Chapter 17, Problem 78E

The standard free energies of formation and the standard enthalpies of formation at 298 K for difluoroacetylene (C₂F₂) and hexafluorobenzene (C₆F₆) are

	$\Delta G_{\text{f}}^{\text{o}}$(KJ/mol)	$\Delta H_{\text{f}}^{\text{o}}$(KJ/mol)
C2F2(g)	191.2	241.3
Hexane	78.2	132.8

For the following reaction:

${\text{C}}_{\text{6}}{\text{F}}_{\text{6}}\left(g\right)\rightleftharpoons 3{\text{C}}_{\text{2}}{\text{F}}_{\text{2}}\left(g\right)$

a. calculate ∆S° at 298 K.

b. calculate K at 298 K.

c. estimate K at 3000. K, assuming ∆H° and ∆S° do not depend on temperature.

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of ${\text{C}}_2{\text{F}}_{\text{2}}$; the value of $\Delta H_f^{°}$ and $\Delta G_f^{°}$ for ${\text{C}}_2{\text{F}}_{\text{2}}$ and ${\text{C}}_6{\text{F}}_{\text{6}}$ is given. The value of $\Delta S°$ and $K$ at the given temperature is to be calculated. The value of $K$ at the given temperature, assuming that $\Delta H°$ and $\Delta S°$ are temperature independent is to be calculated.

Concept introduction: Equilibrium constant, $K$, is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$\begin{array}{l}\Delta G=\text{0}\\ Q=K\end{array}$

The expression for free energy change is,

$\Delta G=\Delta G°+RT\ln \left(K\right)$

The formula of $\Delta H°$ is,

$\Delta H°=\sum n_{\text{p}}\Delta H{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H{°}_{\left(\text{reactant}\right)}$

Explanation of Solution

Explanation

Given

The value of $\Delta G°$ for ${\text{C}}_2{\text{F}}_{\text{2}}$ is $191.2\text{kJ/mol}$.

The value of $\Delta G°$ for ${\text{C}}_6{\text{F}}_{\text{6}}$ is $78.2\text{kJ/mol}$.

The reaction that takes place is,

${\text{C}}_{\text{6}}{\text{F}}_{\text{6}}\left(g\right)\stackrel{}{\rightleftharpoons }{\text{3C}}_{\text{2}}{\text{F}}_{\text{2}}\left(g\right)$

The formula of $\Delta G°$ is,

$\Delta G°=\sum n_{\text{p}}\Delta G{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta G{°}_{\left(\text{reactant}\right)}$

Where,

• $n_{\text{p}}$ is the number of moles of each product.
• $n_{\text{r}}$ is the number of moles each reactant.
• $\Delta G{°}_{\left(\text{product}\right)}$ is the free energy of product at a pressure of $\text{1}\text{atm}$.
• $\Delta G{°}_{\left(\text{reactant}\right)}$ is the free energy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute the values of $\Delta G°$ for ${\text{C}}_6{\text{F}}_{\text{6}}$ and ${\text{C}}_2{\text{F}}_{\text{2}}$ in the above equation.

$\begin{array}{c}\Delta G°=\sum n_{\text{p}}\Delta G{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta G{°}_{\left(\text{reactant}\right)}\\ =\left[\text{3}\left(\text{191}\text{.2}\right)-\left(\text{78}\text{.2}\right)\right]\text{kJ}\\ =\underset{\_}{495.4kJ}\end{array}$

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of ${\text{C}}_2{\text{F}}_{\text{2}}$; the value of $\Delta H_f^{°}$ and $\Delta G_f^{°}$ for ${\text{C}}_2{\text{F}}_{\text{2}}$ and ${\text{C}}_6{\text{F}}_{\text{6}}$ is given. The value of $\Delta S°$ and $K$ at the given temperature is to be calculated. The value of $K$ at the given temperature, assuming that $\Delta H°$ and $\Delta S°$ are temperature independent is to be calculated.

Concept introduction: Equilibrium constant, $K$, is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$\begin{array}{l}\Delta G=\text{0}\\ Q=K\end{array}$

The expression for free energy change is,

$\Delta G=\Delta G°+RT\ln \left(K\right)$

The formula of $\Delta H°$ is,

$\Delta H°=\sum n_{\text{p}}\Delta H{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H{°}_{\left(\text{reactant}\right)}$

Explanation of Solution

Explanation

Given

The value of $\Delta H°$ for ${\text{C}}_2{\text{F}}_{\text{2}}$ is $241.3\text{kJ/mol}$.

The value of $\Delta H°$ for ${\text{C}}_6{\text{F}}_{\text{6}}$ is $132.8\text{kJ/mol}$.

The formula of $\Delta H°$ is,

$\Delta H°=\sum n_{\text{p}}\Delta H{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H{°}_{\left(\text{reactant}\right)}$

Where,

• $\Delta H°$ is the standard enthalpy of reaction.
• $n_{\text{p}}$ is the number of moles of each product.
• $n_{\text{r}}$ is the number of moles each reactant.
• $\Delta H{°}_{\left(\text{product}\right)}$ is the standard enthalpy of product at a pressure of $\text{1}\text{atm}$.
• $\Delta H{°}_{\left(\text{reactant}\right)}$ is the standard enthalpy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute the values of $\Delta H°$ for ${\text{C}}_6{\text{F}}_{\text{6}}$ and ${\text{C}}_2{\text{F}}_{\text{2}}$ in the above equation.

$\begin{array}{c}\Delta H°=\sum n_{\text{p}}\Delta H{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H{°}_{\left(\text{reactant}\right)}\\ =\left[\text{3}\left(\text{241}\text{.3}\right)-\left(\text{132}\text{.8}\right)\right]\text{kJ}\\ =\underset{\_}{591.1kJ}\end{array}$

Given

The formula of $\Delta G°$ is,

$\begin{array}{l}\Delta G°=\Delta H°-T\Delta S°\\ \Delta S°=\frac{\Delta H°-\Delta G°}{T}\end{array}$

Where,

• $\Delta H°$ is the standard enthalpy of reaction.
• $\Delta G°$ is the standard Gibbs free energy.
• $T$ is the given temperature.
• $\Delta S°$ is the standard enthalpy of reaction.

Substitute the values of $\Delta H°,\Delta G°$ and $T$ in the above equation.

$\begin{array}{c}\Delta S°=\frac{\Delta H°-\Delta G°}{T}\\ =\frac{\text{591}\text{.1 kJ}-\text{495}\text{.4 kJ }}{\text{298}}\\ =\underset{\_}{0.3211kJ\text{/}K}\end{array}$

The value of $K$ at $\text{298}\text{K}$ is $\underset{\_}{1.53\times 10^{-87}}$.

Given

The value of $\Delta H°$ is $591.1\text{kJ/mol}$.

The value of $\Delta S°$ is $\text{0}\text{.3211 kJ/K}$.

Temperature is $\text{298}\text{K}$.

Formula

The expression for free energy change is,

$\Delta G=\Delta G°+RT\ln \left(Q\right)$

Where,

• $\Delta G$ is the free energy change for a reaction at specified pressure.
• $R$ is the gas constant $(\text{8}\text{.3145}\text{J}\cdot {\text{K}}^{-\text{1}}\cdot {\text{mol}}^{-\text{1}})$.
• $T$ is the absolute temperature.
• $Q$ is the reaction constant.

Since, the reaction is at equilibrium, the free energy change,

$\begin{array}{c}\Delta G=\text{0}\\ Q=K\end{array}$

Where,

$K$ is the equilibrium constant.

Substitute these values in free energy change expression.

$\begin{array}{c}\Delta G=\Delta G°+RT\ln \left(Q\right)\\ \Delta G°=-RT\ln \left(K\right)\\ \ln \left(K\right)=\frac{-\Delta G°}{RT}\end{array}$

Substitute the values of $R,\Delta G°$ and $T$ in the above expression.

$\begin{array}{c}\ln \left(K\right)=\frac{-\Delta G°}{RT}\\ =\frac{-\text{495}\text{.4}\times {\text{10}}^{\text{3}}\text{J}}{(\text{8}\text{.3145}\text{J/K})\left(298\text{K}\right)}\\ K=\underset{\_}{1.53\times 10^{-87}}\end{array}$

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of ${\text{C}}_2{\text{F}}_{\text{2}}$; the value of $\Delta H_f^{°}$ and $\Delta G_f^{°}$ for ${\text{C}}_2{\text{F}}_{\text{2}}$ and ${\text{C}}_6{\text{F}}_{\text{6}}$ is given. The value of $\Delta S°$ and $K$ at the given temperature is to be calculated. The value of $K$ at the given temperature, assuming that $\Delta H°$ and $\Delta S°$ are temperature independent is to be calculated.

Concept introduction: Equilibrium constant, $K$, is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$\begin{array}{l}\Delta G=\text{0}\\ Q=K\end{array}$

The expression for free energy change is,

$\Delta G=\Delta G°+RT\ln \left(K\right)$

The formula of $\Delta H°$ is,

$\Delta H°=\sum n_{\text{p}}\Delta H{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H{°}_{\left(\text{reactant}\right)}$

Explanation of Solution

Explanation

Given

The value of $\Delta H°$ is $591.1\text{kJ/mol}$.

The value of $\Delta S°$ is $\text{0}\text{.3211 kJ/K}$.

Temperature is $\text{3000 K}$.

The conversion of kilo-joule $\left(\text{kJ}\right)$ into joule $\left(\text{J}\right)$ is done as,

$\text{1}\text{kJ}={\text{10}}^{\text{3}}\text{J}$

Hence,

The conversion of $\text{591}\text{.1}\text{kJ}$ into joule is,

$\begin{array}{c}\text{591}\text{.1}\text{kJ}=\left(\text{591}\text{.1}\times {\text{10}}^{\text{3}}\right)\text{J}\\ =\text{591}\text{.1}\times {\text{10}}^{\text{3}}\text{J}\end{array}$

The conversion of $\text{0}\text{.3211 kJ}$ into joule is,

$\begin{array}{c}\text{0}\text{.3211}=\left(\text{0}\text{.3211}\times {\text{10}}^{\text{3}}\right)\text{J}\\ =\text{3}\text{.2}\times {\text{10}}^{\text{2}}\text{J}\end{array}$

The formula of $\Delta G°$ is,

$\Delta G°=\Delta H°-T\Delta S°$ (1)

And,

$\Delta G°=-RT\ln \left(K\right)$ (2)

Where,

• $\Delta H°$ is the standard enthalpy of reaction.
• $\Delta G°$ is the free energy change.
• $T$ is the given temperature.
• $\Delta S°$ is the standard enthalpy of reaction.

Substitute the value of $\Delta G°$ from equation (2) into equation (1).

$\begin{array}{c}\Delta H°-T\Delta S°=-RT\ln \left(K\right)\\ \ln \left(K\right)=-\left(\frac{\Delta H°-T\Delta S°}{RT}\right)\end{array}$

Where,

• $R$ is the gas constant $(\text{8}\text{.3145}\text{J}\cdot {\text{K}}^{-\text{1}}\cdot {\text{mol}}^{-\text{1}})$.
• $K$ is the equilibrium constant.

Substitute the values of $\Delta H°,R,\Delta S°$ and $T$ in the above equation.

$\begin{array}{c}\ln \left(K\right)=-\left(\frac{\Delta H°-T\Delta S°}{RT}\right)\\ =\frac{\text{591}\text{.1}\text{J}\times {\text{10}}^{\text{3}}-\left(\text{3000}\text{K}\times \text{3}\text{.2}\times {\text{10}}^{\text{2}}\text{J/K}\right)}{(\text{8}\text{.3145}\text{J/K})\left(\text{3000}\text{K}\right)}\\ =\underset{\_}{3.11\times {10}^6}\end{array}$