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Question
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law with t1/2 = 3 hours. What fraction of the sample of sucrose remains after 8 hours?
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Solution
Given: t1/2 = 3 Hours
Now we know that,
k = `0.693/t_(1//2)`
= `0.693/3`
= 0.231 hr−1
Put the above value in the formula of a first-order reaction.
k = `2.303/8 log [R]_0/([R])`
⇒ 0.231 = `2.303/8 log [R]_0/([R])`
⇒ `log [R]_0/([R]) = (0.231 xx 8)/2.303`
= `1.848/2.303`
⇒ `log [R]_0/([R])` = 0.8024
⇒ `[R]_0/([R])` = antilog 0.8024
= 6.3445
`[R]/([R]_0) = 1/6.3445`
= 0.158
∴ The fraction of the sample of sucrose remaining after 8 hours is 0.158.
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