Chapter 12, Problem 33E

The reaction

${\text{I}}^{-}\left(aq\right)+{\text{OCl}}^{-}\left(aq\right)\to {\text{IO}}^{-}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$

was studied, and the following data were obtained:

[I−]0(mol/L)	[OCl−]0(mol/L)	Initial Rate (mol/L · s)
0.12	0.18	7.91 × 10−2
0.060	0.18	3.95 × 10−2
0.030	0.090	9.88 × 10−3
0.24	0.090	7.91 × 10−2

a. What is the rate law?

b. Calculate the value of the rate constant.

c. Calculate the initial rate for an experiment where both I⁻·and OCl⁻ are initially present at 0.15 mol/L.

Interpretation Introduction

Interpretation: A reaction between ${\text{I}}^{-}$ and ${\text{OCl}}^{-}$ certain data related to this reaction is given. The rate law, initial rate and the value of the rate constant are to be calculated.

Concept introduction: The relation between the reaction rate and the concentration of the reactants is stated by the rate law.

Explanation of Solution

Given

The stated reaction is,

${\text{I}}^{-}\left(aq\right)+{\text{OCl}}^{-}\left(aq\right)\to {\text{IO}}^{-}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$

The rate law for the given reaction is calculated by the expression,

$\text{Rate}=k{\left[{\text{I}}^{-}\right]}^m{\left[{\text{OCl}}^{-}\right]}^n$

Where,

• $\left[{\text{I}}^{-}\right]$ and $\left[{\text{OCl}}^{-}\right]$ are the concentrations of ${\text{I}}^{-}$ and ${\text{OCl}}^{-}$.
• $k$ is the rate constant.

The values of $m$ and $n$ are calculated by the comparison of the different rate values from the given table.

The value of $m$ is calculated using the first and second result. Substitute the values of the concentration of $\left[{\text{I}}^{-}\right]$ and $\left[{\text{OCl}}^{-}\right]$ for the first two experiments in the above expression.

$\text{Rate}\text{1}=k{\left[0.12\right]}^m{\left[0.18\right]}^n$

$\text{Rate}2=k{\left[0.060\right]}^m{\left[0.18\right]}^n$

According to the given rate values in the table,

$\frac{\text{Rate}\text{2}}{\text{Rate}\text{1}}=\frac{3.95\times {10}^{-2}\text{mol}/\text{L}\cdot \text{s}}{7.91\times {10}^{-2}\text{mol}/\text{L}\cdot \text{s}}$

Therefore,

$\begin{array}{c}\frac{k{\left[0.060\right]}^m{\left[0.18\right]}^n}{k{\left[0.12\right]}^m{\left[0.18\right]}^n}=\frac{3.95\times {10}^{-2}\text{mol}/\text{L}\cdot \text{s}}{7.91\times {10}^{-2}\text{mol}/\text{L}\cdot \text{s}}\\ {\left(\frac{0.0108}{0.0216}\right)}^m=0.5\\ 0.5={\left(0.5\right)}^m\end{array}$

Simplify the above expression.

$\begin{array}{l}0.5={\left(0.5\right)}^m\\ m=1\end{array}$

The value of $n$ is calculated using the second and third result. Substitute the values of the concentration of $\left[{\text{I}}^{-}\right]$ and $\left[{\text{OCl}}^{-}\right]$ for the second and third experiments in the above expression.

$\text{Rate}2=k{\left[0.060\right]}^m{\left[0.18\right]}^n$

$\text{Rate}3=k{\left[0.030\right]}^m{\left[0.090\right]}^n$

According to the given rate values in the table,

$\frac{\text{Rate}3}{\text{Rate}2}=\frac{9.88\times {10}^{-3}\text{mol}/\text{L}\cdot \text{s}}{3.95\times {10}^{-2}\text{mol}/\text{L}\cdot \text{s}}$

Therefore,

$\begin{array}{c}\frac{k{\left[0.030\right]}^m{\left[0.090\right]}^n}{k{\left[0.060\right]}^m{\left[0.18\right]}^n}=\frac{9.88\times {10}^{-3}\text{mol}/\text{L}\cdot \text{s}}{3.95\times {10}^{-2}\text{mol}/\text{L}\cdot \text{s}}\\ {\left(\frac{0.0027}{0.0108}\right)}^n=0.25\\ 0.25={\left(0.25\right)}^n\end{array}$

Simplify the above expression.

$\begin{array}{l}0.25={\left(0.25\right)}^n\\ n=1\end{array}$

Substitute the values of $m$ and $n$ in the rate law expression.

$\text{Rate}=k{\left[{\text{I}}^{-}\right]}^1{\left[{\text{OCl}}^{-}\right]}^1$

Conclusion

The relation between the reaction rate and the concentration of the reactants is stated by the rate law. The rate law for the given reaction is, $\text{Rate}=k{\left[{\text{I}}^{-}\right]}^1{\left[{\text{OCl}}^{-}\right]}^1$.

Interpretation Introduction

Interpretation: A reaction between ${\text{I}}^{-}$ and ${\text{OCl}}^{-}$ certain data related to this reaction is given. The rate law, initial rate and the value of the rate constant are to be calculated.

Concept introduction: The relation between the reaction rate and the concentration of the reactants is stated by the rate law.

Explanation of Solution

Given

The stated reaction is,

${\text{I}}^{-}\left(aq\right)+{\text{OCl}}^{-}\left(aq\right)\to {\text{IO}}^{-}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$

The rate law for the given reaction is calculated by the expression,

$\text{Rate}=k{\left[{\text{I}}^{-}\right]}^1{\left[{\text{OCl}}^{-}\right]}^1$

The given values for experiment $1$,

$\begin{array}{l}{\left[{\text{I}}^{-}\right]}_{\text{ο}}=0.12\text{mol}/\text{L}\\ {\left[{\text{OCl}}^{-}\right]}_{\text{ο}}=0.18\text{mol}/\text{L}\\ \text{Rate}=7.91\times {10}^{-2}\text{mol}/\text{L}\cdot \text{s}\end{array}$

Substitute the value of rate, the concentration of ${\text{I}}^{-}$ and the concentration of ${\text{OCl}}^{-}$ in the rate expression.

$\begin{array}{c}\text{Rate}=k{\left[{\text{I}}^{-}\right]}^1{\left[{\text{OCl}}^{-}\right]}^1\\ \text{7}\text{.91}\times {\text{10}}^{-2}\text{mol}/\text{L}\cdot \text{s}=k{\left[0.12\text{mol}/\text{L}\right]}^1{\left[0.18\text{mol}/\text{L}\right]}^1\\ k=\frac{\text{7}\text{.91}\times {\text{10}}^{-2}\text{mol}/\text{L}\cdot \text{s}}{{\left[0.12\text{mol}/\text{L}\right]}^1{\left[0.18\text{mol}/\text{L}\right]}^1}\end{array}$

Simplify the above expression.

$\begin{array}{c}k=\frac{\text{7}\text{.91}\times {\text{10}}^{-2}\text{mol}/\text{L}\cdot \text{s}}{{\left[0.12\text{mol}/\text{L}\right]}^1{\left[0.18\text{mol}/\text{L}\right]}^1}\\ =\frac{\text{7}\text{.91}\times {\text{10}}^{-2}\text{mol}/\text{L}\cdot \text{s}}{0.0216{{\text{mol}}^2/\text{L}}^2}\\ \approx \underset{\_}{3.7L/mol\cdot s}\end{array}$

Conclusion

The value of $k$ for the given reaction is $\underset{\_}{3.7L/mol\cdot s}$.

Interpretation Introduction

Interpretation: A reaction between ${\text{I}}^{-}$ and ${\text{OCl}}^{-}$ certain data related to this reaction is given. The rate law, initial rate and the value of the rate constant are to be calculated.

Concept introduction: The relation between the reaction rate and the concentration of the reactants is stated by the rate law.

Explanation of Solution

Given

The stated reaction is,

${\text{I}}^{-}\left(aq\right)+{\text{OCl}}^{-}\left(aq\right)\to {\text{IO}}^{-}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$

The rate law for the given reaction is calculated by the expression,

$\text{Rate}=k{\left[{\text{I}}^{-}\right]}^1{\left[{\text{OCl}}^{-}\right]}^1$

The given concentration values are,

$\begin{array}{l}{\left[{\text{I}}^{-}\right]}_{\text{ο}}=0.15\text{mol}/\text{L}\\ {\left[{\text{OCl}}^{-}\right]}_{\text{ο}}=0.15\text{mol}/\text{L}\end{array}$

The calculated rate constant value is $3.7\text{L}/\text{mol}\cdot \text{s}$.

Substitute the value of the rate constant, the concentration of ${\text{I}}^{-}$ and the concentration of ${\text{OCl}}^{-}$ in the rate expression.

$\begin{array}{c}\text{Rate}=k{\left[{\text{I}}^{-}\right]}^1{\left[{\text{OCl}}^{-}\right]}^1\\ =\left(3.7\text{L}/\text{mol}\cdot \text{s}\right){\left[0.15\text{mol}/\text{L}\right]}^1{\left[0.15\text{mol}/\text{L}\right]}^1\\ =\underset{\_}{0.083mol/L\cdot s}\end{array}$

Conclusion

The initial rate for the given concentration values of ${\text{I}}^{-}$ and ${\text{OCl}}^{-}$ is $\underset{\_}{0.083mol/L\cdot s}$.