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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?
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`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]) = (0.231 xx 8)/2.303`
= `1.848/2.303`
= 0.8024
Taking antilog,
`[R]_0/([R])` = antilog 0.8024 = 6.3445
`[R]_0/([R]) = 1/6.3445`
`[[R]]/[R]_0` = 0.158
Fraction of the sample of sucrose remaining after 8 hours = 0.158
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