A first-order reaction has a rate constant of $k = 6.93\times 10^{-3}\,\text{s}^{-1}$ at a certain temperature. The half-life is closest to

1. First-order half-life: $t_{1/2} = 0.693/k$. 2. Plug in $k = 6.93\times 10^{-3}\,\text{s}^{-1}$. 3. $t_{1/2} = 0.693 / (6.93\times 10^{-3}) = 100\,\text{s}$. 4. Notice the clean cancellation: $0.693 / 6.93 = 0.1$, then $0.1 / 10^{-3} = 100$. 5. Distractor analysis: option B uses $k = 0.0139$ (twice as fast). Option A uses $k = 0.0693$. Option D uses $k = 3.47\times 10^{-3}$ (half as fast). _Source: NCERT Class 12 Chemistry Part 1, Ch 3, §3.3.2 (Example calculations of first-order half-life), p. 13–14._