A first order reaction is 50% completed in 1.26 × 10^14 s . How much…

The correct answer is Option (3) → infinite

For a first order reaction the concentration falls exponentially with time:

    \( \displaystyle [\mathrm{R}] = [\mathrm{R}]_0\,e^{-kt} \qquad \text{or} \qquad t = \frac{2.303}{k}\log\frac{[\mathrm{R}]_0}{[\mathrm{R}]} \)

In every half-life, 0.693/k, the amount left is halved, whatever amount there was. Here the half-life is 1.26 × 10¹⁴ s:

    \( \displaystyle 100\% \xrightarrow{t_{1/2}} 50\% \xrightarrow{t_{1/2}} 25\% \xrightarrow{t_{1/2}} 12.5\% \xrightarrow{t_{1/2}} \cdots \)

For 100% completion, [R] would have to become zero. Then \( [\mathrm{R}]_0/[\mathrm{R}] \) becomes infinitely large, and so does its logarithm and the time. A first order reaction therefore never reaches 100% completion in a finite time; the time needed is infinite.

NCERT (2026-27 edition):
Half-Life of a Reaction – “… 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.”