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The following data were obtained in experiment on inversion of cane sugar.
Time (minutes) 0 60 120 180 360 $\infty$
Angle of rotation +13.1 +11.6 +10.2 +9.0 +5.87 -3.8
(degree)
Determine total time ?
1932 min
None of these
966 min
483 min

A

None of these

B

966 min

C

483 min

D

1932 min

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Solution

Verified by Toppr

The integrated rate law expression for the inversion of can sugar (assuming first order kinetics) is as shown.
$k = \frac{2.303}{t} l o g \frac{r_{0} - r_{\infty}}{r_{t} - r_{\infty}}$
For 60 minutes
$k = \frac{2.303}{60} l o g \frac{13.1 - (- 3.8)}{11.6 - (3.8)} = 0.001549$
For 120 minutes
$k = \frac{2.303}{120} l o g \frac{13.1 - (- 3.8)}{10.2 - (3.8)} = 0.001569$
For 180 minutes
$k = \frac{2.303}{180} l o g \frac{13.1 - (- 3.8)}{9.0 - (3.8)} = 0.001544$
For 360 minutes
$k = \frac{2.303}{360} l o g \frac{13.1 - (- 3.8)}{5.87 - (3.8)} = 0.001551$
Since, the value of k is constant, the reaction follows first order reaction.
The average value of k is $\frac{0.001549 + 0.001569 + 0.001544 + 0.001551}{4} = \frac{0.0062135}{4} = 0.001553 m i n^{- 1}$
To determine the total time, substitute $r_{t} = 0$ in the above expression.
$t = \frac{2.303}{k} l o g \frac{r_{0} - r_{\infty}}{r_{t} - r_{\infty}}$
$t = \frac{2.303}{0.001553} l o g \frac{13.1 - (- 3.8)}{0 - (- 3.8)} = 966 m i n$

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Q1

The following data were obtained in experiment on inversion of cane sugar.
Time (minutes) 0 60 120 180 360

\infty

Angle of rotation +13.1 +11.6 +10.2 +9.0 +5.87 -3.8
(degree)
Determine total time ?

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Q2

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