Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, t1/2=3.00 hours. What fraction of sample of sucrose remains after 8 hours? - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

$C_{12} H_{22} O_{11} + H_{2} O \overset{H^{+}}{\to} C_{6} H_{12} O_{6} + C_{6} H_{12} O_{6}$

It was $1^{st}$ order reaction

$t_{1 / 2} = 3 hrs$

for $1^{st} order,$ $t_{1 / 2} = \frac{0.693}{k}$

$k = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{3} = 0.23 1 {hr}^{- 1}$

$K = \frac{2.303}{t} \log \frac{{[A]}_{0}}{[A]} say {[A]}_{0} = 1$

$[A] = 1 - x$

$0.231 = \frac{2.303}{8 hr} \log [\frac{1}{1 - x}]$

$\log [\frac{1}{1 - x}] = \frac{8 \times 0.231}{2.303} = 0.8024$

$\frac{1}{1 - X} = 10^{0.8024} = 6.3445$

$1 - X = \frac{1}{6.2445} = 0.158 M$

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, $t_{1 / 2} = 3.00$ hours. What fraction of sample of sucrose remains after 8 hours?