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Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsRate of Reactions/Rate LawMedium2 minQB

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Passage / Comprehension

Chlorine dioxide oxidizes iodide ion in an aqueous solution to iodine and get reduced to chlorite ion.

$2 Cl O_{2} (a q) + 2 I^{-} (a q) ⟶ 2 Cl O_{2}^{-} (a q) + I_{2} (a q)$

The order of the reaction with respect to $Cl O_{2}$ was determined by starting with a large excess of $I^{-}$ , so that its concentration was essentially constant. Then, :

$r a t e = k [Cl O_{2}]^{m} [I^{-}]^{n} = k^{'} [Cl O_{2}]^{m}$

where $k^{'} = k [I^{-}]^{n}$ . Given below the time and the concentration of $Cl O_{2}$ taken:

Time (s) $[Cl O_{2}]$ (M)

0.00 $4.77 \times 1 0^{- 4}$

1.00 $4.31 \times 1 0^{- 4}$

2.00 $3.85 \times 1 0^{- 4}$

Determine the rate constant of the reaction

Options:

Answer:

Solution:

$r a t e = k [ClO_{2}]^{1}$

$k = \frac{( 4.31 - 4.77 ) \times 1 0 ^{- 4}}{1} = k [4.77 \times 1 0^{- 4}]^{1}$

∴ $k = 0.1$

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

⏱ 2mℹ️ Source: QB

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