For a reaction 2I^– + 2H⁺ → I₂ (s) + H₂(g), if the initial concentration of I^–1 was 0.80 mol L^–1 and concentration after 20 minutes was 0.68 mol L^–1. Calculate the rate of disappearance of I^– and rate of appearance of I₂.

Solution

Rate = - \frac{1}{2} \frac{∆ {[I]}^{-}}{∆ t} = - \frac{1}{2} \frac{∆ [H^{+}]}{∆ t} = \frac{∆ [I_{2}]}{∆ t} = \frac{∆ [H_{2}]}{∆ t}

We have to calculate the rate of disappearance I^–1 and rate of appearance of I₂, hence

Rate = - \frac{1}{2} \frac{∆ {[I]}^{-}}{∆ t} = \frac{∆ [I_{2}]}{∆ t}

but

\frac{∆ {[I]}^{-}}{∆ t} = \frac{Change in concentration of I^{-}}{Change in time} = \frac{(0.68 - 0.80) mol L^{- 1}}{(20 - 0) minutes} = \frac{0.12}{20} = 0.006 mol L^{- 1} \min^{- 1}

We know that

\frac{∆ [I_{2}]}{∆ t} = \frac{1}{2} \frac{∆ {[I]}^{-}}{∆ t} = \frac{1}{2} \times 0.006 mol L^{- 1} \min^{- 1}

= 0.003 mol L^{- 1} \min^{- 1} .

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