Chapter 12, Problem 122CP

Hydrogen peroxide and the iodide ion react in acidic solution as follows:

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+3{\text{I}}^{-}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{I}}_3^{-}\left(aq\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)$

The kinetics of this reaction were studied by following the decay of the concentration of H₂O₂ and constructing plots of ln[H₂O₂] versus time. All the plots were linear and all solutions had [H₂O₂]₀ = 8.0 × 10⁻⁴ mol/L. The slopes of these straight lines depended on the initial concentrations of I⁻ and H⁺. The results follow:

[I−]0 (mol/L)	[H+]0 (mol/L)	Slope (min−1)
0.1000	0.0400	− 0.120
0.3000	0.0400	− 0.360
0.4000	0.0400	− 0.480
0.0750	0.0200	− 0.0760
0.0750	0.0800	− 0.118
0.0750	0.1600	− 0.174

The rate law for this reaction has the form

$\text{Rate}=\frac{-\Delta [{\text{H}}_{\text{2}}{\text{O}}_2]}{\Delta t}=\left(k_1+k_2[H^{+}]\right){[I^{-}]}^m{[H_2{O}_2]}^n$

a. Specify the order of this reaction with respect to [H₂O₂] and [I⁻].

b. Calculate the values of the rate constants, k₁ and k₂.

c. What reason could there be for the two-term dependence of the rate on [H⁺]?

Interpretation Introduction

Interpretation: The order of the reaction is to be specified with respect to the reactants. The values of the rate constants are to be calculated. The reason corresponds to the fact that two-term dependence of the rate on the concentration of ${\text{H}}^{+}$ ion is to be stated.

Concept introduction: The order of a reaction is the number or the index raise to the concentration term with respect to each component

The rate law or rate equation is a mathematical representation that relates the rate of the reaction with the concentration or pressure.

Rate constant is a proportionality coefficient that relates the rate of chemical reaction at a specific temperature to the concentration of the reactant or the product.

To determine: The order of the reaction with respect to $\left[{\text{H}}_2{\text{O}}_2\right]$ and $\left[{\text{I}}^{-}\right]$.

Answer to Problem 122CP

The order of the reaction with respect to $\left[{\text{H}}_2{\text{O}}_2\right]$ and $\left[{\text{I}}^{-}\right]$ is zero and first order respectively.

Explanation of Solution

Explanation

The order of the reaction with respect to $\left[{\text{H}}_2{\text{O}}_2\right]$ and $\left[{\text{I}}^{-}\right]$ is zero and first order respectively.

Given

The given reaction is stated as,

${\text{H}}_2{\text{O}}_2\left(aq\right)+3{\text{I}}^{-}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{I}}_3^{-}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$

The plot of $\ln \left[{\text{H}}_2{\text{O}}_2\right]$ Vs time is linear.

The concentration of ${\left[{\text{H}}_2{\text{O}}_2\right]}_0=8.00\times {10}^{-4}\text{M}$

The rate law for the reaction,

$\begin{array}{c}\text{Rate}=\frac{\Delta \left[{\text{H}}_2{\text{O}}_2\right]}{\Delta \text{t}}\\ =\left(k_1+k_2\left[{\text{H}}^{+}\right]\right){\left[{\text{I}}^{-}\right]}^m{\left[{\text{H}}_2{\text{O}}_2\right]}^n\end{array}$

The concentration of $\left[{\text{H}}_2{\text{O}}_2\right]$ is very small which is given above. The concentration of $\left[{\text{H}}^{+}\right]$ and $\left[{\text{I}}^{-}\right]$ is significant. Therefore, the reduced form of rate law is represented as,

$\frac{\text{d}\left[{\text{H}}_2{\text{O}}_2\right]}{\text{dt}}=k_{\text{observed}}{\left[{\text{H}}_2{\text{O}}_2\right]}^n$

Where,

• $k_{\text{observed}}=\left(k_1+k_2\left[{\text{H}}^{+}\right]\right){\left[{\text{I}}^{-}\right]}^m$

The concentration of ${\text{H}}_2{\text{O}}_2$ is very small indicates that the reaction is independent of the reactant ${\text{H}}_2{\text{O}}_2$

Therefore reaction is zero order with respect to the reactant ${\text{H}}_2{\text{O}}_2$.

The order of the reaction with respect to reactant $\left[{\text{I}}^{-}\right]$ is determined by comparing the values of slopes given in table by the given formula,

$\frac{\text{Slope}\left(\exp 1\right)}{\text{Slope}\left(\exp 2\right)}=\frac{\left(k_1+k_2\left[{\text{H}}^{+}\right]\right){\left[{\text{I}}^{-}\right]}^m\left(\exp 1\right)}{\left(k_1+k_2\left[{\text{H}}^{+}\right]\right){\left[{\text{I}}^{-}\right]}^m\left(\exp 2\right)}$

Where,

• $k_1,k_2$ are the rate constants.
• Slopes are the experimental determined values.
• $m$ represents the order of the reaction with respect to reactant ${\text{I}}^{-}$

Substitute the values of slopes, concentration of the reactants form the given data, the value of m is determine as,

$\begin{array}{c}\frac{\text{Slope}\left(\exp 1\right)}{\text{Slope}\left(\exp 2\right)}=\frac{\left(k_1+k_2\left[{\text{H}}^{+}\right]\right){\left[{\text{I}}^{-}\right]}^m\left(\exp 1\right)}{\left(k_1+k_2\left[{\text{H}}^{+}\right]\right){\left[{\text{I}}^{-}\right]}^m\left(\exp 2\right)}\\ \frac{-0.360}{-0.120}=\frac{\left(k_1+k_2\left[0.0400\text{M}\right]\right){\left[0.3000\right]}^m}{\left(k_1+k_2\left[0.0400\text{M}\right]\right){\left[0.1000\right]}^m}\\ 3={\left(\frac{0.3000}{0.1000}\right)}^m\\ m=1\end{array}$

The order of the reaction with respect to $\left[{\text{I}}^{-}\right]$ is first order.

Interpretation Introduction

Interpretation: The order of the reaction is to be specified with respect to the reactants. The values of the rate constants are to be calculated. The reason corresponds to the fact that two-term dependence of the rate on the concentration of ${\text{H}}^{+}$ ion is to be stated.

Concept introduction: The order of a reaction is the number or the index raise to the concentration term with respect to each component

The rate law or rate equation is a mathematical representation that relates the rate of the reaction with the concentration or pressure.

Rate constant is a proportionality coefficient that relates the rate of chemical reaction at a specific temperature to the concentration of the reactant or the product.

To determine: The value of the rate constant $k_1$ and $k_2$ for the given reaction.

Answer to Problem 122CP

The value of $k_1$ and $k_2$ are $\underset{\_}{0.82L/mol\cdot min}$ and $\underset{\_}{9.5L^2/mo{l}^2\cdot min}$ respectively.

Explanation of Solution

Explanation

The value of $k_1$ and $k_2$ are $\underset{\_}{0.82L/mol\cdot min}$ and $\underset{\_}{9.5L^2/mo{l}^2\cdot min}$ respectively.

As proved in step 1 that reaction is first order with respect to one reactant while it is zero order with respect to another reactant. The rate law is given as,

$\text{Rate}=\left(k_1+k_2\left[{\text{H}}^{+}\right]\right)\left[{\text{I}}^{-}\right]$

Where,

• $k_1,k_2$ are the rate constant of the given reaction.
• $\left[{\text{H}}^{+}\right],\left[{\text{I}}^{-}\right]$ represent the concentration of the reactants.

To calculate the value of $k_1,k_2$ the slopes of two experiments are considered and then equations are formulated as,

$\text{Slope}=-\left(k_1+k_2\left[{\text{H}}^{+}\right]\right)\left[{\text{I}}^{-}\right]$

Substitute the values of slopes and concentration of the reactants form the given data of experiment 1.

$\begin{array}{c}\text{Slope}=-\left(k_1+k_2\left[{\text{H}}^{+}\right]\right)\left[{\text{I}}^{-}\right]\\ -0.120{\min }^{-1}=-\left(k_1+k_2\left[0.04000\text{M}\right]\right)\left[0.1000\text{M}\right]\\ 1.20{\min }^{-1}=k_1+k_2\left[0.0400\text{M}\right]\end{array}$ (1)

Substitute the values of slopes and concentration of the reactants form the given data of experiment 4.

$\begin{array}{c}\text{Slope}=-\left(k_1+k_2\left[{\text{H}}^{+}\right]\right)\left[{\text{I}}^{-}\right]\\ -0.0760{\min }^{-1}=-\left(k_1+k_2\left[0.0200\text{M}\right]\right)\left[0.0750\text{M}\right]\\ 1.01{\min }^{-1}=k_1+k_2\left[0.0200\text{M}\right]\end{array}$ (2)

Subtracts the equation 1 from equation 2,the value of $k_1,k_2$ are as follows,

$\begin{array}{l}k{}_1=\underset{\_}{0.82L\times mo{l}^{-1}\times mi{n}^{-1}}\\ k{}_2=\underset{\_}{0.82L^2\times mo{l}^{-2}\times mi{n}^{-1}}\end{array}$

Interpretation Introduction

Interpretation: The order of the reaction is to be specified with respect to the reactants. The values of the rate constants are to be calculated. The reason corresponds to the fact that two-term dependence of the rate on the concentration of ${\text{H}}^{+}$ ion is to be stated.

Concept introduction: The order of a reaction is the number or the index raise to the concentration term with respect to each component

The rate law or rate equation is a mathematical representation that relates the rate of the reaction with the concentration or pressure.

Rate constant is a proportionality coefficient that relates the rate of chemical reaction at a specific temperature to the concentration of the reactant or the product.

To determine: The reason for the two-term dependence of the rate on $\left[{\text{H}}^{+}\right]$

Answer to Problem 122CP

The reason for the two-term dependence of the rate of the reaction on $\left[{\text{H}}^{+}\right]$ is the two rate law expressions.

Explanation of Solution

Explanation

The reason for the two-term dependence of the rate of the reaction on $\left[{\text{H}}^{+}\right]$ is the two rate law expressions.

The amount of the acid present in the given reaction is significant. Therefore, the concentration of the given acid takes part in the reaction. The rate law indicates its presence.

On the other hand the given reaction is an acid catalyzed reaction means presence of an acid only enhances the reaction.

The factor which dominates, decide the rate law of the given reaction.