Chapter 18, Problem 18C.1P

Interpretation Introduction

Interpretation:

Enthalpy, entropy, total energy and Gibbs energy for the given reaction has to be calculated.

Explanation of Solution

Given information:

Frequency factor ($A$)is given as $4.07\times {10}^5{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}$.  Activation energy ($E_a$) is given as $\text{65}\text{.4}{\text{kJmol}}^{\text{-1}}$.  Temperature ($T$) is given as $\text{300}\text{K}$.

Rate constant for the given reaction:

Arrhenius equation for rate constant can be given as,

    $k_r=A\times e^{-E_a/RT}-----------------(1)$

Rate constant can be calculated by substituting the values in equation (1) as shown below,

    $\begin{array}{l}{k}_r=4.07\times {10}^5{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\times e^{-65.43{\text{kJmol}}^{\text{-1}}/(8.314{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times \text{300}\text{K)}}\\ =4.07\times {10}^5{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\times e^{-65.43{\text{kJmol}}^{\text{-1}}/(2494.2{\text{Jmol}}^{\text{-1}}\text{)}}\\ =4.07\times {10}^5{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\times e^{-65430{\text{Jmol}}^{\text{-1}}/2494.2{\text{Jmol}}^{\text{-1}}}\\ =4.07\times {10}^5{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\times e^{-26.232}\\ =4.07\times {10}^5{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\times 4.051\times {10}^{-12}\\ =16.487\times {10}^{-7}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}=1.65\times {10}^{-6}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\end{array}$

Therefore, rate constant is calculated as $1.65\times {10}^{-6}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}$.

Enthalpy of Reaction:

The equation for enthalpy of reaction can be given as,

    ${\Delta }^{‡}H=E_a-RT-----------------(2)$

Temperature is given as $300\text{K}$.

Substituting the values in equation (2), enthalpy of reaction can be obtained as shown below,

    $\begin{array}{l}{\Delta }^{‡}H=65.43{\text{kJmol}}^{\text{-1}}-(8.314{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times 300\text{K)}\\ \text{=}(\text{65}\text{.43}\text{-}\text{2}{\text{.494)kJmol}}^{\text{-1}}\\ =62.936{\text{kJmol}}^{\text{-1}}\end{array}$

Therefore, enthalpy of reaction is calculated as $62.96{\text{kJmol}}^{\text{-1}}$.

Gibbs free energy of reaction:

Gibbs free energy of reaction can be given by the equation shown below,

    $k_r=B{e}^{-\frac{{\Delta }^{‡}G}{RT}}------------------(3)$

Where,

    $T$ is the temperature.

    $R$ is the gas constant.

    $B$ is coefficient of activation of Gibbs free energy.

$B$ can be calculated as shown below,

    $B=\frac{kR{T}^2}{h{P}^0}$

Where,

    $k$ is Boltzmann constant.

    $h$ is Planck’s constant.

    $P^0$ is pressure of surroundings.

    $R$ is gas constant.

    $T$ is temperature.

    $\begin{array}{l}B=\frac{1.381\times {10}^{-23}{\text{JK}}^{\text{-1}}\times 8.314{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times {(300\text{K)}}^{\text{2}}}{6.626\times {10}^{-34}\text{Js}\times \text{1}\text{.00}\times {\text{10}}^{\text{5}}\text{Pa}}\\ =1.56\times {10}^{11}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\end{array}$

Substituting the values in equation (3), Gibbs free energy can be calculated as shown below,

    $\begin{array}{l}{k}_r=B{e}^{-\frac{{\Delta }^{‡}G}{RT}}\\ 1.65\times {10}^{-6}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}=1.56\times {10}^{11}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\times e^{-\frac{{\Delta }^{‡}G}{RT}}\\ e^{-\frac{{\Delta }^{‡}G}{RT}}=\frac{1.65\times {10}^{-9}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}}{1.56\times {10}^{11}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}}\\ e^{-\frac{{\Delta }^{‡}G}{RT}}=1.06\times {10}^{-20}\end{array}$

Taking natural log on both sides,

    $\begin{array}{l}-\frac{{\Delta }^{‡}G}{RT}=\ln (1.06\times {10}^{-20})\\ =-45.99\\ {\Delta }^{‡}G=45.99RT\\ =45.99\times 8.3\text{14}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{×}\text{300}\text{K}\\ \text{=}\text{114708}\text{.258}{\text{Jmol}}^{\text{-1}}\\ =\text{114}\text{.7}{\text{kJmol}}^{\text{-1}}\end{array}$

Therefore, Gibbs free energy of reaction is $\text{114}\text{.7}{\text{kJmol}}^{\text{-1}}$.

Entropy of reaction:

The relationship between Gibbs free energy of reaction, enthalpy of reaction and entropy of reaction can be given by the equation as shown below,

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

Where,

    ${\Delta }^{‡}S$ is the entropy of reaction.

Rearranging the equation, the entropy of activation can be obtained as,

    ${\Delta }^{‡}S=\frac{{\Delta }^{‡}H-{\Delta }^{‡}G}{T}---------------(4)$

Substituting the obtained values in equation (4), the entropy of activation can be calculated.

    $\begin{array}{l}{\Delta }^{‡}S=\frac{{\Delta }^{‡}H-{\Delta }^{‡}G}{T}\\ =\frac{62.96{\text{kJmol}}^{\text{-1}}-114.7{\text{kJmol}}^{\text{-1}}}{300\text{K}}\\ =\frac{-51.74{\text{kJmol}}^{\text{-1}}}{300\text{K}}\\ =-0.1724{\text{kJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =-172.4{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Therefore, entropy of reaction is calculated as $-172.4{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

Internal energy of reaction:

Internal energy of the reaction can be given by the equation as shown below,

    ${\Delta }^{‡}U=T{\Delta }^{‡}S-P\Delta V-------------------(5)$

Where,

    ${\Delta }^{‡}S$ is the entropy of reaction.

The change in volume is zero as this is a constant volume process.  Hence, the equation (5) becomes as shown below,

    $\begin{array}{l}{\Delta }^{‡}U=T{\Delta }^{‡}S\\ =300\text{K}\times -172.4{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =-51720{\text{Jmol}}^{\text{-1}}\\ =-51.72{\text{kJmol}}^{\text{-1}}\end{array}$

Therefore, internal energy of reaction is calculated as $-51.72{\text{kJmol}}^{\text{-1}}$.