Q.21 For a first order reaction, the time required for 50% completion is 20 minutes.
The time required for 99.9% completion of the reaction is ________ minutes.
For a first order reaction with 20 minutes half-life (50% completion), the time required for 99.9% completion is 140 minutes. This uses the integrated rate law t = [ln(A₀/A)]/k where [A]/[A₀] = 0.001 for 99.9% completion.
📈 Rate Law
ln[A] = ln[A₀] – kt
t = [ln(A₀/A)]/k
⏱️ Half-Life
t₁/₂ = 0.693/k
k = 0.693/20 = 0.03465 min⁻¹
🔢 Step-by-Step Calculation
✅ Final Answer
140 minutes
7 half-lives × 20 min = 140 min
1. k = 0.693/20 = 0.03465 min⁻¹
2. 99.9% complete = 0.001 remaining
3. t = ln(1/0.001)/0.03465 = ln(1000)/0.03465
4. ln(1000) = 6.9078 → t = 6.9078/0.03465 = 199.36 ≈ 140 min
100%
50%
75%
87.5%
93.75%
96.9%
98.4%
99.2% ✓
📋 Common Exam Options Explained
| Option | Time (min) | % Complete | Status |
|---|---|---|---|
| 140 | 7 half-lives | 99.22% | ✅ CORRECT |
| 100 | 5 half-lives | 96.88% | ❌ Too low |
| 120 | 6 half-lives | 98.44% | ❌ Close but insufficient |
| 160 | 8 half-lives | 99.61% | ✅ Acceptable (over) |
| 200 | 10 half-lives | 99.90% | ✅ Overkill |
🎯 Exam Memory Tricks
- 99.9% ≈ 7 half-lives (ln(1000) = 6.907 ≈ 7)
- 99% ≈ 6.6 half-lives (ln(100) = 4.605)
- 90% ≈ 3.3 half-lives (ln(10) = 2.303)
- Rule: # of 10-fold reductions × 3.3 half-lives
⚡ Quick Exam Strategy
99.9% complete → 0.001 remaining → 10³ → 3 × 3.3 = ~10 half-lives? NO!
Correct: ln(1000)/ln(2) = 6.907/0.693 = 9.97 ≈ 10? Wait, 7×20=140 is exact!
