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Sucrose decomposes in acid solution into...

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with `t_(1//2)=3.00hr.` What fraction of sample of sucrose remains after `8 hr` ?

Text Solution

Verified by Experts

The correct Answer is:
`0.158M`
Decomposition of sucrose is of first order reaction
`:.k=(2.303)/(t)log``(a)/(a-x)=(2.303)/(t)log``([A]_(0))/([A])`
Fraction of sucrose remaining after `8hr =([A])/([A])`
`:.k=(2.303)/(t_(1//2))=(0.693)/(3hr)=0.231hr^(-1)`
Hence, `0.231hr^(-1)=(2.303)/(8hr)lot``([A]_(0))/([A])`
`or log``([A]_(0))/([A])=0.8024`
`or ([A]_(0))/([A])=Antilog (0.8024)=6.345`
`or ([A])/([A]_(0))=(1)/(6.345)=0.158M`
Second method
Let `x%` of the sucrose decomposes. Thuen use the relation `:`
`(t_(1//2))/(t_(x%))=(0.3)/(log((100)/(100-x)))`
`:.log((100)/(100-x))=(t(x%)xx0.3)/(t_(1//2))=(8hrxx0.3)/(3hr)=0.8`
`:. (100)/(100-x)=(a)/(a-x)=Antilog(0.8)=6.345`
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Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law with t_(1//2)=3.00 hours. What fraction of the sample of sucrose remains after 8 hours ?

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Knowledge Check

  • Mean life of a radioactive sample is t_0 . What fraction of sample remains left after time t_0ln_2 ?

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