Time required for $90 \%$ completion of a first order reaction is ' $x$ ' minute. Calculate the time required to complete $99.9 \%$ of the reaction at same temperature.

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Aicharya · Accepted Answer

$\begin{aligned} \mathrm{t}_{90 \%} & =\frac{2.303}{\mathrm{k}} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_t}=\frac{2.303}{\mathrm{k}} \log _{10} \frac{100}{10} \ & =\frac{2.303}{\mathrm{k}} \log _{10} 10 \ \mathrm{t}_{99.9 \%} & =\frac{2.303}{\mathrm{k}} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}=\frac{2.303}{\mathrm{k}} \log _{10} \frac{100}{0.1} \ & =\frac{2.303}{\mathrm{k}} \log _{10} 1000 \ \frac{\mathrm{t}_{99.9 \%}}{\mathrm{t}_{90 \%}} & =\frac{\frac{2.303}{\mathrm{k}} \log _{10} 1000}{\frac{2.303}{\mathrm{k}} \log _{10} 10}=\frac{\log _{10} 1000}{\log _{10} 10}=\frac{3}{1} \ 	herefore \quad \mathrm{t}_{99.9 \%} & =3 	imes \mathrm{t}_{90 \%}=3 x 	ext { minute } \quad 	ext { (since, } \mathrm{t}_{90 \%}=x 	ext { minute) }\end{aligned}$

Time required for $90 \%$ completion of a first order reaction is ' $x$ ' minute. Calculate the time…

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