For a first order reaction, the ratio of the time for 75% completion of a reaction to the time for 50% completion is ____________. (Integer answer)

Solution

Chemistry ma’am · 4 steps

Going to 75% done is just two half-lives in a row, so the time is exactly double.

1. First order time. For a first order reaction, the time to bring the concentration down from [A]₀ to [A] is t = 1/k ln[A]₀/[A]. It depends only on the fraction left, not on the starting amount.
2. Fraction left. At 50% completion, half is left, so [A]₀/[A] = 2. At 75% completion, a quarter is left, so [A]₀/[A] = 4.

   t_(75%)/t_(50%) = (1/k ln 4)/(1/k ln 2) = (2 ln 2)/(ln 2) = 2

3. Check with half-lives. The first half-life takes 100% down to 50%. The second half-life takes 50% down to 25%, which is 75% completion. Two equal half-lives give twice the time.
4. So. The ratio is 2.

The common slip: 1.5. Dividing the percentages, 75/50, gives 1.5. Time is set by the fraction remaining through a logarithm, not by the percentage completed.

The idea to keep. In a first order reaction every half-life is the same length. Count how many halvings get you to the fraction left: 75% done is 2 half-lives, 87.5% done is 3.

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