Chapter 13, Problem 13.103QA

Interpretation Introduction

To find:

a) Write the equation for the overall reaction.

b) Write the rate law predicted by the mechanism for the overall reaction.

c) Identify the catalyst in the reaction.

d) Identify any intermediates formed in the reaction

Answer to Problem 13.103QA

Solution:

a) Overall reaction equation is $2{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{a}\mathrm{q}\right)\mathrm{ }\to 2\mathrm{ }{\mathrm{H}}_2\mathrm{O}\mathrm{ }\left(\mathrm{a}\mathrm{q}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$

b) Rate law predicted by the mechanism for the overall reaction is: $\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}=\mathrm{k}\mathrm{ }\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]\mathrm{ }\left[{\mathrm{I}}^{-}\right]$

c) Catalyst in the reaction is ${\mathrm{I}}^{-}\left(\mathrm{a}\mathrm{q}\right)$

d) The intermediate formed in the reaction is $\mathrm{I}{\mathrm{O}}^{-}\left(\mathrm{a}\mathrm{q}\right)$

Explanation of Solution

1) Concept:

The overall reaction for any given mechanism can be written by adding the two reactions and cancelling out the intermediates. Thus, we add steps 1 and 2 to get the overall reaction for the mechanism. The rate law is written for the overall reaction. A catalyst is a substance added in the reaction, which does not take part in the overall reaction but decrease the activation energy, hence increasing the overall rate of the reaction. Intermediates in a reaction are the stable ions formed during the course of reaction, which are formed in one step and get consumed in the other step. Hence, they do not appear in the overall reaction.

2) Given:

Mechanism for the reaction of hydrogen peroxide and potassium iodide is given in two steps as:

$\mathrm{S}\mathrm{t}\mathrm{e}\mathrm{p}\mathrm{ }1\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{a}\mathrm{q}\right)+{\mathrm{I}}^{-}\left(\mathrm{a}\mathrm{q}\right)\mathrm{ }\to {\mathrm{H}}_2\mathrm{O}\mathrm{ }\left(\mathrm{a}\mathrm{q}\right)+\mathrm{I}{\mathrm{O}}^{-}\left(\mathrm{a}\mathrm{q}\right)\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{s}\mathrm{l}\mathrm{o}\mathrm{w}$

$\mathrm{S}\mathrm{t}\mathrm{e}\mathrm{p}\mathrm{ }2\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{\mathrm{ }\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{a}\mathrm{q}\right)+\mathrm{ }\mathrm{ }\mathrm{I}{\mathrm{O}}^{-}\left(\mathrm{a}\mathrm{q}\right)\mathrm{ }\to {\mathrm{H}}_2\mathrm{O}\mathrm{ }\left(\mathrm{a}\mathrm{q}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)+{\mathrm{I}}^{-}\left(\mathrm{a}\mathrm{q}\right)\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{f}\mathrm{a}\mathrm{s}\mathrm{t}$

3) Calculations:

a) In order to write the overall reaction, we add the two steps as:

$\mathrm{ }\mathrm{ }\mathrm{S}\mathrm{t}\mathrm{e}\mathrm{p}\mathrm{ }1\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{a}\mathrm{q}\right)+{\mathrm{I}}^{-}\left(\mathrm{a}\mathrm{q}\right)\mathrm{ }\to {\mathrm{H}}_2\mathrm{O}\mathrm{ }\left(\mathrm{a}\mathrm{q}\right)+\mathrm{I}{\mathrm{O}}^{-}\left(\mathrm{a}\mathrm{q}\right)\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{s}\mathrm{l}\mathrm{o}\mathrm{w}$

+

$\mathrm{ }\mathrm{S}\mathrm{t}\mathrm{e}\mathrm{p}\mathrm{ }2\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{\mathrm{ }\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{a}\mathrm{q}\right)+\mathrm{ }\mathrm{ }\mathrm{I}{\mathrm{O}}^{-}\left(\mathrm{a}\mathrm{q}\right)\mathrm{ }\to {\mathrm{H}}_2\mathrm{O}\mathrm{ }\left(\mathrm{a}\mathrm{q}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)+{\mathrm{I}}^{-}\left(\mathrm{a}\mathrm{q}\right)\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{f}\mathrm{a}\mathrm{s}\mathrm{t}$

$2{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{a}\mathrm{q}\right)\mathrm{ }\to 2\mathrm{ }{\mathrm{H}}_2\mathrm{O}\mathrm{ }\left(\mathrm{a}\mathrm{q}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{O}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{ }\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}$

b) Rate law for the reaction is written as:

$\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}=\mathrm{k}\mathrm{ }\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]\mathrm{ }\left[{\mathrm{I}}^{-}\right]$

Though ${\mathrm{I}}^{-}$ does not appear in the overall reaction, it is a part of the slow reaction. It is a catalyst, and though it doesn’t affect the rate of the reaction, its concentration can be measured.

c) Species ${\mathrm{I}}^{-}$ is a catalyst as it is getting consumed in the first slow step and regenerated in the second fast step.

d) Species $\mathrm{I}{\mathrm{O}}^{-}$ is an intermediate formed because it is produced in the first step and consumed in the second step. It does not appear in the overall reaction.

Conclusion:

Generally, therate law should be consistent with the overall reaction, but in the case of catalysts, it may not appear in the overall reaction, though it can appear in the slow reaction, which is the rate determining step. And the effect of change of concentration of the rate can be measured. Hence, it may occur in the rate law expression.