t_(87.5) is the time required for the reaction to undergo 87.5 % completion and t₅₀ is the time required for the reaction to undergo 50 % completion. The relation between t_(87.5) and t₅₀ for a first order reaction is t_(87.5)=x × t₅₀ The value of x is ___________. (Nearest integer)

Solution

Chemistry ma’am · 3 steps

87.5% done means 12.5% left, which is 1/8, so three half-lives have passed.

1. What is left. At 87.5% completion, the fraction of reactant remaining is 1 - 0.875 = 0.125 = 1/8.
2. Count half-lives. For a first order reaction the half-life does not depend on concentration, so every t₅₀ halves whatever is present: 1 → 1/2 → 1/4 → 1/8. That is three halvings.

   t_(87.5) = 2.303/k log 8 = 3 × 2.303/k log 2 = 3 t₅₀

3. So. t_(87.5) = 3 × t₅₀, so x = 3.

The common slip: 0. Using 87.5% as the fraction left instead of the fraction reacted gives log(1/0.875)/log 2 ≈ 0.19, which rounds to 0, not 3.

The idea to keep. For first order, turn the fraction left into a power of 1/2. If (1/2)ⁿ is left, then exactly n half-lives have passed.

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