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Surcose decomposes in acid solution into glucose and fructose according to the first order rate law .with t_((t)/(2))=3.00 hours.What fraction of sample of surcose remains after 8 hours? |
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Answer» Solution :`"surcose"_((AQ))overset(H^(+))to"Glucose"_((aq))+"Fructose"_((aq))` The reaction is FIRST order where ,`t_((1)/(2))=3.00` hours `therefore` So, `K=(0.693)/(t)=(0.693)/(300h)=0.231 hr^(-1)` CALCULATE fraction of sample of sucroso REMAINS after 8 hours. Suppose the initial con.of sucrose=`[R]_(0) mol L^(1)` Remaining concentration of sucrose after 8 hours =`[R]_(t) mol L^(-1)` So,`("Sucrose in initial")/("Sucrose after t time")=([R]_(0))/([R]_(t))` k=`(2.303)/(t)` log `([R]_(0))/([R]_(t))` For first order reaction , `therefore 0.231 hr^(-1)=(2.303)/(8 hr) log ([R]_(0))/([R]_(t))` `therefore log ([R]_(0))/([R]_(t))=(0.231 hr^(-1)xx8 hr)/(2.303)=0.8024` Remaining sucrose after 8 hour =`[R]_(t)` and Fraction of remaining sucrose `([R]_(0))/([R]_(t))` `therefore` Fraction of remaining sucrose =`(1)/(([R]_(0))/([R]_(t)))=(1)/(6.3445)` |
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