For a first order reaction A → B, the rate constant, k = 5.5 × 10⁻¹⁴ s⁻¹. The time required for 67% completion of reaction is x × 10⁻¹ times the half life of reaction. The value of x is _____________ (Nearest integer)
(Given : log 3 = 0.4771)

Solution

Chemistry ma’am · 4 steps

67% done means one-third is left, so the time is ln 3/k and k cancels against the half-life.

1. What is left. At 67% completion about one-third of A remains, so [A]₀/[A] = 3. The given log 3 confirms this reading.
2. Two first-order times. For a first-order reaction t = 2.303/k log[A]₀/[A], so t_(67%) = 2.303/k log 3. The half-life is t_(1/2) = 2.303/k log 2.

   t_(67%)/t_(1/2) = (log 3)/(log 2) = 0.4771/0.3010 = 1.585

3. Rate constant drops out. Both times carry the same factor 2.303/k, so the value 5.5 × 10⁻¹⁴ s⁻¹ is never needed.
4. So. t_(67%) = 15.85 × 10⁻¹ t_(1/2), so x = 15.85, which rounds to 16.

The common slip: 6. Taking 0.67 as the fraction remaining gives log(1/0.67)/log 2 = 0.578, so x ≈ 6. At 67% completion only 33% is left.

The idea to keep. For a first-order reaction, any two times are in the ratio of their log terms. The rate constant cancels, so do not let a strange value of k scare you.

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