The half-life for the thermal decomposition of acetone is $80 \mathrm{~s}$ and is independent of initial concentration of acetone. The time required for the reaction to go to $80 \%$ composition is (Given: $\log 2=0.30$ )

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Aicharya · Accepted Answer

The reaction follows first-order kinetics. Hence $k=0.693 / t_{1 / 2}=0.693 /(80 \mathrm{~s})$. For the reaction to be $80 \%$, $[\mathrm{A}]_{t} /[\mathrm{A}]_{0}=0.2 .$ Hence
$$
t=-\frac{2.303 \log \left\{[\mathrm{A}]_{t} /[\mathrm{A}]_{0}ight\}}{k}=-\frac{(2.303)\left(\log 2 \times 10^{-1}ight)}{(0.693 / 80 \mathrm{~s})}=-\frac{(2.303)(0.301-1)}{(0.693 / 80 \mathrm{~s})}=186.1 \mathrm{~s}
$$

The half-life for the thermal decomposition of acetone is $80 \mathrm{~s}$ and is independent of initial…

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