Chapter 18, Problem

Interpretation Introduction

Interpretation:

Enthalpy, entropy, energy and Gibbs energy of activation at $-20\text{°C}$ has to be calculated for the given decomposition of cis- and trans-azoalkanes.

Explanation of Solution

Given information:

    $\begin{array}{l}\theta /°\text{C}\text{-24}\text{.82}\text{-20}\text{.73}\text{-17}\text{.02}\text{-13}\text{.00}\text{-8}\text{.95}\\ {\text{10}}^{\text{4}}\times k_r/s^{-1}1.222.314.398.5014.3\end{array}$

Gibbs Free Energy of Activation:

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

    $k_r=B{e}^{-\frac{{\Delta }^{‡}G}{RT}}------------------(1)$

Where,

    $T$ is the temperature.

    $R$ is the gas constant.

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

Taking log on both sides in equation (1),

    $\ln k_r=\ln B-\frac{{\Delta }^{‡}G}{RT}$

If $1/T$ is plotted against $k_r$, the intercept in B will be obtained as a straight line.  Slope can be given as,

    $\text{Slope}=-\frac{{\Delta }^{‡}G}{RT}--------------(2)$

For $k_r=1.22\times {10}^{-4}{\text{s}}^{\text{-1}}$ and $\theta =-24.82°\text{C}$, the temperature in kelvin is,

    $\begin{array}{l}T=(273+\theta )\text{K}\\ \text{=}(\text{273}\text{-}24.82)\text{K}\\ \text{=}\text{248}\text{.18}\text{K}\end{array}$

From this $1/T$ can be obtained as shown below,

    $\begin{array}{l}1/T=1/248.18\text{K}\\ \text{=}\text{4}\text{.029}\times {10}^{-3}{\text{K}}^{-1}\end{array}$

    $\begin{array}{l}\ln k_r=\ln (1.22\times {10}^{-4})\\ =-9.01\end{array}$

For $k_r=2.31\times {10}^{-4}{\text{s}}^{\text{-1}}$ and $\theta =-20.73°\text{C}$, the temperature in kelvin is,

    $\begin{array}{l}T=(273+\theta )\text{K}\\ \text{=}(\text{273}\text{-}20.73)\text{K}\\ \text{=}\text{252}\text{.27}\text{K}\end{array}$

From this $1/T$ can be obtained as shown below,

    $\begin{array}{l}1/T=1/252.27\text{K}\\ \text{=}\text{3}\text{.964}\times {10}^{-3}{\text{K}}^{-1}\end{array}$

    $\begin{array}{l}\ln k_r=\ln (2.31\times {10}^{-4})\\ =-8.37\end{array}$

For $k_r=4.39\times {10}^{-4}{\text{s}}^{\text{-1}}$ and $\theta =-17.02°\text{C}$, the temperature in kelvin is,

    $\begin{array}{l}T=(273+\theta )\text{K}\\ \text{=}(\text{273}\text{-}17.02)\text{K}\\ \text{=}\text{255}\text{.98}\text{K}\end{array}$

From this $1/T$ can be obtained as shown below,

    $\begin{array}{l}1/T=1/255.98\text{K}\\ \text{=}\text{3}\text{.906}\times {10}^{-3}{\text{K}}^{-1}\end{array}$

    $\begin{array}{l}\ln k_r=\ln (4.39\times {10}^{-4})\\ =-7.73\end{array}$

For $k_r=8.50\times {10}^{-4}{\text{s}}^{\text{-1}}$ and $\theta =-13.00°\text{C}$, the temperature in kelvin is,

    $\begin{array}{l}T=(273+\theta )\text{K}\\ \text{=}(\text{273}\text{-}13.00)\text{K}\\ \text{=}\text{260}\text{.00}\text{K}\end{array}$

From this $1/T$ can be obtained as shown below,

    $\begin{array}{l}1/T=1/260.00\text{K}\\ \text{=}\text{3}\text{.846}\times {10}^{-3}{\text{K}}^{-1}\end{array}$

    $\begin{array}{l}\ln k_r=\ln (8.50\times {10}^{-4})\\ =-7.07\end{array}$

For $k_r=14.3\times {10}^{-4}{\text{s}}^{\text{-1}}$ and $\theta =-8.95°\text{C}$, the temperature in kelvin is,

    $\begin{array}{l}T=(273+\theta )\text{K}\\ \text{=}(\text{273}\text{-}8.95)\text{K}\\ \text{=}\text{264}\text{.05}\text{K}\end{array}$

From this $1/T$ can be obtained as shown below,

    $\begin{array}{l}1/T=1/264.05\text{K}\\ \text{=}\text{3}\text{.787}\times {10}^{-3}{\text{K}}^{-1}\end{array}$

    $\begin{array}{l}\ln k_r=\ln (14.3\times {10}^{-4})\\ =-6.55\end{array}$

Table can be constructed from $1/T$ and $\ln k_r$ as shown below,

    $\begin{array}{l}{\text{10}}^{\text{3}}/T\text{4}\text{.029}\text{3}\text{.964}\text{3}\text{.906}\text{3}\text{.846}\text{3}\text{.787}\\ \ln k_r-9.01-8.37-7.73-7.07-6.55\end{array}$

Slope of the line can be calculated from equation (2) as,

    $\begin{array}{l}-\frac{{\Delta }^{‡}G}{R}=\frac{y_{\text{2}}-y_1}{x_{\text{2}}-x_1}\\ =\frac{-8.37-(-9.01)}{\text{(3}\text{.964}-4.029)\times {10}^{-3}}\text{K}\\ \text{=}\frac{0.64}{-0.065\times {10}^{-3}}\text{K}\\ \frac{{\Delta }^{‡}G}{R}\text{=}\text{9}\text{.86}\times {\text{10}}^{\text{3}}\text{K}\end{array}$

Gibbs free energy of activation is calculated as shown below,

    $\begin{array}{l}{\Delta }^{‡}G\text{=}\text{9}\text{.86}\times {\text{10}}^{\text{3}}\text{K}\times \text{R}\\ \text{=}\text{9}\text{.86}\times {\text{10}}^{\text{3}}\text{K}\times 8.314{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =81.98\times {\text{10}}^{\text{3}}\text{J}{\text{mol}}^{\text{-1}}\\ =81.98\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Therefore, Gibbs free energy of activation is $81.98\text{kJ}{\text{mol}}^{\text{-1}}$.

Enthalpy of Activation:

The equation for enthalpy of activation can be given as,

    ${\Delta }^{‡}H=E_a-RT-----------------(3)$

Temperature is given as $-20°\text{C}$.  Conversion of this into kelvin scale can be done as given below,

    $\begin{array}{l}T=(273+\theta )\text{K}\\ \text{=}(273-\text{20})\text{K}\\ \text{=}\text{253}\text{K}\end{array}$

Substituting the values in equation (3), enthalpy of activation can be obtained as shown below,

    $\begin{array}{l}{\Delta }^{‡}H=81.98{\text{kJmol}}^{\text{-1}}-(8.314{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times 253\text{K)}\\ \text{=}(\text{81}\text{.98}\text{-}\text{2}{\text{.103)kJmol}}^{\text{-1}}\\ =79.88{\text{kJmol}}^{\text{-1}}\end{array}$

Therefore, enthalpy of activation is calculated as $79.88{\text{kJmol}}^{\text{-1}}$.

Entropy of activation:

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

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

Where,

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

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.

    $\begin{array}{l}{\Delta }^{‡}S=\frac{{\Delta }^{‡}H-{\Delta }^{‡}G}{T}\\ =\frac{79.88{\text{kJmol}}^{\text{-1}}-81.98{\text{kJmol}}^{\text{-1}}}{253\text{K}}\\ =\frac{-2.1{\text{kJmol}}^{\text{-1}}}{253\text{K}}\\ =-0.0083{\text{kJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =-8.3{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Therefore, entropy of activation is calculated as $-8.3{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.