anonymous
  • anonymous
The following data were obtained in an experiment on inversion of cane sugar in the presence of 0.1N HCl. Time(min) 0 60 120 180 360 infinity Angle of Rotation 13.1 11.6 10.2 9.0 5.87 -3.8 After what time would you expect a zero reading in Polarimeter........its urgent can someone help me fast!
Chemistry
jamiebookeater
  • jamiebookeater
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chmvijay
  • chmvijay
WOW Very interesting question i Love it :)
anonymous
  • anonymous
@chmvijay will u plz solve it.....
anonymous
  • anonymous
yup it is theorotical just use chemical kinetics and answer this @nincompoop

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chmvijay
  • chmvijay
hope @nincompoop answers this :)
anonymous
  • anonymous
me too hopind the same......lets see...its not supposed to be very difficult .... it is class 12 question....ISC board @chmvijay
anonymous
  • anonymous
*hoping
anonymous
  • anonymous
ok here's the only hint formula\[k =\frac{ 2.303 }{ t }\log_{10}\frac{ r_{0}-r _{\infty} }{ r_{t}-r _{\infty} } \]
anonymous
  • anonymous
@nincompoop @chmvijay @Frostbite i got to go back to work.......if you guys get the answer plz do reply here........thanks! Bye will be back soon
chmvijay
  • chmvijay
ok fine me to leaving to BED will see it hope @nincompoop solves this :)
nincompoop
  • nincompoop
I can't… LOL
nincompoop
  • nincompoop
anonymous
  • anonymous
I guess i will have to ask this to my instructor...!
anonymous
  • anonymous
in the formula \[k={2.303\over t}\log_{10}\frac{r_0-r_\infty}{r_t-r_\infty}\] k = rate of reaction (interms of change in optical activity) t = time r = angles of rotation from this, we calculate the rate of reaction \[ k={2.303\over120}\log_{10}\frac{13.1-(-3.8)}{10.2-(-3.8)}\\ k=1.57\times10^{-3} \] use the equation again, to find "t" when r_t=0 rearranging, the terms, \[ t={2.303\over 0.00157}\log_{10}\frac{13.1-(-3.8)}{0-(-3.8)}\\ \Large\boxed{t=950\,{\rm min}\\ t=15\,{\rm hr}\;50\,{\rm min}} \]
anonymous
  • anonymous
@electrokid Thanks a lot!!! It was great help....
chmvijay
  • chmvijay
@electrokid nice :)

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