More practice questions

4. Q4. Find the half-life of a first order reaction with k = (i) 200 s⁻¹, (ii) 2 min⁻¹, (iii) 4 years⁻¹.

   Show answerHide answer

   t_1/2 = 0.693/k: (i) 3.47 × 10⁻³ s, (ii) 0.347 min, (iii) 0.173 years (NCERT Exercise 3.13, page 86).

5. Q5. The half-life of ¹⁴C is 5730 years. Wood from an old artefact has only 80% of the ¹⁴C of a living tree. Estimate its age, assuming its initial isotope ratio matched the living tree, it has remained a closed carbon sample, and no calibration correction is applied.

   Show answerHide answer

   k = 0.693/5730 = 1.21 × 10⁻⁴ year⁻¹, and t = (2.303/k) log(100/80) = (2.303 × 0.0969)/(1.21 × 10⁻⁴) ≈ 1845 years (NCERT Exercise 3.14, page 86).

6. Q6. A first order reaction has k = 60 s⁻¹. How long does it take for the concentration to fall to 1/16 of its initial value?

   Show answerHide answer

   1/16 = (1/2)⁴, so four half-lives: 4 × 0.693/60 = 0.0462 s, or 4.62 × 10⁻² s (NCERT Exercise 3.16, page 87).

7. Q7. ⁹⁰Sr has a half-life of 28.1 years. If 1 μg is absorbed in a newborn baby’s bones and none is lost by metabolism, how much remains after 10 years and after 60 years?

   Show answerHide answer

   Remaining = 1 μg × (1/2)^t/28.1: after 10 years 0.781 μg, after 60 years 0.228 μg (NCERT Exercise 3.17, page 87).

8. Q8. Show that a first order reaction takes twice as long to go 99% to completion as to go 90%.

   Show answerHide answer

   t_99% = (2.303/k) log 100 = 2 × 2.303/k, and t_90% = (2.303/k) log 10 = 2.303/k, so the ratio is 2 (NCERT Exercise 3.18, page 87).

9. Q9. A first order reaction takes 40 min to go 30% to completion. Find its half-life.

   Show answerHide answer

   k = (2.303/40) log(100/70) = 8.92 × 10⁻³ min⁻¹, so t_1/2 = 0.693/k ≈ 77.7 min (NCERT Exercise 3.19, page 87).

10. Q10. Azoisopropane decomposes to hexane and nitrogen at 543 K, one gas molecule giving two. The total pressure is 35.0 mm Hg at 0 s, 54.0 mm Hg at 360 s and 63.0 mm Hg at 720 s. Find the rate constant.

    Show answerHide answer

    For A → B + C, p_A = 2p_i − p_t: 16.0 mm at 360 s and 7.0 mm at 720 s. k = (2.303/360) log(35.0/16.0) = 2.17 × 10⁻³ s⁻¹ and (2.303/720) log(35.0/7.0) = 2.24 × 10⁻³ s⁻¹; average 2.2 × 10⁻³ s⁻¹ (NCERT Exercise 3.20, page 87).

11. Q11. SO₂Cl₂(g) → SO₂(g) + Cl₂(g) is first order. At constant volume and temperature, starting with pure SO₂Cl₂, the total pressure is 0.5 atm at 0 s and 0.6 atm at 100 s. Find the rate when the total pressure is 0.65 atm.

    Show answerHide answer

    At 100 s, p(SO₂Cl₂) = 2 × 0.5 − 0.6 = 0.4 atm, so k = (2.303/100) log(0.5/0.4) = 2.23 × 10⁻³ s⁻¹. At 0.65 atm, p(SO₂Cl₂) = 1.0 − 0.65 = 0.35 atm, and rate = k × 0.35 = 7.8 × 10⁻⁴ atm s⁻¹ (NCERT Exercise 3.21, page 87). NCERT slip (page 87): NCERT heads the time column “Time/s⁻¹”; the times are in seconds, so it should read “Time/s”.

12. Q12. A → products is first order with k = 2.0 × 10⁻² s⁻¹ and [A]₀ = 1.0 mol L⁻¹. Find [A] after 100 s.

    Show answerHide answer

    [A] = [A]₀e^−kt = 1.0 × e^−2.0 = 0.135 mol L⁻¹ (NCERT Exercise 3.24, page 87).

13. Q13. Sucrose decomposes in acid into glucose and fructose by first order kinetics with t_1/2 = 3.00 hours. What fraction remains after 8 hours?

    Show answerHide answer

    (1/2)^8/3 = 0.157, about 16% (NCERT Exercise 3.25, page 87).

14. Q14. Ammonia decomposes on a platinum surface by zero order kinetics with k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹. Find the rates of production of N₂ and H₂.

    Show answerHide answer

    For 2NH₃ → N₂ + 3H₂, rate = k = d[N₂]/dt = ⅓ d[H₂]/dt: N₂ at 2.5 × 10⁻⁴ and H₂ at 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹ (NCERT Exercise 3.3, page 85). NCERT slip (page 85): NCERT prints k in mol⁻¹ L s⁻¹; a zero order rate constant has the units of rate, mol L⁻¹ s⁻¹.

15. Q15. For a pseudo first order reaction in water, [A] = 0.55, 0.31, 0.17 and 0.085 mol L⁻¹ at 0, 30, 60 and 90 s. Find the average rate between 30 s and 60 s, and the pseudo first order rate constant.

    Show answerHide answer

    Average rate = (0.31 − 0.17)/30 = 4.67 × 10⁻³ mol L⁻¹ s⁻¹. k = (2.303/t) log(0.55/[A]) gives 1.91, 1.96 and 2.08 × 10⁻² s⁻¹ at 30, 60 and 90 s; average 1.98 × 10⁻² s⁻¹ (NCERT Exercise 3.8, page 85, asks for the average rate; the rate constant is added here, reasoned).

16. Q16. For 2N₂O₅ → 4NO₂ + O₂ at 318 K, 10² × [N₂O₅]/mol L⁻¹ is 1.63, 1.36, 1.14, 0.93, 0.78, 0.64, 0.53, 0.43 and 0.35 at t = 0, 400, 800, … 3200 s. Find the order, k and the half-life.

    Show answerHide answer

    log[N₂O₅] against t is a straight line, so the reaction is first order: −d[N₂O₅]/dt = k[N₂O₅]. The slope is about −2.08 × 10⁻⁴ s⁻¹, so k ≈ 4.8 × 10⁻⁴ s⁻¹ and t_1/2 = 0.693/k ≈ 1.44 × 10³ s. The [N₂O₅] curve halves at about 1.5 × 10³ s, which agrees (NCERT Exercise 3.15, pages 86–87).