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(B) For a first-order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.In the first case,initial

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(B) For a first-order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.In the first case,initial concentration $[A]_0 = 0.8 \ mol$ and remaining concentration $[A]_t = 0.8 - 0.6 = 0.2 \ mol$. However,the problem states $0.8 \ mol$ of $A$ produces $0.6 \ mol$ of $B$,implying the ratio of reactant remaining is $\frac{0.8 - 0.6}{0.8} = \frac{0.2}{0.8} = \frac{1}{4}$.Wait,the standard first-order equation is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For the first case: $k = \frac{2.303}{1} \log \frac{0.8}{0.8 - 0.6} = 2.303 \log \frac{0.8}{0.2} = 2.303 \log 4$.For the second case: $t = \frac{2.303}{k} \log \frac{0.9}{0.9 - 0.675} = \frac{2.303}{k} \log \frac{0.9}{0.225} = \frac{2.303}{k} \log 4$.Since $k$ is the same,$t = 1 \ hr$.

The reaction $A \to B$ follows first order kinetics. The time taken for $0.8 \ mol$ of $A$ to produce $0.6 \ mol$ of $B$ is $1 \ hr$. What is the time taken for the conversion of $0.9 \ mol$ of $A$ to produce $0.675 \ mol$ of $B$?

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The reaction $A \rightarrow B$ follows first-order kinetics. If it takes $1 \text{ hour}$ to convert $0.8 \text{ mol}$ of $A$ to $0.6 \text{ mol}$ of $B$,how many hours will it take to convert $0.9 \text{ mol}$ of $A$ to $0.675 \text{ mol}$ of $B$?

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