Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsRate of Reactions/Rate LawMedium2 minQB

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Passage / Comprehension

The inversion of cane sugar was studied in HCl at 298K. The following polarimetric readings were obtained at different intervals of time.

The rate constant of the inversion if it follows first order kinetics will be (Given : $ln \frac{34.83}{28.48} = 0.2$ )

Answer:

Solution:

$C_{12} H_{22} O_{11} + H_{2} O \to C_{6} H_{12} O_{6} + C_{6} H_{12} O_{6}$

Follow first order reaction then we know

$K = \frac{1}{t} ln \frac{Q _{\infty} - Q _{0}}{Q _{\infty} - Q _{t}}$ at $t = 18 min$

$K = \frac{1}{18} ln (\frac{- 10.74 - 24.09}{- 10.74 - 17.74})$

$= \frac{1}{18} ln (\frac{+ 34.83}{+ 28.48})$

$= \frac{1}{18} ln (\frac{34.83}{28.48})$

$= \frac{1}{18} \times 0.2$

$K = 1.1 \times 1 0^{- 2} min^{- 1}$

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:Rate of Reactions/Rate Law

⏱ 2mℹ️ Source: QB

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