An organic compound undergoes first order dec | vedclass

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(A) For a first-order reaction, the time $t$ taken for the concentration to reach $A_t$ from $A_0$ is given by $t = \frac{1}{k} \ln \frac{A_0}{A_t}$.F

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) For a first-order reaction, the time $t$ taken for the concentration to reach $A_t$ from $A_0$ is given by $t = \frac{1}{k} \ln \frac{A_0}{A_t}$.For $t_{1/8}$, the remaining concentration is $\frac{A_0}{8}$, so $t_{1/8} = \frac{1}{k} \ln \frac{A_0}{A_0/8} = \frac{1}{k} \ln 8$.For $t_{1/10}$, the remaining concentration is $\frac{A_0}{10}$, so $t_{1/10} = \frac{1}{k} \ln \frac{A_0}{A_0/10} = \frac{1}{k} \ln 10$.Taking the ratio: $\frac{t_{1/8}}{t_{1/10}} = \frac{\ln 8}{\ln 10} = \frac{\log 8}{\log 10} = \log 8$.Since $\log 8 = \log 2^3 = 3 \log 2 = 3 	imes 0.3 = 0.9$.Therefore, $\frac{t_{1/8}}{t_{1/10}} 	imes 10 = 0.9 	imes 10 = 9$.

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