Determine the order w.r.t. ClO 2 | TestHub

ChemistryMediumsingle choice

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 order w.r.t. $ClO_{2}$

Options:

Answer:

C

Solution:

rate $= k^{'} [Cl O_{2}]^{m}$

$\frac{d [ Cl O _{_{2}} ]}{d t} = k^{'} [Cl O_{2}]_{m}$

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

$= \frac{( 3.55 - 4.31 ) \times 1 0 ^{- 4}}{1} k^{'} [4.31 \times 1 0^{- 4}]^{m} . \dots (2)$

Dividing (1) by (2) we get

$m \approx 1$

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:Experimental determination of order of reaction

⏱ 2mℹ️ Source: QB

Chemistry - Chemical Kinetics Question with Solution | TestHub

Options:

Doubts & Discussion