For this reaction:
\[K = \frac{2.303}{t} \log \frac{[A]_0}{[A]}\]
Given: \( k = 4.6 \times 10^{-2} \, \text{s}^{-1}, \, [A]_0 = 1 \, \text{M}, \, [A] = 0.1 \, \text{M} \)
\[4.6 \times 10^{-2} = \frac{2.303}{t} \log \frac{1}{0.1}\]
\[4.6 \times 10^{-2} = \frac{2.303}{t} \times 1\]
\[t = \frac{2.303}{4.6 \times 10^{-2}} \approx 50 \, \text{sec.}\]