physical chemistry

Question

A person was found to be contaminated by an organic pollutant X, and the level of X in his urine was 1.98 mg/dm3 . Pollutant X is eliminated by the body via a first order reaction and the half-life; t1/2 is 6 days. Calculate the number of days it will take for the concentration of pollutant X in the person’s urine be reduced to 0.035 mg/dm3 .

Answer 1

First-order elimination, so the rate constant comes straight from the half-life:

k = ln2 / t_1/2 = 0.693 / 6 d = 0.1155 d⁻¹

Integrated first-order law:

ln([X]₀ / [X]) = kt  →  t = ln([X]₀ / [X]) / k

t = ln(1.98 / 0.035) / 0.1155 = ln(56.6) / 0.1155 = 4.036 / 0.1155

t ≈ 34.9 days (about 35 days)

Quick sanity check without a calculator: each half-life cuts the level in half, and 1.98 → 0.99 → 0.495 → 0.248 → 0.124 → 0.062 → 0.031 is six halvings, landing just past the target. So the answer has to be a little under 6 × 6 = 36 days. 34.9 fits.

Two things that trip people up here:

1. You never need the volume or the amount of urine. First-order kinetics depends only on the ratio of concentrations, so mg/dm³ cancels and you can plug the numbers in as given.

2. Watch the direction of the log. It is ln(initial/final), not ln(final/initial) — the latter gives −34.9, and a negative time is the usual sign you flipped it.

The same setup handles any "how long until it drops to X" problem — drug clearance, radioactive decay, pollutant elimination. Find k from the half-life first, then use the integrated law once.