A first-order reaction has a half-life of 10 min. How many of the fol…

The correct answer is Option (3) → 3

For a first-order reaction the half-life is constant: it does not depend on the initial concentration, and each half-life halves whatever is left. The time for any fraction comes from \( t = \frac{2.303}{k}\log\frac{[R]_0}{[R]} \).

Correct:

• k = 6.93 × 10⁻² min⁻¹ (A): \( k = \frac{0.693}{t_{1/2}} = \frac{0.693}{10\ \mathrm{min}} = 6.93 \times 10^{-2}\ \mathrm{min^{-1}} \)

• 99.9% complete in about 100 min (B): \( \frac{2.303}{k}\log 1000 = \frac{6.909}{0.0693} = 99.7 \) min, that is 10 half-lives

• 75% complete in 20 min (C): one-fourth is left after two half-lives

Not correct:

• 90% complete in 30 min (D) — one-tenth is left at \( t = \frac{2.303}{0.0693}\log 10 = 33.2 \) min, not 30 min.

• Half-life 20 min when [R]₀ is doubled (E) — the half-life stays 10 min; a half-life that doubles with [R]₀ belongs to a zero-order reaction.

Correct statements: (A), (B) and (C) = 3.

NCERT (2026-27 edition):
3.3.3 Half-Life of a Reaction – “t_1/2 = 0.693/k (3.17)” … “It can be seen that for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species.”

“For zero order reaction t_1/2 ∝ [R]₀. For first order reaction t_1/2 is independent of [R]₀.”

“Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (t_1/2) of the reaction. When reaction is completed 99.9%, [R]_n = [R]₀ – 0.999[R]₀”

    \( \displaystyle t = \frac{2.303}{k}\log 10^{3} = 6.909/k, \qquad t_{1/2} = 0.693/k, \qquad \frac{t}{t_{1/2}} = \frac{6.909}{k} \times \frac{k}{0.693} = 10 \)

3.3.2 First Order Reactions – “The first order rate equation (3.10) can also be written in the form”

    \( \displaystyle k = \frac{2.303}{t}\log\frac{[\mathrm{R}]_0}{[\mathrm{R}]} \qquad (3.15) \)