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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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Answer» Solution :As sucrose decomposes ACCORDING to first order RATE law, `k=(2.303)/(t)log""([A]_(0))/([A])` The aim is to find `[A]//[A]_(0)` As `t_(1//2)=3.0" hour,":.k=(0.693)/(t_(1//2))=(0.693)/(3" hr")=0.231" hr"^(-1)` HENCE, `0.231" hr"^(-1)=(2.303)/(8" hr")log""([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)=0.158.` |
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