Kinetics of an Iodine Clock Reaction How to calculate Rate law
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Step 1
Typically, this reaction involves iodide ions (Iā»), hydrogen peroxide (HāOā), and starch as an indicator. The reaction produces iodine (Iā), which reacts with starch to form a blue complex. Show moreā¦
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How to calculate the rate order of reagents ([H3O+], [I-], [H2O2] and [S2O32-]) in the Iodine Clock Reaction? Also what would the rate law for this reaction be?
Madhur L.
Comment on the order of the Iodine clock reaction , find out rate and rate constant
David C.
In a clock reaction, a dramatic color change occurs at a time determined by concentration and temperature. Consider the iodine clock reaction, whose overall equation is $$ 2 \mathrm{I}^{-}(a q)+\mathrm{S}_{2} \mathrm{O}_{8}^{2-}(a q) \longrightarrow \mathrm{I}_{2}(a q)+2 \mathrm{SO}_{4}^{2-}(a q) $$ As $\mathrm{I}_{2}$ forms, it is immediately consumed by its reaction with a fixed amount of added $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ : $$ \mathrm{I}_{2}(a q)+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \longrightarrow 2 \mathrm{I}^{-}(a q)+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q) $$ Once the $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ is consumed, the excess $\mathrm{I}_{2}$ forms a blue-black product with starch present in solution: $$ \mathrm{I}_{2}+\text { starch } \longrightarrow \text { starch } \cdot \mathrm{I}_{2} $$ The rate of the reaction is also influenced by the total concentration of ions, so $\mathrm{KCl}$ and $\left(\mathrm{NH}_{4}\right) \mathrm{SO}_{4}$ are added to maintain a constant value. Use the data below, obtained at $23^{\circ} \mathrm{C}$ , to determine: (a) The average rate for each trial (b) The order with respect trial (c) The rate constant (d) The rate law for the overall reaction
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