Chapter 14, Problem 14.5QP

(a)

Interpretation Introduction

Interpretation: The rate expression for the given reaction has to be written.

Concept Introduction:

The rate of a reaction is the measure of how fast a reactant is consumed or a product is formed. The rate of a reaction is the ratio of change in concentration of reactant or product to the elapsed time.

Consider a reaction:

$\text{A}\stackrel{}{\to }\text{B}$

In this reaction, the reactant $\text{A}$ has been converted into the product $\text{B}$.

Based on the definition of rate, the rate expression for this reaction can be given as follows:

$\text{Rate}\text{=}\text{-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}$

-ve sign indicates the decrease in concentration of the reactant A.

$\left[\text{A}\right]$-Concentration of the reactant A in molarity.

$\text{Rate}\text{=}\text{+}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}$

+ve sign indicates the increase in concentration of the product B.

$\left[\text{B}\right]$-Concentration of the reactant B in molarity.

$\text{Δt}\text{-}\text{Time}\text{period}$

$\text{Δ}$- Difference between initial and final states.

For a complex reaction such as:

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

The stoichiometric coefficients are considered in the rate expression for the complex reactions.

In this complex reaction:

The stoichiometric coefficients of the reactants are: a and b

The stoichiometric coefficients of the products are: c and d

Therefore, the rate expression for the reaction can be given as follows:

$\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{a}}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{1}}{\text{b}}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{c}}\frac{\text{Δ}\left[\text{C}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{d}}\frac{\text{Δ}\left[\text{D}\right]}{\text{Δt}}$

Explanation of Solution

The given reaction is :

${\text{H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{I}}_{\text{2}}\left(\text{g}\right)\stackrel{}{\to }\text{2HI}\left(\text{g}\right)$

The stoichiometric coefficients are considered in the rate expression for the complex reactions.

In this complex reaction:

The stoichiometric coefficients of the reactants are: 1 and 1

The stoichiometric coefficients of the product is: 2

Therefore, the rate expression for the reaction can be given as follows:

$\text{Rate}\text{=}\text{-}\frac{\text{Δ}\left[{\text{H}}_{\text{2}}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{Δ}\left[{\text{I}}_{\text{2}}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{2}}\frac{\text{Δ}\left[\text{HI}\right]}{\text{Δt}}$

(b)

Interpretation Introduction

Interpretation: The rate expression for the given reaction has to be written.

Concept Introduction:

The rate of a reaction is the measure of how fast a reactant is consumed or a product is formed. The rate of a reaction is the ratio of change in concentration of reactant or product to the elapsed time.

Consider a reaction:

$\text{A}\stackrel{}{\to }\text{B}$

In this reaction, the reactant $\text{A}$ has been converted into the product $\text{B}$.

Based on the definition of rate, the rate expression for this reaction can be given as follows:

$\text{Rate}\text{=}\text{-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}$

-ve sign indicates the decrease in concentration of the reactant A.

$\left[\text{A}\right]$-Concentration of the reactant A in molarity.

$\text{Rate}\text{=}\text{+}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}$

+ve sign indicates the increase in concentration of the product B.

$\left[\text{B}\right]$-Concentration of the reactant B in molarity.

$\text{Δt}\text{-}\text{Time}\text{period}$

$\text{Δ}$- Difference between initial and final states.

For a complex reaction such as:

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

The stoichiometric coefficients are considered in the rate expression for the complex reactions.

In this complex reaction:

The stoichiometric coefficients of the reactants are: a and b

The stoichiometric coefficients of the products are: c and d

Therefore, the rate expression for the reaction can be given as follows:

$\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{a}}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{1}}{\text{b}}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{c}}\frac{\text{Δ}\left[\text{C}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{d}}\frac{\text{Δ}\left[\text{D}\right]}{\text{Δt}}$

Explanation of Solution

The given reaction is :

${\text{2H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)\stackrel{}{\to }{\text{2H}}_2\text{O}\left(\text{g}\right)$

The stoichiometric coefficients are considered in the rate expression for the complex reactions.

In this complex reaction:

The stoichiometric coefficients of the reactants are: 2 and 1

The stoichiometric coefficients of the product is: 2

Therefore, the rate expression for the reaction can be given as follows:

$\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{2}}\frac{\text{Δ}\left[{\text{H}}_{\text{2}}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{Δ}\left[{\text{O}}_{\text{2}}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{2}}\frac{\text{Δ}\left[{\text{H}}_2\text{O}\right]}{\text{Δt}}$

(c)

Interpretation Introduction

Interpretation: The rate expression for the given reaction has to be written.

Concept Introduction:

The rate of a reaction is the measure of how fast a reactant is consumed or a product is formed. The rate of a reaction is the ratio of change in concentration of reactant or product to the elapsed time.

Consider a reaction:

$\text{A}\stackrel{}{\to }\text{B}$

In this reaction, the reactant $\text{A}$ has been converted into the product $\text{B}$.

Based on the definition of rate, the rate expression for this reaction can be given as follows:

$\text{Rate}\text{=}\text{-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}$

-ve sign indicates the decrease in concentration of the reactant A.

$\left[\text{A}\right]$-Concentration of the reactant A in molarity.

$\text{Rate}\text{=}\text{+}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}$

+ve sign indicates the increase in concentration of the product B.

$\left[\text{B}\right]$-Concentration of the reactant B in molarity.

$\text{Δt}\text{-}\text{Time}\text{period}$

$\text{Δ}$- Difference between initial and final states.

For a complex reaction such as:

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

The stoichiometric coefficients are considered in the rate expression for the complex reactions.

In this complex reaction:

The stoichiometric coefficients of the reactants are: a and b

The stoichiometric coefficients of the products are: c and d

Therefore, the rate expression for the reaction can be given as follows:

$\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{a}}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{1}}{\text{b}}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{c}}\frac{\text{Δ}\left[\text{C}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{d}}\frac{\text{Δ}\left[\text{D}\right]}{\text{Δt}}$

Explanation of Solution

The given reaction is :

${\text{5Br}}^{-}\left(\text{aq}\right)\text{+}{\text{BrO}}_3^{-}\left(\text{aq}\right)\text{+}\text{6}{\text{H}}^{+}\left(\text{aq}\right)\stackrel{}{\to }{\text{3Br}}_2\left(\text{aq}\right)+{\text{3H}}_2\text{O}\left(\text{l}\right)$

The stoichiometric coefficients are considered in the rate expression for the complex reactions.

In this complex reaction:

The stoichiometric coefficients of the reactants are: 5,1and 6

The stoichiometric coefficients of the products are: 3and 3

Therefore, the rate expression for the reaction can be given as follows:

$\text{Rate}\text{=}\text{-}\frac{\text{1}}{5}\frac{\text{Δ}\left[{\text{Br}}^{-}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{Δ}\left[{\text{BrO}}_3^{-}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{1}}{6}\frac{\text{Δ}\left[{\text{H}}^{+}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{3}}\frac{\text{Δ}\left[{\text{Br}}_2\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{3}}\frac{\text{Δ}\left[{\text{H}}_2\text{O}\right]}{\text{Δt}}$