A-Level Reaction rates, rate laws and catalysis: worked solution

Question

Theobromine is cleared from the body more slowly than caffeine. Its elimination obeys first-order kinetics with a half-life of 8 hours, and the integrated first-order expression is ln[A]t = -kt + ln[A]0. Suppose an online streamer eats many oversized chocolate bars on air and a blood test taken straight afterwards shows 15.5 mg dm-3 of theobromine. Determine how long it takes for that level to drop below 1.5 mg dm-3.

First order, so k = ln2 / t(1/2) = 0.6931/8 = 0.0866 h⁻¹. Using ln[A]t = -kt + ln[A]0, rearrange: t = (1/k) ln([A]0/[A]t). t = (1/0.0866) x ln(15.5/1.5) = (1/0.0866) x ln(10.33) = (1/0.0866) x 2.335 = 27.0 h. So the level falls below 1.5 mg dm-3 after about 27 hours.