Chapter 6, Problem 6D.9E

(a)

Interpretation Introduction

Interpretation:

The value of activation energy for the reaction has to be calculated where ${10}^{-3}$ of the collisions occur with sufficient kinetic energy for the reaction at the temperature of ${\text{200}}^{\text{ο}}\text{C}$.

Concept Introduction:

Arrhenius equation:

Arrhenius equation gives the temperature dependence of reaction rates.  It can be represented as,

  $\text{k}\text{=}\text{A}{\text{e}}^{\text{-}\frac{{\text{E}}_{\text{a}}}{\text{RT}}}$

Where,

$\text{k}$ is the rate constant

A is the pre-exponential factor

T is temperature

R is the Universal gas constant

${\text{E}}_{\text{a}}$ is the activation energy

Fraction of collision:

According to kinetic theory of gas all the molecules are moving randomly and continuously colliding with each other.  Now according to collision theory reactions occur when the reactant molecules collide with a certain kinetic energy in proper orientation known as activation energy.  Now all the collisions are not effective for reaction to occur.  Only a fraction of the collisions which are done at or above activation energy and in proper orientation is effective.  This fraction is called fraction of collision that depends on the activation energy of the reaction and the temperature.  It can be denoted as,

  $\text{f}\text{=}{\text{e}}^{\text{-}\frac{{\text{E}}_{\text{a}}}{\text{RT}}}$

Where f is the fraction of collision.

Answer to Problem 6D.9E

The activation energy of the reaction for the given collision frequency is $\text{f}\text{=}{10}^{-3}$ is $\text{5}\text{.123}{\text{kJmol}}^{\text{-1}}$

Explanation of Solution

According to Arrhenius equation,

  $\text{k}\text{=}\text{A}{\text{e}}^{\text{-}\frac{{\text{E}}_{\text{a}}}{\text{RT}}}$

The given collision frequency is $\text{f}\text{=}{10}^{-3}$.

Temperature in kelvin,

  $\begin{array}{l}\text{T}\text{=}\text{(200+273)}\text{K}\\ \\ \text{T}\text{=}\text{473K}\end{array}$

Now putting the values in the equation,

  $\begin{array}{c}\text{f}\text{=}{\text{e}}^{\text{-}\frac{{\text{E}}_{\text{a}}}{\text{RT}}}\\ \\ {\text{10}}^{\text{-3}}\text{=}{\text{e}}^{\text{-}\frac{{\text{E}}_{\text{a}}}{\text{8}{\text{.314Jmol}}^{\text{-1}}{\text{K}}^{\text{-1}}\text{×473K}}}\\ \\ \log \left({\text{10}}^{\text{-3}}\right)=2.303\times \left(\text{-}\frac{{\text{E}}_{\text{a}}}{\text{8}{\text{.314Jmol}}^{\text{-1}}{\text{K}}^{\text{-1}}\text{×473K}}\right)\\ \\ -3=2.303\times \left(\text{-}\frac{{\text{E}}_{\text{a}}}{\text{8}{\text{.314Jmol}}^{\text{-1}}\text{×473}}\right)\\ \\ {\text{E}}_{\text{a}}\text{=}\text{5122}\text{.69}{\text{Jmol}}^{\text{-1}}\\ \\ {\text{E}}_{\text{a}}\text{=}\text{5}\text{.123}{\text{kJmol}}^{\text{-1}}\end{array}$

Hence the activation energy of the reaction for the given collision frequency is $\text{f}\text{=}{10}^{-3}$ is $\text{5}\text{.123}{\text{kJmol}}^{\text{-1}}$.

(b)

Interpretation Introduction

Interpretation:

The value of activation energy for the reaction has to be calculated where ${10}^{-6}$ of the collisions occur with sufficient kinetic energy for the reaction at the temperature of ${\text{200}}^{\text{ο}}\text{C}$.

Concept Introduction:

Arrhenius equation:

Arrhenius equation gives the temperature dependence of reaction rates.  It can be represented as,

  $\text{k}\text{=}\text{A}{\text{e}}^{\text{-}\frac{{\text{E}}_{\text{a}}}{\text{RT}}}$

Where,

$\text{k}$ is the rate constant

A is the pre-exponential factor

T is temperature

R is the Universal gas constant

${\text{E}}_{\text{a}}$ is the activation energy

Fraction of collision:

According to kinetic theory of gas all the molecules are moving randomly and continuously colliding with each other.  Now according to collision theory reactions occur when the reactant molecules collide with a certain kinetic energy in proper orientation known as activation energy.  Now all the collisions are not effective for reaction to occur.  Only a fraction of the collisions which are done at or above activation energy and in proper orientation is effective.  This fraction is called fraction of collision that depends on the activation energy of the reaction and the temperature.  It can be denoted as,

$\text{f}\text{=}{\text{e}}^{\text{-}\frac{{\text{E}}_{\text{a}}}{\text{RT}}}$

Where f is the fraction of collision.

Answer to Problem 6D.9E

The activation energy of the reaction for the given collision frequency is $\text{f}\text{=}{10}^{-6}$ is $10.\text{245}{\text{kJmol}}^{\text{-1}}$.

Explanation of Solution

According to Arrhenius equation,

  $\text{k}\text{=}\text{A}{\text{e}}^{\text{-}\frac{{\text{E}}_{\text{a}}}{\text{RT}}}$

Now according to the question the given collision frequency is $\text{f}\text{=}{10}^{-6}$.

Temperature in kelvin,

  $\begin{array}{l}\text{T}\text{=}\text{(200+273)}\text{K}\\ \\ \text{T}\text{=}\text{473K}\end{array}$

Now putting the values in the equation,

  $\begin{array}{c}\text{f}\text{=}{\text{e}}^{\text{-}\frac{{\text{E}}_{\text{a}}}{\text{RT}}}\\ \\ {\text{10}}^{\text{-6}}\text{=}{\text{e}}^{\text{-}\frac{{\text{E}}_{\text{a}}}{\text{8}{\text{.314Jmol}}^{\text{-1}}{\text{K}}^{\text{-1}}\text{×473K}}}\\ \\ \log \left({\text{10}}^{\text{-6}}\right)=2.303\times \left(\text{-}\frac{{\text{E}}_{\text{a}}}{\text{8}{\text{.314Jmol}}^{\text{-1}}{\text{K}}^{\text{-1}}\text{×473K}}\right)\\ \\ -6=2.303\times \left(\text{-}\frac{{\text{E}}_{\text{a}}}{\text{8}{\text{.314Jmol}}^{\text{-1}}\text{×473}}\right)\\ \\ {\text{E}}_{\text{a}}\text{=}10245.\text{39}{\text{Jmol}}^{\text{-1}}\\ \\ {\text{E}}_{\text{a}}\text{=}10.\text{245}{\text{kJmol}}^{\text{-1}}\end{array}$

Hence the activation energy of the reaction for the given collision frequency $\text{f}\text{=}{10}^{-6}$ is $10.\text{245}{\text{kJmol}}^{\text{-1}}$.