For the reaction,$Cl_{2} + 2l^{-} \rightarrow I_{2} + 2Cl^{-}$,the initial concentration of $I^{-}$ was 0.20 mol $L^{- 1}$ and the concentration after 20 min was 0.20 mol $L^{- 1}$.Then the rate of formation of $I_{2}$ in mol $L^{- 1}$ would be

Question

Dr. Abubacker · Accepted Answer

(b)
$Cl_{2} + 2l^{-} ightarrow I_{2} + 2Cl^{-}$
Rate of formation of
$I_{2} = \frac{dI_{2}}{dt} = - \frac{1}{2}\frac{dI^{-}}{dt}$
$= \frac{1}{2} 	imes \frac{0.20 - 0.18}{20}$
$= \frac{1}{2} 	imes \frac{0.02}{20}$
$= 5 	imes 10^{- 4}mol\ L^{- 1}\min^{- 1}$

Q. For the reaction,$Cl_{2} + 2l^{-} \rightarrow I_{2} + 2Cl^{-}$,the initial concentration of $I^{-}$ was 0.20 mol $L^{- 1}$ and the concentration after 20 min was 0.20 mol $L^{- 1}$.Then the rate of formation of $I_{2}$ in mol $L^{- 1}$ would be

Explanation

Reviewed by Dr. Abubacker

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