Time required for completion of 99.9 % of a First order reaction is ________ times of half life (t_(1 / 2)) of the reaction.

Solution

Chemistry ma’am · 4 steps

Only 0.1% is left, that is 1/1000, and halving ten times gives almost exactly that.

1. Fraction left. After 99.9% completion, 0.1% of the reactant remains, so [A]₀/[A] = 100/0.1 = 1000.

   t_(99.9%) = 2.303/k log 1000 = (2.303 × 3)/k = 6.909/k

2. Divide by half-life. For a first order reaction t_(1/2) = 0.693/k. The k cancels, so t_(99.9%)/t_(1/2) = 6.909/0.693 = 9.97.
3. Quick check. Each half-life halves what is left. After 10 half-lives the fraction left is 1/2¹⁰ = 1/1024, which is almost exactly 1/1000.
4. So. t_(99.9%) ≈ 10 t_(1/2), so the answer is 10.

The common slip: 7. Treating it as 99% completion uses log 100 = 2 and gives a ratio of about 6.6, which rounds to 7. Here 99.9% means log 1000 = 3.

The idea to keep. For first order, the time to reach any fraction left depends only on that fraction, never on the starting amount. Count half-lives: 1/1000 ≈ 1/2¹⁰, so about 10 half-lives.

Answer from the NTA final answer key. Solution by PrepWiser.