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Question
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law with `"t"_(1/2)`= 3 hours. What fraction of the sample of sucrose remains after 8 hours?
Solution
`"t"_(1/2)` = 3 Hours
Now we know that,
k = `0.693/"t"_(1/2)`
= `0.693/3`
= 0.231 hr−1
Put above value in the formula of first order reaction,
k = `2.303/8log [R]_0/([R])`
So,
`log [R]_0/([R])=(0.231xx8)/2.303` = 0.8024
Taking antilog on both sides,
`[R]_0/([R])` = 6.3445
`[[R]]/[R]_0` = 0.158
Fraction of the sample of sucrose remaining after 8 hours = 0.158
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