Question

Sucrose hydrolysis in acid solution into glucose and fructose following first order rate law with a half-life of $3.3\text{h}$ at ${25}^{o}C$. After $9\text{h}$, the fraction of sucrose remaining is $f$. The value of $\log \frac{1}{f}$ is $\times {10}^{-2}$. (Round off to the nearest integer).
[Assume: $\ln 10=2.303,\ln 2=0.693$]

Answer 1

$\text{Sucrose}\stackrel{hydrolysis}{\to }\text{Glucose + Fructose}$
$t_{\frac{1}{2}}=3.3\text{h}=\frac{10}{3}\text{h}$
$C_t=\frac{\frac{C_o}{T}}{2^t\frac{1}{2}}$
Fraction of sucrose remaining =$f=\frac{C_t}{C_0}=\frac{1}{\frac{t}{\frac{2T_1}{t}}}$
$\frac{1}{f}=2^{\frac{t}{t_{\frac{1}{2}}}}$
$\log \frac{1}{f}=\log (2{)}^{\frac{t}{t_{\frac{1}{2}}}}=\frac{t}{t_{\frac{1}{2}}}\log 2$
$\frac{9}{\frac{10}{3}}\times 0.3-\frac{8.1}{10}-0.81=x\times {10}^{-2}$
$x=81$