Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a

Question

Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at $25^{\circ} \mathrm{C}$. After 9 h , the fraction of sucrose remaining is $f$. The value of $\log _{10}\left(\frac{1}{f}ight)$ is _______ $	imes 10^{-2}$. (Rounded off to the nearest integer)\ [Assume $: \ln 10=2.303, \ln 2=0.693]$\

Professor PC Thomas Classes · Answer

(81) : For first order reaction,\ $t_{1 / 2}=\frac{0.693}{k}$ So, $k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{3.33} \mathrm{~h}^{-1}$ $$ \left(	ext { As } t_{1 / 2}=3.33 \mathrm{~h}ight) $$ $k=\frac{2.303}{t} \log \frac{a}{a-x}$\ Given, $a-x=f \quad a=1$\ $k=\frac{2.303}{t} \log \left(\frac{1}{f}ight)$\ $\Rightarrow \log \left(\frac{1}{f}ight)=\frac{0.693}{3.33} 	imes \frac{9}{2.303}$\ $\log \left(\frac{1}{f}ight)=0.8132=81 	imes 10^{-2}$\