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(A) For a first-order reaction,the rate constant $K$ is given by $K = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t}$.For reaction $A$,$94\%$ has reacted,so

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$4.06$

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(A) For a first-order reaction,the rate constant $K$ is given by $K = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t}$.For reaction $A$,$94\%$ has reacted,so the remaining amount is $100 - 94 = 6\%$. Thus,$K_1 = \frac{2.303}{t} \log \frac{100}{6}$.For reaction $B$,$50\%$ has reacted,so the remaining amount is $100 - 50 = 50\%$. Thus,$K_2 = \frac{2.303}{t} \log \frac{100}{50}$.Taking the ratio: $\frac{K_1}{K_2} = \frac{\log(100/6)}{\log(100/50)} = \frac{\log(16.67)}{\log(2)} = \frac{1.2219}{0.3010} \approx 4.06$.

For two first-order reactions:
$A \rightarrow \text{Products}$
$B \rightarrow \text{Products}$
If at the same time,$50\% \, B$ has reacted and $94\% \, A$ has reacted,calculate the ratio $K_1/K_2$.

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For two first-order reactions:$A \rightarrow \text{Products}$$B \rightarrow \text{Products}$If at the same time,$50\% \, B$ has reacted and $94\% \, A$ has reacted,calculate the ratio $K_1/K_2$.