Chapter 14, Problem 78GQ

(a)

Interpretation Introduction

Interpretation:

The rate law has to be determined for the given reaction.

Concept introduction:

Rate law: It is an equation that related to the dependence of the reaction rate on the concentration of each substrate (reactants).

For a reaction,

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

  Where,

  A and B are reactants

  a and b are stoichiometric coefficients

$\text{Rate}\text{=}\text{-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{k}{\left[\text{A}\right]}^{\text{x}}{\left[\text{B}\right]}^{\text{y}}{\left[\text{C}\right]}^z$

  Where,

  k is the rate constant

Rate constant: Rate constant is an expression used to relate the rate of a reaction to the concentration of reactants participating in the reaction.

Answer to Problem 78GQ

The rate law for the given reaction is $\text{Rate}\text{=}\text{k}\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]$.

Explanation of Solution

Given information,

The reaction is $\text{2}{\text{H}}_2{O}_2{}_{\left(aq\right)}\underset{}{\to }\text{2}{\text{H}}_2{O}_{\left(l\right)}+O_2{}_{\left(aq\right)}$

  $\begin{array}{cc}\begin{array}{l}\left[H_2{O}_2\right]\\ \left(\text{mol/L}\right)\end{array}& \begin{array}{l}\text{Initial}Reaction\text{Rate}\\ \left(\text{mol}{\text{O}}_{\text{2}}\text{/L}\text{.}\min \right)\end{array}\\ 0.0500& 5.30\times {10}^{-5}\\ 0.100& 1.06\times {10}^{-4}\\ 0.200& 2.12\times {10}^{-4}\end{array}$

Calculate the value of $x$ from experiment 1 and 2

  $\begin{array}{c}\frac{Rate1}{Rate2}\text{=}\frac{k{\left[H_2{O}_2\right]}_1^x}{k{\left[H_2{O}_2\right]}_2^x}\\ \\ \frac{\left(5.30\times {10}^{-5}\text{mol/L}\text{.min}\right)}{\left(1.06\times {10}^{-4}\text{mol/L}\text{.min}\right)}\text{=}{\left(\frac{0.0500\text{mol/L}}{0.100\text{mol/L}}\right)}^x\\ \\ 0.50=0.50\\ \\ n=1\end{array}$

Therefore, the rate law of the given reaction is $\text{Rate}\text{=}\text{k}\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]$.

(b)

Interpretation Introduction

Interpretation:

The value of rate has to be determined for the given reaction.

Concept introduction:

Rate law: It is an equation that related to the dependence of the reaction rate on the concentration of each substrate (reactants).

For a reaction,

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

  Where,

  A and B are reactants

  a and b are stoichiometric coefficients

  $\text{Rate}\text{=}\text{-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{k}{\left[\text{A}\right]}^{\text{x}}{\left[\text{B}\right]}^{\text{y}}{\left[\text{C}\right]}^z$

  Where,

  k is the rate constant

Rate constant: Rate constant is an expression used to relate the rate of a reaction to the concentration of reactants participating in the reaction.

Answer to Problem 78GQ

The value of rate constant is $1.06\times {10}^{-3}{\min }^{-1}$.

Explanation of Solution

Given information,

The reaction is $\text{2}{\text{H}}_2{O}_2{}_{\left(aq\right)}\underset{}{\to }\text{2}{\text{H}}_2{O}_{\left(l\right)}+O_2{}_{\left(aq\right)}$

  $\begin{array}{cc}\begin{array}{l}\left[H_2{O}_2\right]\\ \left(\text{mol/L}\right)\end{array}& \begin{array}{l}\text{Initial}Reaction\text{Rate}\\ \left(\text{mol}{\text{O}}_{\text{2}}\text{/L}\text{.}\min \right)\end{array}\\ 0.0500& 5.30\times {10}^{-5}\\ 0.100& 1.06\times {10}^{-4}\\ 0.200& 2.12\times {10}^{-4}\end{array}$

Therefore, the rate law of the given reaction is $\text{Rate}\text{=}\text{k}\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]$.

Calculate the value of rate constant from experiment 1

  $\begin{array}{c}\text{Rate}\text{=}\text{k}\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]\\ \\ \text{Rate}\text{constant, k}\text{=}\frac{\text{Rate}}{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}\text{=}\frac{\left(5.30\times {10}^{-5}\text{mol/L}\text{.min}\right)}{\left(0.0500\text{mol/L}\right)}\frac{4.\text{0}\text{×}{\text{10}}^{\text{-4}}\text{M/s}}{}\\ \\ \text{=}1.06\times {10}^{-3}{\min }^{-1}\end{array}$

Therefore, the value of rate constant is $1.06\times {10}^{-3}{\min }^{-1}$.

(c)

Interpretation Introduction

Interpretation:

The value of rate has to be determined for the given reaction.

Concept introduction:

Rate law: It is an equation that related to the dependence of the reaction rate on the concentration of each substrate (reactants).

For a reaction,

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

  Where,

  A and B are reactants

  a and b are stoichiometric coefficients

  $\text{Rate}\text{=}\text{-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{k}{\left[\text{A}\right]}^{\text{x}}{\left[\text{B}\right]}^{\text{y}}{\left[\text{C}\right]}^z$

  Where,

  k is the rate constant

Rate constant: Rate constant is an expression used to relate the rate of a reaction to the concentration of reactants participating in the reaction.

Rate constant for a particular reaction is always constant. It does not depend on the concentration of the reactant.

Answer to Problem 78GQ

The value of rate constant is $1.06\times {10}^{-3}{\min }^{-1}$.

Explanation of Solution

Given information,

The reaction is ${\text{H}}_2{O}_2{}_{\left(aq\right)}\underset{}{\to }{\text{H}}_2{O}_{\left(l\right)}+\text{1/2}{O}_2{}_{\left(aq\right)}$

Rate constant for a particular reaction is always constant. It does not depend on the concentration of the reactant.

Hence the rate constant of the given reaction is same the rate constant of $\text{2}{\text{H}}_2{O}_2{}_{\left(aq\right)}\underset{}{\to }\text{2}{\text{H}}_2{O}_{\left(l\right)}+O_2{}_{\left(aq\right)}$ since the reaction is same and only concentration differs.

Therefore, the value of rate constant is $1.06\times {10}^{-3}{\min }^{-1}$.