The decomposition of benzene diazonium chlori | vedclass

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Question

(A) For a first order reaction,$k = \frac{2.303}{t} \log \frac{a}{a-x}$.For $t_{1/4}$,$x = \frac{a}{4}$,so $t_{1/4} = \frac{2.303}{k} \log \frac{a}{3a

Vedclass · Accepted Answer

$272$

Vedclass · Answer

(A) For a first order reaction,$k = \frac{2.303}{t} \log \frac{a}{a-x}$.For $t_{1/4}$,$x = \frac{a}{4}$,so $t_{1/4} = \frac{2.303}{k} \log \frac{a}{3a/4} = \frac{2.303}{k} \log \frac{4}{3}$.For $t_{1/10}$,$x = \frac{a}{10}$,so $t_{1/10} = \frac{2.303}{k} \log \frac{a}{9a/10} = \frac{2.303}{k} \log \frac{10}{9}$.Taking the ratio: $\frac{t_{1/4}}{t_{1/10}} = \frac{\log(4/3)}{\log(10/9)} = \frac{\log 4 - \log 3}{\log 10 - \log 9} = \frac{2 \log 2 - \log 3}{1 - 2 \log 3}$.Using $\log 2 = 0.3$ and $\log 3 = 0.477$: $\frac{t_{1/4}}{t_{1/10}} = \frac{2(0.3) - 0.477}{1 - 2(0.477)} = \frac{0.6 - 0.477}{1 - 0.954} = \frac{0.123}{0.046} \approx 2.67$.Thus,$\frac{t_{1/4}}{t_{1/10}} 	imes 100 \approx 267$. The closest option is $272$.

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