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 $K = \frac{1}{t} \ln \left( \frac{C_0}{C_t} \right)$.For $t_{1/8}$,the concentration

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$9$

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(B) For a first-order reaction,the rate constant $K$ is given by $K = \frac{1}{t} \ln \left( \frac{C_0}{C_t} ight)$.For $t_{1/8}$,the concentration $C_t = C_0 / 8$,so $K t_{1/8} = \ln(8) = 3 \ln(2)$.For $t_{1/10}$,the concentration $C_t = C_0 / 10$,so $K t_{1/10} = \ln(10)$.Taking the ratio: $\frac{t_{1/8}}{t_{1/10}} = \frac{3 \ln(2)}{\ln(10)} = 3 \log_{10} 2$.Given $\log_{10} 2 = 0.3$,we have $\frac{t_{1/8}}{t_{1/10}} = 3 	imes 0.3 = 0.9$.Therefore,$\frac{t_{1/8}}{t_{1/10}} 	imes 10 = 0.9 	imes 10 = 9$.

An organic compound undergoes first-order decomposition. The time taken for its decomposition to $1/8$ and $1/10$ of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. What is the value of $\frac{t_{1/8}}{t_{1/10}} \times 10$? (Given: $\log_{10} 2 = 0.3$)

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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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