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Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, witht_(1//2)=3.00 hours. What fraction of the sample of sucrose remains after 8 hours? |
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Answer» Solution :First order rate equation is : `k=(2.303)/(t)"log"([A]_(0))/([A])""…(i)` As `t_(1//2)=3.0` hours `therefore k=(0.693)/(t_(1//2))=(0.693)/(3H)=0.231h^(-1)` SUBSTITUTING the values in equation(i), we GET `0.231 H^(-1)=(2.303)/(8h)"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`. `([A])/([A]_(0))` represents the fraction of SUCROSE left after 8 hours. |
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