Sucrose hydrolyses in acid solution into gluc | vedclass

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Question

(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{3.33 \ h} = \frac{0.693}{10/3 \ h} = 0.2079

Vedclass · Accepted Answer

$81$

Vedclass · Answer

(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{3.33 \ h} = \frac{0.693}{10/3 \ h} = 0.2079 \ h^{-1}$.The integrated rate law for a first order reaction is $\ln \frac{[A]_0}{[A]_t} = kt$,which can be written as $\log_{10} \frac{[A]_0}{[A]_t} = \frac{kt}{2.303}$.Given that the fraction of sucrose remaining is $f = \frac{[A]_t}{[A]_0}$,we have $\frac{1}{f} = \frac{[A]_0}{[A]_t}$.Substituting the values: $\log_{10} (\frac{1}{f}) = \frac{0.2079 	imes 9}{2.303} = \frac{0.693 	imes 3 	imes 9}{10 	imes 2.303} = \frac{0.693}{2.303} 	imes 2.7 = 0.3009 	imes 2.7 = 0.81243$.Thus,$\log_{10} (\frac{1}{f}) = 81.24 	imes 10^{-2}$.Rounding to the nearest integer,we get $81$.

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