The decomposition of a substance R takes place according to Ist order…

The correct answer is Option (3) → (ln2)/8 s⁻¹

For a first order reaction, the half-life does not depend on the starting concentration:

    \( \displaystyle t_{1/2}=\frac{0.693}{k}=\frac{\ln 2}{k} \)

So in every half-life the concentration is halved, whatever it was at the start. After n half-lives the fraction left is (1/2)ⁿ.

Here the concentration falls to 1/8 of its initial value:

    \( \displaystyle \frac{1}{8}=\left(\frac{1}{2}\right)^3\ \Rightarrow\ 3\ \text{half-lives} \)

    \( \displaystyle 3\,t_{1/2}=24\ \mathrm{s}\ \Rightarrow\ t_{1/2}=8\ \mathrm{s} \)

    \( \displaystyle k=\frac{\ln 2}{t_{1/2}}=\frac{\ln 2}{8}\ \mathrm{s^{-1}} \)

The integrated rate equation gives the same value: k = (2.303/24) log 8 = (2.303 × 0.903)/24 = 0.0866 s⁻¹ = 0.693/8 s⁻¹.

So the rate constant is (ln 2)/8 s⁻¹.