Chapter 18, Problem 18C.5AE

(i)

Interpretation Introduction

Interpretation:

Entropy of activation has to be calculated for second order gas-phase ozone decomposition which has frequency factor as $\text{4}\text{.6}\text{x}{\text{10}}^{\text{12}}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}$ and activation energy of $\text{10}\text{.0}{\text{kJmol}}^{\text{-1}}$.

Explanation of Solution

Given information:

    $\begin{array}{l}A=4.\text{6}\text{x}{10}^{12}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\\ \text{T}\text{=}\text{298}\text{K}\\ E_a=\text{10}\text{.0}{\text{kJmol}}^{\text{-1}}\end{array}$

Entropy of activation:

In the problem statement it is said that the reaction takes place at low pressure.  Hence, the reaction is assumed to be bimolecular and the rate constant is second-order.

The rate constant equation can also be given as shown below by use of Eyring equation.

    $\begin{array}{l}{k}_r=B{e}^{{\Delta }^{‡}S/\mathrm{R}}{e}^{-{\Delta }^{‡}H/RT}\\ Where,\\ B=\left(\frac{kT}{h}\right)\left(\frac{RT}{p^{\sout{\text{O}}}}\right)=\frac{kR{T}^2}{h{p}^{\sout{\text{O}}}}\end{array}$

Comparing the above equation with equation (1),

    $k_r=B{e}^{{\Delta }^{‡}S/\mathrm{R}}{e}^{-E_a/RT}\text{e}=A{e}^{-(E_a/RT)}$

Therefore.

  $A=eB{e}^{{\Delta }^{‡}S/\mathrm{R}}$

The above equation implies that,

    ${\Delta }^{‡}S=R\left(\ln \frac{A}{B}-2\right)----------------(1)$

$B$ can be calculated as shown below,

    $\begin{array}{l}B=\frac{kR{T}^2}{h{p}^{\sout{\text{O}}}}\\ =\frac{\left(\text{1}\text{.381}\text{x}{\text{10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\right)\text{x}\left(\text{8}\text{.314}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\text{x}{\left(\text{298}\text{K}\right)}^2}{\text{6}\text{.626}\text{x}{\text{10}}^{\text{-34}}\text{Js}\text{x}{\text{10}}^{\text{5}}\text{Pa}}\\ =\frac{\text{1019615}\text{.025}\text{x}{\text{10}}^{\text{-23}}{\text{J}}^{\text{2}}{\text{mol}}^{\text{-1}}}{\text{6}\text{.626}\text{x}{\text{10}}^{\text{-34}}\text{Js}\text{x}1\text{atm}}\\ =\text{1}\text{.538}\text{x}{\text{10}}^{\text{16}}{\text{Jatm}}^{\text{-1}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\end{array}$

It is known that $1J$ is equal to $\text{9}\text{.87}\text{x}{\text{10}}^{\text{-3}}{\text{dm}}^{\text{3}}\text{atm}$.  Therefore, the value of $B$ can be expressed as shown below,

    $\begin{array}{l}B=\text{1}\text{.538}\text{x}{\text{10}}^{\text{16}}\text{x}\text{9}\text{.87}\text{x}{\text{10}}^{\text{-3}}{\text{dm}}^{\text{3}}\text{atm}{\text{atm}}^{\text{-1}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\\ =15.180\text{x}{\text{10}}^{\text{13}}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\\ =1.518\text{x}{\text{10}}^{\text{14}}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\end{array}$

Substitution of the obtained values in the equation (1), entropy of activation can be calculated as shown below,

    $\begin{array}{l}{\Delta }^{‡}S=R\left(\ln \frac{A}{B}-2\right)\\ =R\left(\ln \frac{4.\text{6}\text{x}{10}^{12}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}}{1.518\text{x}{\text{10}}^{\text{14}}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}}-2\right)\\ =\text{8}\text{.314}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\left(-5.496\right)\\ =-45.69{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

The entropy of activation is calculated as $-45.69{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

(ii)

Interpretation Introduction

Interpretation:

Enthalpy of activation has to be calculated for second order gas-phase ozone decomposition which has frequency factor as $\text{4}\text{.6}\text{x}{\text{10}}^{\text{12}}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}$ and activation energy of $\text{10}\text{.0}{\text{kJmol}}^{\text{-1}}$.

Explanation of Solution

Given information:

    $\begin{array}{l}A=4.\text{6}\text{x}{10}^{12}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\\ \text{T}\text{=}\text{298}\text{K}\\ E_a=\text{10}\text{.0}{\text{kJmol}}^{\text{-1}}\end{array}$

Enthalpy of activation:

Enthalpy of activation can be given by the equation shown below,

  $\begin{array}{l}{E}_a={\Delta }^{‡}H+RT\\ {\Delta }^{‡}H=E_a-RT\\ =10.0{\text{kJmol}}^{\text{-1}}-\left(\text{8}\text{.314}\text{x}{\text{10}}^{\text{-3}}{\text{kJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{x}298\text{K}\right)\\ =10.0{\text{kJmol}}^{\text{-1}}-\left(\text{8}\text{.314}\text{x}{\text{10}}^{\text{-3}}{\text{kJmol}}^{\text{-1}}\text{x}298\right)\\ =10.0{\text{kJmol}}^{\text{-1}}-2.477{\text{kJmol}}^{\text{-1}}\\ =7.523{\text{kJmol}}^{\text{-1}}\end{array}$

Therefore, the enthalpy of activation is calculated as $\text{7}\text{.523}{\text{kJmol}}^{\text{-1}}$.

(iii)

Interpretation Introduction

Interpretation:

Gibbs energy of activation has to be calculated for second order gas-phase ozone decomposition which has frequency factor as $\text{4}\text{.6}\text{x}{\text{10}}^{\text{12}}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}$ and activation energy of $\text{10}\text{.0}{\text{kJmol}}^{\text{-1}}$.

Explanation of Solution

Given information:

    $\begin{array}{l}A=4.\text{6}\text{x}{10}^{12}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\\ \text{T}\text{=}\text{298}\text{K}\\ E_a=\text{10}\text{.0}{\text{kJmol}}^{\text{-1}}\end{array}$

Enthalpy of activation (${\Delta }^{‡}H$) is calculated as $\text{7}\text{.523}{\text{kJmol}}^{\text{-1}}$.

Entropy of activation (${\Delta }^{‡}S$) is calculated as $-45.69{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

Gibbs energy of activation:

Gibbs energy can be calculated using the equation given below,

    ${\Delta }^{‡}G={\Delta }^{‡}H-T{\Delta }^{‡}S---------------(1)$

Substitution of the obtained values in equation (1), Gibbs energy of activation can be calculated as shown below,

    $\begin{array}{l}{\Delta }^{‡}G=\text{7}\text{.523}{\text{kJmol}}^{\text{-1}}-\left(298\text{K}\text{x}(-45.69{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}})\right)\\ =\text{7}\text{.523}{\text{kJmol}}^{\text{-1}}+13.61562{\text{kJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =\text{21}\text{.138}{\text{kJmol}}^{\text{-1}}\end{array}$

Therefore, the Gibbs energy of activation is calculated as $\text{21}\text{.138}{\text{kJmol}}^{\text{-1}}$.