Question

Surcrose a hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of $3.33hat{25}^{\circ }C.$ After $9h$ the fraction of sucrose remaining is $f$. the value of $lo{g}_{10}\left(\frac{1}{f}\right)$ is __________ $\times {10}^{-2}$. (Rounded off to the nearest integer)
[Assume $:ln10=2.303,ln2=0.693]$

• A. 81.00
• B. 81.0
• C. 81

Answer 1

Answer: (A), (B), (C)

$\begin{array}{cccccccc}& Sucrose& +& H_{2}O& \stackrel{H^{+}}{\to }& Glucose& +& Fructose\\ t=0& a& & & & 0& & 0\\ t=9hr& a-x& & & & x& & x\end{array}$

For first order reaction,
$t_{1/2}=\frac{0.693}{k}$

$k=\frac{0.693}{3.33}h{r}^{-1}$

$k=\frac{2.303}{t}log\frac{a}{a-x}$
$(a-x)=f$
$a=1$
$\frac{0.693\times 9}{3.33\times 2.303}=lo{g}_{10}\left(\frac{1}{f}\right)$

$lo{g}_{10}\left(\frac{1}{f}\right)=0.81$
$lo{g}_{10}\left(\frac{1}{f}\right)=81\times {10}^{-2}$