A non-steroidal anti-inflammatory drug is metabolized with a rate constant of 3.25 day–1. What is the half-life for this metabolism reaction?

Question

BlackTom AI · Accepted Answer

The half-life for first-order processes is given by t1/2 = ln 2 / k, so apply that formula directly.

1) Relevant concepts
- A rate constant with units day−1 indicates first-order kinetics. For first-order decay/metabolism, t1/2 = ln(2)/k (ln(2) ≈ 0.693147).

2) Step-by-step reasoning
- Given k = 3.25 day−1 and the first-order half-life formula, substitute k into the formula.

3) Calculations (all intermediate steps)
- ln(2) = 0.693147...
- t1/2 = ln(2) / k = 0.693147 / 3.25
- To compute: 0.693147 × 4 / 13 (since 3.25 = 13/4) = 2.772588 / 13 ≈ 0.21328 day
- Rounding to three significant figures gives t1/2 ≈ 0.213 day.

4) Why this is correct
- The formula t1/2 = ln(2)/k is derived from integrating the first-order rate law; using the given first-order rate constant 3.25 day−1 yields the computed half-life. The numeric division shown above directly yields 0.213 day, matching the correct calculation.

5) Why the other options are incorrect
- a. 2.25 day would correspond to k = ln(2)/2.25 ≈ 0.308 day−1, not the given 3.25 day−1.
- b. 0.308 day is the half-life for k ≈ 2.25 day−1 (k = ln(2)/0.308 ≈ 2.25), so it mismatches the given k.
- d. 1.63 day corresponds to k = ln(2)/1.63 ≈ 0.425 day−1, again not 3.25 day−1.

6) Conclusion
- The correct half-life for k = 3.25 day−1 is approximately 0.213 day, which confirms choice c.

A non-steroidal anti-inflammatory drug is metabolized with a rate constant of 3.25 day–1. What is the half-life for this metabolism reaction?单项选择题

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