Chapter 11, Problem 30E

The reaction

${\text{2I}}^{\text{-}}\left(aq\right){\text{+S}}_{\text{2}}{\text{O}}_{\text{8}}{}^{\text{2-}}\left(aq\right)\to {\text{I}}_{\text{2}}\left(aq\right){\text{+2SO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)$

was studied at 25°C. The following results were obtained where

$\text{Rate}=-\frac{\Delta \left[{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}{}^{\text{2-}}\right]}{\Delta t}$

[I−]0(mol/L)	[S2O82−]0(mol/L)	Initial Rate (mol/L · s)
0.080	0.040	12.5 × 10−6
0.040	0.040	6.25 × 10−6
0.080	0.020	6.25 × 10−6
0.032	0.040	5.00 × 10−6
0.060	0.030	7.00 × 10−6

a. Determine the rate law.

b. Calculate a value for the rate constant for each experiment and an average value for the rate constant

Interpretation Introduction

Interpretation: A reaction between ${\text{I}}^{-}$ and ${\text{S}}_{\text{2}}{\text{O}}_8^{2-}$ and its rate expression is given. The rate law 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.

Answer to Problem 30E

The rate law for the given reaction is stated as follows.

Explanation of Solution

Given

The stated reaction is,

$2{\text{I}}^{-}\left(aq\right)+{\text{S}}_2{\text{O}}_8^{2-}\left(aq\right)\to {\text{I}}_2\left(aq\right)+2{\text{SO}}_4^{2-}\left(aq\right)$

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

$\text{Rate}=k{\left[{\text{I}}^{-}\right]}^m{\left[{\text{S}}_2{\text{O}}_8^{2-}\right]}^n$

Where,

• $\left[{\text{I}}^{-}\right]$ and $\left[{\text{S}}_2{\text{O}}_8^{2-}\right]$ are the concentrations of ${\text{I}}^{-}$ and ${\text{S}}_2{\text{O}}_8^{2-}$.
• $k$ is the rate constant.

The values of 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 as only the value of $\left[{\text{I}}^{-}\right]$ changes in these results. Substitute the values of the concentration of $\left[{\text{I}}^{-}\right]$ and $\left[{\text{S}}_2{\text{O}}_8^{2-}\right]$ for the first two experiments in the above expression.

$\text{Rate}\text{1}=k{\left[0.080\right]}^m{\left[0.040\right]}^n$

$\text{Rate}2=k{\left[0.040\right]}^m{\left[0.040\right]}^n$

According to the given rate values in the table,

$\frac{\text{Rate}\text{2}}{\text{Rate}\text{1}}=\frac{6.25\times {10}^{-6}\text{mol}/\text{L}\cdot \text{s}}{12.5\times {10}^{-6}\text{mol}/\text{L}\cdot \text{s}}$

Therefore,

$\begin{array}{c}\frac{k{\left[0.040\right]}^m{\left[0.040\right]}^n}{k{\left[0.080\right]}^m{\left[0.040\right]}^n}=\frac{6.25\times {10}^{-6}\text{mol}/\text{L}\cdot \text{s}}{12.5\times {10}^{-6}\text{mol}/\text{L}\cdot \text{s}}\\ {\left(\frac{0.040}{0.080}\right)}^m=0.5\\ 0.5={\left(0.5\right)}^m\end{array}$

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 first and third result. Substitute the values of the concentration of $\left[{\text{I}}^{-}\right]$ and $\left[{\text{S}}_2{\text{O}}_8^{2-}\right]$ for the first two experiments in the above expression.

$\text{Rate}\text{1}=k{\left[0.080\right]}^m{\left[0.040\right]}^n$

$\text{Rate}3=k{\left[0.080\right]}^m{\left[0.020\right]}^n$

According to the given rate values in the table,

$\frac{\text{Rate}3}{\text{Rate}\text{1}}=\frac{6.25\times {10}^{-6}\text{mol}/\text{L}\cdot \text{s}}{12.5\times {10}^{-6}\text{mol}/\text{L}\cdot \text{s}}$

Therefore,

$\begin{array}{c}\frac{k{\left[0.080\right]}^m{\left[0.020\right]}^n}{k{\left[0.080\right]}^m{\left[0.040\right]}^n}=\frac{6.25\times {10}^{-6}\text{mol}/\text{L}\cdot \text{s}}{12.5\times {10}^{-6}\text{mol}/\text{L}\cdot \text{s}}\\ {\left(\frac{0.020}{0.040}\right)}^m=0.5\\ 0.5={\left(0.5\right)}^m\end{array}$

Simplify the above expression.

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

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

$\text{Rate}=k{\left[{\text{I}}^{-}\right]}^1{\left[{\text{S}}_2{\text{O}}_8^{2-}\right]}^1$

Conclusion

The relation between the reaction rate and the concentration of the reactants is stated by the rate law. The arte law for the given reaction is, $\text{Rate}=k{\left[{\text{I}}^{-}\right]}^1{\left[{\text{S}}_2{\text{O}}_8^{2-}\right]}^1$.

Interpretation Introduction

Interpretation: A reaction between ${\text{I}}^{-}$ and ${\text{S}}_{\text{2}}{\text{O}}_8^{2-}$ and its rate expression is given. The rate law 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.

Answer to Problem 30E

The value of $k$ for all experiments is $\underset{\_}{3.9\times 10^{-3}mo{l}^{-1}\times L^{-1}\times s^{-1}}$.

Explanation of Solution

Given

The stated reaction is,

$2{\text{I}}^{-}\left(aq\right)+{\text{S}}_2{\text{O}}_8^{2-}\left(aq\right)\to {\text{I}}_2\left(aq\right)+2{\text{SO}}_4^{2-}\left(aq\right)$

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

$\text{Rate}=k{\left[{\text{I}}^{-}\right]}^1{\left[{\text{S}}_2{\text{O}}_8^{2-}\right]}^1$

Where,

• $\left[{\text{I}}^{-}\right]$ and $\left[{\text{S}}_2{\text{O}}_8^{2-}\right]$ are the concentrations of ${\text{I}}^{-}$ and ${\text{S}}_2{\text{O}}_8^{2-}$.
• $k$ is the rate constant.

Substitute the values of rate and the concentration of   ${\text{I}}^{-}$ and ${\text{S}}_2{\text{O}}_8^{2-}$, for different experiments, from the given table, in the above expression.

For experiment 1,

$\begin{array}{l}\text{Rate}=k{\left[{\text{I}}^{-}\right]}^1{\left[{\text{S}}_2{\text{O}}_8^{2-}\right]}^1\\ \text{12}\text{.5}\times {\text{10}}^{-6}\text{mol}/\text{L}\cdot \text{s}=k{\left[\text{0}\text{.080}\text{mol}/\text{L}\right]}^1{\left[0.040\text{mol}/\text{L}\right]}^1\\ \text{12}\text{.5}\times {\text{10}}^{-6}\text{mol}/\text{L}\cdot \text{s}=k\left[0.0032{{\text{mol}}^2/\text{L}}^2\right]\end{array}$

Simplify the above expression.

$\begin{array}{c}k=\frac{\text{12}\text{.5}\times {\text{10}}^{-6}\text{mol}/\text{L}\cdot \text{s}}{\left[0.0032{{\text{mol}}^2/\text{L}}^2\right]}\\ =\underset{\_}{3.9\times 10^{-3}mo{l}^{-3}\times L^1\times s^{-1}}\end{array}$

For experiment 2,

$\begin{array}{l}\text{Rate}=k{\left[{\text{I}}^{-}\right]}^1{\left[{\text{S}}_2{\text{O}}_8^{2-}\right]}^1\\ \text{6}\text{.25}\times {\text{10}}^{-6}\text{mol}/\text{L}\cdot \text{s}=k{\left[\text{0}\text{.040}\text{mol}/\text{L}\right]}^1{\left[0.040\text{mol}/\text{L}\right]}^1\\ \text{6}\text{.25}\times {\text{10}}^{-6}\text{mol}/\text{L}\cdot \text{s}=k\left[0.0016{{\text{mol}}^2/\text{L}}^2\right]\end{array}$

Simplify the above expression.

$\begin{array}{c}k=\frac{\text{6}\text{.25}\times {\text{10}}^{-6}\text{mol}/\text{L}\cdot \text{s}}{\left[0.0016{{\text{mol}}^2/\text{L}}^2\right]}\\ =\underset{\_}{3.9\times 10^{-3}mo{l}^{-3}\times L^1\times s^{-1}}\end{array}$

For experiment 3,

$\begin{array}{l}\text{Rate}=k{\left[{\text{I}}^{-}\right]}^1{\left[{\text{S}}_2{\text{O}}_8^{2-}\right]}^1\\ \text{6}\text{.25}\times {\text{10}}^{-6}\text{mol}/\text{L}\cdot \text{s}=k{\left[\text{0}\text{.080}\text{mol}/\text{L}\right]}^1{\left[0.020\text{mol}/\text{L}\right]}^1\\ \text{6}\text{.25}\times {\text{10}}^{-6}\text{mol}/\text{L}\cdot \text{s}=k\left[0.0016{{\text{mol}}^2/\text{L}}^2\right]\end{array}$

Simplify the above expression.

$\begin{array}{c}k=\frac{\text{6}\text{.25}\times {\text{10}}^{-6}\text{mol}/\text{L}\cdot \text{s}}{\left[0.0016{{\text{mol}}^2/\text{L}}^2\right]}\\ =\underset{\_}{3.9\times 10^{-3}mo{l}^{-3}\times L^1\times s^{-1}}\end{array}$

For experiment 4,

$\begin{array}{l}\text{Rate}=k{\left[{\text{I}}^{-}\right]}^1{\left[{\text{S}}_2{\text{O}}_8^{2-}\right]}^1\\ \text{5}\text{.00}\times {\text{10}}^{-6}\text{mol}/\text{L}\cdot \text{s}=k{\left[\text{0}\text{.032}\text{mol}/\text{L}\right]}^1{\left[0.040\text{mol}/\text{L}\right]}^1\\ \text{5}\text{.00}\times {\text{10}}^{-6}\text{mol}/\text{L}\cdot \text{s}=k\left[0.0013{{\text{mol}}^2/\text{L}}^2\right]\end{array}$

Simplify the above expression.

$\begin{array}{c}k=\frac{\text{5}\text{.00}\times {\text{10}}^{-6}\text{mol}/\text{L}\cdot \text{s}}{\left[0.0013{{\text{mol}}^2/\text{L}}^2\right]}\\ \approx \underset{\_}{3.9\times 10^{-3}mo{l}^{-3}\times L^1\times s^{-1}}\end{array}$

For experiment 5,

$\begin{array}{l}\text{Rate}=k{\left[{\text{I}}^{-}\right]}^1{\left[{\text{S}}_2{\text{O}}_8^{2-}\right]}^1\\ \text{7}\text{.00}\times {\text{10}}^{-6}\text{mol}/\text{L}\cdot \text{s}=k{\left[\text{0}\text{.060}\text{mol}/\text{L}\right]}^1{\left[0.030\text{mol}/\text{L}\right]}^1\\ \text{7}\text{.00}\times {\text{10}}^{-6}\text{mol}/\text{L}\cdot \text{s}=k\left[0.0018{{\text{mol}}^2/\text{L}}^2\right]\end{array}$

Simplify the above expression.

$\begin{array}{c}k=\frac{\text{7}\text{.00}\times {\text{10}}^{-6}\text{mol}/\text{L}\cdot \text{s}}{\left[0.0018{{\text{mol}}^2/\text{L}}^2\right]}\\ \approx \underset{\_}{3.9\times 10^{-3}mo{l}^{-3}\times L^1\times s^{-1}}\end{array}$

The value of rate constant is same for all the reactions.

Conclusion

The value of the rate constant $k$ is $\underset{\_}{180mo{l}^{-2}\cdot L^2\cdot mi{n}^{-1}}$.