75 g of 100 g decomposes, which is 75%. So t(75%) = 100 minutes.
2. In the second case 150 g of 200 g is again 75%, and for a first order reaction the time for a given percentage does not depend on the initial amount. The time is 100 minutes.

Teacher's Note:
1. Convert both cases to percentages.
2. First order: the time for a fixed fraction is independent of the initial concentration.

Answer: (a) mol L⁻¹ h⁻¹

Explanation:
1. The same amount (10% of the initial amount) decomposes every hour, so the rate is constant.
2. A constant rate means a zero order reaction, and k then has the unit of rate, mol L⁻¹ h⁻¹.

Teacher's Note:
1. Constant rate means zero order.
2. Zero order: unit of k = unit of rate.

Answer: (d) 0.15 mol dm⁻³

Explanation:
1. For a zero order reaction [A] = [A]₀ − kt.
2. t = 30 min = 0.5 h, so [A]₀ = [A] + kt = 0.05 + 0.2 × 0.5 = 0.15 mol dm⁻³.

Teacher's Note:
1. Convert 30 minutes into hours so that the units agree with k.
2. [A]₀ = [A] + kt for zero order.

Answer: (c) 1

Explanation:
1. 10 g → 5 g takes 30 min, and 5 g → 2.5 g → 1.25 g takes two further periods of 30 min, which is 60 min in all (total 90 min).
2. The time for each halving is constant (30 min), so the reaction is of first order.

Teacher's Note:
1. A constant half-life means first order.
2. Check that each halving takes the same time.

Answer: (b) 0.63 g

Explanation:
1. 96 h ÷ 24 h = 4 half-lives.
2. Mass left = 10 ÷ 2⁴ = 10 ÷ 16 = 0.625 g ≈ 0.63 g.

Teacher's Note:
1. Number of half-lives = time ÷ t½.
2. Divide the initial mass by 2 for each half-life.

Answer: (d) 23.1 min

Explanation:
1. Rate = k[A], so k = 1.5 × 10⁻² ÷ 0.5 = 3.0 × 10⁻² min⁻¹.
2. t½ = 0.693/k = 0.693 ÷ 0.03 = 23.1 min.

Teacher's Note:
1. Get k from the rate law first.
2. Then use t½ = 0.693/k.

Answer: (a) 300 s

Explanation:
1. t½ = 0.693 ÷ 6.9 × 10⁻³ ≈ 100 s.
2. One-eighth means three half-lives, so t = 3 × 100 = 300 s.

Teacher's Note:
1. One-eighth is (½)³, so three half-lives.
2. t½ comes from t½ = 0.693/k.

Answer: (b) an increase in the number of molecules with energy equal to or greater than the activation energy

Explanation:
1. The collision frequency rises by only a few per cent for a 10 K rise.
2. The fraction of molecules with energy ≥ E_a (the factor e^−E_a/RT) increases greatly, so the rate roughly doubles.

Teacher's Note:
1. Collisions increase slightly; effective collisions increase a lot.
2. Temperature does not change E_a.

Answer: (d) Plot (d)

Explanation:
1. An endothermic reaction has products at a higher energy than the reactants: plots (c) and (d).
2. A high activation energy means a tall barrier above the reactants, which is plot (d); plot (c) has only a small barrier.

Teacher's Note:
1. First decide exothermic or endothermic from the product level.
2. Then compare the height of the peak above the reactants.

Answer: (d) (i) and (iii) both

Explanation:
1. For a first order reaction log (a − x) = log a − kt/2.303, so a plot of log (a − x) against t is a straight line with a negative slope: graph (i).
2. The half-life of a first order reaction is independent of the initial concentration a, so a plot of t½ against a is a horizontal line: graph (iii).
3. A plot of log (a − x) against 1/t is not a straight line for any simple order.

Teacher's Note:
1. Use the linear forms of the integrated equations.
2. Constant half-life means a horizontal t½ against a line.

Answer: (a) 100 kJ mol⁻¹

Explanation:
1. ΔH = E_a(forward) − E_a(reverse), and ΔH = −30 kJ mol⁻¹.
2. −30 = 70 − E_a(reverse), so E_a(reverse) = 100 kJ mol⁻¹.

Teacher's Note:
1. In an exothermic reaction the reverse reaction has the higher barrier.
2. Draw the energy profile if you are not sure of the signs.

Answer: (c) Plot (c)

Explanation:
1. Taking logarithms, ln k = ln A − E_a/RT, which has the form y = c + mx with y = ln k and x = 1/T.
2. So a plot of ln k against 1/T is a straight line of negative slope −E_a/R, which is plot (c).

Teacher's Note:
1. Compare the equation with y = mx + c.
2. The slope of the Arrhenius plot gives E_a/R.