A drug undergoes first-order elimination with a half-life of 6 hours. What percentage of the drug will be eliminated after 18 hours?

Explanation

## First-Order Elimination and Half-Life **Key Point:** In first-order kinetics, the drug concentration is reduced by **50% every half-life**. ### Calculation: - t½ = 6 hours - Total time = 18 hours - Number of half-lives = 18 ÷ 6 = **3 half-lives** ### Remaining Drug After Each Half-Life: | Time (hours) | Half-Lives | % Remaining | % Eliminated | |---|---|---|---| | 0 | 0 | 100% | 0% | | 6 | 1 | 50% | 50% | | 12 | 2 | 25% | 75% | | 18 | 3 | 12.5% | **87.5%** | **Formula:** % Remaining = (1/2)^n × 100, where n = number of half-lives - After 3 half-lives: (1/2)³ = 1/8 = 12.5% remaining - Therefore: 100% − 12.5% = **87.5% eliminated** **High-Yield Mnemonic:** "**3 half-lives = 7/8 gone**" (87.5% eliminated)