## leena1996 2 years ago 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!

1. chmvijay

WOW Very interesting question i Love it :)

2. leena1996

@chmvijay will u plz solve it.....

3. leena1996

yup it is theorotical just use chemical kinetics and answer this @nincompoop

4. chmvijay

hope @nincompoop answers this :)

5. leena1996

me too hopind the same......lets see...its not supposed to be very difficult .... it is class 12 question....ISC board @chmvijay

6. leena1996

*hoping

7. leena1996

ok here's the only hint formula$k =\frac{ 2.303 }{ t }\log_{10}\frac{ r_{0}-r _{\infty} }{ r_{t}-r _{\infty} }$

8. leena1996

@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

9. chmvijay

ok fine me to leaving to BED will see it hope @nincompoop solves this :)

10. nincompoop

I can't… LOL

11. nincompoop

@electrokid

12. leena1996

I guess i will have to ask this to my instructor...!

13. electrokid

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, $t={2.303\over 0.00157}\log_{10}\frac{13.1-(-3.8)}{0-(-3.8)}\\ \Large\boxed{t=950\,{\rm min}\\ t=15\,{\rm hr}\;50\,{\rm min}}$

14. leena1996

@electrokid Thanks a lot!!! It was great help....

15. chmvijay

@electrokid nice :)