Q.60 A first order reaction is 87.5% complete at the end of 30 minutes. The half-life of the
reaction is __________ minute(s).
Problem Solution
For a first-order reaction, the integrated rate law is t = (2.303/k) log₁₀([A]₀/[A]), where [A]₀ is initial concentration and [A] is concentration at time t.
87.5% completion means 12.5% remains, so [A]₀/[A] = 8. Thus, log₁₀(8) = 0.9031, and at t = 30 min, k·30 = 2.303 × 0.9031 = 2.0794.
The half-life is t₁/₂ = 0.693/k = (0.693 × 30)/2.0794 = 10 min (exact, as 2.0794 = 3 × 0.693).
Half-Life Method
First-order half-life is concentration-independent: one half-life reduces to 50% remaining, two to 25% (75% complete), three to 12.5% (87.5% complete).
30 min for three half-lives gives t₁/₂ = 10 min per half-life.
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First order reaction 87.5% complete 30 minutes half-life problems test core chemical kinetics understanding for CSIR NET Life Sciences exams. The half-life calculation reveals 10 minutes exactly through simple half-life counting or rate law integration.
Key Concepts
- Half-Life Definition: Time for reactant concentration to halve, t₁/₂ = 0.693/k for first-order reactions. Independent of initial concentration.[web:6][web:9]
- 87.5% Completion Insight: Remaining fraction 1/8 = (1/2)³, so exactly 3 half-lives.
Step-by-Step Derivation
- Note 87.5% complete means [A] = 0.125 [A]₀, ratio = 8.
- Time equation: 30 = (2.303/k) log(8), where log(8) = 3 log(2) = 0.9030.
- k = (2.303 × 0.9030)/30 = 0.0693 min⁻¹.
- t₁/₂ = 0.693/0.0693 = 10 min.
CSIR NET Exam Tips
- Recognize powers of 1/2: 50% (1), 75% (2), 87.5% (3), 93.75% (4) half-lives.
- Avoid zero/second-order confusion—first-order t₁/₂ is constant.
| Half-Lives | % Complete | Fraction Remaining | Time (at 10 min/half-life) |
|---|---|---|---|
| 1 | 50% | 1/2 | 10 min |
| 2 | 75% | 1/4 | 20 min |
| 3 | 87.5% | 1/8 | 30 min |
