In the following first order reactions A xrig | vedclass

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(A) For the first order reaction $A \xrightarrow{K_1} B$,the rate constant $K_1$ is given by: $K_1 = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} \

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(A) For the first order reaction $A \xrightarrow{K_1} B$,the rate constant $K_1$ is given by: $K_1 = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} ight)$.Given $90\%$ of $A$ has reacted,$[A]_t = 10\% 	ext{ of } [A]_0$,so $K_1 = \frac{2.303}{t} \log \left( \frac{100}{10} ight) = \frac{2.303}{t} 	imes 1 = \frac{2.303}{t}$.For the reaction $B \xrightarrow{K_2} 	ext{Product}$,the rate constant $K_2$ is given by: $K_2 = \frac{2.303}{2t} \log \left( \frac{[B]_0}{[B]_{2t}} ight)$.Given $99\%$ of $B$ has reacted,$[B]_{2t} = 1\% 	ext{ of } [B]_0$,so $K_2 = \frac{2.303}{2t} \log \left( \frac{100}{1} ight) = \frac{2.303}{2t} 	imes 2 = \frac{2.303}{t}$.Therefore,the ratio $\frac{K_1}{K_2} = \frac{2.303/t}{2.303/t} = 1$.

In the following first order reactions $A$ $\xrightarrow{K_1} B$ $\xrightarrow{K_2} \text{Product}$,find the ratio $K_1/K_2$ if $90\%$ of $A$ has been reacted in time $t$ while $99\%$ of $B$ has been reacted in time $2t$.

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