Question

The thiosulfate ion is oxidized by iodine as follows

$\mathbf{2}{\mathit{S}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}{\mathit{I}}_{\mathbf{2}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\to }{\mathit{S}}_{\mathbf{4}}{\mathit{O}}_{\mathbf{6}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathit{I}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$

In a certain experiment, $\mathbf{7}\mathbf{.}\mathbf{05}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{\text{mol/L}}$of ${\mathit{S}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}$is consumed in the first \(11.0\) seconds of the reaction. Calculate the rate of consumption of ${\mathit{S}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}$, calculate rate of production of iodine ion.

Short answer

Rate of consumption of${\text{[S}}_{\mathrm{2}}{\mathrm{O}}_{\mathrm{3}}^{-}]$ is equal to rate of production of$[{\mathrm{I}}^{-}\mathrm{]}$ is equal to .$6.41\times {10}^{-4}{\text{Ms}}^{-1}$

Step-by-step solution

Understanding the meaning of the term rate of consumption and production in a chemical reaction:

Rate of consumption and rate of production in a chemical reaction refers to the average rate of reaction. Average rate of reaction is defined as the change in concentration of a reactant or product divided by the time interval over which the change occurs.

Calculating rate of production of iodine ion and rate of consumption of [S2O3-] in the given reaction:

Rate$=\frac{∆\left[{\mathrm{S}}_2{\mathrm{O}}_3^{-}\right]}{∆\mathrm{t}}=\frac{7.05\times {10}^{-3}\mathrm{M}}{11.0\mathrm{s}}=6.41\times {10}^{-4}{\mathrm{Ms}}^{-1}$

Due to the reaction coefficients in this chemical reaction, rate of consumption of$\left[{\text{S}}_2{\text{O}}_3^{-}\right]$ is equal to the rate of production of $\left[{\text{I}}^{\text{-}}\right]$.