A patient was given a 200 mg dose of a drug IV, and 100 mg was eliminated during the first 2 hours. If the drug follows first-order elimination kinetics, how much of the drug will remain 6 hours after its administration

Okay, let's tackle this question. The patient was given 200 mg IV, and 100 mg was eliminated in the first 2 hours. The question is about first-order elimination kinetics and how much remains after 6 hours.

First, I need to recall what first-order kinetics means. First-order means the elimination rate is proportional to the concentration. The half-life is constant, right? So if 100 mg is eliminated in 2 hours, that's half the dose. So the half-life here is 2 hours because 200 mg to 100 mg is one half-life.

Now, after 6 hours, how many half-lives have passed? 6 divided by 2 is 3 half-lives. Each half-life reduces the amount by half. Starting with 200 mg:

After 1st half-life (2h): 100 mg
After 2nd (4h): 50 mg
After 3rd (6h): 25 mg

So the answer should be 25 mg remaining. Let me check the options. The correct answer is D, 25 mg. The other options might be numbers like 50, 100, or 12.5 mg, which are common mistakes if someone miscalculates the half-lives or misunderstands the kinetics.

Wait, first-order kinetics uses the formula N = N0 * (1/2)^(t/t1/2). Plugging in the numbers: 200*(1/2)^(6/2) = 200*(1/2)^3 = 200*(1/8)=25. Yep, that's right. So the correct answer is 25 mg. The other options probably represent errors in calculating the number of half-lives or using zero-order kinetics by mistake.

**Core Concept**
First-order elimination kinetics describes drug elimination where the rate depends on the concentration. The half-life (t½) is constant and the drug concentration decreases exponentially over time.

**Why the Correct Answer is Right**
In first-order kinetics, the half-life is the time required to eliminate half the drug. Here, 200 mg → 100 mg in 2 hours defines a half-life of 2 hours. After 6 hours (3 half-lives), the remaining drug is:
200 mg → 100 mg (2h) → 50 mg (4h) → **25 mg (6h)**.
The formula **N = N₀ × (1/2)^(t/t½)** confirms this, yielding 200 × (1/2)³ = 25 mg.

**Why Each Wrong Option is Incorrect**
**Option A:** Likely represents zero-order kinetics (fixed amount eliminated per time), which contradicts the given first-order scenario.
**Option B:** May reflect miscalculating 2 half-lives instead of 3 (200 → 50 mg at 4h).
**Option C:** Could stem from doubling the half-life (e.g., assuming t½ = 6h instead of 2h).

**Clinical Pearl / High-Yield Fact**
First-order kinetics is the most common elimination pattern in clinical

✓ Correct Answer: D. 25 mg