Chapter 13, Problem 1PE

The iodate ion reacts with sulfite ions in the reaction

${\text{IO}}_{\text{3}}^{\text{-}}{\text{+3SO}}_{\text{3}}^{\text{2-}}\stackrel{}{\to }{\text{I}}^{\text{-}}{\text{+3SO}}_{\text{4}}^{\text{2-}}$

At what rate are the iodide and sulfate ions being produced if the sulfite ion is disappearing at a rate of $\text{2}\text{.4 }\times {\text{ 10 }}^{\text{-4}}{\text{ mol L}}^{\text{-}}{}^{\text{1}}{s}^{\text{-1}}$? (Hint: Pay attention to the stoichiometry.)

Interpretation Introduction

Interpretation:

The rate of production of iodide and sulfate ion, for the given equation, if the sulfite ion is disappearing at a rate of $2.4\times {10}^{-4}\text{ mol}\cdot {\text{L}}^{-1}{\text{s}}^{-1}$, is to be determined.

Concept Information:

For a generalized reaction:

$\text{aA}+\text{ bB }\to \text{ cC }+\text{dD}$

$\begin{array}{l}\text{Rate of reaction }\\ =-\left(\text{1}/\text{a}\right)*\left(\Delta \left[\text{A}\right]/\Delta \text{t}\right)=-\left(\text{1}/\text{b}\right)*\left(\Delta \left[\text{B}\right]/\Delta \text{t}\right)\\ =-\left(\text{1}/\text{c}\right)*\left(\Delta \left[\text{C}\right]/\Delta \text{t}\right)=-\left(\text{1}/\text{d}\right)*(\Delta \left[\text{D}\right]/\Delta \text{t}\end{array}$

Stoichiometry refers to the branch of chemistry that determines the relationship between reactants and/or the products involved in the chemical reaction that determines desired quantitative data.

The rate of reaction may be defined as the speed at which a chemical reaction proceeds or the concentration of reactant that is transformed into the product at a particular unit of time.

Answer to Problem 1PE

Solution:

$8\times {10}^{-5}\text{ mol}\cdot {\text{L}}^{-1}{\text{s}}^{-1}$

$2.4\times {10}^{-4}\text{ mol}\cdot {\text{L}}^{-1}{\text{s}}^{-1}$

Explanation of Solution

Given information:

The iodate ion reacts with sulfite ions in the reaction,

${\text{IO}}_3^{-}+3{\text{SO}}_3^{2-}\to {\text{I}}^{-}+3{\text{SO}}_4^{2-}$

The rate of disappearance of sulfite ion is $\frac{d\left[{\text{SO}}_3\right]}{dt}=2.4\times {10}^{-1}\text{ mol}\cdot {\text{L}}^{-1}{\text{s}}^{-1}$

Recalling the above reaction,

${\text{IO}}_3^{-}+3{\text{SO}}_3^{2-}\to {\text{I}}^{-}+3{\text{SO}}_4^{2-}$

The rate law of the above reaction is written as:

$r=-\frac{d\left[{\text{IO}}_3\right]}{dt}=-\frac{1}{3}\frac{d\left[{\text{SO}}_3\right]}{dt}=\frac{d\left[\text{I}\right]}{dt}=\frac{1}{3}\frac{d\left[{\text{SO}}_4\right]}{dt}$

The negative sign in the above equation describes the disappearance of the reactants.

The rate of formation of sulfate ion is calculated as:

Substitute the value of $\frac{d\left[{\text{SO}}_3\right]}{dt}$ in the above equation of rate law, and write the equation for the formation of the sulfate ion,

$\begin{array}{l}r=-\cancel{\frac{1}{3}}2.4\times {10}^{-4}\text{ mol}\cdot {\text{L}}^{-1}{\text{s}}^{-1}=\cancel{\frac{1}{3}}\frac{d\left[{\text{SO}}_4\right]}{dt}\\ r=\frac{d\left[{\text{SO}}_4\right]}{dt}=2.4\times {10}^{-4}\text{ mol}\cdot {\text{L}}^{-1}{\text{s}}^{-1}\\ \frac{d\left[{\text{SO}}_4\right]}{dt}=2.4\times {10}^{-4}\text{ mol}\cdot {\text{L}}^{-1}{\text{s}}^{-1}\end{array}$

The negative sign is ignored as this determines only the magnitude of the reaction.

The rate of formation of sulfate ion is $2.4\times {10}^{-4}\text{ mol}\cdot {\text{L}}^{-1}{\text{s}}^{-1}$.

The rate of formation of iodide ion is calculated as:

Substitute the value of $\frac{d\left[{\text{SO}}_3\right]}{dt}$ in the above equation of rate law, and write the equation for the formation of the iodide ion,

$\begin{array}{l}r=-\frac{1}{3}2.4\times {10}^{-1}\text{ mol}\cdot {\text{L}}^{-1}{\text{s}}^{-1}=\frac{d\left[\text{I}\right]}{dt}\\ r=\frac{d\left[\text{I}\right]}{dt}=\frac{2.4\times {10}^{-4}\text{ mol}\cdot {\text{L}}^{-1}{\text{s}}^{-1}}{3}\\ \frac{d\left[{\text{SO}}_4\right]}{dt}=8\times {10}^{-5}\text{ mol}\cdot {\text{L}}^{-1}{\text{s}}^{-1}\end{array}$

The negative sign is ignored as this determines only the magnitude of the reaction.

Conclusion

The rate of formation of sulfate ion is $2.4\times {10}^{-4}\text{ mol}\cdot {\text{L}}^{-1}{\text{s}}^{-1}$ and the rate of formation of iodide ion is $8\times {10}^{-5}\text{ mol}\cdot {\text{L}}^{-1}{\text{s}}^{-1}$.