For the first order reaction,half life is 14 | vedclass

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(B) For a first order reaction,the half-life is given by $t_{1/2} = \frac{\ln 2}{k} = 14 \, sec$. Thus,$k = \frac{\ln 2}{14}$. The time $t$ required f

Vedclass · Accepted Answer

$42$

Vedclass · Answer

(B) For a first order reaction,the half-life is given by $t_{1/2} = \frac{\ln 2}{k} = 14 \, sec$. Thus,$k = \frac{\ln 2}{14}$. The time $t$ required for the concentration to reduce to $\frac{1}{8}$ of its initial value $(a_t = \frac{a_0}{8})$ is given by the integrated rate equation: $k = \frac{1}{t} \ln \left( \frac{a_0}{a_t} ight) = \frac{1}{t} \ln \left( \frac{a_0}{a_0/8} ight) = \frac{1}{t} \ln 8$. Since $\ln 8 = \ln(2^3) = 3 \ln 2$,we have: $k = \frac{3 \ln 2}{t}$. Equating the two expressions for $k$: $\frac{\ln 2}{14} = \frac{3 \ln 2}{t}$. Solving for $t$: $t = 14 	imes 3 = 42 \, sec$.

For the first order reaction,half life is $14 \, sec$. The time required for the initial concentration to reduce to $\frac{1}{8}$ of its value is .......... $sec$.

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