# CORRECT OPTION?

The following information is given:

Half life = 150 s

Initial concentration = 0.1 M

Final concentration = 0.01 M

$Forafirstorderreactions:\phantom{\rule{0ex}{0ex}}k=\frac{0.693}{{t}_{{\displaystyle \frac{1}{2}}}}=\frac{0.693}{150}=0.00462{s}^{-1}\phantom{\rule{0ex}{0ex}}Now,\phantom{\rule{0ex}{0ex}}\mathrm{ln}A=-kt+\mathrm{ln}{A}_{o},A=finalconcentration,{A}_{o}=Initialconcentration\phantom{\rule{0ex}{0ex}}t=\frac{\mathrm{ln}{A}_{o}-\mathrm{ln}A}{k}=\frac{\mathrm{ln}0.1-\mathrm{ln}0.01}{0.0462}=498.395\approx 500$

Thus, it takes 500 seconds.

Thus, option (c) is the correct answer

Hope it helps.

Regards

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