Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law,with $t_{1/2} = 3.00 \ h$. What fraction of sample of sucrose remains after $8 \ h$ $?$
(N/A) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 3.00 \ h$,we have $k = \frac{0.693}{3.00 \ h} = 0.231 \ h^{-1}$.
The integrated rate law for a first order reaction is $\ln \frac{[R]_0}{[R]} = kt$ or $\log \frac{[R]_0}{[R]} = \frac{kt}{2.303}$.
Substituting the values: $\log \frac{[R]_0}{[R]} = \frac{0.231 \ h^{-1} \times 8 \ h}{2.303} = \frac{1.848}{2.303} \approx 0.8024$.
Taking the antilog,$\frac{[R]_0}{[R]} = 10^{0.8024} \approx 6.3445$.
The fraction remaining is $\frac{[R]}{[R]_0} = \frac{1}{6.3445} \approx 0.1576$.
Thus,the fraction of sucrose remaining after $8 \ h$ is approximately $0.158$.
Given $t_{1/2} = 3.00 \ h$,we have $k = \frac{0.693}{3.00 \ h} = 0.231 \ h^{-1}$.
The integrated rate law for a first order reaction is $\ln \frac{[R]_0}{[R]} = kt$ or $\log \frac{[R]_0}{[R]} = \frac{kt}{2.303}$.
Substituting the values: $\log \frac{[R]_0}{[R]} = \frac{0.231 \ h^{-1} \times 8 \ h}{2.303} = \frac{1.848}{2.303} \approx 0.8024$.
Taking the antilog,$\frac{[R]_0}{[R]} = 10^{0.8024} \approx 6.3445$.
The fraction remaining is $\frac{[R]}{[R]_0} = \frac{1}{6.3445} \approx 0.1576$.
Thus,the fraction of sucrose remaining after $8 \ h$ is approximately $0.158$.
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