UPSC Combined Geo-Scientist, Mains · 2025 · Mains Paper II
Q12(a). t₀.₇₅ = 3 t₀.₅ for a second-order reaction
Question
For a second-order reaction A → P, show that the time required for 75% of the reaction (t₀.₇₅) is equal to 3 × t₀.₅.
Answer in brief
From 1/[A] − 1/[A]₀ = kt, t₀.₅ = 1/k[A]₀ and t₀.₇₅ = 3/k[A]₀, so t₀.₇₅ = 3 t₀.₅.
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Coverssecond-order reactionhalf-lifeintegrated rate lawt0.75 = 3 t0.5chemical kinetics
All parts of Question 12
- 12(a) t₀.₇₅ = 3 t₀.₅ for a second-order reaction5 marks
- 12(b) Orders in sucrose and H⁺ from half-lives10 marks
- 12(c) Pressures for a given coverage on a Langmuir surface5 marks