The reaction A to B follows first-order kinet | vedclass

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

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(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.In the first case,initial amount $[A]_0 = 0.80 \ mole$ and amount of $B$ formed is $0.60 \ mole$,so remaining amount $[A]_t = 0.80 - 0.60 = 0.20 \ mole$.$k = \frac{2.303}{1} \log \frac{0.80}{0.20} = 2.303 \log 4$.In the second case,initial amount $[A]_0 = 0.90 \ mole$ and amount of $B$ formed is $0.675 \ mole$,so remaining amount $[A]_t = 0.90 - 0.675 = 0.225 \ mole$.$k = \frac{2.303}{t} \log \frac{0.90}{0.225} = \frac{2.303}{t} \log 4$.Since $k$ is constant,$2.303 \log 4 = \frac{2.303}{t} \log 4$.Therefore,$t = 1 \ hr$.

The reaction $A \to B$ follows first-order kinetics. It takes $1 \ hr$ for $0.60 \ mole$ of $B$ to be formed from $0.80 \ mole$ of $A$. How much time (in $hr$) will it take for $0.675 \ mole$ of $B$ to be formed from $0.90 \ mole$ of $A$?

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