Chapter 14.6, Problem 12E

(A)

Interpretation Introduction

Interpretation:

The equilibrium constant using the shortcut van’t Hoff approach.

Concept Introduction:

The expression of the equilibrium constant is,

$K_T=\exp \left(-\frac{\Delta {\underset{\_}{G}}_T^0}{RT}\right)$

Here, standard change in Gibbs free energy is $\Delta {\underset{\_}{G}}_T^0$, gas constant is $R$, and temperature is $T$.

The short-cut van’t Hoff equation is,

$\frac{\Delta {\underset{\_}{G}}_T^0}{RT}=\frac{\Delta {\underset{\_}{G}}_R^0}{R{T}_R}+\left(\frac{\Delta {\underset{\_}{H}}_R^0}{R}\right)\left(\frac{1}{T}-\frac{1}{T_R}\right)$

Here, standard change in Gibbs free energy at reference temperature is $\Delta {\underset{\_}{G}}_R^0$, standard enthalpy of a reaction at reference temperature is $\Delta {\underset{\_}{H}}_R^0$, and reference temperature is $T_R$.

The expression of the standard change in Gibbs free energy at reference temperature is,

$\Delta {\underset{\_}{G}}_R^0=\Delta {\underset{\_}{G}}_{f,1-\text{pentanol}}^0-\Delta {\underset{\_}{G}}_{f,1-\text{pentene}}^0-\Delta {\underset{\_}{G}}_{f,\text{water}}^0$

Here, standard Gibbs free energy of $1-\text{pentanol}$ is $\Delta {\underset{\_}{G}}_{f,1-\text{pentanol}}^0$, standard Gibbs free energy of $1-\text{pentene}$ is $\Delta {\underset{\_}{G}}_{f,1-\text{pentene}}^0$, and standard Gibbs free energy of $\text{water}$ is $\Delta {\underset{\_}{G}}_{f,\text{water}}^0$.

The expression of the standard enthalpy of reaction at reference temperature is,

$\Delta {\underset{\_}{H}}_R^0=\Delta {\underset{\_}{H}}_{f,1-\text{pentanol}}^0-\Delta {\underset{\_}{H}}_{f,1-\text{pentene}}^0-\Delta {\underset{\_}{H}}_{f,\text{water}}^0$

Here, standard enthalpy of reaction of $1-\text{pentanol}$ is $\Delta {\underset{\_}{H}}_{f,1-\text{pentanol}}^0$, standard enthalpy of reaction of $1-\text{pentene}$ is $\Delta {\underset{\_}{H}}_{f,1-\text{pentene}}^0$, and standard enthalpy of reaction of $\text{water}$ is $\Delta {\underset{\_}{H}}_{f,\text{water}}^0$.

(B)

Interpretation Introduction

Interpretation:

The equilibrium constant using the rigorous approach.

Concept Introduction:

The expression of the standard change in Gibbs free energy as a function of temperature is,

$\frac{\Delta {\underset{\_}{G}}_T^0}{RT}=\left\{\begin{array}{l}\frac{\Delta {\underset{\_}{G}}_R^0}{R{T}_R}+\frac{J}{R}\left(\frac{1}{T}-\frac{1}{T_R}\right)-\frac{\Delta A}{R}\ln \left(\frac{T}{T_R}\right)-\frac{\Delta B}{2R}\ln \left(T-T_R\right)-\frac{\Delta C}{6R}\ln \left(T^2-T_R^2\right)\\ -\frac{\Delta D}{12R}\ln \left(T^3-T_R^3\right)-\frac{\Delta E}{20R}\ln \left(T^4-T_R^4\right)\end{array}\right\}$

Here, $J$, $A$, $B$, $C$, $D$, and $E$ are constants.

The expression of the constant $J$ is,

$J=\Delta {\underset{\_}{H}}_R^0-\Delta A{T}_R-\frac{\Delta B}{2}{T}_R^2-\frac{\Delta C}{3}{T}_R^3-\frac{\Delta D}{4}{T}_R^4-\frac{\Delta E}{5}{T}_R^5$

The expression of the constant $\Delta A$ is,

$\begin{array}{c}\Delta A=A_{1-\text{pentanol}}-A_{1-\text{pentene}}-A_{\text{water}}\\ =-A_{\text{water}}\end{array}$