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In a reaction of acidified hydrogen peroxide with potassium iodide, the concentration of iodine formed rises from 0 to $10^{- 5} mol {dm}^{- 3}$ in 10 seconds. What is the rate of reaction?

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$10^{- 6} mol {dm}^{- 3} s^{- 1}$

$10^{- 5} mol {dm}^{- 3} s^{- 1}$

$10^{6} mol {dm}^{- 3} s^{- 1}$

$10^{4} mol {dm}^{- 3} s^{- 1}$

answer is A.

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Detailed Solution

The reaction is $2 I^{-} + H_{2} O_{2} + 2 H^{+} \to I_{2} + 2 H_{2} O$
$Rate = + \frac{d (I_{2})}{dt} = \frac{10^{- 5}}{10} = 10^{- 6} mol {dm}^{- 3} s^{- 1}$

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