For the reaction I2 - 2S2O32- - S4O62- - 2I-I- -d-I2-dt - - 1-2 d-S2O32-dtII- -d-I2-dt - - 2 d-S2O32-dtIII- -d-I2-dt - -2 d-I-dt d-S2O32-dtIV- d-S4O62-dt - 1-2 d-I-dtThe correct option is-

The correct option is B I and IV I_2 + 2S_2O_3^2 → S_4O_6^2 + 2I^⊖ Rate = d[I_2]/dt = nbsp; d[S_2O_3^2 ]/2dt = nbsp;+d[S_4O_6^2 ]/ ...

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Byju's Answer

Standard XII

Chemistry

Nucleophilic Substitution in Alkyl Halides

For the react...

Question

For the reaction $I_{2} + 2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 I^{⊖}$
I. $\frac{- d [I_{2}]}{d t} = - \frac{1}{2} \frac{d [S_{2} O_{3}^{2 -}]}{d t}$
II. $\frac{- d [I_{2}]}{d t} = - 2 \frac{d [S_{2} O_{3}^{2 -}]}{d t}$
III. $\frac{- d [I_{2}]}{d t} = - 2 \frac{d [I^{⊖}]}{d t} \frac{d [S_{2} O_{3}^{2 -}]}{d t}$
IV. $\frac{d [S_{4} O_{6}^{2 -}]}{d t} = \frac{1}{2} \frac{d [I^{⊖}]}{d t}$
The correct option is:

Only I

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I and IV

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II and IV

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Only III

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Solution

The correct option is B I and IV
$I_{2} + 2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 I^{⊖}$
Rate $= \frac{- d [I_{2}]}{d t} = \frac{- d [S_{2} O_{3}^{2 -}]}{2 d t}$
$= \frac{+ d [S_{4} O_{6}^{2 -}]}{d t} = \frac{+ d [I^{⊖}}{2 d t}$
Hence, correct options are I and IV.

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For the reaction I2+2S2O2−3→S4O2−6+2I⊖I. −d[I2]dt=−12d[S2O2−3]dtII. −d[I2]dt=−2d[S2O2−3]dtIII. −d[I2]dt=−2d[I⊖]dtd[S2O2−3]dtIV. d[S4O2−6]dt=12d[I⊖]dtThe correct option is:

The correct option is B I and IVI2+2S2O2−3→S4O2−6+2I⊖Rate =−d[I2]dt=−d[S2O2−3]2dt=+d[S4O2−6]dt=+d[I⊖2dtHence, correct options are I and IV.

The correct option is B I and IV
$I_{2} + 2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 I^{⊖}$
Rate $= \frac{- d [I_{2}]}{d t} = \frac{- d [S_{2} O_{3}^{2 -}]}{2 d t}$
$= \frac{+ d [S_{4} O_{6}^{2 -}]}{d t} = \frac{+ d [I^{⊖}}{2 d t}$
Hence, correct options are I and IV.