Question

n. Sucrose decomposes in acid solution into glucose and fructose according to a first-order rate law,
with a half-life of 3.33 h at 25 °C. What fraction of a sample of sucrose remains after 9.00 h?

Answer 1

Answer:

The fraction of a sample of sucrose remains after 9.00 h is 0.155.

Explanation:

According to the fundamentals for a first order reaction,

k=  $\frac{2.303}{t} \log \frac{\{\mathrm{R} 0\}}{\mathrm{R}}$

it is given that 0.5 t = 3.33 hours;

k = $\frac{0.693}{0.5 t}$ =    $\frac{0.693}{3.33}$ = 0.208  

then 0.208 =   $\frac{2.303}{9} \log \frac{\{\mathrm{R} 0\}}{\mathrm{R}}$

$\log \frac{\{\mathrm{R} 0\}}{\mathrm{R}}$  =   $\frac{0.208 \times9}{2.303}$

              = 0.812

taking antilog on both sides we get

   $\frac{[\mathrm{R} 0]}{[\mathrm{R}]}$ = antilog(0.812) =    $\frac{[\mathrm{R} 0]}{[\mathrm{R}]}$= 6.486  

$\frac{[\mathrm{R}]}{[\mathrm{R} 0]}$= $\frac{1}{6.486}$  = 0.155

Hence the fraction of sample of sucrose that remains after 9 hours is 0.155.