An organic compound undergoes first order dec | vedclass

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(B) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $t_{1/8}$,the remaining concentrat

Vedclass · Accepted Answer

$0.9$

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(B) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $t_{1/8}$,the remaining concentration is $\frac{1}{8}$ of the initial concentration $[A]_0$,so $[A]_t = \frac{[A]_0}{8}$.$k t_{1/8} = 2.303 \log \frac{[A]_0}{[A]_0/8} = 2.303 \log 8 = 2.303 	imes 3 \log 2$.For $t_{1/10}$,the remaining concentration is $\frac{1}{10}$ of the initial concentration $[A]_0$,so $[A]_t = \frac{[A]_0}{10}$.$k t_{1/10} = 2.303 \log \frac{[A]_0}{[A]_0/10} = 2.303 \log 10 = 2.303 	imes 1$.Dividing the two equations:$\frac{t_{1/8}}{t_{1/10}} = \frac{3 \log 2}{1} = 3 	imes 0.30 = 0.9$.

An organic compound undergoes first order decomposition. The time taken for its decomposition to $\frac{1}{8}$ and $\frac{1}{10}$ of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. What is the value of $\frac{t_{1/8}}{t_{1/10}}$? $[\log 2 = 0.30]$

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