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Cane sugar is gradualaly converted into ...

Cane sugar is gradualaly converted into dextrose and laevuloe by dilute acid. The rate of inversion is observed by measuring the polarisation angle, at various times, when the following results are obtained:
`{:("Time (min)",0,10,20,30,40,100,oo),("Angle",32.4,28.8,25.5,22.4,19.6,-6.1,-14.1):}`
Show that the reaction is of first order. Calculate the value of t, when the solution is optically inactive.

Text Solution

Verified by Experts

in case, the inversion of cane sugar is a first order change, then
`k=(2.303)/(t)log_(10).([A]_(0))/([A])=(2.303)/(t)log_(10).(r_(0)-r_(oo))/(r_(t)-r_(oo))`
When `t=10, r_(0)=32.4, r_(t)=28.8, r_(oo)=-14.1`
`k=(2.3030)/(10)log_(10).(32.4-(-14.1))/(28.8-(-14.1))`
`k=(2.3030)/(10)log_(10).(46.5)/(42.9)=0.008 "min"^(-1)`
When `t=20mr_(0)=32.4,r_(t)=25.5,r_(oo)=-14.1`
`k=(2.3030)/(20)log_(10).(32.4-(-14.1))/(22.5-(-14.1))=(2.3030)/(20)log.(46.5)/(39.6) 0.008"min"^(-1)`
When `t=30,r_(0)=32.4,r_(t)=22.4,r_((oo)=-14.1`
`k=(2.3030)/(30)log_(10).(32.4-(-14.1))/(22.5-(-14.1))=(2.3030)/(30)log.(46.5)/(36.5) =0.008 "min"^(-1)`
Thus, the reaction is of first order as the value of k is constant.
Solution will be optically inactive when half of the came sugar in inverted.
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