For the first order reaction A → B, the half life is 30 min. The time taken for 75% completion of the reaction is _________ min. (Nearest integer)
Given : log 2 = 0.3010
log 3 = 0.4771
log 5 = 0.6989

Solution

Chemistry ma’am · 3 steps

75% done means 25% left, which is exactly two half-lives for a first order reaction.

1. Count half-lives. In a first order reaction every half-life cuts what remains to half, whatever the starting amount. After one half-life 50% of A is left, and after two half-lives 25% is left.
2. Check with the formula. The integrated rate law gives t = 2.303/k log[A]₀/[A] with k = 0.693/t_(1/2). For 75% completion, [A]₀/[A] = 100/25 = 4.

   t_(75%)/t_(1/2) = (log 4)/(log 2) = (2 × 0.3010)/0.3010 = 2

3. So. t_(75%) = 2 × 30 min = 60 min, so the answer is 60.

The common slip: 12. Using 75% as the amount left gives [A]₀/[A] = 4/3, so t = 30 × log(4/3)/(log 2) ≈ 12 min, not 60.

The idea to keep. For first order, the fraction left after n half-lives is (1/2)ⁿ. Read 75% done as 25% left, and you get two half-lives straight away.

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