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

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=
`(2.303)/(t) \log \frac{\{\mathrm{R} 0\}}{\mathrm{R}}`

it is given that 0.5 t = 3.33 hours;

k =
`(0.693)/(0.5 t)` =
`(0.693)/(3.33)` = 0.208

then 0.208 =
`(2.303)/(9) \log \frac{\{\mathrm{R} 0\}}{\mathrm{R}}`

`\log \frac{\{\mathrm{R} 0\}}{\mathrm{R}}` =
`(0.208 *9)/(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]}`=
`(1)/(6.486)` = 0.155

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