VKC ScienceChemistry for written exams Telegram

UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2020 · Paper I

Q7(a)(ii). First order: t_(99.9%)≈10 t_(50%)

  • 8 marks
  • Answered
Question

Show that for a first order reaction, the time required for 99.9% completion is almost 10 times the time for 50% completion.

Answer in brief

t=(2.303)/(k)log([A]₀)/([A]): t_(99.9%)=(2.303×3)/(k) and t_(50%)=(2.303×0.301)/(k), so the ratio is 3/0.301=9.97≈10.

Key points

  • Integrated first-order law: t=(2.303)/(k)log([A]₀)/([A])
  • 99.9 % completion: [A]=0.001[A]₀: t_(99.9%)=(2.303)/(k)1000=(2.303×3)/(k)=(6.908)/(k)
  • 50 % completion: [A]=0.5[A]₀: t_(50%)=(2.303)/(k)2=(2.303×0.3010)/(k)=(0.693)/(k)
  • Ratio: t_(99.9%)t_(50%)=(1000)/(2)=(3)/(0.3010)=9.97≈10

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

  1. Integrated first-order law: t=(2.303)/(k)log([A]₀)/([A])
  2. 99.9 % completion: [A]=0.001[A]₀: t_(99.9%)=(2.303)/(k)1000=(2.303×3)/(k)=(6.908)/(k)
  3. 50 % completion: [A]=0.5[A]₀: t_(50%)=(2.303)/(k)2=(2.303×0.3010)/(k)=(0.693)/(k)
  4. Ratio: t_(99.9%)t_(50%)=(1000)/(2)=(3)/(0.3010)=9.97≈10
  5. Meaning: the ratio does not depend on k or on [A]₀; 1000≈2¹⁰, so 99.9 % completion takes about ten half-lives.

Application. in radioactivity and pharmacokinetics the same rule is used: after about ten half-lives less than 0.1 % of a nuclide or a first-order-eliminated drug remains.

Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram

Model answer to Q7(a)(ii), 2020 Paper I: solution page 35
Solution page 35 of 45
Read the whole 2020 solution2020 question paper

All parts of Question 7